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Full text of "Lecture notes in mathematics: rudiments of Riemann surfaces"

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ERSITY LIBRARY 



Lecture Notes in Mathc^maties 
Number 2 




RUDIMENTS 
OF RIEMANN SURFACES 



B. Frank Jones, Jr. 



Houston, Texas 
1971 



RUDIMENTS OF RIEMANN SURFACES 



B . Frank Jones , Jr 

QS 

3 

HO. 1^ 



Rice University 
1971 



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3 1272 00076 4025 






Digitized by the Internet Archive 

in 2011 with funding from 

LYRASIS Members and Sloan Foundation 



http://www.archive.org/details/lecturenotesinma19712rice 



PREFACE 

In the spring semester, 1969, I taught a course at 
Rice University on Rieraann surfaces. The students were 
primarily seniors who had taken one semester of complex 
variables and had been exposed at least to the language 
of general topology. I made detailed lecture notes at 
the time, and this volume contains those notes with minor 
changes . 

The purpose of the course was to introduce the 
various ideas of surfaces , sheaves , algebraic functions , 
and potential theory in a rather concrete setting, and 
to show the usefulness of the concepts the students had 
learned abstractly in previous courses. As a result, I 
discussed the material carefully and leisurely, and for 
example did not even attempt to discuss the notions of 
covering surface, differential forms, Fuchsian groups, 
etc. Therefore, these notes are quite incomplete. For 
comprehensive treatments of the subject, please consult 
the bibliography. 

I gratefully acknowledge some of the standard books 
which I consulted, especially M. H. Heins ' Complex Function 
Theory , G. Springer's Introduction to Riemann Surfaces , and 
H. Weyl's The Concept of a Riemann Surface . Also, I relied 
heavily on L. Bers ' lecture notes, Riemann Surfaces, and 
especially on his Lectures 15-18. 



One of the students was Joseph Becker, to whom I ow« 
special thanks. He helped and prodded me over and over 
and gave me tremendous encouragement. 

Thanks also go to the typists, Janet Gordon, Kathy 
Vigil, and Barbara Markwardt , and to Rice University for 
publishing the notes. 



Houston, July 12, 1971 



CONTENTS 

Chapter I . INTRODUCTION 1 

Chapter II . ABSTRACT RIEMANN SURFACES 16 



Chapter III . THE WEIERSTRASS CONCEPT OF A RIEMANN 

SURFACE 46 



Chapter IV . BRANCH POINTS AND ANALYTIC 

CONFIGURATIONS 77 



Chapter V . ALGEBRAIC FUNCTIONS 110 

Chapter VI . EXISTENCE OF MEROMORPHIC FUNCTIONS 149 



Chapter VII . CLASSIFICATION OF SIMPLY CONNECTED 

RIEMANN SURFACES 221 



Chapter VIII . THE TORUS 241 



APPENDIX 

Final examination 270 

Solutions to problems 6 and 7 271 

Solutions to final exam problems 283 



REFERENCES 293 




is an extension of f. But this is not the kind of dif- 
ficulty that we wish to consider. 

Rather, the basic problem is that of multiple- 
valued "functions." Phrased in terms of continuations, 
there is not always a largest region to which a holo- 
morphic function can be extended. As an example let 
D = [z: |z-l|<l] and f(z) = principal determination of 

log z 

1 J n=1 

defined on D . Of course, we also 

have f(z) = logjzj 4- i arg z, 

where arg z is between - S and -j. Now f can be extended 

to a holomorphic function on the plane C with the negative 

real axis removed, the extension being log|z| + i arg z, 

where -rr < arg z < rr. But there are other regions which 

can be considered as largest regions of extension; e.g., 

the plane C with the positive imaginary axis removed and 

the extension being log|z| + i arg z, where - -J^-arg z<5. 

It is admittedly frequently useful to "cut" the plane 
C along a line from to » as visualized in the above 
cases, and to consider there a single-valued "branch" 
or "determination" of log z, and such a technique is ex- 
ploited e.g. in contour integrals. 

But from the point of view of this course the cut- 
ting of C really enables one to evade the issue, which 
is namely how can one speak of log z and face up to its 



multiple-valuedness in a fearless way. And the same 
question for other functions. The answer given by Rie- 
mann is that the plane C is too deficient to admit such 
functions, so we consider other surfaces where functions 
can be defined which are single-valued and still exhibit 
the essential behavior of (in our example) log z. 

Let us now consider an explicit method for building 
such a surface for log z. Take an infinite sequence of 
planes minus the origin, which are to be considered as 

distinct: call them C', where n is any integer. On C' 

' n' J b n 

define a function f by 

n J 

f (z) = log|z| + i arg z + 2nrri , 

where -tt < arg z <, n. Now we "glue" the planes C' in a 

reasonable way. This "gluing" is tantamount to defining 

a topology on the union of the (disjoint) sets C ; . To 

define this topology we shall describe a neighborhood 

basis of each point. For a point z € C' which does not 
r r n 

lie on the negative real axis a neighborhood basis shall 

consist of all open disks in C' with center at z. If 

r n 

z 6 C' and z is a negative real number, a neighborhood 
basis shall consist of all sets 

fw€C': |w-z|<c, Im wsO) u fw€C', ,: |w-zj<e, Im w<0] , 
n ' n+1 ii' > 

where < e < |z | . 



C n 



c' 

n+1 



JL 



It is then easily checked that the set S = U C becomes 

n=-» n 
a topological space with a neighborhood basis for each 

point of S being described as above. Also, if f is the 

function from S to C which equals f on C' for each n, 

' n n 

then f becomes a continuous function on S. Indeed, it 

suffices to check continuity at points z € C' which are 

J r n 

negative real numbers. In the semidisk in C' depicted 

above f = f takes values close to logJz| + in + 2nni, 

and in the semidisk in C',, f = f ,, takes values close 

n+1 n+1 

to log | z | - in + 2(n+l)rri, so in the whole neighborhood 
of z f is close to log|z| + irr + 2nni = f(z). Thus, f 
is continuous. 

Thus , we have succeeded in defining a set S which 
carries a single-valued function f which obviously is 
closely related to log z. We shall later point out the 
essential feature of S which allows us to call it a Rie- 
mann surface (definition to be given in Chapter II) . 

We remark that it is easy to visualize S as a col- 
lection of planes glued together as indicated and forming 

3 
in R an infinite spiral. The next surface we construct 

will not have so simple a form. 



For this construction consider the function z ' , 
where m is an integer >2. Since each z £■ has m dis- 
tinct m roots, this is a multiple- valued "function." 
In order to treat this function consider distinct 
copies of the plane minus the origin, C-/ ,0/ , • • ■ >C ' . 

Define a function f on C ' by the formula: 

n n J 

i fi 
if z = re , r>0, -tt<9^tt , 



f (z) = r l/n» e i9/m e 2iT7(n-l)/r 



m 



Let T = lJ C ' and define a topology on T exactly as be- 
n=l n 

fore, except that a neighborhood basis of a negative 

real number z £ C ' is treated a little differently. The 

m J 

same situation obtains as in the figure on p . 4, with 

C' , replaced by £' . Note that an attempt to visualize 

T as a spiral in R is doomed, since the "top" level C ' 

m 

has to be glued to the "bottom" level C-,' along their 

negative real axes , and this without crossing any of the 

intermediate levels C^,..., C* -* and also without crossing 

the seam where C-/ is joined to C^ (in case m = 2) . As 

before, define a function f on T by the formula f = f 

* n 

on C'. As in the figure on p . 4 , if z € C ' is a negative 
n o r- j n ° 

real number, then in the semidisk in C ' f takes values 

' n 

, fc , il/m in/m 2irr(n-l)/m , . ., ..... 

close to |z| e e v a an ° in tne semidisk in 

C^ f takes values close to | z | lAvin/m^iTrn/i^ so f 
stays close to fz 1 1 /m e ±Tr (2n- 1) /m = f ^ in a neighborhood 
of z. And this holds even if n = m, in which case C ' , 



is replaced by C,'. Thus, f is continuous on T and gives 
a reasonable representation of z ' m . 

Now a very interesting addition can be made to T. 
Namely, consider each C' to have its origin replaced, 
but with the origins in each C representing a single 
point to be added to T. Thus, consider TufO] (the orig- 
inal set with one point added) and let a neighborhood 
basis of consist of sets of the form 

m 
{0} U U [z€C^: |z|<e] 
n=l 

for 0<e< m . Extend f by f (0) =0. Then f is again con- 
tinuous on TU{0}. In the very same way, the point <*> can 
be added. Let 

t = Tu{0}ur»] , 

let a neighborhood basis of OT consist of sets of the form 

m 

i>] u u [ z ^ c n : I Z IH:} > 
n=l 

and let f( m ) = co . Then f is a continuous function from 
T to the extended complex plane (Riemann sphere) C. Ob- 
viously the points and m are in some sense different 
from the other points in T. They are called branch 
points , and are said to have order m-1. 

Although T is somewhat difficult to visualize as 

3 
situated in P , we shall now easily see that it is homeo- 

A ~ A 

morphic to the sphere C.' In fact, the mapping f: T • C 
is a homeomorphism. We have shown that it is continuous; 



it is onto since every complex number is an m root; 
it is 1-1 since different complex numbers definitely 
have different m roots and also the same complex 
number z * has m distinct m roots. General topol- 
ogy then shows f is continuous since T is compact and 

A 

C is Hausdorff; but it is quite easy to see directly 
that f is continuous. Indeed, f (z) is essentially 
z (positioned on the correct C ') . 

The nature of this homeomorphism and the geometry 
involved in the construction of T are perhaps better 
seen when one considers the Riemann sphere C instead of 

C as the basic region from which f is to be built. If 

a 2 2 2 

one regards C as the Euclidean sphere {(x,y,z): x +y +z =1} 

3 
in R by means of s tereographic projection and uses m 

distinct copies C,,...,C with the gluing described 
t- 1 ' ' m & o 

above to be done along the meridians corresponding to 
the negative real axis , then an essentially equivalent 
surface T is obtained. Now consider the action of the 

A 

function f. On C it is given by the determination 

f of the m root and maps C onto a portion of C cut 
n r n r 

off by two meridians which correspond to rays in the 
plane with an included angle of 2rr/m. In other words , 
it "spreads open" the cut in C from a hole with open- 
ing to a hole with (1 - — )2tt opening. Here is a picture, 
a "top" view looking "down" on the north pole, <*>: 



A A 

C „ . ^rrr. C 

n 



f 

> 





image of 

u under f 
n 



Thus, the image of T under f consists of m "slices" of 
C, and the gluing in T shows that these slices of C are 
pieced together in such a way that T is mapped homeo- 
morphically onto C. 

If one is interested only in the topological prop- 
erties of T, then the procedure discussed in the above 
paragraph can be considerably shortened by ignoring 
the specific nature of the cuts and of the function f. 
We illustrate with the case m = 2. Since we shall only 
discuss topological properties , we replace the cut along 
a meridian by any old cut on the sphere which looks 
reasonable, and take two copies of the sphere: 

. - - • A „ A 

• Cl N c 2 



The gluing is to be done in such a way that the shaded 
areas are to be attached, as are the unshaded areas. 
The action of the function f is now replaced by a con- 
tinuous opening of the two holes : 




/■ 





It is then obvious how to attach these spheres with holes; 
the resulting figure looks like 
a figure which is obviously homeo- 
morphic to a sphere. 

Now we shall briefly indicate the construction of 
some other Riemann surfaces. For example, suppose a 
and b are distinct complex numbers and consider the 
multiple-valued "function" ,/(z-a) (z-b) . The same pro- 
cedure which works for Jz can be applied here if C or 
C is cut between a and b. In trying to define this func- 
tion one finds that the sign changes when a circuit is 
made around either a or b, so two copies of C can be 
joined along the cut as before to provide a surface on 
which a function which is single- valued and has the prop- 
erties of V(z-a) (z-b) can be defined; the figure is 
exactly that which appears at the bottom of p. 8, where 
the two slits go from a to b on each sphere. Here it 
should be remarked that either branch of y(z-a) (z-b) is 
meromorphic at », since one branch is approximately z 
at oo and the other branch approximately -z. The branch 
points on the surface we have constructed are a and b, 



10 



A 

and the surface is again homeomorphic to C. However 
note that the function ,/(z-a) (z-b) is not the homeo- 
morphism in this case. Indeed, this function assumes 

A 

every value in C exactly twice. A natural homeomorphism 
in this case is the function on this surface corresponding 



A 

function and two copies of C cut from a to b , we obtain 
the same surface. 

Using the same process, we shall now construct a 
Riemann surface which is not homeomorphic to a sphere. 
For this consider the expression ./(z-a) (z-b) (z-c) , where 
a,b,c are distinct complex numbers. In order to attempt 
to define a single- valued function from this formula, 
consider two copies of the sphere each having two 
cuts , say from a to b and from c to »; these cuts should 
not intersect : 





In defining continuously the square root in this case, a 
change of sign results in going around a, or b, or c, or 
00 . The cuts we have provided prohibit this, and we 
also see just how to glue in order to obtain a continuous 
function: the shaded areas along the cuts from a to b 



11 



are to be attached, and likewise along c to oc . Now let 
S denote the resulting surface with the four branch 
points ajbjC, 00 included, the topology being defined in 
the by now usual manner. This surface is not homeo- 
morphic to a sphere. To see this we will exhibit a 
closed curve on S which does not separate S into two 
components . This is the curve shown on the left sphere 





which encircles the cut from a to b. To see that this 
curve does not disconnect S consider the typical example 
of the curve (shown by a dotted line) which connects two 
points which at first glance might be separated by the 
given closed curve. 

Probably the best way to see this topological prop- 
erty is to apply the method sketched on p. 8. After 
the first step we obtain the following spaces to be 
glued: 








12 



After the gluing, the resulting figure appears as shown: 




This figure is clearly homeomorphic 

to a torus or a sphere with "one f 

handle." The same topological 

type of surface arises from the function 




y<fe-a) (z-b) (z-c) (z-d) , where a,b,c,d are distinct. The 
only difference is that the cuts on C go from a to b 
and from c to d. 

This same argument allows the treatment of the 
function v '(z-a.j ) (z-a„) . . . (z-a ) , where a, ,. . . ,a are 
distinct. Two copies of C are used with cuts from a. 



to a j , a~ 



to a , , etc. If m is even, the last cut is 

from a , to a , and if m is odd, from a to <=. The 
m-1 m' ' m 

same gluing procedure gives a topological type as il- 
lustrated: 

there are -^ connecting 

. . . c . m+1 

tubes if m is even, — ■*— 

if m is odd. 




This is homeomorphic to a sphere with "handles" 



13 




. , m- 2 m- 1 
there are — y- or — k— 

handles if m is even or 

odd, respectively. This 

is said to be a surface 

having genus equal to the 

number of handles . 



Local coordinates. 
In preparation for the definition of abstract Rie- 
mann surfaces to be given in the next chapter, we shall 
now examine a common property of all the surfaces we 
have constructed. Namely, each point on the surface 
has a neighborhood homeomorphic to an open subset of 
C--the essential defining property for a surface. This 
assertion is of course completely trivial except where 
we have made cuts and where we have inserted branch 
points , for outside these exceptional points the neigh- 
borhoods can just be taken to be disks on the various 
copies of C and the homeomorphism essentially the 
identity mapping onto the same disk, now regarded as 
lying in some other fixed copy of C. The situation 
for points on the cuts which are not branch points is 
not much more involved. Refer to the neighborhood de- 
fined and depicted on pp. 3,4; call this neighborhood 
U(z) and let h be the. disk [w€C: jw-z|<e}. Then define 



cp: U(z) 



14 I 

by the obvious relation 

cp(w) = w. 

The effect of cp is obviously to attach the two semi- 
disks used to make up U(z). It is now trivial to check 
that each point which is not a branch point has a neigh- 
borhood homeomorphic to an open set (a disk) in C, and 
this is true for all the surfaces we have constructed. 
If =° is not a branch point and does not lie on a cut, 
a neighborhood can be taken to be the complement of a 
large closed disk in the appropriate copy of C and the 
mapping into C the function ^(z) = z 

Now for the branch points. It should be no surprise 
that the branch points can be treated, for we have 
pointed out how the surface with branch points added 
is homeomorphic to a sphere or a sphere with handles 
(in the cases we have considered) , making the neighbor- 
hoods of the branch points look not very special at all. 
Now we write down this homeomorphism explicitly in the 
case of the Riemann surface for z , since all the 
other branch points we have considered have the same 
behavior as is exhibited in this case (for m = 2) . In 
fact, the homeomorphism is exactly the "function" z 
(which has been made s ingle- valued) . In terms of the 
notation of p. 5, this is the function f. A similar 
construction works when the branch points at » are 
considered. 



15 



Finally, consider how these various homeomorphisms 
are related. That is, suppose given two overlapping 
neighborhoods U-, and U2 on the surface with correspon- 
ding homeomorphisms cp, and cp~ . Then the function 
X'.;, :?-, is defined on an open subset of C and has values 
in another open subset of C, and is clearly a homeo- 
morphism. The thing to be noted is that it is holo- 
morphic . Except where U-, or U- involves a branch point 
this is trivial, as the map cp^ocp, is the identity where 
it is defined. If U-, involves a branch point with m 

— 1 t~V-> 

sheets, then x>, is essentially the m power, and 

- 1 m 
ep^ocp, (z) = z , which is holomorphic. If U^ involves 

a branch point, then cp^ocp, is a holomorphic determi- 
nation of the m root. 

The observation made above will be used to give a 
definition of Riemann surface in the next chapter. 



16 II 

Chapter II 
ABSTRACT RIEMANN SURFACES 



In the introduction we have considered one method 
of constructing Riemann surfaces and have pointed out 
various properties. In the rest of the course several 
other methods will be given, especially the extremely 
important sheaf of germs of meromorphic functions in 
Chapter III and its generalization, the analytic con - 
figuration , in Chapter IV. Other examples will be con- 
sidered in the present chapter. All of these Riemann 
surfaces have one feature that cries out for attention, 
so before coming to the concrete examples we shall 
define this characteristic feature and call any object 
which possesses it a Riemann surface. 

DEFINITION 1 . A surface is a Hausdorff space S 
such that V p€S 3 an open neighborhood U of p and an 
open set W c C and a homeomorphism cp: U - W. Such a 
mapping cp is called a chart or a coordinate mapping . 

DEFINITION 2. Let S be a surface. An atlas for 



S is a collection of charts {cp } , where a runs through 
some index set, such that every point of S belongs to 

the domain of some co . If cp : U - W , then we are 

a a a a ' 



II 



17 



saying that 



s = u u 



Note that if U and U meet, then both cp and cp are 
defined on the intersection U fl U and these mappings 
provide homeomorphisms between this intersection and 

the open sets cp (U flu* ) and cp, (U fflj ) in C , respectively. 

r a v a 3 3 a 3 r J 

Therefore, there is defined the function 



V*B : VW - ^a< U a nU ,) 



a\ 




i W. 



For brevity we shall frequently speak of cp ocp without 

Q. p 

mentioning that it is defined only on cp (U HU ) . The 

functions cp ocp^ are called coordinate transition func- 

v a ' 9 

tions of the atlas , because if cc and cp are thought of 
as defining coordinates on U HU the mapping cp c cp 
determines how to change from one coordinate system to 
another. 



18 II 



DEFINITION 3. An atlas fcp } is analy tic if each 

====== l+ a J l 

coordinate transition function cp ocp is analytic. 

Just note that this definition makes good sense, 
as cp ocp a is a complex- valued function on an open set 
in C and thus the usual meaning of analytic function 
is what is meant. 



DEFINITION 4 . Two charts cp, and cp 2 on a surface S 
are compatible if the functions cp, °<p" and cp„ocp are 
analytic. A chart cp is compatible with an analytic 
atlas {ep } if cp and cp are compatible for all a. 



DEFINITION 5 . An analytic atlas is complete if it 
contains every chart compatible with it. 

We are now almost ready to define a Riemann surface 
as a surface together with an analytic atlas . But there 
is a slight technical problem which must be overcome. 
Namely, there is almost never a convenient canonical 
atlas , and we therefore either need to define some 
sort of canonical atlas or need to define an equivalence 
relation between analytic atlases. Since these ap- 
proaches are really the same, we arbitrarily pick the 
former possibility. This is the reason for Definition 5, 
Now we give a lemma which actually relates these con- 
cepts . 



II 19 



LEMMA 1 . For any analytic atlas [cp } on a sur - 
face S, there exists exactly one complete analytic 
atlas containing it . This complete analytic atlas is 

the collection of all charts compatible with (cp } . 

ex 



Proof: Let G be the set of all charts compat- 
ible with {cp }. We first prove that G is an atlas, 
then that it is complete. Suppose then that cp,cp'€ C 
Thus, cp:U - W and cp ' : u' - W are homeomorphisms 
from open sets in S to open sets in C. Suppose U 

and u' meet and let p 6Unu'. Since {cp } is an 

r o ^a 

atlas, there exists cp :U - W such that p €U . 

^a a a r o a 

Then 



cp'ocp = (cp'°cp a )°(;o a °cp ) 



and cp'°cp and cp °cp~ are both holomorphic since 

cp,cp' are compatible with cp . Thus, cp'om" is 

ex 

holomorphic. Thus, G is an analytic atlas. To prove 
that G is complete, suppose i|i is compatible with 
G. Since G contains {cp } (since {cp } is itself 
an analytic atlas), i|i is compatible with [cp }. 
That is, iKG. Thus, G is complete. 

Finally, to prove that G is unique, suppose ft 
is a complete analytic atlas containing icp } . If 



20 II 



eoSB, then cp is compatible with {cp }, and thus 
cp€G. This proves BOG. Now suppose cp£G. Let 
•J16S. Arguing as above, we find 



CP° Uf = (iP CP a )°(cP a °1tF~ ) , 

l)l°Cp = (ty°CD )o (cp a °ip ) , 



and thus tp and $ are compatible. Thus, cp is 
compatible with 3. As 8 is complete, cp?8. Thus 
GcS. Hence, G = 8. 

QED 



As a result of this lemma, we see that two analytic 
atlases are contained in the same complete analytic 
atlas if and only if each chart from one atlas is com- 
patible with each chart from the other atlas, or if 
and only if the union of the two atlases is itself an 
analytic atlas. 

DEFINITION 6. A Riemann surface is a surface to- 



gether with a complete analytic atlas. 

Thus, to specify an abstract Riemann surface, we 
must specify a surface and a complete analytic atlas. 



21 



The effective purpose of Lemma 1 is to enable us to 
forget about the rather cumbersome completeness assump- 
tion. So when we wish to construct a Riemann surface, 
we will be satisfied to exhibit one analytic atlas, 
keeping in the back of our minds that Lemma 1 implies 
the existence of a unique larger complete analytic 
atlas. This is quite helpful, as it will usually be 
more or less obvious what can be chosen to be an 
analytic atlas, 

It is most important for beginners in this sub- 
ject not to be beguiled by Definition 6. The crux 
of the theory of Riemann surfaces is not this definition. 
This definition just gives a convenient term in a book- 
keeping sense to keep track of the structure implied 
in the definition of complete analytic atlas. Thus, 
this chapter has been called " abstract Riemann surfaces." 
It will be up to us to verify for the many concrete 
Riemann surfaces we find that the above definition obtains 
Now we pass to some examples. 

Examples. 



1. This is by far the most trivial example. Let 
S be any open subset of C; the atlas con- 
sists of the single chart cp which is the 



22 II 



identity mapping on S. In this case cp 
is obviously a homeomorphism and the only 
transition function is cpocp" = identity on 
S. 

A most important example is the Riemann 
sphere . We take this to be the topological 
space C = CUf™}, where points in C have 
their usual neighborhoods and a neighborhood 
basis of 0= consists of the sets {z:|z|>a} 
U{°°3 f° r 0<a<°=. This is clearly a topo- 
logical space and stereographic projection 

is a homeomorphism of C onto the unit 

3 
Euclidean sphere in R . The atlas we pick 

will consist of two charts. Let Ui = W, = C 
and cp-itU-i "* W-. be the identity. Let 
U 2 = C - {0}, w~2 = C, and cp2 : U2 "* W 2 be 
given by cpo(z) = z , ^(oo) = 0. These are 
clearly charts, and cp2 or P i ( z ) = cpo(z) = z , 
cp-iocpo (z) = z~ , which shows the coordinate 
transition functions are holomorphic. 

As we mentioned above, c is homeomorphic 

3 
to the unit sphere in R . It is a fact that 

any topological space homeomorphic to a Riemann 
surface can itself be made into a Riemann sur- 
face. To see this, suppose S is a Riemann 



II 23 



surface with analytic atlas {3 } and that 

J ■ a J 

T is a topological space and I : T -> S a 
homeomorphism. Then the maps {cp °?} form 
an analytic atlas for T with transition 
functions 



(cp a o§)o (cPgol ) = CP a oCp g 



4. All the surfaces constructed in Chapter I are 
Riemann surfaces. The verification was briefly 
indicated on pp. 13-15. 

5. Any open subset of a Riemann surface can be 
made into a Riemann surface in a natural way: 
If T is an open set in the Riemann surface 

S, then for a chart cp :U -* W on S let 

^a a a 

the mapping ty be the restriction of cp 
to U flT. Then an analytic atlas fcp } on 
S gives rise to an analytic atlas { \Ji } on 
T. 

6. The torus. Of course, the examples mentioned 
in 4 include a Riemann surface homeomorphic 
to a torus; cf. p. 12. Here is another way 
to make a torus into a Riemann surface. 



24 II 



Problem 1 . Let u^ and ou be nonzero complex 
numbers whose ratio is not real. Let 
- {n-^uu^ + n2U)2: n^r^ integers}, 
and for any z£C let [z] = z + Q. 
Prove that 3 6 > such that 
In-juu, + n~ui)o | £ 6 if n, , n~ are 
integers which are not both zero. Let 
c/0 be the set of all [z] for zee, 
noting that [z] = [z'] » z - z'GO* For 
any [z] define a neighborhood basis 
of [z] to consist of all sets 



UgCfzl) = {[wl: jz - w|<e) 



for e>0. Prove that C/Q becomes a 
Hausdorff space. For es^/2 let 
r :U ([zi) - A. = (C C C: j C | <e] be defined 
by co([w]) = w-z. Prove that these form 
charts in an analytic atlas for c/n. 

The relation to a torus is that c/n is 
homeomorphic to a torus in a natural way. 
This can perhaps best be seen by considering 
the set A = [t^ + t 2 0) 2 : 0st 1 <l,0^t 2 <l}cc, which 
is obviously in one-to-one correspondence with 

c/n. 



II 



25 




JJ-J+UU2 







The topology in A is 
determined in a natural 
fashion: a neighborhood 
basis of a point t, uu-. + 
t 2 oi2 with < t, < 1, 
< t« < 1, can be taken 
to be sufficiently small 
disks centered at that 
point. For a point p 
as indicated in the figure, 
a neighborhood basis can 
be taken to be sets 



{z^A: ]z-p|<e}u{zeA: ] z-p- a j 1 1 < e } 



for all sufficiently small e. And a neighbor- 
hood basis of can be described in a similar 
fashion, corresponding to the four smaller 
sectors in the figure. Of course, this top- 
ology just corresponds to a gluing in the sense 
of Chapter I and one easily sees that now A 
is homeomorphic to C/Q, the homeomorphism 
being the mapping A - c/n which sends z 
to [z]. Finally, if one imagines this gluing 
carried out with a strip of paper the shape 
of A, it becomes clear that A is homeomorphic 



26 



II 



to a torus . 
7 . The sheaf of germs of moromorphic functions 
to be discussed at length in Chapter III will 
be a Riemann surface in a natural way. 

DEFINITION 7 . A path in a topological space S 
is a continuous function y from I = [0,11 into S. 
The initial point of y is y(0) and the terminal 
point of y is y(l). And y is said to be a path 
from y(0) to y(l). 

DEFINITION 8 . A topological space S is disconnected 
if 3 open sets A,BcS such that S = AUB, A and 
B are disjoint, and neither A nor B is empty. A 
topological space S is connected if it is not dis- 
connected . 

PROPOSITION 1 . A Riemann surface S is connected 
if and only if for any points Pq and p-, in S there 
exists a path in S from Pq to_ p-,. 

Proof : Suppose S is disconnected, and let A 
and B be the corresponding sets of Definition 8. 
Let Pq£A and p-i c B. If there is a path y in S 



IT 



27 



from p n to p-, , then y(I) is connected (it is a 
general result that a continuous image of a connected 
space is connected). However, the sets A-, = Y(l)nA 
and B-, = y(I)hB show that in the sense of Definition 
8 y(I) is disconnected. 

Conversely, suppose S is connected and let p^, 
p-.cS. Let A = {p$S:3 path in S from p~ to p}. 
Then A contains p~ and is thus not empty. Also, 
A is open: if p^A then using an open neighborhood 
U of p and a chart cp '• U - A from U onto a disk 
A, then U^A. For if p 'gu and if y is the path 
from p.-. to p, then a path Yi from p„ to p' is 



Yl (t) = 



(Y (2t), 0*ts%, 
( p" 1 ((2-2t)cp(p) + (2t-l)cp(p')), ^t 5 l 





28 II 



Thus, A is open. A similar proof shows that A is 
closed: if p' is a limit point of A, then we can 
use the same picture as above, except that U is 
now picked to be a neighborhood of p' homeomorphic 
to a disk A. Since p' is a limit point of A, 
there is a point p<=unA. Then the same construction 
as above shows that there is a path in S from p~ 
to p' ; i.e., p'€A. Thus, A contains all its limit 
points and is therefore a closed set. Since A is 
open and closed and nonempty, and S is connected, 
we have A = S. Thus, Pi€A. 

SID 
Remark . Note that the above proof is entirely 

topological. In general topology this theorem states 
that a connected, locally arcwise connected space is 
arcwise connected. 

Now we turn to the important concept of analytic 
functions. 

DEFINITION 9 . Let S, and S 2 be Riemann sur- 
faces, U an open subset of S^ and f a continuous 
function from U to S 2 . Then f is analytic if 
for every chart cp,:!^ - W^ on S 1 and everv chart 
cp 2 -U 2 - W 2 on S 2 , the function cp 2 °f°cp 1 is holo- 
morphic (Here and elsewhere when we use a phrase 



II 29 



like "every chart cp," we mean every chart cpi in 
the complete analytic atlas for S,.) 

Remark . Since the coordinate transition functions 
are holomorphic, to check the analytic ity of f in a 
neighborhood of a point Pq€U it is sufficient to 
check the analyticity of cp2°f°<?i f° r some chart cp, 
in a neighborhood of p Q and some chart cp 2 ^ n a neighbor- 
hood of f(p ). This remark also immediately leads to 

PROPOSITION 2 . In the notation of Definition 9 
f is analytic on U if and only if f is analytic 
in some neighborhood of each point of U. 

Proof is left to the reader. 
PROPOSITION 3. If f:S 



If f:S, - S2 is analytic and 



g:S2 - S~ is analytic , then g°f:S, -» S~ is analytic 

Proof : Let p^gS,. Choose a chart cpo:U, - W-, 
in a neighborhood of gof(p~). Choose a chart 
cpy'.^n ~* ^9 i- n a neighborhood of f(Pr\) such that 
g(U2) c Uo- Choose a chart cp-pU-j - W, in a neighbor- 
hood of p Q such that f(U-j)<=u 2 . Then 



30 II 



cp 3 °S ofo cpi = (cp 3 °gocp2 )°(cp2 ofo cp]_ ) 



is a composition of holomorphic functions and is thus 
holomorphic. Thus, g°f is analytic in a neighborhood 
of p~ and Proposition 2 shows this suffices. 

QED 
Examples . 

1. If S-, is an open subset of the Riemann 
surface C and S 2 = C, then f:S-, - r 
is analytic according to Definition 9 « f 
is analytic in the usual sense (satisfies 
the Cauchy-Riemann equation) . 

2. If S-, = t and f is continuous from a 
neighborhood of <= into So, then f is 
analytic in a neighborhood of <= » the 
function z - f(z~ ) is analytic in a 
neighborhood of 0. This follows because a 
chart near » on € is the mapping 

co(z) = z~ . 

3. Likewise, if S 2 = t and f:S-, - c is 
continuous in a neighborhood of Pq and 
f(Pn) = °°3 then f is analytic in a neighbor- 
hood of Pn " T -*- s ana ly t; i- c from a neighbor- 



II 31 



; 

4. An analytic function from a Riemann surface 

to C is said to be holomorphic ; an analytic 
function from a Riemann surface to C is 
said to be meromorphic . 

5. Any chart in the complete analytic atlas of 
a Riemann surface is holomorphic. 

6. Consider the torus C/fi as discussed in 6 
on p. 24. Let tt:C - C/q be the canonical 
mapping n(z) = [z] . Then n is analytic. 
To see this consider <p:U ([z]) - A„ as in 
Problem 1. Then in a neighborhood of the 
fixed point z we have ep°n(w) = cp([w]) = w-z, 
a holomorphic function of w. 

7. Again for the torus c/fi considered in 6, 
we show that if S is a Riemann surface 
and f:C/n - S, then f is analytic 

» 3 F:C - S analytic such that 

F = fo-rr. 

First, if f is analytic and F is defined 
this way, then F is a composition of 
analytic functions and is thus analytic. 



32 II 



Now suppose F is analytic and F = fon. 
We shall then prove that f is analytic in 
a neighborhood of any point fz]^C/Q. Take 
cp:U ([z]) - A as in Problem 1. Then for 
p€U,([z]) we can write p = [w], where 
jw-z| < e and cp(p) = w-z. Thus 

f(p) = f(rr(w)) = F(w) - F(z +~,(p)), 

and we have exhibited f as a composition 

of analytic functions, so that f is analytic 

near p Q = [z] . 

This example really indicates the importance 
of the notion of analytic functions, since 
we see that there is a natural identification 
of analytic functions on c/fi with analytic 
functions F on C which are doubly periodic , 
i.e., which satisfy 

F(z+uj 1 ) = F(z), 
F(z+a) 2 ) = F(z). 

When S = c these are the elliptic functions. 
For the Riemann surfaces constructed in the 



II 33 



introduction there are corresponding analytic 
functions. For example, consider the Riemann 
surface S for log z and the function f 
on S corresponding to log z (pp. 3-4). 
Then f is holomorphic on S. Likewise, 
consider the Riemann surface T for z 
and the corresponding function f (pp. 5-6). 
Then f is meromorphic on T. This really 
follows from 5 above since near the branch point 
the function f is a chart and likewise 
near the branch point =>, and away from the 
branch points the verification is obvious. 
9. The analytic functions from c to £ are 
the rational functions. 
10. The analytic functions from c to C are 

the constant functions (Liouville's theorem). 

Now we shall develop some general properties of 
analytic functions. The main thing to note is the 
fact that local properties of analytic functions of 
a complex variable usually go over to corresponding 
properties in the general case in an obvious and trivial 
fashion. For example, we have 



34 II 



PROPOSITION 4 . An analytic function f:S 1 - S 2 
which is not constant on any neighborhood is an open 
mapping . 

Proof : We must show that if Pn€S-, and u, is 
a neighborhood of p^, then f(U-i) contains a neighbor- 
hood of fCpj-j). We can assume cp-, : II, - W-, is a chart 
for S, and cp2 : U? " w ? a cnart f° r ^9 anc ^ f(U-i)cU 9 . 

Then ep2 c f°CD-j is a nonconstant holomorphic function 
on W-, and by the known property that a holomorphic 
function of a complex variable is open if not constant 
we see that cpoofocpn (W-, ) contains a neighborhood G 
of cp„of(p„). As cp„ is a homeomorphism, this implies 
f(u\) contains a neighborhood cp" (G) of f(p n ). 

OED 

also, global topological properties of Riemann 
surfaces can be combined with local properties of analytic 
functions in a decisive manner. 

PROPOSITION 5 . If S, is a connected Riemann 
surface and if 

f :S X - S 2 , g:S L - S 2 , 



II 3S 



are analytic functions such that f and g coincide 
on some set which has a limit point in S-. , then 



f = 



Proof : Let A = {peS-, : f and g coincide in a 
neighborhood of p}. Clearly, A is open by its very 
definition. Also, A t 0, for if f(p ) = g(p ) 
with p n - p Q (p n jt p ), then Pq^A; to see this let 
cp j _:U i - W i be charts for S ± , PgCUp f(p Q ) = g(p Q )€U 2 . 
Then cp2°f°cp^ and cp^ogocp-, are holomorphic in w\ 
and agree on a sequence in W, tending to cp-i (p n )<EW, , 
and thus by the known property for holomorphic functions 
of a complex variable, cp 2 °f°cp^ and cp2°g°cpT coin- 
cide in a neighborhood of cp 1 (p Q ). Thus, f and g 
coincide in a neighborhood of p~, and we see that 
Pq(=A. A similar proof shows that A is closed; 
just use the previous argument with p ft taking the 
role of a limit point of A. As S-. is connected, 
A = S, . 

QED 

PROPOSITION 6 . If S is a connected Riemann 
surface and if f:S - C is holomorphic , then Jf! has 
no relative maximum in S unless f is constant. 



36 II 



Proof : Suppose |f| has a relative maximum at 

P : ! f (p)| £ l f (P )| for P near Po- Then the 
maximum principle for holomorphic functions of a 
complex variable implies f is constant in a neigh- 
borhood of Pq. Proposition 5 implies f is constant 
on S. 

QED 

PROPOSITION 7 . If f is a holomorphic function 
on a Riemann surface minus a point , S - {Pq}j and 
if f is bounded in a neighborhood of p~, then f 
has a unique extension to a holomorphic function on 
S. 

Proof: Apply the usual theorem on removable 
singularities to show that if cp:U - W is a chart in 
a neighborhood of p^, then there is a holomorphic 
function g on W such that f°cp = g on 
W - (cp(Pfl)}. The extension of f near p« is then 
g°cp • 

QED 

PROPOSITION 8 . If S is a compact connected 
Riemann surface , the only holomorphic functions on 
S are constants . 

Proof : Suppose f:S - C is analytic. Since S 
is compact, the continuous function |f| assumes its 



II 37 



maximum at some point of S. Since S is connected, 
Proposition 6 implies f is constant. 

QED 
Now let us examine in some detail the local pro- 
perties of meromorphic functions. Let f be mero- 
morphic in a neighborhood of p^. in a Riemann surface 
S. If ;p : U - W is a chart in the complete analytic 
atlas for S and u is a neighborhood of p~, then 
a translation of the set W in C allows us to 
assume cp(p n ) = 0. Thus, fo^ is meromorphic in 
a neighborhood of in C. Thus, focp" has a 
Laurent expansion 



-Ik 
foco (z) - Z a w z , a N $ 0. 

k=N k W 



If i|cU-j -• W-, is another chart in the complete analytic 
atlas for S, Ui a neighborhood of p~, t(Pr)) = u > 
then cpo^, and its inverse are holomorphic and map 

to 0, and thus near w = 



<Poi|r "(w) = E c, w , c-. £ 0, 
k=l k I 

Therefore, 



38 II 



f o lj! (w) = focp o(po\|( (w) 



N N , 

■ a N c l w + 



where the additional terms involve higher powers of 
w. Therefore, 



-1 k 

fo-k (w) ■ £ b,w , b„ ^ 0. 

k=N K N 



Thus, the number N does not depend on the particular 
chart used, but depends only on the function f. It 
is called the divisor of f at p~ and is written 

n = a f (p ). 

There is another integer associated with f 
which is perhaps more important. Suppose f is not 
constant near p~. If the divisor N of f at p^ 
is negative, then the multiplicity of f at p« is 
said to be -N. Now suppose 3^(p ) £ 0. Then the 
multiplicity of f at p~ is the divisor of f - f(Pn) 
at Pq. Thus, we have for m^(p ), the multiplicity 
of f jat p , the formula 



II 39 



m f (p Q ) = -o f (P ) if 3 f(P(P < °» 

m f (p ) = s f _ f( p o )(p ) if a f (p > * °- 

Thus, rn^Cpj-.) is a positive integer which is completely 
determined by f. 

In terms of m^p^) we can obtain a simple 
representation for f by choosing an appropriate 
chart near p„. Thus, let m = m^(p n ) and consider 
two cases: 

9f(P(-)) s . In this case the Laurent expansion 
appears in the form 



foco _1 (z) = f(p n ) + a z m + . . . , a 7* 0. 
v ' VK m ' m 



Let a be one of the m — roots of a and note that 

m 



m . m+1 . m m/T , _ k k>. 

a z + a m , -, z +...=az(l + E — z ) . 

m m+1 v , -. a 

k=l m 

oo fl 

Let h(z) be the principal m — root of 1 + E — z 

k=l a m 
near z = 0, so that 



foq,- 1 (z) = f(p n ) + (azh(z)) m 



40 II 



Now define a new chart near p« by the equation 

Hp) " acp(p)h(cp(p)), p near p Q . 

Then f is a chart in the complete analytic atlas 
for S since the mapping z - azh(z) is a conformal 
equivalence near 0; and 

f° 1 lf~ 1 (w) = f°cp" 1 °cp il(~ 1 (w) 

-1/ N L / ,-1/ nxnUI 



- f(P ) + (acp i|f (w)h(cp°ilf" (w))) 
0' 

o- 



= f(P ) + UU~ l (»))) m 

= f ( Pn ) + w ra . 



df(Pf>) < 0. Now the Laurent expansion is 



j- -1/ \ -m . 1-m . , f, 

*°V (z) = a_ m z + ai _ m z + ..., a_ m t 0, 

a, 
-m,, 1-m , >. 

= a z (1 + z + . . . ) • 

-m v a 

-m 



In this case choose a such that a = a and h 

-m a l-m 

holomorphic near with h(0) = 1, h(z) = 1 + - — — z + 



a 
-m 



Then 



f°cp" (z) = (azh(z)) _tn 



II 41 



so a similar argument shows that there is a chart ^ 
at p n such that 



f o ijr (w) = w 



Summarizing, if m = m^Cp^), then there is a 
chart y in a neighborhood of p such that 



fo*" 1 (w) = f(p ) + w m if a f (p ) > o , 

fo^-i( w ) = w - m if a f (p ) < o. 



We note that it is easy to prove that 
m gof (p ) = m g (f(p ))m f (p ). 

DEFINITION 10 . Two Riemann surfaces S, and 
S^ are equivalent if there are analytic functions 

f:S l "* S 2 and §:S 2 "* S l sucn tnat f0 § = the identity 
on S2 and gof = the identity on S,. Thus, each 
mapping f and g is bijective, analytic, and h&s analytic 
inverse . 

It is routine to check that we have defined an 
equivalence relation. Note that equivalent Riemann 
surfaces are homeomorphic. The converse is not valid. 
We shall see that among the tori c/Q constructed in 



42 II 



6 on pp. 23-25 there are infinitely many nonequivalent 
Riemann surfaces. However, if a Riemann surface is 
homeomorphic to fc, then it is equivalent to t 
with its usual complete analytic atlas. This will 
be proved in Chapter VII. 

Here is perhaps the simplest example of two 
homeomorphic nonequivalent Riemann surfaces. Let 
S-, be C with the usual analytic atlas. Let 
A = {z:|z| < 1} and define a homeomorphism cp : C - A. 
E.g., 

cp(z) = 



A+|z| 2 



Let A have the usual analytic atlas and define S^ 
to be the Riemann surface induced on C by the homeo- 
morphism cp, as in 3 on p. 22 • In other words, 
S~ has an analytic atlas consisting of the single 
chart cp . Then S-, and S~ are not equivalent. For 
suppose f:S-, - S ? is analytic. Then by definition 
"o°f is a holomorphic function from C to A and 
is therefore constant by Liouville's theorem. Thus, 
f is constant. 



II 43 



We wish to consider a final feature of analytic 
functions. Suppose S and T are Riemann surfaces 
and that f:S -* T is analytic. If qcT we say that 
f takes the value q n times if f~ ({q^) = (p-i»-.-P, 1 
is finite and 



I 

I m f (p k ) = n 



k=l 



(thus we are counting "according to multiplicity"). 
If f ({q}) =0 we have n = 0. If this situation 
occurs, then there are charts :pi,:U, ~ W, in the 
complete analytic atlas for S with p, pU, and a 
chart cp:U - W in the complete analytic atlas for 

T such that the collection of sets fU, } is disioint 

-1 m f^ p k' 
and cp°f°cp (z) = z for zcW, . By diminishing 

k k 

the sizes of the U, (if necessary) we can also 

assume that each U. is contained in a compact set 

in S. Also, the explicit form for cpofo^" given 

above shows that there exists a neighborhood V of 

q such that the restriction of f to U\ takes 

each value in V exactly m.p(p, ) times. Therefore, 

I 
the restriction of f to U U, takes each value in 

k=l R 
V exactly n times. Using this background information, 



44 II 



we can prove 



PROPOSITION 9. Let S and T be Riemann surfaces 



and f:S - T an analytic function which is not constant 
on any neighborhood , 

1 . If S is compact and T is connected , then 

f takes every value in T the same number 
oft ime s . Also , it follows that T is compact . 

2 . If f takes every value in T the same (finite) 



number of times , then f is proper . I.e . , 

^-1 

f of any compact set is compact . In parti 

ular , if also T is compact , then S is compact . 



Proof : The rest of the proof is just topology. 
Assume the hypothesis of 1. Since f(S) is a 
continuous image of a compact set, f(S) is compact. 
Since T is Hausdorff, f(S) is closed. Proposition 
4 implies f(S) is open. Since T is connected, 
f(S) = T. (Thus, T is itself compact.) If f takes 
any value infinitely often, then since S is compact 
there is a limit point in S of the set where f 
takes this value, and Proposition 5 implies f is 
constant in a neighborhood of this limit point. Thus, 



II 45 



f takes every value q in T a finite number N(q) 
times. Now we use the argument just preceding this 
proposition with no change in notation. Since f 

does not take the value q on the compact set 

I 
S - U U, , there exists a neighborhood G of q such 

k=l K i 

that the compact and thus closed set f(S - u U, ) is 

k=l K 
disjoint from G. Thus, the restriction of f to 

I 

U U takes each value in VnG exactly N(q) 

times, and outside u U, f takes no value in 

k-1 k 
VnG. Thus, N(q') = N(q) for q'^VflG. Thus, the 

integer-valued function q -• N(q) is continuous on 

T. Since T is connected, N is constant and part 

1 ic proved. 

Now we prove 2. Let f take every value in 

T n times. If q^T, the analysis preceding the 

proposition again shows that f takes each value 

I 
in V exactly n times in U \J.. Therefore, by 

■ = k=l 
hypothesis f -1 (V) c U U v . Therefore, f _1 (V) is 

k=l k 
contained in a compact set in S since the U. 's 

are contained in compact sets in S. If FcT is 

m i 

compact, then Fcyv., where f _i (V.) is contained 

j=l J , J 

in a compact set in S. Thus, f" (F) is contained in 

a compact set S, and since f~ (F) is closed, it is 

compact. 

QED 



46 III 



Chapter III 
THE WE IERS TRASS CONCEPT OF A RIEMANN SURFACE 

In this chapter we shall consider the process of 
analytic continuation and obtain the Weierstrass defi- 
nition of analytic function. As we shall see, a concise 
language can be given to this process which brings us to 
construct a Riemann surface as a replacement for 
("multiple-valued") analytic function. 

The basic idea and the basic difficulty have been 
indicated on p . 1. Given a holomorphic or even a mero- 
morphic function defined on an open set, we want to extend 
it as far as possible and somehow take care of the multi- 
ple-valuedness that arises . 

What we shall basically consider is analytic con- 
tinuation along paths. First, we shall describe the clas- 
sical concept and then we shall fix everything up with 
lots of notation so that we end up with a Riemann sur- 
face and so that the problems of analytic continuation 
go over into statements about topological and other prop- 
erties of Riemann surfaces. Analytic continuation is 
classically considered in the following way: suppose f 
is meromorphic in a disk A and has a Laurent expansion 
about the center a of A converging in A: 



Ill 47 



f(z) = I a (z-a) k , z 6 A, 
k=-» K 



If b € A, it then makes sense to consider the Laurent 
expansion of f about b. This can be done in at least 
two ways ; we can write 

(z-a) k = (z-b + b-a) k 

- (b-*) k (l + B^) k 

n 



= (b-a) k E & &=& 



n=0 ' n (b-a) n 



by the binomial theorem, and then we insert this into 
the formula for f, rearrange terms (permissible if z is 
near b) , and thus obtain for b * a a Taylor series ex- 
pansion for f in a neighborhood of b. The other proce- 
dure would be simply to write 



°° f( n )rM n 
f(z) = S r , ^ (z-b) 11 , z near b 

n=0 n - 



If it so happens that the new series has radius of con- 
vergence larger than the distance from b to the boundary 
of A, we then have what is termed a direct or an immediate 
analytic continuation of f. Taking the new function in 
the new disk, we can again apply this process, etc. We 
can thus arrive at a "sequence A-. ,Ao,... of disks and cor- 
responding meromorphic functions f,,f ,-,,... such that 



48 III 

f , is meromorphic in A, , 

the center of A, belongs to A, , , 

f, is a direct analytic continuation of f^-i • 

Our first adjustment of this process will be to ignore 
a definite procedure for direct analytic continuation. 
Thus, instead of considering f, to be constructed from 
f by a definite process, we shall just require 
f, , = f, in A, , fl A, - Also, there is then no reason 
to require the center of A. to be in A, -. . 

DEFINITION 1 . Let y: [0,1] - C be a path. An 

analytic continuation along y 1S a collection of disks 

A-, ,A , . . . ,A and meromorphic functions f, , f „ , . . . ,*f 
i z n i z n 

such that 



f, is meromorphic in A, , 

f k-i H f k in Vi n A k' 

and such that there exist = t n <t,<...<t = 1 with 

1 n 



Y([t k-l 5t k ]) c A k 5 k = lj2 



,n. 



We clearly wish to consider all possible analytic 

continuations along paths , usually starting with a given 

meromorphic function f, in a given disk A-, . This will 

define a meromorphic function f in A , but the value 
r n n' 

f (y(l)) is not in general independent of y, so that we 
cannot in general define a meromorphic function in C 



Ill 



^9 



which extends f -, . For example, if 

A ] _ = {z: lz-lj < 1} 
and f, is the principal determination of z 2 in A, 



f (z) = (l+(z-l))^ = A(z-l)' 
1 n=0 n 



then analytic continuations along paths from 1 to -1 
definitely depend on the path. Consider the figure 




analytic continuation along 
Y-, yields a holomorphic func- 
tion near -1 whose value at 
-1 is i; but analytic con- 
tinuation along Yo yields 
the value -i. 



These statements are trivial to justify since we can 
write z = re 1 - with -rr < 6 < n (except on the negative 
real axis) and then z 2 = r 2 e . Along y-i 8 increases 
to n and along Yo B decreases to -rr , and thus the two 
different values at -1 result. 

Therefore, if we wish to define some meromorphic 
function which is a largest possible analytic continua- 
tion of f, , or which is derived from f-, by analytic 
continuation on all paths for which continuation is 



50 III 

possible, we shall have to have something other than C 
on which to define the extended function. So we now 
begin to introduce the Riemann surface on which these 
continuations will be defined. 

By the principle of the uniqueness of analytic con- 
tinuation (p. 1) , it suffices to know the original func- 
tion in an arbitrarily small open neighborhood. Such a 
"germ" of a function uniquely will determine the function 
everywhere. So we make the following definitions. 

DEFINITION 2 . Let a € C and suppose f and g are 
functions which are meromorphic in neighborhoods of a. 
Then f is equivalent to g, written f ~ g, if f = g in 
some neighborhood of a. 

Clearly this is an equivalence relation, and we 
then make the following 

DEFINITION 3 . Let a £ C. Then M a is the collection 
of equivalence classes of functions meromorphic in a 
neighborhood of a. Any element of M is called a germ 
of a meromorphic function . If f is a meromorphic function 
in a neighborhood of a, then [f! a is the germ to which f 
belongs. We say Tf] is the germ of f a_t a. 

Bv definition, [f]_ = f g 1 *» f = g near a. 



Ill 51 

DEFINITION 4. M = U M . We also define the 
=== a€C a 

obvious mapping tt: M - C by n([f] ) = a. 

Below we shall make M into a topological space in 
a natural way and then M will be called the sheaf of germs 
of meromorphic functions , and M the stalk over the 
point a. 

We shall define a topology on M by exhibiting a 
neighborhood basis for each point in M. Simple con- 
siderations then show that if we define a set in M to 
be open if it contains one of these special neighbor- 
hoods of each of its points , then the class of open sets 
in M forms a topology for which each point has the given 
neighborhood basis as a basis of open neighborhoods in 
this topology if the given neighborhood bases satisfy 
the following conditions: 

1 . any two neighborhoods of a point contain 

a third neighborhood of that point; 

2 . any neighborhood contains a neighborhood 

of each of its points . 
Furthermore, the topology of M is Hausdorf f if also 

3. any two distinct points of M are contained 
in disjoint neighborhoods. 



The topology on M . Suppose [f]„ € M. Then there 
exists a disk A centered at a such that f is meromorphic 



Ill 



on A. Define 



U(a,f,A) « {Cf] b : b€A}. J 



A neighborhood basis of [f!L is defined to be all sets 
U(a,f,A) such that f is meromorphic on A. Although 
the definition is quite simple, we have already in- 
corporated into it the notion of direct analytic con- 
tinuation, for the definition states that the germs "close" 
to [f] are just the germs of the function f itself at 
points close to a. Now we check the various require- 
ments for neighborhood bases: 

1. U(a,f,A 1 ) n U(a,f ,A 2 ) z>u(a,f,A 3 ) 

if a c A-, n Ao ; 

2. suppose [f], € U(a,f,A). Then let A' be 

a disk centered at b such that A ' c A , and 
note that 

U(b,f ,A') c U(a,f,A) ; 




3. now we check that M is Hausdorff . Suppose 
that Tf]„ and [g]_ t are distinct points in 
M. If a 4 a ' , then take L and A' to be 



Ill 53 

disjoint disks centered at a and a', 
respectively, and note that U(a,f,A) and 
U(a',g,A') are obviously disjoint. If 
a = a', then choose a disk A such that f 
and g are meroraorphic in A. Then U(a,f ,a) 
andU(a,g,A) are disjoint. For otherwise 
there would exist a point r f]^ = _gJw f° r 
some b c A, and thus f = g in a neighbor- 
hood of b. By the uniqueness of analytic 
continuation, f = g in A , contradicting 
[f] a - [gl a . 

Now M is a topological space. Note how much information 
is contained in the statement of the validity of the 
Hausdorff separation axiom--namely , this property re- 
flects the uniqueness of analytic continuation. 

M is a surface. The charts are almost obvious. 



Just use the mapping tt restricted to the various neigh- 
borhoods U(a,f,A). Suppose we call cp the restriction of 
tt to U(a ,f ,A) . Then 

cp: U(a,f,A) - A 

is given by cp([f],) = b, and cp" (b) = [f]v. Thus, cp is 
a bijection. Also, if U(b,f,A') cTJ(a,f,A), then clearly 

cp(U(b,f,A')) - o'. 



54 III 

Thus, rp induces a one-to-one correspondence between a 
neighborhood basis of [flL and a neighborhood basis of 
-u(rf], ) = b. Thus, cp is a homeomorphism. This proves 
that M is a surface. 

Moreover, we now have a nice atlas on M and we 
claim it is an analytic atlas and thus 

M is a Riemann surface . Suppose cp is the restriction 
of tt to U(a,f,A) and ■,) is the restriction of rr to 
U(b,g,A')« If z € A and ^p" 1 (z) £ U(a,f,A) n U(b,g,A'), 
then x~ (z) = [f] = [gl , and thus vlr (cp~ (z)) = z. Thus, 
where it is defined we have 

il/o cp = identity! 

The coordinate transition functions are thus trivially 
holomorphic and M is a Riemann surface. 

The mapping rr:-M - C is holomorphic. This really 
needs no checking at all, since n restricted to any neigh- 
borhood u(a,f,A) is a chart in the analytic atlas we have 
constructed, and such charts are always holomorphic 
(p. 31, no. 5). 



Problem 2 . Define V: M - C by the formula 



V([f] a ) = f(a). 



Prove that V is meromorphic . 



Ill 55 

Thus , we have two meromorphic functions n and V on 
M which are quite natural and simple functions to con- 
sider. We shall in the next chapter define an extension 
of M which is quite a bit more complicated, and again 
will be able to single cut two natural meromorphic func- 
tions , which we shall again designate n and V. In 
that context these functions will appear very much alike, 
although on M the function rr seems to be somewhat simpler 
than V. 

In terms of M we can give a characterization of 
analytic continuation along paths (see Definition 1) . 

PROPOSITION 1 . Let [f] fl € M be given , and let y 
be a path in C starting at a . A necessary and sufficient 
condition that there exists an analytic continuation 
along y with f, = f in a disk A, containing a (using 
the notation of Definition 1) is that there exists a 
path y in M such that 

TToy = Y, 

y(0) = [f] a . 

Proof : The necessity is quite clear. Using the 
notation of Definition 1 we define 



<*> = [f k ] v(t) ■ Vl * * * tj 



56 III 

Since y(t k _ ± ) £ ^.-^ and f fc-1 , f fc in ^_^\, we 

have [f k . 1 3 Y(tk _ i) = Cf k 3 Y (Vl>'. ThUS ' Y ^ Unam " 
biguously defined, and clearly n(y(t)) = y(t), 

y(0) = [fi],/ fi \ = Lfl a - The continuity of y is im- 
mediate from the definition of the topology of M and 
the continuity of y. 

The proof of sufficiency relies on a compactness 
argument. The continuity of y and the definition of the 
topology of M show that for each t € [0,1] there exists 
an open interval I (open relative to [0,1]) containing 
t and a meromorphic function f defined on a disk ^ 
centered at y(t) such that 

y(l t ) c U(y(t),f t ,A t ) = U t c M. 

The compactness result we need is that there exists 
<■: > such that any interval in [0,1] of length not 
exceeding z is contained in some one of the intervals 
I . The proof proceeds in the following manner. For 
each s € [0,1] there exists r(s) > such that 

r0,l]n(s-r(s) ,s+r(s)) c l g . 

As [0,1 J is compact, there exist points s,,...,s, such 

that 

k 
[0,1] c U (s.-%r(s.),s.+%r(s.)). 

T=l J J J J 

Let e = minfr(s.): lsj^k}. Then if x € [0,1] choose j 
such that |x-s . | < %r(s.). If |y-x] £ \ rZ } then 



Ill 57 



y-s. | £ |y-x| + |x-Sj| < %s+%r(Sj) < r( Sj ) 



so 



y € (s -r(s ),s +r(s )) c I . 

J j J J ° A 

Thus , 

[0,1] n [x-% e ,x+% e ] c I , 

j 

as required. 

Now choose points t«,...,t such that = t <t,<. . ,<t 
= 1, t,-t, , ^ e, 1 ^ k ^ n. By choice of e, the inter- 
val r t, i ,t, ] is contained in some set I constructed 

" k-1' k T k 

above. Thus, we are given a collection of disks A and 

T k 
meromorphic functions f on A , and we have to check 

T k T k 
that we have thereby obtained an analytic continuation 

along y- Since v(lt, _, ,t, ]) c y(I ) <= U , we obtain 

Y([t k . ls t k 3) = n(Y([ Vl ,t k ])) c nCU^) - A^ . 

If z € A HA , then the corresponding points in 

T k-1 T k 
U and U are [f ] and [f ] , respectively. 

T k-1 T k T k-1 z T k z 

In particular for z = y(t, _, ) we have 

k-1 k 

so that f = f in a neighborhood of y(t, -, ) , and by 

T k-1 T k • tc ~ i 

analytic continuation f s f in A P, A Finally, 

Ik-1 T k T k-1 T k 
it is clear that since y(0) = [f] , we have f = f in 



58 III 

A, f. A , so the analytic continuation along y which 
we have constructed begins with the given meromorphic 
function f in a neighborhood of a. 

QED 

DEFINITION 5 . If y and y are paths into C and M, 
respectively, such that y = tto y > then v is said to be a 
lifting of y . 

PROPOSITION 2 . "The uniqueness of analytic con- 
tinuation" (Topologically speaking, "The unique lifting 
theorem") . I_f y-i and y? are paths in M such that 

He y-, = no \„ , then either 

Y x (t) = Y 2 ( t ) for every t € [0,1] , 
or 

Y 1 (t) = Y 2 ( t ) for no t € [0,1] . 

Proof : Let A = [t€[0,l]: Y x (t) = Y 2 ( t )}- B Y con_ 
tinuity of y-, and v-, A is a closed set. It is also an 
open set, for consider any t~ € A. Let a = noy-(tn) and 

y.(t n ) = [f]„. Then f is meromorphic on a disk A centered 
l u a 

at a and a neighborhood U(a,f,A) of [f] a is defined. By 
continuity of y • , Y-(t) i- s contained in U(a,f,A) for t 
sufficiently near t^ and thus for those values of t 



viw - r f v; l(t ) 



Ill 59 

and since tto y-, - tt°Y 2 we obtain y-,(t) = Y 2 (t), or, t€A. 

Since A is both open and closed and [0,1] is con- 
nected, we have either A = [0,1] or A is empty. 

QED 

Thus , although in the sense of Definition 1 
analytic continuation is not a uniquely defined con- 
struction (since different choices could be made for 
the t, ' s and the disks A,), yet viewed as a lifting 
problem we do have a strong uniqueness statement. 
Moreover, we see now that the natural choice we have 
used in the proof of necessity of Proposition 1 was 
really forced upon us . There was no other way to 
choose y(t) . 

This discussion definitely does not imply that 
the unique continuation property along paths leads to 
a germ at the end point of the path which is uniquely 
determined by the end point. The discussion on p . 49 
makes this clear. In terms of the example of the 

square root mentioned there, observe that if f-, is the 

J. 

principal determination of z 2 near 1 and if y is the 

path y(t) = e , 0^t<L } then y can be lifted to a 
path y such that y(0) = [f-jL, but y(l) = [-£-,]•,• Thus, 
y(0) i v(l) , although y(0) = y(1). Another way of 
stating this is that rf-,]-, and [-f, ], are "far apart" 
in the topology of M, and yet both lie in M-, and can 
be connected by a path in M. 



60 III 

Now we begin to prove the famous "monodromy 
theorem," which essentially states that the phenomenon 
just discussed cannot occur on simply connected regions. 
A consequence will be the fact that on any simply con- 
nected region in C which does not contain the origin 
one can define a (single-valued) analytic determination 

1 /m 

of z ,l°g z > etc. First we introduce the notation 

I = [0,1] , 

I 2 = [0,l]x[0,l] . 

2 
LEMMA 1 . Let f: I - C be continuous , and let 

2 

P: I - M satisfy nor = 7. 

Assume that for each fixed u t I, r(t,u) is a continu- 
ous function of t; and also that r(0,u) is a continuous 
function of u. Then T is continuous . 

Proof : This follows in a purely topological man- 
ner from the unique lifting theorem (Proposition 2) and 
the description of lifting in terms of analytic continua- 
tion given in Proposition 1. Let € I. We shall then 
prove that there exists e > such that T is continuous 
on I x (In(a-.;,H;)), and the lemma will then be proved. 
For each fixed u define for t € I 

v u (t) = :(t,u) , 
v u (t) = r(t,u) . 

Then v and v are continuous and no v = v • Now we 
u u u u 



Ill 61 

apply Proposition 1, which guarantees the existence of a 

collection of disks A-.,..., A , meromorphic f, defined in 

A, , l^k^n, and points t, such that = t A <t-,< . . .<t = 1 
k' ' r k 1 n 

and 

Ys^Vl'^ C A k » 
f k-l s f k ° n A k-1 n A k > 
and 

7 3 <t) = Cf k ] YB ( t) , Vi^^k • 

the latter choice being forced as follows from the proof 
of Proposition 1 and the unique lifting theorem. Note 
that we have fixed 3 and applied Proposition 1 to the 
paths v a nd y„. 

Since r is uniformly continuous, there exists e-. > 
such that 

Y u^ t k-l' t k^ c A k if ' u ~ 6 ' < G l ' u€I 

(recall that Ai is an open disk). Also, since r(0,u) is 
a continuous function of u, there exists e„ > such that 

7(0, u) 6 U(a,f 1 ,a 1 ) if | u- 3 ' < e 2 ' u€I 

(a = center of A-,). Thus, if e = min(<r-| , s ? ) , 

v u (0) « [f^ (0) if |u-p| < , , u€I. 



62 III 

The unique lifting theorem now implies that 

Y U < t > = [f k 3 Y„(t) Vl it$t k |U " 9 I° ' U€I 
That is , 



ii 



r(t,u) = [f k ] r(tjU: 



> t. , ^t^t. , |u-3 |<€ , u6I, 



i) ' k-1 k ' 

But the continuity of I"(t,u) in the indicated range im- 
plies that of r(t,u) in the same range. 

QED 

DEFINITION 6 . If T is a topological space and 
Yp : I - T and y-i : I -» T are paths in T having the same 
end points , then Yq and y-i are homo topic with fixed end 
p oints if there exists a continuous 

2 
r: I - T 

such that 

r(t,o) = Y (t) , 

r(t,i) = v 1 (t) , 

r(o,u) = y (0) = Yl (0) , 

r(i,u) = y (i) = y x (i) • 

The function T is called a homotopy between Yn an< ^ y, . 
If there is possibility of confusion we will say v and 
Y-i are T-homotopic with fixed end points . 



Ill 63 

DEFINITION 7 . A connected topological space T is 
simply connected if each pair of paths Yn an d y-i i- n T 
having the same end points are homotopic with fixed end 
points . 

Now we can state various trivial consequences of 
Lemma 1 . 

2 
Covering Homotopy Theorem . Let F: I -* C be a 

homotopy in C and let p € M such that rr(p) = r(0,u) 

(O^u^l) , and suppose for each u € I the path t - ^(t,u) 

can be lifted to a path in M starting at p , say F(t ,u) , 

so that 

nor = r . 

Then r is a homotopy in M. 

Proof : For each u t I the function t -. r(t,u) is 
continuous, by hypothesis. And 7(0, u) = p is constant 
and thus a continuous function of u. Therefore, Lemma 1 
implies F is continuous. Finally, consider the two 
paths 

Y(u) = ?(l,u) , 
Y '(u) = f(l,0) . 

We then have 

Y(0) = y'(0) 



64 III 

and 

ttoy(u) = r(l,u) = r(l,0) = uoy'(u). 

Thus Proposition 2 implies y = y'- That is, r(l,u) 
= 7(1,0), and thus r is a homotopy. 

QED 

Monodromy Theorem . Let D be a simply connected 
region in C , a € D , and f a meromorphic function in a 
neighborhood of a . Assume that f has an analytic con - 
tinuation along every path in D which starts at a . Then 
there exists a meromorphic function F on D such that 
f = F in a neighborhood of a . 

Proof : The hypothesis means that for every path y 
in D such that y(0) = a, there exists a path y in M such 
that r,o y = y and y(0) = Lf] a . If Yq and Yn are paths 
in D from a to z , then y- and y-i ar e D-homotopic with 
fixed end points , and by the covering homotopy theorem 
the paths Yn an< 3 y-i a re homotopic with fixed end points; 
in particular, v n (l) = y-, (1) . Thus, we can define un- 
ambiguously 

F(z) = V(y(1)) , 



where y is a path in M such that y(0) = [f] a and no y is 
a path in D from a to z . Now we must check the properties 



Ill 65 

of F. First, suppose y(1) = l"g] , where g is holomorphic 
in a disk A centered at z. For w € A we use the path 
which goes from a to z along y and then from z to w 
along a line segment. The lifting from a to z is y 
and the lifting from z to w is just the germ of g at 
points on the segment from z to w. Since F is unambig- 
uously defined, F(w) = V([g] ) = g(w) . Thus, F is mero- 
morphic in A and this proves F is meromorphic in D. In 
particular, if z = a we can take g = f and we obtain 
F(w) = f(w) for w near a. 

QED 
One application of the monodromy theorem has al- 
ready been mentioned. Namely, on a simply connected 
region D c C - {0} , there exists a holomorphic deter- 
mination of log z. The only hypothesis which needs to 
be checked is that log z can be analytically continued 
along all paths in D. This can be verified in a simple 
manner, but we omit the proof now since a slightly dif- 
ferent version of the same result will be given in the 
discussion of algebraic functions in Chapter V. 

We next want to give an example pertinent to the 
monodromy theorem, but we shall first give a rather 
simple but important theorem on analytic continuation, 
the so-called "permanence of functional relations." 
This is a generalization of a familiar result on single- 
valued functions , an example of which is the fact that 
the identity sin 2z = 2 sinz cosz follows from its 



66 III 

validity for real z and the analyticity of all the 
functions involved. The theorem we shall give is 
really a generalization of usual theorems on unique 
analytic continuation because we are not here dealing 
with single-valued functions. Also, a more general 
theorem could be stated. 

Permanence of Functional Relations . Let A(z,w) 
be a holomorphic function for z in a region D c C and 
all w f C . Let y be a path in M such that y = no y is 
a path in D and each t t I yields y ( t ) = [ f t ] , t v , 
where f is a holomorphic function in a neighborhood 
of v(t). I_f A(z ,f„(z)) = in a neighborhood of y(0), 
then for each t <E I, A(z,f (z)) = in a neighborhood 
of Y (t). 

Remark . We have not given a definition for a 

function A to be holomorphic in two complex variables. 

One definition states that for any (z ,w ) in the do- 

J N o o 

main of definition of A , A has a power series expansion 

i k 
A(z,w) = i a ,(z-z ) J (w-w ) 

j ,k=0 JK ° ° 

converging absolutely for z near z and w near w . The 
° ° J o o 

important property we need is that if f is a holomorphic 
function of one variable near z , then A(z,f(z)) is holo- 
morphic for z near z . For the most important case we 
shall consider this is quite obvious; namely, the case 



Ill 67 

in which the function A(z,w) is a polynomial in w with 
coefficients holomorphic functions of z: 

A(z,w) = a Q (z)w n +. . .+a n _ 1 (z)w+a n (z) . 

Proof : Since f is holomorphic near y(t), the 
function A(z,f (z)) is holomorphic near y(t) and thus 
defines a germ at v(t) which we denote 

Y x (t) = [A(z,f t (z))] v(t) . 

Since «(t) = [f^] /. N , it follows that v n is a path in 
v ' <- t y(t) ' 1 

M (i.e., y-i i s continuous) and obviously n y-i = Y- 
Also define 

VZW = [ ° ] Y(t) • 

Then Yo is a path in M with ^=Yo = Y- By hypothesis, 
Y-i (0) = Yo(0). Thus, Proposition 2 implies y-, = Yo and 
this implies the result. 

QED 

Before giving the example, let us make one important 
observation about analytic continuation. This is the 
fact that if two germs at a point a are different, then 
they remain different under analytic continuation along 
any fixed path. This is another consequence of Proposi- 
tion 2, which in this" case would read that if 
Y x (l) = y 2 (1), then y x (0) = y 2 (0). Also, if [f\ & is a 



68 III 

germ at a and if f can be continued analytically along 
every path in a region D and if the continuation of f 
depends only on the terminal point of the path and not 
on the path itself, then there is a meromorphic F 
defined in D such that F = f near a. The proof of 
this is exactly like the proof of the monodromy the- 
orem was once we knew that analytic continuation did 
not depend on the path (see pp. 64-65). 

The monodromy theorem has of course two critical 
hypotheses. We have already indicated the reason for 
assuming D is simply connected, and now we shall ex- 
amine the other main hypothesis, that f has an analytic 
continuation along every path in D. Note especially 
that the hypothesis does not state that f can be con- 
tinued analytically to each point of D along some path 
in D. We shall now give an example to refute such a 
possibility for a weakening of the hypothesis of the 
theorem. 

This example will be the Riemann surface for the 
"inverse" of the function G(w) = w -3w, and the analytic 

continuation process will reduce to finding paths on 

2 
the surface. As G'(w) = 3w -3, the inverse function 

theorem of complex analysis will apply if w = 1 and 

w i -1. Since G(l) - -2 and G(-l) = 2, we conclude 

that if G(w ) = z f ±2 , then there exists a unique 

holomorphic function f in a neighborhood of z such 



Ill 69 

that G(f(z)) = z near z and f(z ) = w . But for each 

z ^ ±2 there are three distinct corresponding values 

of w and thus three distinct solutions f of G(f(z)) = z 

defined near z . We shall make this multiple-valued 
o r 

correspondence z - w into a single- valued function on 
an appropriate Riemann surface by the technique of the 
introduction, even though we no longer possess an 
explicit formula for w in terms of z. Thus, we select 
three copies of the z-plane cut along the real axis 
from 2 to °= and from -2 to »: 

Each of these slit planes is 
simply connected, so the mono- 
dromy theorem applies to show 

^ '_y k a that in each plane we can 

define a global solution f 
to the equation G(f(z)) = z 
and f is holomorphic in the 
slit plane. 
In order to accomplish the corresponding gluing we must 
see what happens to these functions at the slits. So 
we wish to examine carefully the values of w corresponding 
to real z such that 2<|z|<=°. To do this we introduce 
coordinates z = x+iy, w = u+iv and compute from 

3 
(u+iv) - 3 (u+iv) = x+iy . 

We find 



70 



III 



u - 3uv - 3u = x , 
3u v-v -3v = y . 



1) = 0. 



Along the slits we have y = 0, or 3v(u - -^ 

2 v 2 
Thus, v = or u - -«- = 1. This locus in the w-plane 

looks like the real axis and a hyperbola: 



x<-2 




x>2 f 



;X<-2 




3 
For v = we have x = u -3u. Thus, x>2 « u>2 and 

x<-2 « u<-2, as one easily sees by considering the 

graph of u -3u. For u - ->r- = 1 we have 

3 2 -3 

x = u -3u(3u -3)-3u = -8u +6u. Again, it is easily 

seen that x>2 <=> u<-l and x<-2 » u>l. 

Now we distinguish three regions in the w-plane: 

? 2 
A = {(u,v): u-^>l, u>0] - f(u,0): 2^u<»] , 

9 2 
■o r / \ 2 V n , 

B = i(u,v) : u -^ <1] , 



2 v 



C = {(u,v): u -^ >1, u<0] - {(u,0): -<^u^-2] 



Ill 



71 



Then one easily sees that the function G maps A ,B , and C 
each onto a copy of the z-plane, cut as described. Sup- 
pose we use three copies of the z-plane, labeled C» , C R , 
and C p . In order to see how these should be glued along 
the cuts, we just need to check the sign of y near the 
boundaries of A,B, and C in the w-plane. This is in- 
dicated in the figure. 




/ 



/ y>0 



y<0 



o / 2 v 
y = Jv(u 



T 



D 



w-plane 



Now we can easily indicate the method of gluing the 
planes C A ,C B ,C C : 



-2 2 



\ 



\ 



•2 2 



■2 2 



72 III 

Note in particular that the cuts from 2 to °° in C. and 
from -2 to m in C r can now be erased. This is the 
basic reason this example has been introduced . "Over" 
the point z = 2 lie two points of our Riemann surface, 
one a branch point, the other not. Likewise for z = -2. 

Now we have a function f defined on this Riemann 
surface which represents all the solutions of G(w) = z 
for any z. Now suppose we start at z = with the 
solution f of the equation G(f (z)) = z near z = 0, 
f (0) = 0, f holomorphic . Given any complex number 
a, there is some path y from to a along which f has 
an analytic continuation. If a * ±2, one can indeed 
go along any path from to a which does not pass through 
±2. If a = 2, use the path: 



Y 



Here is the reason. The start- 
ing point corresponds to z = 0, 
( \ "w = and thus to the origin in 

\ -2 "'*2 c r* I n order to get to the point 

^.^^"-^ 2 in C. (where this is not a 

branch point) , we pass through 
the cut joining C R to C . . 

Likewise, if a = -2, use the path 






V 



Ill 73 

But the conclusion of the monodromy theorem fails. 
Otherwise, by the permanence of functional relations 
there would exist a function F holomorphic in all of 
C such that G(F(z)) = z, z € C. It is rather clear 
that this cannot happen since by its very nature the 
relation z - w must be multiple valued. A direct 
proof would be this . Since F - 3F = z , we have 
F(z) - » as z - <=. Thus, F has a pole at °> and so 
the Laurent expansion of F at « shows that F(z) = az 
+ . . . (smaller powers of z) , where a 4 and n is a 
positive integer. But then F - 3F = a z + ... 
and there is no way this can behave like z near ». 

We shall return to this example in Chapter V, 
where algebraic functions in general are treated. But 
it should even be noted here that the branch point 2 
lying in C and C„ and the branch point -2 lying in 
C. and C R can be added to the surface in the way 
described in Chapter 1, and likewise * in C«,C R , and 
C r can be added, all three sheets being joined there. 
The resulting surface is a Riemann surface and the 
function f on it corresponding to the mapping z - w is 
meromorphic. Also, f is easily seen to be one-to-one 
since the inverse mapping w - z is single- valued. There- 
fore, since f is also onto, f is an analytic equivalence 

A A 

with f, so this Riemann surface is equivalent to C. 



74 III 



Theorem of Poincare and Volterra. Let S be a 



connected open subset of M. Then for any a € C the 
set 



{[f] a : [f] a € S] = S n TT -1 (a) 



is countable or finite 



Proof : Since S is connected we can consider some 
fixed [gJ-u £ S and then note that each element of 
S n T7 (a) can be connected to [g!L by a path y in S , 
by Proposition 1 of Chapter II. That is, if y = rr° y j 
then analytic continuation of g along v results in f , 
if [f]„ is the point we are considering. By Propo- 

3. 

sition I it follows that if y' is another path such 
that for a sufficiently small e>0 



lY'OO - Y (t)|< e , 0^1 , 

then [g] //q\ can be analytically continued along y ' and 
the resulting germ is [f] / ,-. x. We have again appealed 
to the unique lifting theorem and the argument used in 
the proof of Lemma 1. Now there is such a \ ' with 
initial point b and terminal point a, such that y' is 
a polygon with vertices (except for a and b) at rational 
complex numbers (i.e. , complex numbers whose real and 
imaginary parts are both rational) . Thus , S " n (a) 
consists of germs [f]_ which come from analytic con- 
tinuation from [g]v along paths which are polygons with 



Ill 75 

rational vertices . There are only countably many such 
paths so the theorem is proved. 

QED 

Of course, the example which is immediately sug- 
gested by this theorem is the Riemann surface for log z, 
which has countably many sheets. In the language of 
germs, we have over a point a 4 0,°=, the germs 
[log z + Zn^ri] , where log z represents an arbitrary 
determination of the logarithm near a , and n is any 
integer . 

DEFINITION 8 . Let f be a meromorphic function in a 
neighborhood of a point a € C. The Riemann surface ( in M) 
of f is the component of [f] in M. 

Here we have used a topological word "component," 
which by definition is a maximal connected set--a con- 
nected set contained in no strictly larger connected set. 
Since M is a surface, in this case the component containing 
[f] (the component of ^f] ) is the collection of germs 

a. ~ ~~ a. 

which can be joined to [f] by a path (in M) . 

For example, the Riemann surface of any determination 
of z near a point a ■£ consists of all germs [f]^ 
such that f(z) = z near b, b ± 0. By the permanence of 
functional relations "all the germs in this Riemann sur- 
face must satisfy this identity, and we thus need only 



76 III 

verify that any germ satisfying the identity can be 
joined to any other such germ. This can of course be 
easily checked directly, but an argument will be given 
in Chapter V for a general theorem along these lines. 

There is an obvious deficiency in the Riemann 
surface for z . Namely, the branch points and ro 
are missing. This situation is true in general for M-- 
it has been constructed without branch points (a phrase 
which we haven't even yet defined), and also it does not 
contain germs of functions meromorphic at °°. The latter 
is not a serious omission and indeed we could have con- 
sidered from the start germs of meromorphic functions 
on any fixed Riemann surface. But in the next chapter 
we shall construct a Riemann surface which contains M 
in a very precise sense and has all the branch points 
and also the germs at ». Then we shall give a satisfac- 
tory definition of the Riemann surface of a meromorphic 
function, replacing Definition 8. 



IV 77 



Chapter IV 

BRANCH POINTS AND ANALYTIC CONFIGURATIONS 

Before going to the definitions we give some 
motivating thoughts. The basic thing we want to do 
is give up the special role played by the independent 
variable. So consider [f] . This germ of course is 
determined by a meromorphic function f defined 
near a, the correspondence being written z -♦ f(z). 
We could also consider z as depending on a complex 
parameter t and write for example a + t - f(a + t) 
as the correspondence, where now t is near zero. 

But also we could write a + sin t - f(a + sin t), 

3t 3t 

or a + e - 1 - f(a + e - 1), etc. All these 

would be legitimate representations of f because 

the correspondence t - z indicated in each case 

is a conformal equivalence of a neighborhood of t = 

onto a neighborhood of z = a. Thus, in general we 

could consider a pair of meromorphic functions 

P(t) = a + p(t), 
Q(t) = f(a + p(t)), 

where t is a conformal equivalence of a neighborhood 
of onto a neighborhood of 0. Thus, each small 



78 IV 

parameter value t corresponds uniquely to a value 

of z (=P(t)) near a and the corresponding value 

Q(t) of f. We would not like to allow a representation 

of the form 



P(t) = a + t 2 , 
Q(t) = f(a + t 2 ), 



however. The reason is basically because two different 
values of t can give the same value of P. However, 
the thing that is really wrong here is that two dif- 
ferent values of t can give the same value both of 
P and of Q. This will be an important observation 
in our preparation for the definition. 

Now consider the Riemann surface in M for the 
function z ' m . This consists of germs [f] , a^O, 
such that f is some determination of z near a. 
So we have a representation 

P(t) = a + t, 

Q(t) = (a + t) (some determination), 



for t near 0. Suppose Q(0) = a so that a is 
one of the m — roots of a. Then 
a new parameter t by the equation 



one of the m — roots of a. Then we can introduce 



IV 79 



a + t = (a + T) m 



In fact, -t— (t=0) = ma ^ 0. Thus, we can also re- 
present [f] by the pair of functions 



P^t) - (a + r) m 
Qt(t) - a + t 



In our desire to obtain a representation near the 
branch point, we would like to use a pair P(t) = t, 
Q(t) = t . Of course, this is not allowed, but the 
answer to the dilemma is obtained by just formally 
setting a = in the above formulas to obtain the 
pair 



Pi<T) = ^ , 
Q X (t) = T . 



Note how useful such a pair is. We obtain all the 
values of z just by using the m different 
solutions of t = z. These yield the same value of 
P, (regarded as the independent variable) for the 
m different corresponding values of Q-, . Thus in a 
very real sense we have introduced a point corresponding 
to the branch point 0, and it fits in very well with 



80 IV 

the regular points near 0. Of course, we again 
would allow parameter changes as before, so that the 
pair 

P(t) = oCt)* 1 , 

Q(t) = P (t) 

is regarded as equivalent to the pair P,, Q-, if D 
is a conformal equivalence, with p(0) = 0. And as 
before we do not allow a pair such as 



P(t) = t 2m 
Q(t) = t 2 



because different values of t can yield rhe same values 
cor both P and Q. 

Finally, we exhibit pairs which we want to 
imagine as germs at ». If f is meromorphic in a 
neighborhood of <=o 5 then we use the parameter t near 
and let the independent variable be z = — . Thus 
we have 



P(t) = t" 1 , 
Q(t) = f(t _1 ) , 



defined for t near 0. More generally, we can also 



IV 81 



consider a as a branch point, yielding for the 
Riemann surface for z ' the pair of functions 



P(t) = t" m 
Q(t) = t" 1 



Now we are ready for the formal development. 

DEFINITION 1 . A parameter change is a function 
p holomorphic in a neighborhood of such that 

p(0) = , 
p'(0) jt . 

Equivalently, we could say that o(0) = and D 
is one-to-one in a neighborhood of 0. 

DEFINITION 2 . A pair is an ordered couple of 
functions P, Q meromorphic in a neighborhood of 
such that in a sufficiently small neighborhood of 

1. P is not constant, 

2. the mapping t - (P(t),Q(t)) is 

one-to-one . 

Examples of pairs have already been given. Here 



82 IV 

are some other examples. First, (sin t, sin t) is 
a pair, although the points t = and t = n give 
the same value to both P and Q. Second, (t m ,t n ) 
is a pair if and only if m t , and either n = 
and m = +1 or n ^ and m and n are relatively 
prime. The only thing which really needs checking 
here is that if m and n are relatively prime, then 
(t ,t ) is a pair. This follows because the Euclidean 
algorithm (see Chapter V) shows there exist integers 
m' and n' such that mm' + nn ' = 1. Now suppose 



f . m jn*. /.m . n% 
(t p t ± ) = (t 2 , t 2 ) 



t,, mm ^mm , nn ^_nn M , . . i 

Then t, = t ? and t-, = t 2 Multiplying, we 

obtain 



mm '+nn ' mm '+nn ' 
C l * C 2 



tl - t 2 



DEFINITION 3 . Let (P,Q) and (P 1 ,Q ] _) be pairs 
Then (P,Q) is equivalent to (P -, , Q-, ) if there exist! 
a parameter change p such that the equations 

P x = P°P , 

Q x - Qc p , 



IV 83 

are valid in a neighborhood of 0. If (P,Q) is 
equivalent to (P-pQ-i), this will be written 
(P,Q)~(P 1 ,Q 1 ). 

PROBLEM 3 . Suppose (P,Q) and (PpQ^ are 
pairs. Prove that if there exists a function p 
holomorphic in a neighborhood of such that 
p(0) = and ? ± = P° p , Q x = Qc P near 0, then p 
must be a parameter change. Also, p is uniquely 
determined (near 0) by these equations. 

LEMMA 1 . ~ is an equivalence relation . 

Proof : Reflexive : (P,Q) ~ (P,Q) since D (t) = t 

works . 
Symmetric : If (P,Q) „ (PpQp and p 
satisfies 



P x = Po 



Qi = Q°P , 



then also 



P = P 1 op~ 1 
Q = Q^o -1 



near and o~ is holomorphic, 



84 IV 

proving (P 1 ,Q 1 ) ~ (P,Q). 
Transitive : If (P,Q) ~ (PpQi) and 

(PpQ-,) ~ (PojQo) an d we have 
parameter changes p and p, 
satisfying P 1 = Po p , P 2 = P;l Pi> 
likewise for the Q's, then 

?2 = P ]/ °i = P ° P" Pi ' 
Q 2 - Q^Px - Q°p° Pl , 

and pc-p-i is also a parameter 
change, showing that 

(P,Q) . (P 2 ,Q 2 ). 

(2ED 

DEFINITION 4 . An equivalence class of pairs is 
a meromorphic element . The meromorphic element containing 
a pair (P,Q) is designated e(P,0). Thus, 

e(P,Q) = {(PpQ^rCPpQi) is a pair and 
(P,Q) „ (PpQ^j. 

Define M to be the collection of all meromorphic 
elements . 

DEFINITION 5 . The two functions n:M - £, 

V:M £, 



IV 85 

are given by the formulas 

TT(e(P,Q)) = P(0), 
V(e(P,Q)) = Q(0). 

We simply remark that tt and V are well defined 
since if (P,Q) „ (P ] _,Q 1 ), then clearly P(0) = P-,(0) 
and Q(0) = Q-j^ (0) . The number n(e(P,Q)) is sometimes 
called the center of e(P,Q), and V(e(P,Q)) is 
called the value of e(P,Q). 

Another remark which is simple but useful is that 
if (P,Q) is a pair and p is a parameter change, 
then (P°p,Q°p) is also a pair and is therefore 
equivalent to (P,Q). 

As has been indicated in the motivation for M, 
we definitely wish to consider M c M in a natural 
manner. Of course, the way we do this is to define a 
function on M with values in M and prove this 
function is one-to-one. This means that each element 
of M is identified with an element of M in a one- 
to-one fashion, and the identification is this: to 
a germ [f] we associate the meromorphic element 
e(a + t,f(a + t)). Now we prove this is a one-to- 
one function. Suppose [g!v is another germ and 
that e(a + t,f(a + t)) = e (b + t,g(b + t)). This 
means that there exists a parameter change p such 



86 IV 

that for t near 

a + t = b + (t) , 

f(a + t) = g(b + p(t)) . 

The first equation implies a - b and p(t) = t, and 
then the second equation implies f(a + t) = g(a + t) 
for t near 0. Thus, [f] = [g] fe . 

We now begin to topologize M, then make M a 
surface, then a Riemann surface. We remark that as 
sets the inclusion M c M is an isomorphism of M 
onto its image in M (this we have just proved), 
and we will eventually see that as Riemann surfaces 
this is still true: the mapping of M onto its image 
in M will be seen to be an analytic equivalence. 

Before beginning this program we wish to spell 
out a notational convenience. Frequently we shall 
write 

e(P,Q) = e(P(t),Q(t)) 

to designate a meromorphic element. We have already 
used this type of notation in the discussion of M c M, 
where we wrote e(a + t,f(a + t)). Of course, this 
means e(P,Q), where P(t) = a + t, Q(t) = f(a + t), 
but it would seem pedantic to be so strict with the 



IV 87 



notation and certainly would be confusing. We couldn't 

2 
even use notation such as e(t , t) . In order to 

attempt to be consistent we shall try to use t for 

the dummy variable in an expression such as the above. 

Thus, for example, 



e(P(t Q + t), Q(t o + t)) 



stands for the meromorphic element e(P-,,Q-,), where 
for small t 



P x (t) = P(t Q + t) 
Q 1 (t) = Q(t Q + t) 



DEFINITION 6 . Let (P,Q) be a pair and assume 
P and Q are both meromorphic on a disk A. Then 
let U(P,Q,A) be the collection of meromorphic elements 
according to the formula 



U(P,Q,A) = {e(P(t Q + t),Q(t Q + t)):t Q €A). 



We have assumed A sufficiently small that the mapping 
t - (P(t),Q(t)) is one-to-one on A (cf. Definition 2) 

Note that by this latter assumption each couple 
(P(t + t), Q(t Q + t>) for t £A is indeed a pair, 
and thus U(P,Q,a) makes sense. 



88 IV 

The sets U(3?,Q,A) will form a neighborhood basis 
of e(P,Q) when A is allowed to vary over all suffi- 
ciently small disks centered at 0. Since U(P,Q,A) 
is not defined in terms of the equivalence class 
e(P,Q) but rather in terms of the particular pair 
(P,Q)^e (P ,Q) , we shall need a lemma comparing two 
neighborhoods constructed with different but equivalent 
pairs . 

LEMMA 2 . Suppose (P,Q) ^ (P ]L ,Q 1 ). If U(P,Q,a) 
is defined , then there exists a disk A, centered at 
such that 

U(P 1 ,Q 1 ,A 1 ) c U(P,Q,A). 

Proof : By definition there exists a parameter 
change p such that P-, = P° p , Q-i = Q° o in a disk 
A-i centered at 0. We choose A-i sufficiently small 
that p' never vanishes in A-, and p(Ai)cA, and 
also that U (P,,Q,, A-, ) is defined. Now let 
e€U(P 1 ,Q 1 ,A 1 ). Then e = e(P 1 (t £) + t), Q 1 (t Q + t)) 
for some t . £A-.. Now 

P 1 (t Q + t) = P(p(t Q + t)) = P ( (t o ) + 0l (t)) 

where 



IV 89 

Pl (t) = p (t + t) - D (t ) . 

Note that p^CO) = and p-^O) = p '(t ) ± and thus 
p, is a parameter change. Since also 

i 

Q 1 (t Q + t) = Q(p(t Q ) + Pl (t)) , 

we conclude that (P x (t o + t),Q 1 (t Q + t)) „ (P(p(t Q ) + t), 
Q(p(t Q ) + t)). Thus, 



e = e(P( P (t o ) + t),Q(p(t Q ) + t)) 6 U(P,Q,£) 



and this proves the lemma. 

PROPOSITION 1 . The collection of sets U(P,Q,A)is 
a system of basic neighborhoods for a topology on M. 

Proof: Clearly any point e(P,Q) in M belongs 
to U(P,Q,zO, and just as on p. 52 we have two things 
to check: 

1. Suppose there are given U(P,Q,a) and 

U(P, ,Q, , £,), basic sets defined in terms 
of pairs (P,Q) ^ (P- L ,Q 1 ). By Lemma 2 
there exists- a disk A« centered at 
such that A 2 c=a 1 and U(P 1 ,Q 1 ,^ 2 ) cU(P,Q,A). 
Thus, 



90 IV 



U(P,Q,A) .? U(P 1 ,Q 1 ,A 1 ) 3 U(P 1 ,Q 1 ,i 2 ) 



2. Suppose e(P,Q)€ U(P,Q,A). Then for a point 
t Q €A, (P,Q) ~ (P(t Q + t),Q(t o + t)). If 
A ' is the disk centered at whose radius 
is the radius of A minus It I , then it 
is clear that 



U(P(t o + t) ,Q(t Q + t) ,A') c U(P,Q,A) 



By Lemma 2 there exists a disk A centered at 
such that U(P,Q,A) c U(P(t +t) ,Q(t +t) ,A ') , 
Thus , 

U(P,Q,Z) c U(P,Q,A). 

Before proving that M is a Hausdorff space, we 
introduce some normal representations for meromorphic 
elements. Suppose we consider an element e(P,Q). 
The discussion of pp. 37-41 defines the multiplicity 
m of P at and shows a particularly simple form 
P has in terms of a judiciously chosen chart for the 
Riemann surface (a neighborhood of in this case) . 
Thus, in the present framework we conclude that there 
exists a parameter change q such that near t = 

P(t(t)) - P(0) + t m if P(0) ^ CD , 

P(p(t)) = t" m if P(0) = « . 



Thus, if Q-, = Q ? , we see that 



IV 91 

e(P,Q) = e(P(0) + t m ,Q 1 ) if P(0) j- », 
e(P,Q) = e(t" m ,Q 1 ) if P(0) = -. 

Note that the integer m is well defined, being the 
multiplicity of P. For if P, is derived from P 
by means of any parameter change, then P, has the 
same multiplicity m. 

A point of M of the form e(a + t ,Q) or 
e(t ,Q) is called a branch point of order m-1. 

It should be remarked that the normal form is not 
unique if m>l. In fact, if uu is any root of id = 1, 
then for example 



(a + t ra ,Q(t)) „ (a + t m ,Q(wt)) 



as the parameter change t - ujt shows. Thus, e.g. 

2 2 
e(t ,t) = e(t ,-t) . This is the only possible type of 

ambiguity. 

PROPOSITION 2 . M is a Hausdorff space . 

Proof ; Compare pp. 52-53. The fact that M is 
Hausdorff is an obvious and immediate consequence of 
the uniqueness of analytic continuation. The present 
proof is surprisingly more involved. Suppose that 
e(P,Q) and e(P-,,Q-,) are not contained in disjoint 



92 IV 

neighborhoods. We can assume both these elements to 
be in normal representation , so that 



ti t x. \ i *.m x.~m 

P(t) = a + t or t 



and 



P^Ct) = b + t n or t" n 



Let A k be the disk centered at with radius k~ . 
Then for any sufficiently large k the neighborhoods 
U(PjQ,Av_) anc * U(P-.,Q-.,^ ) have a common point, say 

e(P(s k + t),Q(s k + t)) = e(P ] _(t k + t),Q 1 (t k + t)), 

where s u st k^k* "'" n P art i cu l ar n an( ^ V have the 
same value at these two points, so 

P ^"W- Q(s k ) = Q x (t k ) . 

If ever s k = t k = 0, then e(P,Q) - e(P ;L ,Q 1 ), which 
is what we're trying to prove. Thus, we can assume 
s, or t, ^ 0. Now letting k - « implies first 
P(0) = P^O), so we have either 



P(t) = a + t m and P^t) = a + t n , 



IV 93 



or 



P(t) = t" m and P^t) = t" n 



We shall eventually prove that m = n, so then P = P, , 
Also, we see immediately that in either case s, = tj\ 
Choose arbitrary n— roots of s, , say 



Then 



n 

a k = s k ■ 



m \ ~ ~ m 

c k J t? 



m 

a k 
Since there are only n choices for each number — . 

r k 

we can choose a subsequence of k's such that these 

numbers are all equal to a common n — root of 1, 

say uu . Renaming this subsequence, it follows that 

we can assume 



^k = fc k 



Then 



Q(aJ) - Q 1 (u.a") 



94 IV 

Since this equation is valid for o\ - 0, a-. ^ 0, we 
can now apply the uniqueness of analytic continuation 
to conclude 



Q(s n ) = Q 1 (-jus m ), s small . 



Thus, note that 



n N ~ , n 



(P(s ll ),Q(s u )) s (P 1 (u U s ul ),Q 1 (u)S m )), s small. 

Since the mapping t - (P(t),Q(t)) is one-to-one (small 
t), then the mapping s -(P(s ),Q(s )) is exactly , 
n-to-one (small s). Likewise, the mapping s - (P, (uus ), 
Qi (jus )) is exactly m- to-one. Since these mappings 
are identical, we must have m = n ! 

Thus, Q(s ) = Q-, (uus ) and we conclude 

Q(t) = Q 1 (uut). 
Since P(t) s P-. (uut) (as uu =1), we have 

(P,Q) „ (P 1 ,Q 1 ), 

the parameter change being just p(t) = uut. Thus, 

e(P,Q) = e(P 1 ,Q 1 ). 

QED 



IV 95 

Now we have certain obvious charts for M. Namely, 
we define the mapping 

cp:U(P,Q,A) - A 
by the formula 



co(e(P(t o + t),Q(t o + t))) = t Q 



or 



cp" 1 (t n ) = e(P(t + t),Q(t + t)) 



The definition of cp is of course clear enough, 

but for cp to be well defined something must be checked. 

Namely, if two points in U(P,Q,a) are the same, then 

they correspond to the same t . Another way of saying 

this is that cp~ is one-to-one. But if cp~ (t ) = rp~ (t'), 

then Tr(<p" 1 (t )) = rrCcp" 1 ^)) and V^" 1 ^)) = Vfo -1 ^) ) , 

so that P(t Q ) = P(t^) and Q(t Q ) = Q(t^). Since the 

mapping t - (P(t),Q(t)) is one-to-one for tfA, this 

implies t = t ' . This shows that at least cp maps 

U(PjQ>A) to A in a bijective fashion (one-to-one 

and onto). 

It is now easy to see that cp is a homeomorphism. 
In fact, if e Q = e(P(t Q + t),Q(t Q + t)) is any point 
in U(P,Q,A), then a neighborhood basis for e consists 



96 IV 

of the sets U(P(t + t),Q(t + t),A') = [e(P(t + t-, + t) 

Q(t + t-^ + t)):t,€A'}, where A' is a sufficiently 
small disk centered at 0. The image of a set like 
this under the mapping cp is precisely {t + t-, : t-|£A'}, 
and these sets form a neighborhood basis for the point 
t -A. Thus, x induces a one-to-one correspondence 
between a neighborhood basis for e and a neighbor- 
hood basis for cp (e ). Thus, cp is a homeomorphism. 
Thus, M is a surface. 



PROPOSITION 3 . The given charts form an analytic 
atlas for M . Thus , M is a Riemann surface . 

Proof : Suppose two neighborhoods U(P,Q,a) and 
U(P-i ,Qi , A-i ) meet. Let cp and cpi be the respective 
charts. Let e be a common point in these neighbor- 



hoods and cp (e ) = t , cpi(e )= t-,. We need to check 
the analyticity of cp-i°cp in a neighborhood of t . 

Now by definition 

e Q = e(P(t Q + t),Q(t o + t)) = e(P 1 (t 1 + t),Q 1 (t ] _ + t)). 

Therefore there exists a parameter change _ such that 
for t near 

P(t Q + t) = P ] _(t 1 + p(t)) , 

Q(t Q + t) = Q 1 (t 1 + p(t)). 



IV 97 



For z near t we have 



cp _1 (z) = e(P(z + t),Q(z + t)) 



We need tc express this in terms of P, and Q, , rather 
than P and Q. So we compute as follows: 

P(z + t) = P(t Q + (z-t Q +t)) = P 1 (t 1 + p(z-t Q +t)) 

= P 1 (t 1 + p(z-t Q ) + [p(z-t +t) - o(z-t o )l). 

The function 

Pl (t) = p(z-t Q +t) - p(z-t ) 

satisfies p-i(O) = and p-f(O) = p'(z - t )„ Thus, 
p^(0) ^'0 if z - t is sufficiently small, so p, 
is also a parameter change. Since the same computation 
is valid for Q and Q, , we obtain 



cp _1 (z) = e(P 1 (t 1 + p(z - t Q ) + t),Q 1 (t 1 + D (z - t Q ) + t)) 



Therefore, 



cp^cp" (z) = t 1 + p (z - t Q ), 



which holds for all z sufficiently near t . Since 



98 IV 

P is holomorphic, we have now proved that cpi°cp 
is holomorphic near t . 

QED 
Now we list various properties of M. 

PROPOSITION 4. M - M is a discrete set. That 



is , each point of M has a neighborhood consisting 
only of itself and points of M. 

Proof : Suppose e(P,Q)eM and that A is a 
disk centered at such that U(P,Q,A) is defined 
and for tea - [0], P' (t) ? 0, P(t) ^ ». Then 
U(P,Q,A) is a neighborhood of e(P,Q) having the 
required properties. Indeed, if e €U(P,Q,a) but 
e £ e(P,Q), then for some t gA - {0} 



e Q = e(P(t Q + t),Q(t Q + t)). 



Now t - P(t + t) is holomorphic and one-to-one near 
t = 0, so 

p(t) = P(tL + t) - P(t n ) 



is a coordinate change. Thus, if a = P(t ) 



e Q = e(P(t Q ) + t,Q(t Q + p^Ct))) 



IV 99 



= CQ(t Q + P -1 (z - a)l a , 



that is, e is the germ of the meromorphic function 
z - Q(t + p" (z - a)) at a. Thus, e o ?M. 

QED 

PROBLEM 4 . Prove that tt and V are meromorphic 
functions on M. Prove that the mapping which identifies 
M as a subset of M is an analytic equivalence of 
M onto its image in M. 

The second half of this problem completely 
justifies regarding M as a subset of M. Of course, 
we previously could only consider M c M as sets, but 
now also as Riemann surfaces. Also, M is open in 
M, as is implied by Proposition 4. 

PROPOSITION 5 . _If efE, then e is a branch 
point of order m - 1 if and only if m (e) = m 
(definition on p. 38). 



Proof : Suppose e = e(a + t ,Q), and cp: U(a+t ,Q,A) - A 
is a related chart. Then 



rrocp (t ) = a + t , 
v . o o ' 



so that the multiplicity of n°cp~ at is m. A 



100 IV 



similar computation applies if e = e(t~ m ,Q) 



QED 



PROPOSITION 6 . Any two points in the same component 
of M can be joined by a path in M every point of 
which except the initial and terminal points lies in 
M. 

P roof : This is a topological consequence of 
Proposition 4. Suppose y is a path. Since y(I) 
is compact and M - M is discrete and closed, 
y(I)n(M - M) is finite. Let e be a point in this 
set which is not an initial or terminal point of y. 
Let tn be the smallest and t-, the largest numbers 
t in (0.1) such that y(t) = e . Choose a neighbor- 
hood u of e and a chart cp:U - A, where AeC 



is a disk 



/>-,' 



e 
o 



Choose < ti < t Q «£ t-^ < ti < 1 
such that v(t£)cu, y(tpfU. 

Choose a path 6 in a joining 
cp°Y(t(}) anc * cp°y(tj) and missing 
tp(e ). Then let 






Y (t), Ostst^, 

! -l ft-ta \ 
Yl (t) =V p ^Ep^j' *&****{> 

y '.y(t), t^stsl. 



IV 101 

Then yi i- s a path in M having the same end points 
as y> but yi does not pass through e . Since we 
need to remove only finitely many points like e , the 
theorem follows. 

QED 
PROPOSITION 7. There is a natural one-to-one 



correspondence between components of M and components 
of M. Namely , if S is a component of M, S is 
contained in a unique component of M, which is the 
closure of S _in M; conversely , if S is a component 
of M, then S contains a unique component of M, which 
is SnM. 

Proof: Let S be a component of M. Certainly 
S is contained in a unique component S of M. 
Since components are closed, S contains the closure 
of S. But also any eeS can be joined to a fixed 
e Q gS by a path y such that y(0) = e Q , y(1) = e, 
y([0,1))cM, b y Proposition 6. Since y([0,1)) is 
connected and y(0)pS, it follows that v([0,l))cS 
(since S is a component). Thus, e is a limit 
point of S and thus belongs to the closure of S, 
showing S is contained in the closure of S. 

Conversely, let. S be a component of M. If a 
component of M is contained in S, this component 
is also contained in Sp,M. Thus, it suffices to show 



102 IV 

SnM is a component of M, Proposition 6 shows that 
SfiM is connected, for if e and e are in SfiM, 
they can be joined by a path in M. Since the end 
points are in S, the entire path is in S (S is 
a component ) and thus is in SnM. And if a point of 
M can be joined by a path in M to a point in SnM, 
that point must be in S and thus in SnM. Thus, 
SrM is a component. 

QED 

DEFINITION 7. A component of M is called an 
analytic configuration . This is a translation of the 
term "analytische Gebilde" used by Weyl. Another 
term is analytic entity . 

DEFINITION 8 . Let f be a meromorphic function 
in a neighborhood of a point a<=C. The Riemann sur - 
face of f is the analytic configuration containing 

rf] • 

J a 

This definition is finally the complete idea which 

was begun in Definition 8 of Chapter III. We have now 

included the branch points in the surface and nothing 

else needs to be added. 

It is important to observe that the nice analytic 

continuation or lifting properties of M do not hold 

in M. For example, Proposition 2, the unique lifting 



IV 103 

theorem, of Chapter III would be false if phrased for 
M. Just consider a neighborhood of a branch point 
to see this. For example, let 



V-.(s) = e((s+t) 2 , s+t), -lsssl 



2 
Y 2 (s) = e(|s|+t) , |s|+t), -lsssl 



2 
Then tt°Yi( s ) = n Yo( s ) = s > but 



Y 1 (s) = y 2 ( s ) for 0ss<l , 
Yi(s) jL Yo( s ) for -lss<0 

We thus are led to a pictoral idea of branch point: 
two liftings of a given path in C which begin at a 
common point in M must be the same until a branch 
point is reached. But then the liftings can branch 
into several different paths in M. 

If f is a meromorphic function which is not 
one-to-one, it of course has no inverse. But we can 
easily consider the Riemann surface inverse to its 
Riemann surface, as follows. 

PROPOSITION 8 . -Let S be an open connected sub - 
set of M such that V is not constant on S. Then 
the mapping 



104 IV 

i:S - M (i for "inverse") 

defined by 

i(e(P,Q)) = e(Q,P) 

is an analytic equivalence of S with i(S). 

Proof : First, note that i is well defined. 
This depends on the obvious fact that if (P,Q) ^ (P, ,Q-, ) 
then (Q,P) ~ (Q-. ,P-i). Now we prove i is analytic. To 
do this we introduce charts in the canonical way: 

cp:U(P,Q,A) - li , 
♦ :U(Q,P,A) - & , 

where 



cp" 1 (t ) = e(P(t +t),Q(t +t)), 
f l (t Q ) = e(Q(t o +t),P(t o +t)). 



Then 



i^cp" 1 (t o ) = v(e(Q(t o +t),P(t o +t))) = t Q , 



so that trivially i|/°i°cp" is analytic. Thus, i is 



IV 105 

analytic . 

Since i is analytic, i(S) is an open connected 
subset of M, and n is not constant on i(S) since 

V is not constant on S. Furthermore, i:i(S) -. S 
is analytic by what we have already proved and 

is i = identity. Thus, the inverse of i is analytic. 

QJED 

We now give an interesting and rather surprising 
application of some of these ideas. 

DEFINITION 9 . Let f be meromorphic on an 
open set Dec and let w?u. Then w is an asymptotic 
value of f if there exists a path of the form 

y:[0,a) -D (where 0<as») 

such that 



lim f(y(s)) = w 
s-o 



and y - SD, meaning that for every compact set K c D , 
there exists s such that y(( s o>°)) c D_K - 

THEOREM 1 . If f is holomorphic on an open set 
Dec, then there exists an asymptotic value of f. 



106 



IV 



Proof : Clearly it suffices to treat the case 
in which D is connected and f is not constant on 
D. Now define 



S = {Tf1 a :acD} 



Clearly, S is an open subset of the sheaf of germs 

M, and the mapping rr.S - D is an analytic equivalence 

Now we complicate the situation by regarding ScM 

and letting T = i(S) in the sense of Proposition 8. 

Then V:T - D is an analytic equivalence since 

V = tto i . According to the definition of McM, we 

have 

T = (e(f(a+t),a + t):a<=D} . 

Thus, foV = tt on T. That is, we have a commutative 
diagram 



X 



A 



IV 107 

and V represents the multiple-valued inverse of f. 
Note that a point e(f(a+t),a + t) is a branch point 
(order at least 1) if and only if f'(a) = 0, and this 
holds only for a discrete and thus countable set of 
points a€D. Let E = {a€D:f'(a) = 0} and note 
that f(E) is a countable subset of f. Choose arbi- 
trarily a £D - E. Since f(E) is countable and there 
are uncountably many rays from f(a ) to a> t it 
follows that there exists a ray from f(a ) to ~ 
which contains no point of f(E). Let this ray be 
represented by a path a:[0,») - C, so that a(0) = f(a ), 
a(s)(^f(E), lira a(s) = ». 

Now we consider the process of lifting a to 
T by the mapping n. Note that if e<=T and n(e)^f(E), 
then V(e)<£E and thus e^M. Thus, lifting a is a 
problem of lifting to M, not merely M, and the 
unique lifting theorem obtains. Let s be the 
supremum of all numbers s, such that there exists a 
lifting a on [O.s,) with a(0) = e(f(a +t), a + t). 
Then there is a unique path d corresponding to the 

maximal s , 
o ' 



a:[0,s o ) - T 



such that ttocl = a and a(0) = e(f(a +t), a + t), 
Then define y = Voa } so that y I s a path in D 



108 IV 



such that 



lim f(y(s)) = lim f°V°a(s) = lim rroa(s) 
s-s„ s-s_ s-s„ 



= lim a(s) - a(s Q ) . 
s-s 



Here ct(s Q ) = « if s = » . Thus, the theorem follows 

if we Know that y leaves every compact set in D. 

If s Q = » this is perfectly clear since lim. f( Y (s)) = ». 

S-»« 

Suppose s Q < co, and suppose that for some compact KcD , 

y does not eventually leave K. By the Bolzano- Weierstrass 

theorem, there exists a point z Q c K- and a sequence 

s, < Sy < ..., s - s , such that y(s ) - z . Since 

V:T • D is a homeomorphism , a(s ) - V (z Q ) = e(f(z +t),z +t) 

Since tt is continuous, f( z Q ) = li- m n°a(s ) = lim a(s n ) 

= a(s o )^f(E), so e Q = e(f(z Q + t),z Q + t)*TnM. This 

contradicts the maximality of s , for the topology 

of M implies that a neighborhood U of e is homeo- 

morphic by a homeomorphism cp to a disk A centered 

at f( z ) an d cp is just the restriction of tt to U . 

Therefore, if n is chosen such that a (s )ca, . then 

for sufficiently small e > we can define 



a(s) = cp~ oa(s), s n s s < s Q + e 



and we obtain a lifting of the required sort past the 



IV 109 

supposedly maximal s . This contradiction shows that 
leaves every compact set in D. 

QM 

A comprehensive reference to questions of this 
sort can be found in MacLane, G. R., "Asymptotic values 
of holomorphic functions ," Rice University Studies 
49 (No. 1) 1963, pp. 1-83. The example we have just 
treated can be found on page 7 of MacLane 's monograph. 



110 



Chapter V 
ALGEBRAIC FUNCTIONS 

What we are going to study in this section is 
solutions of an algebraic equation in two complex 
variables; i.e., equations of the form 

A(z,w) = 0, 

where A is a polynomial in z and w. The viewpoint 
is that we want to regard w as a function of z satis- 
fying A(z,w(z)) =0. Of course, we expect w to be 
multiple-valued and then we construct a Riemann surface 
on which a function like w can be defined. Examples 
of this procedure were given in the introduction. There 
we treated the following examples of A: 

m 
w - z, 

2 
w - (z-a) (z-b) , 

(z-b)w - (z-a) , 

w - (z-ap (z-a 2 ) . . . (z-a m ) ; 

also on pp. 68 f f. we discussed the polynomial 

w - 3w - z . 

All the Riemann surfaces associated with these examples 
can be easily visualized as subsets of M and as such enjoy 
the topological property of compactness . The main fact 



Ill 



to come out of this section is that algebraic equations 
always lead to compact surfaces and that, conversely, 
every compact analytic configuration has a unique alge- 
braic equation associated with it. 

It follows from general topological considerations 
that every compact orientable surface (as Riemann surfaces 
are) is homeomorphic to a sphere with a certain number 
g of handles and g is called the genus of the surface; 

cf. p. 13. Before analyzing algebraic equations, 
we shall discuss heuristically a remarkable formula 

involving the genus, the number of sheets, and the 

branching of a compact Riemann surface. 

The Riemann-Hurwitz formula . Consider a compact 
analytic configuration S. We first discuss its Euler 
characteristic . This can be defined in terms of a 
"triangulation" of S. We do not wish to pause to 
define triangulation, but if f is the number of tri- 
angles (faces), e the number of edges, and v the number 
of vertices, then the Euler charactersitic is v-e+f . 
A theorem of topology is that this number is a topo- 
logical invariant of the surface and equals 2-2g: 

v-e+f = 2-2g. 

Now S has certain branch points e,,...,e of orders 
b-,,...,b , respectively, b. s 1. Define 



112 



V = lb. 
J-l J 



The number V is called the ramification index or 
total branching order of S. Also S has a certain 
number n of sheets when viewed as spread over C; 

this is the number such that n takes every value 

a 
in C n times ... see pp. 43-44. The Riemann-Hurwitz 

formula is 



n 



v = „ + 



i. 



To prove this formula consider a triangulation of 
the sphere C such that every point rr(e.) is a vertex. 
Let f,e, and v be the number of faces, edges, and 
vertices. Since C has genus 0, we have the Euler 
formula 

v-e+f = 2. 



Now consider the preimage by tt of these triangles. 
By lifting the triangulation of C to S we obtain nf 
faces and ne edges in the triangulation of S, since 
S has n sheets. And each vertex which is not a rr(e.) 
is lifted to n new vertices. But each vertex rr(e.) 
does not get lifted to n new vertices. Rather, if 
z Q is one of these values, then rr (i z }) consists of 
exactly 



113 



n - I b. 

distinct points. Thus, the number of vertices in the 
triangulation of S is 

nv - V. 

Therefore, 

(nv-V) - ne + nf = 2-2g. 

Since v-e+f = 2 we can write this relation as 

In - V = 2-2g, 
and the assertion is proved. 

Let us test this formula on some of the cases we 
have considered. For example, on d. 12 we treated 



w - (z-a, ) . . .(z-a), 



a,,..., a distinct. If m is even there are branch 
1 m 

points of order 1 at each a. and nowhere else, so 
V = m and thus 



m -=n + g-l = 2 + g- 1, 
m-2 



If m is odd then also. «> is a branch point of order 1 

rn t 

and so V = m+1 and g = — ^— . These results agree with 
p. 13. 



114 



3 
Next, consider the example w -3w-z discussed on 

pp. 68 - 73. There the points 2 and -2 are 

branch points of order 1 and a> is a branch point of 

order 2, so V = 4. Since n = 3, we find g = and 

the Riemann surface is homeoraorphic to a sphere. This 

again agrees with our earlier findings, for on p. 73 

we discussed an analytic equivalence of the surface 

. , A 

with C. 

Of course, we have not rigorously derived the 
formula, but we have given a sketch of a rigorous 
proof. But the formula should prove useful as a 
check in working out other examples. Every time one 
sees a compact Riemann surface, he should try out 
this formula on it. Two things in the formula deserve 
special attention. One is that V is always an even 
integer. The other is that a purely topological number 
g is equal to the number -j - n + 1 which depends very 
much on features of the surface which are not purely 
topological . 

Now we proceed to the analytical discussion of 
algebraic equations. 

Problem 5 . In the spirit of pp. 68-73, 
discuss the algebraic equation (w -1) - z = 0. 

Lemma 1 . Let a, , . . . ,a be holomorphic on an open 

set D c £ and A(z,w) =? w n + a, (z)w n ~ +. . .+a , (z)w +a (z) 
1 n-1 n v J 



115 



Suppose z € D and 



A(zo,w ) = 0, 

"A, 



dw 



(Zo^o> * ° 



Then there exists a function f holomorphic in a neighbor- 
hood of z c such that 

A ( z , f ( z ) ) =0, z near z , 
f(z ) - w 0J 

A(z,w) = 0, z near z Q , w_near_w Q => w = f (z) . 

Proof : This is merely an implicit function theorem 
and could be derived from the general implicit function 
theorem of differential calculus - we would just have 
to check the validity of the Cauchy-Riemann equation. 
However, the proof is much simpler in the present case 
than the proof of the general theorem and is even almost 
elegant, so we present it. 

Since A is not constant in w, the zeros of A(z ,w) 
are isolated. Thus, there exists e > such that 
A(z ,w) ^ for < |w-w | < e. Let y be the path 
Y(t) =w +ee TTl ,0<;t5l. Since the image of y 
Y is compact and A(z ,w) ^ 

there, there exists 6 > such 
that 



w 

o 



A(z,w) + for |z-zj <{ 



w-w 
o 



116 



Therefore, the residue theorem implies that for each 
fixed z, |z-z j < 6, 



SW dw 



Zni y A(z,w) 

is equal to the number of zercs of A(z,w) (minus the 
number of poles of A(z,w)) for |w-w | < e- And it is 
clear that this number is a continuous function of 
z for jz-z | < 6, and is therefore constant. For 
z = z Q we are counting the number of zeros of A(z ,w) 
in |w-w Q ]< e. Since A(z Q ,w) -• only at w = w Q and 
since w is a first order zero (- — (z ,w ) ^ 0), we have 
proved that the above integral is equal to 1 for |z-z |<c 
Thus, j z-z j < 6 implies there exists a unique f(z) 
such that |f(z)-w j < e and A(z,f(z)) = 0. Again, the 
residue theorem implies 

i lr< 2 >"> 

f(2) = si' w T77^T du ' |z " z ° ! <6 ' 

Y A(.z,w; 

From this formula it follows immediately that f is holo- 
morphic. Of course, f(z ) = w_. 

To prove uniqueness, suppose g is holomorphic near 
z Q and A(z,g(z)) = 0, g(z ) = w_ . Then by continuity of 
g it follows that there exists < 6, <. 5 such that for 
lz-% | < 6 1 , |g(z)-w | < e. Therefore, g(z) = f(z) for 



117 



|M e l < H- 

QED 

COROLLARY . Suppose z € D and chat there exists 
no w satisfying 

A(z OJ w) = 0, 

Then there exist unique holomorphic functions f -,,..., f 
in a neighborhood of z such that 



A(z.,f, (z)) = near z , 1 s k < n , 

for each z near Zq , the numbers f , (z) , . . . 

f (z) are distinct, 
n 



Proof ; Since A(Zq,w) is a polynomial in w of 
degree n, it has n zeros. By hypothesis these zeros 
are distinct, say A(z ,w,) =0, 1 <. k s n, w, , . . . ,w dis- 
tinct. Apply Lemma 1 to w = w, to obtain the holomorphic 
solutions f, . Since f, (z ),..., f (z Q ) are distinct, it 
follows by continuity that for z sufficiently near z , 
f, (z), . . . ,f (z) are distinct. 

QED 

LEMMA 2 . Let A be defined as in Lemma 1 . Assume 
that for every z € D there exists no w satisfying 



118 



A(z,w) - 0, 

$w v ' J 



Let f be a holomorphic function in a neighborhood of 
z c e D satisfying 



A(z,f(z)) s near z 0> 



Then f can be analytically continued along any path in 



D starting at z Q . 

Proof : Let y. [0,1] - D be a path with Y (0) = z - 

We are trying to prove the existence of a path v: [0,1]->m 

such that y(0) = [f] and n° > = \ • By the general 

z o 

discussion of analytic continuation we know that -, exists 
on some interval [0,t o ]j t > 0, and that , is uniquely 
determined (Proposition 2 of Chapter III) . Let s be the 
supremem of such t Q . Then y exists on the interval [0,s ) 
Now we apply the above corollary to the point v (s ). 
obtaining holomorphic functions f -.,...,£ in a neighbor- 
hood of v(s ) satisfying the conclusion of the corollary 

on a disk A centered at v(s )• Choose any s, < s such 

p J 1 o 

that y( s i) 6 A. Then y( s t) = 
^X [g] , *., where g is holomorphic 

- ^v( Sl ) \ vCs x ; 

""-» )~ > ~ in a neighborhood of y( s -i) an & 

v(s ) / Y 

° / by the permanence of functional 

L relations (p. 66) 



119 



A(z,g(z)) s near y(s x ). 

Thus, g(z) is one of the n zeros of the polynomial A(z,w) 
and must therefore be equal to one of the f, (z). Thus, 
by Lemma 1 and its corollary we find that for a unique k, 
g(z) = f, (z), z near y(s-i ) • By the uniqueness of analytic 
continuation , 

Y (s) = [f k ] , s 1 s s < Sq . 
v(s) 

This formula serves to define y for s = s as well and 
even for s > s , s-s sufficiently small, if s Q < 1. 
Thus we conclude that s = 1 and that y exists on [0,1]. 

QED 

COROLLARY . In addition to the hypothesis of Lemma 
2, assume that D is a simply connected region. Then 
there exist holomorphic function s f -,,..., f on D such 
that 

A(z,f,(z)) = for z € D, 1 g k «s n., 
for each z € D, the numbers f-,(z),..., 

f (z) are distinct. 

n v ' 

Proof ; Use the corollary of Lemma 1 to obtain 

f -,,..., f near some point in D, say z . Use Lemma 2 

In r J 

and the monodromy theorem (p. 64) to obtain holomorphic 
extensions on all of D, noting that A(z,f, (z)) = on D 
follows from the permanence of functional relations. If 
for some z £D, f.(z) = f, (z), Lemma 1 implies f . = f , 



120 



near z and then f. = f, in D by analytic continuation, 
contradicting f.(z ) 4 fv.Cz ) if j ^ k. Thus, j = k. 

JO K O 

QED 



The above corollary is about as far as we can 
go without really analyzing what happens near points 
z such that A(z,w) has a multiple zero. To carry out 
such an analysis will require a little algebraic 
background, which we now begin. 

First of all, what we shall be considering is 
functions A which are polynomials in z and w. It is 
always possible and frequently useful to arrange A 
according to powers of w or according to powers of z. 
Thus, we write 

A(z,w) = a Q (z)w n -I- a- L (z)w n ~ +...+a n (z)w + a n (z), 

where a,-,, a,,..., a are polynomials in z, and we assume 
u 1 n 

a~ ^ 0. We then say that A has degree n with respect 
to w. We say that a polynomial B is a factor of A if 
there exists another polynomial C such that A = BC . If 
A has no factors other than constants or constant multi- 
ples of A, we say that A is irreducible . It will 
also frequently be useful to factor a~ from A, writing 

A(z,w) = a Q (z) [w n +a 1 (z)w n " 1 +. . .+a n _ 1 (z)w+a n (z)], 



121 



a, 

where <x = — is a rational function of z. Conversely, 
R a 

given rational functions of z, a v ...,a , we can let a n 
° 1 n u 

be the least common multiple of the denominators of 
a,,..., a , and use the above formula to define a poly- 
nomial A. This innocent statement will prove to be 
extremely useful in constructing polynomials. We shall 
frequently be able to construct holomorphic functions 
a. on c minus a finite set, and by some argument show 
that a.K. has no essential singularities in c. Then we 
use the fact that a meromorphic function a, on C must 
be rational; cf. p. 33, no. 9. 

LEMMA 3 . Let A and B be polynomials in z and w which 
have no common nontrivial factor , and assume A, B ^ 0. 
Then there are at most finitely many z such that there 
exists w such that 

A(z,w) = 0, 
B(z,w) = 0. 

Proof : We shall use the Euclidean algorithm. To 
do this it is most convenient to regard A and B as 
polynomials in w. Then we employ the factorization 
mentioned above to write 



A = a Q (z)A 
B = b Q (z)B 



where 



122 



A'(z,w) = w 11 + a 1 (z)w n_1 + ...+ a n (z), 
B'(z,w) = w m + p^z)^ -1 +...+ 3 m (z), 

and a-i , • • • , a > 3..,...,j3 are rational functions of z. 

We rely heavily on the fact that the rational functions 
of z form a field. Also, we write for short deg A' = n 
and deg B = m. By long division we have uniquely 

A' = B'Q 1 + Rp deg R, < deg B'. 

Here Q-, and R-, are polynomials in w with coefficients 

in the field of rational functions of z, and if R, = 

we set deg R, = -co. If R, ^ 0, we apply this again 
to obtain 

B' - R ] _Q 2 + R 2 , deg R 2 < deg R ± . 
Continue this division process: 

R, = R..Q- + R,, deg R < deg R. ; , 



\-2 = \-l\ + V de § R k < de § R k-1 
R k-1 = \ Q k+l ' 



As indicated in this scheme, the process eventually 
terminates (R^ji = 0) since the degrees of the R.'s 
keep decreasing. We assume of course that R, ^ . 
Note that if R, = 0, then B' is a factor of both A' 
and B ' , thus B is a polynomial in z alone and the con- 
clusion of the lemma is trivial. Thus, we can assume 



123 



R-. ^ 0. Working up through the above scheme, we see 
successively that R, is a factor of R, , , thus R, .,,..., 
and finally R, is a factor of B', and thus of A'. By 
hypothesis, R, must have degree in w, so R, is just a 
rational function of z. Now we eliminate finitely many 
z by requiring that a n (z) 4 0, b n (z) 4 0, and z is not 
a pole of any of the coefficients of any of the poly- 
nomials Q, , . . . , (X, and R^(z) 4 0. Then we claim that 
there does not exist w such that A(z,w) = 0, B(z,w) = 0. 
For suppose such w exists. Then also A'(z,w) = 0, 
B'(z,w) = 0, since a~ (z) 4 0, b fl (z) 4- 0. Since Q, (z,w)4°°, 
the first equation in our division scheme implies R-, (z,w) 
= 0. Likewise, R^(z,w) = 0, and on down the line until 

we reach the contradiction R^.(z) = 0. 

QED 

Remark. Perhaps a cleaner way of giving this argu- 
ment is to work up through the above equations to write 

R k = CA' + DB', 

where C and D are polynomials in w with coefficients which 
are rational function of z. By clearing all the fractions 
out of this expression, we obtain 

R = EA + FB, 
where R is a not identically vanishing polynomial in z 
alone, and E and F are polynomials in z and w. Then if 
A(z,w) = and B(z,w) = 0, it follows that R(z) = 0. Since 
R has only finitely many zeros, this proves the lemma. 



124 



For our purposes the most important applications 
of this lemma occur when the polynomial A is irreducible . 
Then A and B have no common nontrivial factor except 
perhaps A itself. Thus, if A is not a factor of B, 
Lemma 3 is in force. The most important example is 
the case in which the degree of B with respect to w is 
lower than that of A. 

DEFINITION 1 . Let A be a polynomial, 

A(z,w) = a (z)w + a-, (z)w +. . .+a (z) , a-, £ 

Then a point z € C is a critical point for A if one of 
the following conditions holds: 

•1- . Z = CD J 

2. a Q (z) = 0; 

3. there exists w € C such that 

A(z,w) = 0, 

If z is not critical, then z is a regular point for A. 
PROPOSITION 1. If A is irrreducible . then there 



are only finitely many critical points for A . 

Proof : Since a.-, has only finitely many zeros, 
there are only finitely many z satisfying 1 or 2. 

Since the degree of — with respect to w is less than 

fc ^w K 

dA 
n, A and -r— have no nontrivial factor in common, and 

ow 



125 



Lemma 3 implies that at most finitely many z satisfy 
condition 3. 

Of course^ what we are aiming for is an analytic 

description of the solutions of A(z,w) = 0. If we wish 

to do this in a neighborhood of a regular point z , the 

corollary to Lemma 1 contains all the information we need, 

namely that there are n distinct holomorphic solutions 

f, , . . . ,f nearz : A(Zjf.(z)) = . Viewed as points in 
In o k r 

M, we have found 

e, = e(z +t, f, (z +t)) 
k v o k v o 

such that 

A(z +t, i' k (z +t)) s 0, t near 0. 

Another way of expressing this relation is that 

A(n(e) ,V(e)) - for e near e, . 

Likewise, for the simplest example of critical point we 
have 

A(z,w) = w -z 



and the element 



e - e(t n ,t) 



satisfies A(t n ,t) = near 0, or 



A(rr(e), V(e)) s for e near e . 

o 



126 



Thus, we make the following definition. 

DEFINITION 2 . The Riemann surface of the polynomial 
A(z,w) is the largest open subset of M on which A(rr,V) = 
Thus, a meromorphic element e(P,Q) belongs to the Riemann 
surface of A if and only if A(P(t), Q(t)) = for t near 
0. 

This latter assertion follows from the fact that 
if m is a chart defined on \](?,Q,L) in the canonical 
way indicated on p. 95, then 



P(t ) = nor/ 1 (t ), 
Q(t ) = Vocp" 1 (t^) J 



so that 



e(P,Q) = e(n «" 1 , Voce" 1 ) 



Notation . S. is the Riemann surface of A. 

The first main result we shall obtain is that if A 
is irreducible and has degree n in w, then S. is compact, 
connected, and rr restricted to S. takes every value in 

A 

C n times. First, we need a lemma on polynomials and 
their zeros. 

LEMMA 4 . I_f w, a-, , . . . , a are complex numbers such 
that 



127 



n , n-1 

n-1 n 



w ~ + ai« + • • •+ i_ i« + a_ = 0. 



then 



wl < I a.-, | +• • •+ | a | + 1. 
1 ' 1 ■ n ' 



Proof : If | w | < 1, the result holds. If |w| ;> 1, 
then 

|w| n * |a 1 !| W | n " 1 +...+|a n _ 1 || W | + |a n | 

£ (laj +...+ |a n |)M n -\ 
so that 



w| £ h 1 | + ... + |a n 



QED 



THEOREM 1 . If A is irreducible , then S A is an 

analytic configuration . 

Proof : By Proposition I, if D is the set of reg- 
ular points for A, then T-D is finite. We shall first 
prove that S» tt (D) is connected; this assertion 
forms the main point of the proof. Note that 

s A n tt -1 (D) c M. 

For suppose e(P,Q) € S A (1 tt (D), and let z = 7(0), 



w = Q(0) . Then z Q € D and A(z ,w o ) = 0. Since z Q is 
a regular point, — (z ,w ) ^ 0. Thus, Lemma 1 implies 

there is a unique holomorphic f near z such that 



128 



A(z,f(z)) = 0, f(z D ) = w Q . Since A(P(t), Q(t)) s 

and P(t) is near z , Q(t) near w Q for small t, we then 
have Q(t) = f(P(t)). If the mapping t- (P(t), Q(t)) 
is to be one-to-one (as it must), then the mapping 
t - P(t) must be one-to-one, showing that P has mul- 
tiplicity 1 at 0. Thus, e(P,Q) f M. 



So we must now prove that if z,, and z..€D and [f j 
and [g] € S. , then there is a path in S. H rT~ (D) 



'0 



A 



connecting these two germs. Since C-D is finite, D 

is connected, and thus there is a path v in D with 

initial point z, and terminal point z„. By Lemma 2 

there exists a (unique) path \ in M such that r y = v 

and /(Q) = [gj . By the permanence of functional 
Z l 

relations, ;(t) is in S for every t. In particular, 

v(l) € S. and thus is represented by a holomorphic 

function near z, which forms zeros of A. By the 

corollary to Lemma 1, there are unique holomorphic 

functions f ,,.... f in a neighborhood of z n such that 
In ° 

[f, ] € S A and f-,(z),...,f (z) are the distinct zeros 

of the function A(z,w), if z is near z,-,. Thus, 

[£] = [f.l and Y (l) = [f,]_ for some j and k. 
z J 

To finish the proof that S. "■ tt (D) is connected, 
it suffices to prove that for any j and k there exists 
a path Y in D from Zq to Zq such that analytic continu- 
ation of f. along v leads to f , . Let us suppose that in 



V 129 

all such analytic continuations f, can be analytically 

continued to f, , f . -,,..., f , but not to f ,,,..., f (where 
r /.' ' m m+1' n v 

we have renumbered the f.'s). Here 1 <. m <, n. and we 

J 

want to prove m = n . 

Now consider the function 

m 

B(z,w) = Tf (w-f k (z)) 

k=l 

defined for all w £ C and for z in a neighborhood of z„ . 

For each fixed w the function B(z,w) can be analytically 

continued along all paths in D with initial point z~ 

(Lemma 2), and analytic continuation along a closed path 

of this nature must simply lead to a permutation of 

f ,,..., f : such a continuation could not lead to any of 

I'm J 

the f . . f . and two different f 's could not be 

m+1 n j 

continued to the same f, , by the unique lifting theorem. 
Therefore, B(z,w) is analytically continued into itself 
along any closed path in D from z^ to z n , since B is a 
symmetric function of f ,,..., f . Another way of looking 
at this is to perform the indicated multiplication in B 
and write near z^ 

B(z,w) = w m + a 1 (z)w m " + ... + 9- m ( z )> 

whe re 

a k (z) = (-1) v f f ... f . 
i 1 <i 2 <. ..<i k H L 2 L k 

By the same reasoning, each ex, is symmetric in f n f . 

K 1 m 

so a k has the property that it can be analytically 



130 



continued along all paths in D and analytic continuation 
along closed paths leads back to a, • Thus each <x, can 
be extended to a single - valued holomorphic function in D 

Now for a trick that will be used over and over. 
The function a, is holomorphic in C except at finitely 
many points. We shall now estimate the growth of a. 
at these points to conclude a, does not possess any 
essential singularity. Suppose now that a is one of the 
critical points( oneof the points in C-D) . Then for 
some sufficiently large integer N we have near a 

|a (z)| > |z-aj N , |a k (z)| < C (1 <; k g n) 

(C is some constant) if a ^ »; if a = » we have near a 

|aQ.(z)| > c, |a (z)| s |z| (1 s k ^ n) 

(c is some positive constant). Thus, for z near a 
and A(z,w) = 0, Lemma 4 implies: 

if a ^ oo, |w| < nC|z-a| L + 1, 

if a = co. w < — z + 1 . 
11 c ' ' 

Since A(z,f, (z)) = 0, we thus obtain for z near a, 

i c t \ I j i -N | iN 

| f , (z) I <, const|z-a| or const |z| 

if a ^ - or a = oo, respectively. Thus, the formula 
for o.^ shows that for z near a. 



131 



-Nk Nk 

| ou (z) | <, const |z-a| or const |z| 



in the two cases. Thus, a, has either a pole or a 
removable singularity at a. Since this is true at 
each critical point, a, is meromorphic in C ana is 
thus a rational function. 

Let t>Q be the least common multiple of all the 
denominators of the a. 's expressed as fractions without 
common factors, and let 



B(z,w) = b (z)B(z,w) 

= b Q (z)w m + b 1 (z)w m * + ...+ b m (z), 

a polynomial in z,w of degree m in w. Since for z 



near z~ 



B(z,f,(z)) = A(z,f,(z)) - 0, 



the conclusion of Lemma 3 does not hold for the poly- 
nomials A and B. Thus, A and B possess a common non - 
trivial factor. Since A is irreducible, this factor 
must be A itself. Thus, the degree of B must be at 
least the degree of A, so m = n . 

We have now completed the proof that S. (1 rr (D) is 
connected. The rest is easy. Suppose e(P,Q) £ S. . Then 
for 4 sufficiently small disk A centered at 0, U(P,Q, A) 
consists only of points in S. D rr (D) with the possible 
exception of e(P,Q), since c-D is finite. Thus, e(P,Q) 



132 



-1 • — 

can be joined to a point in S, D rr (D) by a path in M. 

Thus, S. is connected. 

To prove S. is a component we show it is both open 

and closed in M. It is trivially open by Definition 2. 

Suppose e(P-,Q) is in the closure of S. and let cp: U(P,Q, A) 

A be a canonical chart. Then there exists t € A such 

o 

that cp" (t o ) € S A . Thus, since co" (t Q ) = e(P(t +t), 
Q(t +t)), we have 

A(P(t Q +t),Q(t o +t)) h for t small. 

Thus, since A(P(t),Q(t)) is a meromorphic function for 
t € A which vanishes for t near t , A(P(t),Q(t)) = in 
A. That is, e(P,Q) € S A , proving S. is closed. 

£ED 

WARNING It is tempting to think that if z is a 

critical point of the type 3, that is, if the equation 

A(z Q ,w) = has a double root; and if e(P,Q) e S., 

P(0) = z , and Q(0) is a double zero, then e(P,Q) is a 

branch pdint of order at least 1. This is not true in 

general. For example, let 

k( \ 2 2 3 

A(z,w)=w -z -z. 



Then A is irreducible and z = is a critical point, the 
zeros of A(0,w) - w both vanishing. There are two point 
in S. lying near z = 0, and these are given by 



133 



e(t, t/l+t), e(t, -tv'l+t), 

where JT+t is the principal determination of the square 
root for t small. Clearly, neither of those meromorphic 
elements is a branch point of order s 1. 

THEOREM 2 . S. is compact . 

Proof: We again write 



A(z,w) = a Q (z)w n + ... + a n (z). 

Consider the function tt: S a -. X, . By Proposition 9.2 of 
Chapter II, it suffices to prove that the restriction of 

tt to S, takes everv value in c n times. Of course, 
A 

it suffices to consider the case in which A is irreducible 



Let D be the set of regular points for A; by Proposition 
1 the set C-D is finite, 
sufficiently small that 



1 the set C-D is finite. Let a € C and choose e > 



A = [z: | z-a | < ej (A = {z: |z| > e" } if a = 0=) 



contains only points of D except possibly for a itself 
Let a' be the set A with a line from a to the circum- 
ference removed; for def initeness, let 

A' = [z: z € A, z-a not a nonnegative real 
number } . 

Since A' is simply connected and contains 
no critical points for A, the corollary to 




134 



lemma 2 implies that there are functions £.,...,£ 

In 

holomorphic in A' such that for each z € A', 
f,(z)j...,f (z) are the distinct solutions of A(z,w) = 0. 
Likewise, there are functions g, ,...,g holomorphic 
in the region A" as illustrated: 

,. a Now just below the slit in A' 



the function f, must coin- 
k 

cide with a unique g.. In 



/ 

v a i 

\ / turn, g. must coincide with 

a unique f just above the 

slit in A'. Let us denote I = o(k) . Thus, f ,, N is the 

a(k) 

result of analytically continuing f, in a counterclock- 
wise manner around A'. By the unique lifting theorem, 
the function a is a permutation of the integers l,2,...,n 
This permutation has a unique decomposition into cycles. 
Let us consider a cycle of length m and let us renumber 
the functions f, so that this cycle is represented by 
-(1) = 2, cr(2) = 3, ...,o(m-l) = m, a(m) = 1. Define 
for small t 

f ] _(a+t ), < arg t < — , 

f.,(a+t m ), ^ < arg t < 2 ^, 
Q(t) = \ 2 v ' ' m & m ' 



f (a+t m ), (m-l)^i < arg t < 2 



-m 



TT. 



(If a = =o replace a + t by t throughout.) By the 
definition of j and the particluar enumeration of this 



135 



cycle, Q has an obvious extension to a holomorphic 
function defined for < |t| < e . Also, since 
each f, (a + t ) is a solution of A(a + t , w) = 0, 
Lemma 4 can be applied exactly as on p . 130 to 
show that | f , (a + t )| <; const |t| as t - 0, for 
some positive integer N. Therefore, Q cannot have an 

essential singularity at 0, and thus Q is mero m orphic 

.. i | 1/m 
for I t| < e . 

Now we prove that (a+t , Q(t)) is a pair . Suppose 

that for small s and t, a + t = a + s , Q(t) = Q(s) . 

Then t m = s™ . If (k-1)— <; arg t < k— and 
v y m & m 

(j-l)~ s arg s < j-^, then 

Q(t) = f,(a+t m ) (= lim f (a + t m e i6 )), 
K 6^0+ k 

Q(s) - f^a+s 111 ). 

Since t = s , f, (a+t ) = f.(a+t ). Since the functions 

f, ,....£ represent distinct solutions (and likewise 
1 n r 

,g ), we must have j = k. By the inequalities 
for arg t and arg s, the equation t = s now implies 
t = s. Thus, the mapping t - (a+t , Q(t)) is one-to-one. 

Since A(a+t , Q(t)) = 0, this argument finds an 



element 



e(a+t m , Q(t)) 



belonging to S. ■ 



136 



If the permutation a is decomposed into cycles of 

lengths m-, , ..., m , then by the same argument we produce 

elements in S. of the forms 

m 
e(a + t \ Q,(t)), 



m 
e(a + t ^, Q^(t)). 

m. 
Since the multiplicity of n at each point e(a+t L , Q.(t)) 

is m. (Proposition 5 of Chapter IV), it follows that tt 

takes the value a a_t least m-,-f. . .+m = n times. The same 

is true if a = a, though we of course need to use slightly 

different notation. 

An obvious remark shows that ri takes each value at 

most n times... of course, we speak of the restriction of 

tt to S. . In fact, if a is a regular point for A, then 

the points in S. D n ((a)) are in M (cf. p. 128 ) and 

these elements are exactly the germs of the n holomorphic 

solutions near a by the corollary to Lemma 1. Thus, tt 

takes on the value a exactly n times in S. . By the 

argument on p. 43, if tt takes some value (a 

critical value) more than n times in S., then n takes 

every neighborhing value more than n times in S. , which 



implies n takes some regular value more than n times in 

QED 



S., a contradiction 



137 



Remarks. 1. One sees finally the reason for 
discussing M - it contains precisely enough points to 
discuss branch points in general, and in particular to 
discuss all the solutions of algebraic, equations. If 
e(P,Q) € S. and cp: U(P,Q,A) - A is the canonical chart, 
the function & is called a uniformizer for A near the 
point P(0). It replaces the multiple-valued solutions 
of A = by two single-valued meromorphic functions. 
It is of course only defined locally. 

m . 

2. The elements e(a + t x , Q. (t)) produced in the 

above proof are obviously different. The only possi- 
bility for two of them to coincide is for two of the 
multiplicities m. and m. to coincide and for Q.(t) = Q.(uut) 
for some root of unity uu . But this would force the 
corresponding cycles to overlap, as can be easily 
checked. 

3. The function Q on p . 134 is meromorphic and 
thus has a Laurent expansion: 

Q(t) = I a t\ 
k=N K 

1/m 
Substituting formally z=a+t,ort= (z-a) } gives 

a series 



1 a, (z-a) 
k=N K . 



with a similar series 



k/m 



138 



-k/m 



S a z 
k=N K 



in case a = «. These are called Puiseaux series, and 
have the property that for any determination of z 
the sum of the series gives a solution of A(z,w) = 0, 

1 /m 

and differeint determinations of z yield different 
solutions. Of course, all this information is contained 
in the idea of the corresponding meromorphic element. 

4. It is almost amazing how easy it was to find the 

m. 
elements e(a+t 1 , Q,(t)) in S. . However, when one 

observes what had to be known, it is quite obvious that 

it should be easy. Namely, we had to have completely 

solved the equation A(z,w) = away from the critical 

points, and then it was a simple matter of checking 

what S. looks like above these finitely many critical 

points. But this sort of procedure can almost never 

be carried out in practice for rather obvious reasons. 

We can't even usually hope to solve the equation near 

a critical point and observe how the zeros behave under 

analytic continuation around the critical point. 

5. Even without knowing Proposition 9.2 of Chapter 
II, it is almost obvious why S. is compact. For S. consists 
essentially of n copies of the (compact) sphere C branched 
above certain finitely many points. The only way S. 
could fail to be compact would be for certain of these 
branch points not to be included in S. . Essentially the 



139 



proof shows they are indeed all included and this state- 
ment is phrased in the perhaps deceptive statement that 

the restriction of n to S. takes every value n times. 

A J 

3 3 
Problem 6 . Let A(z,w) = w -3zw+z . Prove that A 

is irreducible. Find its critical points and discover 

the types of meromorphic elements which belong to S. . 

Compute the genus of S. by the Riemann-Hurwitz formula 

(p. 112). 

3 a 
Problem 7 . Same for A(z^w) = zw -3w+2z , where a 

is any integer (positive or negative) . Of course, if 

a < then this is interpreted to be the problem for 

the polynomial 

1-a 3 „ -a , ., 
z w - 3z w + I . 

Now we pass to the converse of Theorem 2. This 
states that every compact analytic configuration is the 
Riemann surface of a unique (to within a constant factor) 
irreducible algebraic function. In Chapter VI this state- 
ment will be improved considerably and will state that 
any compact connected Riemann surface is analytically 
equivalent to a compact analytic configuration (and thus 
has an associated irreducible polynomial) . 

Before stating this converse of Theorem I, we 
make a useful observation about S. . First, divide out 
the leading coefficient ap(z) to write A(z,w) = a^ (z )A ' (z,w) > 
where 



140 



A'(z,w) = w n + a, (z)w +. . ,+a (z) 
i n 



and n.,...,a are rational functions of z . We assume 
A (and thus A') to be irreducible. If a is a regular 
point for A, then the corollary to Lemma 2 shows the 
existence of the holomorphic zeros f ,,..., f as usual. 
Thus, 

e k = e(a+t, f k (a+t)) 

is a point in S., 1 5 k g n, and the elements e, are 
the only ones in S fl it" ({a}). Also, V(e ) = f k (a), 
so the numbers V(e, ) are the n solutions of A'(a,w)=0. 

Thus, we obtain a factorization 

n 
A'(a,w) = J[ (w-V(e k )) 

k=l 

= IT (w-V(e)). 
e€S A 

n(e)=a 

THEOREM 3 . Let S be a compact analytic configuration 
Then there exists a unique (up to constant factor) irre - 
ducible polynomial A such that S = S. . 

Proof : Since S is compact and n: S - C is analytic, 
Proposition 9.1 of Chapter II shows that the restriction 
of tt to S takes every value the same number n of times. 
Let D be the subset of C defined by 



£-D = irr(e): e € S, m (e) > 1 or V(e) = a,} 



141 



Thus, if e € S and n(e) € D, then m (e) = 1 and V(e) 4- »• 
Since S is compact, the set of elements e such that e € S 
and m (e) > 1 or V(e) = » is finite . A fortiori , C-D 

TT 

is finite. We now take our clue from the discussion on 
p. 140 and define for z € D, w € C 



= TT 



A'(z,w) = I \ (w-V(e)). 

ees 

n(e)=z 

That discussion implies that if S = S» for some A, then 
this must be the formula for A'(z,w) for regular points 
z, since all the regular points must be contained in D. 
Thus, the uniqueness assertion of the theorem is estab- 
lished. Moreover, we have an explicit formula for A' 
and we now just have to check various details. 

First, if z € D then there are exactly n elements 

o J 

e,,...,e € S with n(e, ) = z , since n takes the value 
z n times and rr must have multiplicity 1 at each e, . 
Suppose e, = e(P(t) , Q,(t) ) , where P(t) = z +t if z f » 

and P(t) = t~ if z - ». Let cp,: U(P,Q k ,£ i ) - A be 

a canonical chart. For small t , 

o 

snn" 1 ({P(t o )})= t^ 1 (t o ),...,^ 1 (t o )}, 

so that 

n 

A'(P(t o ),w) =7]'( w -Q k (t o )) 
k=l 

since V(r D ~ (t )) = Q, (t ). This equation shows that if 
k ° K ° 



142 



we expand 

A'(z,w) = w + a, (z)w +...+ a (z), 

then a 1 ,...,a are holomorphic on D. 
in 

Now we examine the behavior of cu at the isolated) 
points of C-D. Suppose a € £-D . Let e(a+t m , Q(t)) 
be one of the points in S (1 tt (j_a})5 in case a = co this 
must be replaced by e(t ,Q(t)). Then for z near a but 
not equal to a, there are m points in S Pi tt ({z}) 
determined by this one element, namely, 

e(a+(t k + t) m , Q(t k + t)), where t™ = z-a. 

(We now discuss the case a 4- m ', the case a = co is 
handled entirely similarly.) The corresponding values 
of V(e) are Q(t, ), 1 < k s m. Thus, for some N we have 

|V(e)| s |t k |" N - Iz-a|- N/m 

for these m points e £ S n n (z) . Treating the other 

points in S fl tt (z ) in a similar fashion, we obtain for 

some integer M 

-M -1 

|V(e)| <. | z-a | if e € S A n (z), z near a. 

Thus 

-Mk 

|a k (z)| < const |z-a| if z is near a; 

if a = co this estimate should read 

Mk 



| a k (z) | < const | z 



143 



Therefore,, ctv. is meromorphic on C and thus a-i, is rational . 



Now that we have produced a polynomial A we must show 
that A is irreducible and that its Riemann surface is S. 
This will essentially be done all at once. Suppose that 
there exists a factorization of A in the form A = BC, 
where B and C are polynomials and B is irreducible ... in 
fact; there is always such a factorization with a poly- 
nomial B of degree at least one in w (perhaps C is constant) 
Then B has a Riemann surface S R which is a compact analytic 

configuration. Let e be an element in S,, such that m (e) = l 

d rr 

and V(e) 4- <*>; this includes all but finitely many points 

in S„ . We also assume n(e) £ D, eliminating again at 

most finitely many points. Then if rr(e) = z we let 

P(t) =z +tifz ^co and P(t) = t" 1 if z = ». Then 
o o v o 

e = e(P,Q), and B(P(t), Q(t)) = for t near 0. Thus, 
since A = BC we have 

A'<P(t),Q(t)) = for t near 0. 

The formula for A' at the bottom of p. 141 implies 

n . 

TT (Q(t)-Q (t)) =0 for t near 0. 
k=l R 

Since each factor Q-Q^ which is not identically zero can 
have only isolated zeros, it follows that for some k, 

Q = Q k - 

Thus, e = e(P,Q k ) = e, 6 S. Thus, except for finitely many 



144 



points S R c S . Since S is compact in the Hausdorff space 
M, S is closed and thus 

S„ - S. 



Since S„ is a component of M and since S is connected, 

D 

it follows that S„ = S. 

15 



Now it is all done. For., n assumes (when restricted 
to S_) every value the same number of times, this number 
being the degree of B as a polynomial in w. But n assumes 
(when restricted to S) every value n times. Thus, B has 
degree n in w . Thus, C has degree in w and thus is just 
a polynomial in z. Therefore, if we discard all the 
common polynomial factors in z from the polynomial A(z,w), 
we must have C s const. This shows that A is irreducible, 
its only possible nontrivial factor turning out to be 



itself. And S. = S = S . 



QED 



We thus see that on any compact analytic configuration 
S the two meromorphic functions n and V are related by 
an algebraic equation. These two functions of course 
allow us to construct other meromorphic functions on S; 
namely any rational func tion of tt and V is meromorphic 
on S. The amazing fact is that there are no other mero- 
morphic functions on S . In fact, we have 



145 



THEOREM 4 . Let S be a compact analytic configuration 

on which n assumes every value n times ■ Let f be any 

meromorphic function on S . Then there exist unique rational 

functions " ns -'-, r i i such that 
n-1 

n-1 . 
f = I a. n-V J . 
j=0 J 

Proof : Suppose z is a regular point for A, the 
polynomial is such that A is irreducible, and S = S. • If the 
the formula for f is to hold, then we must have 

f(e) = I a.(z)V(e) J if e € S, n(e) = z. 

Now S n ({z}) = [e ,...,e ] has exactly n points and 
the numbers V(e, ) are distinct. The above equations read 

n-1 . 

f(e, ) = L a.(z) V(e, ) J , 1 £ k < n. 
k j-0 J k 

These are n equations in n "unknowns", a (z),...,a -i(z), 
and the determinant of the system is 



det 




146 



This is a so-called Vandermonde determinant and its 
value is well known and easily seen to be 

7T (V(e k )-V(e^)), 

ls-t<k<n 

which is not zero. Thus, a f ,(z),...,a -> (z) are uniquely 
determined. It is also clear that these numbers a-(z) 
really depend only on z and not on a particular ordering 
e,,...,e of the points in n ({z}). Thus, a ,...,a _-■ 

are uniquely determined at the regular points for A, and 

thus are unique since they are to be rational functions. 

Knowing what a. must be, we now prove that they 

exist. By Cramer's rule, we can write down a formula 

for a-(z) in terms of a determinant involving f(e, ) and 
J K 

V(e, ), divided by the Vandermonde determinant. Near a 
fixed regular point we can choose the e, in terms of 
charts to be analtyic functions and thus f(e,) and V(e, ) 
become analytic functions, proving a. is holomorphic on 
the set of regular points. As usual, we now prove that 
a. cannot have any essential singularities. Since we 
obtain upper bounds for f(e, ) and V(e, ) in the standard 
manner we are used to by now, it remains to obtain a 
lower bound for the Vandermonde. 

Suppose then that a is a critical point. We assume 
in the following that a # »; the case a = co is treated by 
mere formal changes in the analysis. The points in 
S n rr ([a]) have the forms 



147 



m , J 

e(a+t J , Q.(t)), 1 < j s J, S m. = n, 

J i = l J 



where Q. is meromorphic near 0. Let the positive 



integer m be the least common multiple of the integers 
m. • If z is a number sufficiently near a but not equa' 
to i, choose an arbitrary s € C such that 



m 
z-a = s . 



Let 

2ni/m 

. = € 
J 



a 4 - e J 



Then for <. I <, m -1, 

m/m. m. 
z-a = (uj. s 3 ) 3 , 

„ m/m. 

and the numbers id ."'s J are different for <, I s m.-l. 
J J 

Thus, S fl rr ((zj) consists of the points 

m/m. m. m/m. 

e., = e(a+(uur s J + t) J . Q. («)? s J + t)) 



for < I < m.-l, 1 < j < J. Thus, V(e. ,) = Q. (ojIs J) . 

The Vandermonde contains terms of the form 

. m/m . / m/m , 

V(e )-V(e., ,) = Q ( ffl *s J ) - Q.,(tD*,s J ) ; 

J "w J "^ JJ JJ 

which is a meromorphic function of s, not vanishing for 
small s ^ since z is regular for a. Thus, there exist: 
an integer N such that 



148 



N N/m 
|V(e ) - V(e,, ,)| * |s| = |z-a| 

so the Vandermonde has modulus bounded below by 

N n(n-l) 

I i m "2 

z-a 



Thus, we have proved that each a- is rational and 

by definition 

n-1 . 

f(e) = E a .(TT(e))V(e) J 
j=0 J 

for all but finitely many e <E S (those such that rr(e) is 
a critical point for A). Thus, these two meromorphic 
functions on S coincide. 

QED 



VI 149 

Chapter VI 
EXISTENCE OF MEROMORPHIC FUNCTIONS 

The thrust of this chapter is the proof that 
there exist nonconstant meromorphic functions on any 
Riemann surface. It will take a tremendous amount 
of machinery to achieve this result; in particular, 
we will need to give a careful and fairly complete 
discussion of harmonic functions on Riemann surfaces. 
But before beginning this topic, we shall exhibit one 
problem which can be solved using the existence of 
meromorphic functions. 

First, we introduce a lemma which really logic- 
ally belongs in Chapter IV, but has not been needed be- 
fore now. 

LEMMA 1 . Let m be a positive integer and Q a_ 
meromorphic function near having Laurent expansion 

CO 

Q(s) = £ a-s j , 
3-— J 

and assume that no positive integer except 1 is a common 
factor of all j such that a. ?* 0. Let n be an integer 

relatively prime to m. Then (t m , Q(t n )) is a pair . 

Proof : We have to prove that the mapping t - (t ,Q(t )) 
is one-to-one near 0. If this is not the case, then 
there exist s, - and t, -» such that s, ^ t, and s, = t, , 



150 VI 

Q(s£) - Q(t{J). Thus, (s k /t k ) m - 1, and by taking a 
subsequence we can assume that there exists a fixed uu 
such that uu ^ 1, uu = 1, s, = uut, (cf. p. 93). 
Therefore, Q(uo n t k ) = Q(t k ), so that the two functions 

Q(iu s) and Q(s) agree on a sequence s = t, - 0. Since 
they are both meromorphic , they must be identical: 

l, a.'ju s J = L a- s , s near 0. 

Therefore the coefficients must agree: a-«J -* = a. for 
all j. This means that a. f implies u; -* = 1 . It 

follows easily that ju = 1. For the set i j : uj n -^ = 1} 
is an additive subgroup of the integers and the Euclidean 
algorithm implies that any subgroup equals the integer 
multiples of a fixed positive integer j . Thus, a- ^ 
implies j contains j as a factor. By hypothesis, 

i = 1 and therefore u = 1. Since n and m are relatively 

J o J 

prime, the Euclidean algorithm again implies there 
exist integers p and q such that pm + qn = 1 . Thus, 

_ pm+qn _ / m,p. n q 

a contradiction. 

^ED 

THEOREM 1 . Let S be any connected Riemann surface . 
Let f and g be meromorphic functions on S such that f ^ 
constant . Then there exists a unique analytic $ : S - M 



VI 



151 



such that 

f = TTo $ } 
g = Vof. 

It is convenient to draw a diagram to indicate 
these two equations: 



^M 



f,g 



TT,V 



A A 
C X C 



The statement of the theorem is then exactly that there 
exists an analytic $ making this diagram commutative. 

Proof : Uniqueness : Suppose p 6 S and that m.p(p) 
= 1. Since f = rtoi, it follows that m^p) = 1. Let 
i|i be any compatible chart in a neighborhood of p. 
Suppose |(p) = e(P,Q) and let ®: U(P,Q.,A) - A be a 
canonical chart. We assume iji(p) = 0. 

,u(p,q,a; 

' §(p) 




/ 



. 



152 VI 

Recall from p. 126 that 



p - - 1 

r - Uocp , 

Q = Voce" 1 . 

Now the mapping -ooSoH-' is a parameter change , since it 
is one-to-one near and maps to 0. Thus, we consider 



P ° ( cp ° § ° <jr ) = TT° $ ° f ■ f o \|l , 

Q°(» f°i|(~ ) = V°$°ili" = g°v" 



and we thus have 



(P,Q) ~ (f°y 1 , go Mr" 1 ) . 



Therefore, if m f (p) = 1 we have 



(p) = e(fo f , gof ) . 



This proves that £ is uniquely determined except on the 
discrete set where the multiplicity of f is greater 
than 1. Since § is continuous on S, then | is also 
uniquely determined everywhere. 

Existence : We already know how to define $ at 
points where m f = 1 . Therefore, we so define $ at those 
points, just noting that the definition §(p) = e(f°u; } 
g°'Ji ) really is independent of the particular chart 
o; a different selection of the chart merely gives a 



VI 153 

parameter change. 

Now suppose p € S and m f (p ) = m. Choose a 
chart $ near p such that $(p ) = and 

fot" 1 (t) - f(p Q ) + t m 

if f (p ) 4- ». As usual, if f (p ) = <» we have instead 

fof" 1 (t) = t _m . 

Then consider the Laurent expansion of go ty : 



goi|f (t) = £ a,t 



Let n be the largest positive integer which is a factor 
of all k such that a, ^ 0; in case a, = for all k, let 
n = in. Then we can let k = nj in the above series and 
we obtain 



-1 (t) • S a nj t nj = Q(t n ), 



oo 



where 

00 

Q(s) = £ a.s J 

and r, . = a .. Thus, either n ■ = for all i or there 

is no common factor of all j with a- ^ except 1 (and 

-1). Let a be the positive integer which is the 

greatest common divisor of m and n and define 

m n 
$(p ) = e(f(p o ) + t», Q(t^)) 



154 VI 



(replace f(p ) + t w by t ^ if f(p ) * «>) . We have to 

check that this is really a meromorphic element, i.e., 
that 



is a pair. If a. = for all j, then — = 1 and it is 

J u 

obvious. Otherwise, Lemma 1 applied to the relatively 

prime integers — and — shows that we do have a pair. 

U u 

Thus, $(p ) makes sense; we do not need to check that 
we have defined it independently of the choice of f 
(there are only m choices to make) since we can regard 
the choice of f to be an arbitrary "function" of p • 

We now observe that this definition of *(p ) works 
even when m = 1; then ^ = 1 and the definition agrees 
with the earlier definition of § at points where f has 
multiplicity 1. Note that obviously 

TT°*(P Q ) ■ f (P )> 

Vo?(p o ) - Q(0) = gov _1 (0) = g(p Q ). 

Thus, the required commutativity of the diagram is 
proved. We thus need to check the analyticity of $ 
in order to finish the proof. We prove that $ is 
analytic in a neighborhood of p , using the above 
notation . 

Let z be near 0, z ^ 0, and let p = v (z) . Then 
for sufficiently small z, p is a point where f has multi- 



VI 155 



plicity 1. Define the chart f, = \|r-z, so that iiu (p) = 
and 

*i (t) = f (z+t) • 

Therefore^ according to our first definition of §, 



$(v" (z)) = e(fo*" , go^ ) 

= e(fo^" 1 ( z +t), goif 1 (z+t)) 
= e(f(p o ) + (z+t) m , Q((z+t) n )) 

Now we introduce the canonical chart near $ (p ): 

r o 

call it cp: U - A, where 

m n 

.f 1 (t o ) = e(f(p o ) + (t Q +t) U , Q((t o +t) U )) 



We introduce next the parameter change p defined by 

o(t) = (z+t) W - z^; 

since z 7^ 0, this is. a parameter change. And we have 

m 
m u 

(z+t) = (z +p(t)) (and a similar formula 

w i th n ) , 

showing that 

m ri 

$(f 1 (z)) = e(f(p )+(z^+p(t))^, Q((zM+ p ( t ))U)) 

m n. 

= e(f(p )+(zW+t)^ Q((z"+t)")) 

=.^ _1 (z^). 
Thus, 



156 VI 



30 $oy (z) = Z^, 



_1 

so c°^'°i is holornorphic near 0. This proves that ? 

is analytic near p . 
r o 

QED 

COROLLARY . Let S be any compact connected Riemann 
surface and f , g meromorphic functions on S such that 
f ^ constant . Then there exist a unique compact 
analytic configuration T and analytic function ? from 
S onto T such that the diagram commutes : 

S * > T 

f >g \ / n,V 

1 x 1 



Proof : This is trivial. We just let T = §(S). 
Since S is compact and connected and § is continuous,. T 
is also compact and connected. Since § is analytic and 
nonconstant, Proposition 4 of Chapter II implies T is 
an open subset of M. As T is thus closed and open and 
connected, it is an analytic configuration. 

QED 

COROLLARY . Under the hypothesis of the previous 
corollary, there exis ts a unique irreducible polynomial 
A(z,w) such that 



VI 157 

A(f( P ). g(p)) = for p € S. 

Proof: We first prove uniqueness. If A has the 
required properties, then A(rr($(p)), V(§(p))) = for 
p £ S. Since 5 is onto, this implies A(rr(e), V(e)) = 
for e ST. Therefore the Riemann surface for A satisfies 
S A => T. Since T is a component of M and S is connected, 

Pi A 

S. = T. Thus, Theorem 3 of Chapter V shows A is unique. 

Existence is trivial. Simply let A be chosen by 
Theorem 3 of Chapter V such that S A = T. The above 

argument worked the other direction proves A(f,g) = 0. 

QED 

We are most interested in the possibility that the 
mapping $ of S onto T is also one-to-one. For then we 
will have an analytic equivalence of the compact Riemann 
surface S with an analytic configuration. The next theorem 
gives some equivalent conditions. 

THEOREM I . Let S be a compact connected Riemann 
surface and f , g meromorphic functions on S such that 
f £ constant . Let a, T, A be the objects of the two 
previous corollaries . Assume that f takes every value 
n times . The the following conditions are equivalent . 



1 . $ is an analytic equivalence of S onto T 

2 . There exists 
f(p) = z} has n points 

3 . For all excep 
f(p) = zj has n points 



2, There exists a point z € C such that tg(p): 



3. For all except finitely many z £ r , ig(p): 



158 VI 

4 i The polynomial A has degree n _in w . 

Proof : 1 => 4 : Since § is an analytic equivalence 
and f = no? ; rr also takes every value n times. The results 
of Chapter V, especially Theorem 3, imply that A has 
degree n in w . 

4 => 3 : If z is a regular point for A (and this 
is true for all but the finitely many critical points), 
then T n ( {z}) = {e, , . . . ,e } and the numbers 
V(e, ) , . . . , V(e ) are distinct, being the solutions of 
A(z,w) = 0. Since § is onto, there exist p , ...,p € S 

such that |(p k ) = e k> Then g(P k ) - v ( e k ) and f ^k^ = 
Tr(e k ) = z, so 

ig(p) : f (p) = z) 

has at least n points V(e, ) , . . . , V(e ). Since f takes 
every value n times, this set can contain no more than 
n points. 

3^2 : Trivial. 

2 =■> 1 : Finally we come to an interesting 

proof. By hypothesis there are n points p,,...,p such 

that f(p^.) = z and the numbers g(p-, ),..., g(p ) are distinct. 

In particular, since f takes the value z n times, m f (p,) = 1 

Since f = r^f, m (p, ) = 1. Let e = $(p,). We shall show 
$ l 1 

that $ takes the value e one time. Suppose then that 
$(p) - e. Then f(p) - -r(e) = noiCp^ = f(p x ) = z, so 



VI 159 

p = p for seme k. Then g(p k ) = V(e) = V°§(p 1 ) = g(p 1 ), 
so p, = p, . Thus $(p) = e if and only if p = p, . Since 

moreover m (p, ) = 1, we have now proved that $ takes the 

§ 1 

value e one time. But Proposition 9.1 of Chapter II 
implies $ takes every value one time. That is, $ is one- 
to-one. Thus, <£ is an analytic equivalence. 

QED 

Let us comment on 4. Suppose that the mapping § 
takes every value k times, and that rr: T -> C takes every 
value m times . Then since f = n *, it is easy to see 
that f takes every value mk times. In the notation of 
Theorem 2. this means n = mk and A has degree m In w. 
Thus, in general the degree of A is a factor of n. For 



example, consider the trivial case in which S = C . f(z) = z 
and g(z) - z . Tl 
corollary implies 



and g(z) ■ z . The uniqueness assertion of the previous 



A(z,w) = w - z . 

2 
For, A is irreducible and A(f(z),g(z)) = g(z) - f(z) = 

4 ^ 
z -z ' = 0. Thus, f takes every value 4 times, the 

degree of A is 2, so we conclude that $ takes every value 

2 times. In fact, our explicit construction shows that 

for z ^ 0,oo, 

4 2 
$(z) - e((z+t) , (z+t) ). 

s(-z) = e((-z+t) 4 , (-z+t) 2 ) 

= e((z-t) 4 , (z-t) 2 ) 



160 



vr 



= e((z+t) 4 , (z+t) 2 ) 
= $(z) 



by the parameter change t -> -t. 

Now we state the main theorem of this section and 
show how it can be used to produce functions f and g 
which satisfy the criteria of Theorem 2. Note that we 
must at least produce a nonconstant meromorphic 
function f on S; the following theorem allows us to do 
even better. 

THEOREM 3 . Let S be any connected Riemann surface 
and let p , q € S . p --}=■ q • Then there exists a meromorphic 
function f oin S such that f(p) ^ f(q)- 

We are nowhere near being able to prove this yet. 
But assuming its validity for the moment we prove 

COROLLARY . Let S be a compact connected Riemann 
surface . Then S is analytically equivalent to an analytic 
configuration . 

Proof ; First apply Theorem 3 to find a nonconstant 
meromorphic f on S . Now we show how to construct 
another meromorphic g on S which satisfies criterion 2 
of Theorem 2. Assume that f takes every value n times. 



VI 16! 



If n = 1, take g = 0. Suppose n > 1. Since the points 
of S where the multiplicity of f is greater than 1 are 
isolated., there exists z € C such that f Qz}) consists 
of n distinct points p,,...,p . Theorem 3 implies that 
if j ^ 1, there exists a meromorphic h on S such that 
h(p.) 4- h(p-, ) . Choose a complex number a 4 l n (Pi ) , • • • , 
h(p )1. Then there exists a Mobius transformation 

„ t v aw + b 
F(w) = -r— t 

v ' cw + d 

such that F(h(p 1 )) = 1, F(h(p.)) = 0, F(a) = ». Thus, 
there exists a meromorphic h. = Foh such that 

h j (p 1 ) = 1, 

h.( P .)=0 3 

h.(p, ) is in C for 1 <. k <. n. 
J p k y 

n 
Define g, = i\ h.. Then g, is meromorphic on S and 
1 j=2 J l 

gl ( Pl ) = 1, 

g 1 (p,) =0, 2 s j <; n. 

Repeating this construction, there exist meromorphic 
functions g,, ...,g on S such that 

g k (P k ) - 1, 

g k ( P .) = if j + k. 



Now define 



n 

£ kg,. 
k=l R 



162 VI 

Then g is meromorphic on S and g(p^) = k, 1 < k s n. There- 
fore, 

g(p) : f (p) = z = 1,2 n 

has n points. So criterion 3 of Theorem 2 is satisfied 
and therefore S is analytically equivalent to an analytic 
configuration (criterion 1 of Theorem 2). 

QED 

Now we shall begin to introduce the machinery needed 
to prove Theorem 3. The basis is the idea of ha rmonic 
functions on Riemann surfaces. First, we recall that a 

function u on an open set in C is said to be harmonic if 

2 5 2 u ?/u 
u is of class C and — j + —n =0. A convenient way of 

3x dy 
discussing this is to define the differential operators 



»„ _ 1 Su 1 du 
Su = 7 lx" + 71 ^ ' 

- 1 du 1 du 
Su = 7 "5x • 7T 17 • 



Then 

3lu = lau = k(r-% + -^-7 ) • 

5x =>,y 

Now the equation If = is exactly the Cauchy-Riemann 
equation. Thus, f is holomorphic if and only if If = 0; 



moreover, in this case af = £', the ordinary complex 
derivative of f. Thus, if u is of class C , then u is 
harmonic if and only if du is holomorphic. In particular, 



VI 163 



if u is harmonic then au has derivatives of all orders. 

Likewise, u is harmonic so that Su = du has derivatives 

of all orders. Thus, — = ou + au has derivatives of 

ox 

all orders, and the same is true for -r— . Thus, harmonic 

Ay 

functions have derivatives of all orders. 



Chain rule . There is a chain rule for these 
differential operators, which we now describe. Suppose 
V and W are open sets in C and h: V - W is a class C 
mapping of V into W. 

Let u: W - C be of class C . Then u<?h is a] so of 
class C and if we let D-, denote partial differentiation 
with respect to the first argument and D., with respect to 
the second argument, the usual chain rule reads 

D-^uch) = (D 1 u)ch D-^Reh) + (D 2 u)rh D-^Imh). 
D 2 (uoh) = (D u)oh D z (Reh) + (D 2 u)?h D^Clmh). 



Therefore . 



d(u°h) = (D-,u)oh a(Reh) + (D^u^h d(Imh) 

/■^ \ i. Bh+ah , /t\ \ u oh- dh 
= (D-,u)oh ■, + (D 2 u)oh —j-. 

1 1 1 I 

= (7D u + ^HD 2 u)oh ah + (2D ; ,u j-,- D 2 u)oh3h 

= (Su)oh dh + (Bu)eh ah. 



The corresponding formula for o(ush) follows the same way 
We thus obtain 



164 VI 

a(uoh) = (dh)oh dh + (du)oh 3h, 
"3~(u°h) = (Su)oh ~bh 4- (3u)ohdh~. 

As a special case, suppose h is holomorphic . Then 3h = h', 
dh = 0, so we obtain 

d(u°h) = (3u)oh h', 
(1) 

S(uoh) = (3u)oh F . 

2 2 

a ft 

Now define A = — j + — j (the Laplacian ) ; as we 

3x fty 

have seen, A = 4 33 - 433. Thus, if h is holomorphic, 
the above chain rule implies 

A(uoh) = 4l[(3u)°h h'] 

= 4"i[(ftu)oh]h' + 4(*u)oh "ih' (Leibnitz' 

rule) 

= 4(3ftu)oh"h 7 h / + 0, 

so we obtain 

2 
(2) A(uoh) = (Au)oh |h'| . 

We need one more formula involving 3. Suppose f is 
holomorphic. Then 

5Ref =il+-d". f/ + ° , 

so we have 



VI 165 



(3) 3 Ref = \1' 



DEFINITION 1. Let u be a real-valued function 



defined on a Riemann surface S. Then u is harmonic 
if for every chart ■■-; u - W in the complete analytic 
atlas for S,, u >$ is harmonic on W. 



PROPOSITION 1. Let S be a Riemann surface and 



u: S - ?,. Then the following conditions are equivalent . 

1 . u is harmonic . 

2 . For each p € S there exists a chart r D : U -• W 
in the complete analytic atlas for S such 
that p € U and u°cp is harmonic in a 
neighborhood of cp(p) . 

3 . In a neighborhood of each point of S 
there exist holomorphic functions f and 
g such that u = f + g. 

4 . In a neighborhood of each point of S there 
exists a holomorphic function F such that 
u = ReF. 

Proof ; 1 => 2 : Trivial. 

2 =» 1 : If u°cp is harmonic near rc(p) as 
in condition 2, and if i|i is any compatible chart near p, 
then 

UO\|) =UO!p 0((£0\|j ) , 

so uof is also harmonic by formula (2) on p. 164 . This 



166 VI 



proves 1 



2 =» 3 : Since uo^ is harmonic, d(u° c ) is 



holomorphic. Locally, any holomorphic function has a 
primitive, so there exists a holomorphic function f 
near p such that near <s(p) 



aCuoof 1 ) = (foa," 1 )' 



Define g = u-f. Then 



d(go^) ) =5>(g°cp ) - 5(uocp" ) - d(f°cp~ ) 



= o(uo, 4 ) - (fo _1 ) ' 



= 0. 
Thus, gc- is holomorphic, proving g is holomorphic 



3 =■ 4 : Using u = f + g, we have since u is 
real, u = Reu = Ref + Reg = Ref + Reg, so we merely 
take F = f + g- 

4 => 2 : We have u ^ = Re(Fo^ ) is harmonic 

near tp(p) . 

QED 

PROPOSITION 2 . Let S and T be Riemann surfaces , 
F : S - T an analytic mapping . If u is a harmonic function 

on T, then uoF is harmonic on S. 



VI 167 



Proof : If 33 is a chart on S and v a chart on T. 
then we must investigate (uoF)oc • This is 

(uoF)o-; = ue,' o(\jJoFocp ) 

and we know uo i) is harmonic and tyoFog is holomorphic 
Therefore,, formula (2) of p. 164 obtains. 

3£D 



PROPOSITION 3. Let S be a connected Riemann surface 



and u a harmonic function on S . If u van ishe s on a 
neighborhood of some point of S , then u = . 

Proof : Define A = L p £ S: u = in a neighborhood 

of p}. Then A is open by definition and A £ by hypothesis 

We now demonstrate that A is closed: suppose p is a 

limit point of A. By criterion 4 of Proposition 1. there 

exists a holomorphic function F near p such that 

u = ReF near p . Thus, ReF vanishes on some open set 

near p , namely on the intersection of A with any 

neighborhood of p where F is defined. But then F must 

be constant on this open set and by the uniqueness 

of analytic continuation F is constant. Thus, u is 

constant near p and thus p € A. Since A is open and 
*o r o v 

closed and not empty, and since S is connected, A = S. 

SED 

The fundamental Theorem 3 actually follows from a 
theorem on the existence of harmonic functions , which we 
now state . 



168 VI 

THEOREM 4 . Let S be any connected Riemann surface 
and let p £ S . Let ©: U - W be a chart in the complete 
analytic atlas for S with p € U and cp(p) = 0. Let n be 
a positive integer . Then there exists a harmonic function 
u on S - [p j such that for z near 

uocrj (z) = c logjzj + Ref(z), 

where c is some real constant and f is meromorphic in 
a neighborhood of and has a pole of order n at 0. 

Thus., Theorem 4 guarantees the existence of a 
harmonic function on S - {p} with prescribed singularity 
at p. For emphasis., we repeat that the order of the 
pole of f at is exactly n: in the notation of p. 38,. 
& f <0) = -n. 

Now we shall indicate how the knowledge of Theorem 

4 leads to a proof of Theorem 3. Let p ,q be the 
r *o ^o 

distinct points on S mentioned in the hypothesis of 
Theorem 3. Let u be harmonic on S - ipj with repre- 
sentation near p as prescribed by Theorem 4 with 
(say) n = I: 

uonj (z) = c log|z| + Ref(z), 

f (z) = — + ... (Laurent expansion 
near 0) , a # 0. 

Let i|i be a chart near q , i|f(q ) = 0, and let v be a 
harmonic function on S - {q } with expansion near 



VI 



of the form 



169 



voiji" (z) = d logjzj + Reg(z), 



g(z) = ^ + ..., p t 



Using these two harmonic functions we shall construct 
the meromorphic function required in Theorem 3. 
Here is how it is done: let p, € S and let o be a 
chart near p, (in the complete analytic atlas for S) . 
Near p, we define 



F(p) 



B(voa" )(a(p)) 



First, we show this definition to be independent of a. 
Let a-, be another chart near p 1 and let h = aoa, , so 
that h is holomorphic and has a holomorphic inverse. 
Then formula (1) of p. 164 implies 

a(uo a J 1 )(a 1 (p)) = S(uoa" 1 oh)(a 1 (p)) 

= &(uo a " 1 )(h(a 1 (p)))h / (a 1 (p)) 
- B(u=a" 1 )(a(p))h / (a 1 (p)). 

Therefore, 

3(uoa^ 1 )(a 1 (p)) a(uoa _1 )(a(p)) 

- — _ } 

d(v°c^ 1 )(a 1 (p)) 5(voa" 1 )(a(p)) 

since the common nonzero factor h'(a-,(p)) cancels after 
division. Thus, the definition of F is independent of 



170 VI 



the choice of chart 



Next, since uo<t and voo are harmonic, the 
functions d(u°j ) and $(voa ) are holomorphic, and 



not identically zero since otherwise e.g. v»a would 
be holomorphic (Cauchy-Riemann equation) and thus 
constant (since it is real-valued) . But then Proposition 
3 would imply that v is constant on S - [q }., which 
manifestly contradicts its singular behavior near q . 
Thus, the zeros of d(v°o ) are isolated, so the for- 
mula for F exhibits F as the quotient of two holomorphic 
functions near p, , the denominator not vanishing iden- 
tically, and thus F is meromorphic near p., . Thus, F 
is meromorphic on S - [p } - (q }. 

Finally, we must examine the behavior of F near 
p and q . Near p we use the chart & and compute 
according to (3) of p. 165 

SCuosf 1 ^)) = J^ + £f'(z) 



2z 

so that F°co has a pole of order at least 2 at 0. 

Thus, F(p ) = oo. Likewise, near q we use the chart 
vt o n o 

is and compute 

d(vov (z)) S_ + . . . , 

2z Z 

so that F° ty has a zero of order at least 2 at 0. 



VI 171 

Thus, F(q ) = 0. This concludes the proof of Theorem 
3. 

We have therefore finally reduced the problem 
to that of demonstrating Theorem 4. It will take a 
considerable amount of machinery and technique in the 
area of harmonic function theory to accomplish this, 
so we now begin a discussion of the relevant properties 
we need. 

Proposition 4 . Let u be continuous on &, the 
closure of an open disk a c C, and harmonic in £. 
Suppose l has center z and radius r. Then 

1 2rT in 

u(z ) = ■+- r u(z + re b )dQ. 
o 2tt Jq o 

Proof : Let < p < r. Then the divergence 
theorem implies 

o = r (-£-£ + -^-§)dxdy = j |a dS, 

lz-z o i< P * x *y |z-z o i= P v 

where dS is the element of arc length on the circle 
I z-z I = p and ■— = is the directional derivative in the 

direction of the outer normal. Another way of writing 
this is 

- lJ^^ z o + P ei9 ))P d e. 



172 VI 

Dividing by p and then moving -r— outside the sign of 

o p 

integration implies 

a 2 n i a 
= -I- f u(z + pe lb )de. 
dp oq o 

Therefore, the continuous function of p fe [0, r] given 

by 

1 ^ n i9 

p - -sr- f u(z + pe )da 
2n J o 

is constant. Since its value at p = is u(z ), the 
result follows. 

QED 

Now we show how to apply this simple property of 
harmonic functions to obtain a representation of u in 
all of A, not just at the center. First, we take A 
to be the unit disk, [z: |z| < 1} for simplicity of 
computations. Let a € a and consider the Mobius 
transformation 



T(z) = 



1-az 



T maps A onto a conformally, A onto A, and T(a) = 
Thus uoT is harmonic on , 
that Proposition 4 implies 



Thus uoT is harmonic on A, continuous on A, so 



uoT _1 (0) = 4- f n uoT _1 (e ir5 )d K . 



2 

Now we introduce the change of variable 



VI 



173 



e ie = T -i (e ic ) 



Then e c = T(e ), so that a simple computation yields 

iG - i9 

dcp _ e ae " 

e - a 1-ae 



10 

+ 



io -ie - 

e - a e -a 



, - ie.- i9 - 
1-ae +ae -aa 

, in w -ie - N 

(e -a)(e -a) 



l-|«| 2 



ie |2 • 

e -a 



Therefore, 



2 2 

u < a > = 2T J n ^ll* 1 ^ u < ei9 ) d 9 
Ztt J e l8 -ar 



Define 

2 

(4) P(z,e ie ) - * ^ | z | 2 , |z| < 1; 

| e -z | 

this is the so-called Poisson kernel. We want to observe 



certain things about it 

1. P a 0; 

•2tt„/_ ie. 



2. r^PCz^^^de = 1, |z| < 1; 

i 9. 

3. P(z.,e ) is a harmonic function of z; 

4 . for any 6 > 0, 



174 VI 



lim T 
ie i6 i9 P'(z,e 1B )de = o 
z-e ° |e -e °| ^6 



The first property is obvious and the second follows 
from formula (4) applied to the harmonic function u = 1. 

The third follows from the formula for 7?. which reads 

d9 

i-9 - 16 



i9v e ° ze 



2rrP(z,e J ) = -f— + 



19 , - 11 

e -z 1-ze 



exhibiting P as a sum of two harmonic functions of z. 

ie 

To prove the fourth, assume j z-e °| < 6/2. Then 

ifl ie ie ie 

|e -z| s |e -e °|-|e °-z| > 6-6/2 = 5/2, so that 

2 
I P(z,e l9 )d9 g j~ 1- I Z L • 2 n < 4^(1-1 zl) 

ie ie 2tt u727 



and this clearly tends to zero as z - e 



These four properties are all we need to establish 
the following converse to formula (4) . 

PROPOSITION 5. Let f be a continuous function on 



the circle z = 1. Define 



f 2n P(z,e ie )f(e i9 )d9, |z| < 1, 
« <( ° 

f(z), |z| = 1. 
Then u is harmonic for | z | < 1 and continuous for \z\ < 1 • 



VI 175 



Proof : The fact that u is harmonic for jz| < 1 fol- 
lows from 3 by differentiation under the integral sign. 

i9 
Clearly., we need only prove that lim u(z) = f(e °) for 

i9 
|z| < 1, z - e , in order to finish the proof. Let 

18 

e > 0. By continuity of f at e , there exists 6 > 

19 c ie ie 

such that IfCe 1 ^) - f(e °) | < j- if | e -e °| < 5. 

Now 2 implies 

19 2 tt 1a ia 19 

u(z) - f(e °) = f P(z,e l9 )[f(e ld )-f(e °)]di 

Choose a constant C such that |f(e "') | <; C for all 9- 
Then 

ie . 

|u(z)-f(e °)| £ f J P(z,e l8 )d9 

id ie 
Ie -e °|<5 



+ 2C J P(z,e lB )d6 

ie ie 

ie -e °bfi 



Since the first integral is bounded by 1, and property 

4 implies there exists 6' > such that the second inte- 

i6 
gral is bounded by -f^ if | z-e °| < 5 ' ' , we obtain 

i8 



|u(z)-f(e °)| < c 



18 
if I z-e °| < 6' 



QED 



Of course, it is not necessary to restrict our 
attention to the unit disk. If we consider functions 



176 VI 



in the disk A of center z and radius r. the formula 

o 

analogous to that of Proposition 5 is 

, 2tt r - j z-z | . 

u(z) = ^r- f ■ ■*-* — 2 f( z + re lb )d6. 

Z * J !re l6 -(z-z o )| Z ° 

This can be derived in the same manner,, or merely 
by considering the change of variable z-z = rw and 
using Proposition 5 as it stands. 

The Poisson integral formula we have just derived 
has several immediate applications which will be of 
great importance to us. For example, we have 

PROPOSITION 6 . Let D be an open set in C and K 
a compact subset of D . Then there exists a constant C 
which depends only on K and D such that if u is har - 
monic in D then 

sup|~| * C sup |u| . 
K ax D 

A similar result holds with — replaced by any deri - 



vative of any order 



Proof : For any z € K there exists a disk A of 

center z and radius r such that the closure of A is 
o 

contained in D. For |z-z | < -jr , the Poisson integral 



formula implies 



where 



l-^(z) | 5 c sup|u| , 



VI 177 



2 I I 2 
. r - z-z 

I ' O ' I 
C = SUP -r— — r 

-i dx | 13/ viz ' 
I I ^ 1 re - (z-z ) 
z-z < -?rr ' v o x ' 
' o 2 

is easily seen to be finite. Since K can be covered by 

finitely many such disks as [z: j z-z | <— r], the result 

follows . 

QED 

PROPOSITION 7 . Let D be an open set in C and UpU^, . . . 
a sequence of harmonic functions in D which converge uni - 
formly on compact subsets of D to a function u . Then u 

cu . 

is harmonic in D and the sequence n converges to — . 

also uniformly on compact sets in D . 

Proof : If a is a disk whose closure is contained in 
D, then u has a Poisson integral representation in A of 

n or 

the form 

2 

, 2 n r -z-z ia 

u (z) = 4- f ' 2 u (z + re Ld )d9. 

re - (z-z ) 

For fixed z <= A let n - co in this formula and use the 

uniform convergence to pass the limit under the integral 

sign to obtain 2 ? 

u < z > - h L r: — : »<« +» 18 >«. 

|re 10 -(z-z o )| 2 

Therefore, u is harmonic in is. Therefore, u is harmonic in D 

Now suppose K is a compact subset of D. Choose an 
open set D-, such that K c D, and the closure of D, is a 
compact subset of D. Let C be the constant of Proposition 
6 relative to K and D-, . Then 



178 VI 

OU -, 
SUP 'lE "SF * C SU P l U n" U 

K a ?X D n 
By hypothesis, sup J u -u| -> as n - <*,. and therefore 

D i n 

-^r- - -^7 uniformly on K. As K is arbitrary, the 
result follows. 

PROPOSITION 8 . Let D be an open set in C and 

u n • u ,, . . . a sequence of harmonic functions in D which 

i ^ — — ' ■ ■ 

are uniformly bounded on every compact subset of D . 
Then there exists a subsequence n-, < n ? < ... such that 

lim u 

exists uniformly on compact subsets of D . 

Proof : If A is a disk such that its closure is a 
compact subset of D, then there exists a constant C 
depending only on a such that |u (z) j <, C for z € k, 
n s 1. Therefore, Proposition 6 implies that if \l 
is the concentric disk with half the radius of L, 
then for some other constant C 

du 3u 

l-srl * c v l-#l * c i on **■ 

Now we apply the mean value theorem on the disk \L (details 
omitted) to conclude that for z, z' € \k> 



VI 179 

|u n (z)-u n (z')| s 2C 1 |z-z'| • 

This proves that the family of functions u,,u...... is 

equicontinuous on %A. Since A was arbitrary, it follows 
that the family u. , u^,... is equicontinuous on each 
compact subset of D. By the Arzela - Ascoli theorem, 
there exists a subsequence with the required property 
that u converges uniformly on compact subsets of D. 

QED 

DEFINITION I . An open subset D of a Riemann 
surface is an analytic disk if there exists a chart 
cp: U - W in the complete analytic atlas such that 
c(D) is a disk whose closure is a (compact) subset 
of W. 

Notation . If A is a subset of a topological space, 
A denotes the closure of A and BA denotes the 
boundary of A. 

PROPOSITION 9 • Let D be an analytic disk in a 
Riemann surface S and let f : 3D - n be continuous . 
Then there exists a unique function P^ on D such that 
P.p is continuous on D , harmonic in D, and P^ = f 
on ^D. 

Proof : Let $: U - W be a chart in the complete 
analytic atlas for S satisfying the condition of 
Definition 2. If : (D) = [z: |z-z | < r], then 



180 VI 



PfoiiT must be continuous on r(D) • harmonic on <£>(D), 

and P f °c&"" = foqT on 5qj(D) (= ep(dD)). Thus, if 

z € a(D), then 

2 r 2 -lz-z [ 2 
P f 8 T _1 (z) - JZ J" " , r i9 ' Z Z ° ,2 f^ _1 (2 +re ie )d9 



Therefore, P f is uniquely determined and Proposition 5 
implies that P as defined by th 
the conditions of Proposition 9. 



implies that P as defined by this formula satisfies 



QED 

DEFINITION 3 . If D is an analytic disk in a 
Riemann surface S and if u: S -» R is continuous, u^ 
is the unique continuous function on S which agrees 
with u on S-D and is harmonic in D. The existence 
and uniqueness of u^ are guaranteed by Proposition 9 

LEMMA 2 . Let S be a Riemann surface and p £ S 
Let <c > . Then there exists a neighborhood U of p 

such that for all functions u which are harmonic and 



nonnegative on S , and for all p , q g U 
u(p> < (l+e)u(q). 



Proof ; There exists a chart cc: U - W in the com- 
plete analytic atlas for S such that W contains (z: |z|<l} 
and cp(p ) = 0. This can obviously be achieved by com- 
posing an arbitrary chart with a suitable linear trans- 
formation of C onto itself. Let v = u°rp . Then 



VI 



181 



according to p. 173., 



2rr 



v(z) = i" P(z,e l9 )v(e l0 -)de J \z\ < 1 
"0 

Now l-|z| <, | e 1 -z| < 1 + |zl so we obtain 



1- z __ 1- z 



1 ' (l+|zj) |e -z| (l-|z|) 



= i±Ui 

l-|z| 



i-j. 



Therefore, since v(e ) > 0, 



1-lzl 1 ,2rr 



1+1 zl 1 „2tt , i8. 



1+ z Iji 



— I v(e )d6 < v(z) < ' — L -5— f v(e )de 



1- z 







By Proposition 4 this pair of inequalities can be 
written in the form 



I V (0) < V (z) ,™.v(0) 



1+ z 



1- z 



If < 6 < 1 and |zj < 6, we obtain 



■^ v(0) < v(z) ^ ±±± v(0) . 
1+6 1-6 



Therefore, if jz| < 6 and |w| < 5, 

v<0) £ (-±±V 
1-5 1-6 



v(z) < ^v<0) s (±±^) v(w) 



182 VI 

Pick 5 such that 

(1T5) £ 1 + «• 

Then let U = cc~ ({z: |z| < 6}). This is a neighborhood 
of p = '5 (0) and for p,q f U, 

u(p) = v(cp(p)) s (l+e)v(ro(q)) = (l+c)u(q). 

QED 

Harnack's Inequality . Let S be a connected Riemann 
surface and K a compact subset of S. Then there exists 

a constant C depending only on K and S such that for 

all nonnegative harmonic functions u on S and all p,q € K^ 

u(p) • Cu(q). 

Proof : It obviously suffices to consider the class 
,! of functions which are harmonic and positive on S ; if 
u is harmonic and u > 0, then for every e > 0, u+e £ H 
and if the inequality is true for functions in u then 
u(p) + e < C(u(q) + e) . Then let e-*0. Of course, we 
are debating a triviality anyw.ay, because Harnack's 
inequality implies that if u a and u is harmonic, 
then either u = or u > 0. Now choose some fixed 
point p £ S and define 



o 



u(p) U (P„) 
F(p) = sup max (^_ } . -tfjj) : u € I 



o' 



VI 183 

We are going to prove F is continuous. Let p, f S and 
let e > 0. Let U be a neighborhood of p, satisfying the 
condition of Lemma 2. Then for u € d and p,q € U, 



$1, , (l +e) Hgij . (1+0F(q) . 



u(p n ) u (pJ 

_ °- s (1+s) _°_ s (l+ e )F(q) 

u(p) u(q) 



Therefore, 

(5) F(p) < (1+e) F(q) for p,q 6 U. 

In particular, if F(p,) < « we choose q = p, to conclude 
that F(p) < m for all p 6 U; if F(p.,) = « we choose 
p = p-, to conclude that F(q) = a. for all q £ U. Therefore, 
the sets 

[p € S: F(p) < ■}, [p € S: F(p) - »} 

are open. As they are obviously disjoint and their union 
is obviously S, the connectedness of S implies one of 
these sets is S, the other empty. Since F(p ) = 1, p 
belongs to the first of the sets, and thus we have 
proved that F < a everywhere on S. 

Now we obtain the continuity. Taking q = p, in 
(5), 



taking p = p^, 



F(p)-F( Pl ) < eF(p T ) if p € U; 
-eF( Pl ) s (l+e)(F(q)-F( Pl )) if q 6 U. 



184 VI 

Thus, we obtain 

" 1+7 F(p l } " F (P)- F (P!) * eF( Pl ) if p € U, 

and since e is arbitrary, this proves that F is 
continuous at p, . 

Since F is continuous on S and K is compact, there 
exists a constant c such that F(p) < c for p 6 K. There- 
fore, if p, q £ K 

u(p) < F(p)u(p o ) s F(p)F(q)u(q) <c 2 u(q). 

Harnack's Convergence Theorem . Let S be a connected 
Riemann surface and 3 a nonvoid family of harmonic 
functions on S which is directed upwards , i.e., if 
u,v € 0, there exists w £ 3 such that w 2 u, w 2 v. 
Let U = sup 3, i.e., for p € S 

U(p) = sup[u(p) : u € 3} • 

Then there exists a sequence u-. < u.. <. u„ «j . . . such that 
u € 3 for all n and u - U uniformly on compact subsets 
of S . Moreover , either U = 00 o_r U is harmonic on S . 

Proof : Let u be an arbitrary function in 3 and 
let 

0' = [u e 3: u s u } . 

Then sup 3' = sup 3. Indeed, since 3' c 3 the inequality 
sup 3' <. sup 3 is obvious, and if u € 3 then there exists 
v € 3 such that v s u and v s u ; therefore, v 6 3' and 



VI 185 

v s u, proving that sup 3' 2 sup r*. 

For any compact set K c S let C„ be the corres- 
ponding constant in the conclusion of Harnack's inequality 
Then for u € 3 ' , u-u is a nonnegative harmonic function, 
so for all p,q € K it follows that 

u(p) ~ U ^P) * c K (u(q) - u (q)) 
* C R (U(q) - u o (q)). 
Taking the supremum over all u € $' implies 



(6) U(p) - U o (p) g C K (U(q) - U Q (q)) 



It fellows that if U(q) < =0, then U(p) < <*, and here p,q 
can be any points in S (just take K to be the compact set 
Cp,q)). Therefore, either U s » or U < ». 

Let p € S be fixed and choose a sequence u/.u^u^, .. 

from 3 such that u ' (p ) -• U(p ). By hypothesis, we 

can let u-, = u-,' and then find inductively u € such that 
li J n 

u 2 u ' , u s u , . 
n n n n-1 

Then we have a sequence u, ^Uo <u.,s. . . from 3 such that 

u (p ) - U(p ). Note that (6) holds for an arbitrary 

u € 3, so we can take u = u in (6) . If U(p ) = », 
o on vr o 

then for any compact set K containing p Harnack's in- 
equality implies 

U n (p o } ' U l (p o } " C K (u n (q) " u l (q)) > q € K > 



186 VI 



and therefore u (q) -» « uniformly on K. This proves 

the result in case U = ». If U(p ) < », then (6) implies 



U(p) - u n (p) < c K (u(p o ) - u n (p Q )), p eK, 

and therefore u (p) - U(p) uniformly for p € K. Finally, 
Proposition 7 shows that U is harmonic in this case. 

QED 

Now we need to introduce the basic building block 
other than the Poisson integral, which is subharmonic 
functions. The basic theory is contained in the following 
proposition . 

PROPOSITION 10. Let u be a continuous real-valued 



function on a connected Riemann surface S . Then the 
following conditions are equivalent. 

1. For every analytic disk D c S, u <. u~. (Cf. 
Definition 3 .) 

2. If s is a proper open subset of S, if & is 
compact, if h is continuous on © and harmonic 
in <s, and if u <. h on S^, then u & h in &. 

3. For each p € S there exists a chart ®: U - W 
in the complete analytic atlas for S such 
that p £ U and 



1 -1 i 9 
u(p) s -j — f uojp (ep(p)+re )de 



for small positive r , 



VI 187 



4. For each p € S and every chart -: U - W 
in the complete analytic atlas for S 
such that p € U and every z sufficiently 
close to co(p)j 

-1 1 2rr -1 ie 
uo :i (z) s o— i u-2 (z+re )d9 

2TT J Q 

for small positive r . 

Proof : We are going to establish four implications, 
three of which are absolutely trivial. 

2=1 ; Assume that 2 holds and let D be an 
analytic disk. Use 2 with & = D and h = u_. restricted 
to & . Then u = h on ^D so we obtain u sh in D, i.e., 
u < u n in D . Since u = u„ outside D, 1 follows. 

1=4 : Assume that 1 holds and consider the 
analytic disk 

D = cp~ ( {w: | z-w| <r}) 

for sufficiently small r. Then u < u„, so in particular 

UOep (z) < UpOcr, (z) . 

Since u^or is continuous on j_w : |z-w| s r} and har- 
monic in the interior, the mean value property of 
Proposition 4 implies 

V^" 1(z) = -7^ i u D o,' 1 (z+re ie ) de 
= i- f 2n uo cp " 1 (z+re i6 )d6 



188 VI 

since Up. = u on dD . Thus, 4 follows. 

4=>3 : Completely trivial: we allow e very chart in 
4 and moreover 3 is just 4 at the single point z = >.o(p) • 

3=>2 : Finally here is something which requires 
thought. Assume that 3 holds and assume we have the 
hypothesis of 2. Define v = u-h in 9 . Let M = sup _v. 
Since Q is compact and v is continuous, the supremum 
is attained, so the set 

A = {p e <$: v(p) = M} 

is either nonvoid or v = M somewhere on S&. In the 
latter case, since v < on fiS we obtain M < and the 
result follows. So we assume v < M everywhere on B", 
in which case A is not empty. Since v is continuous, 
the set A is closed relative to &. We use 3 to show that 
A is open : suppose p £A. Pick a chart co according to 
3 with respect to the point p. Then for small positive r 

u(P> < TZ r TT u r " 1 (co(p) + re i9 )de. 

Since h is harmonic, it satisfies the similar relation 
with equality instead of inequality (Proposition 4) and 
thus we obtain by subtraction 

v(p) *T-[ 2ri vo,/ 1 ( cp (p) + re i9 )de. 
But the left side is v(p) = M and the integrand 



VI 189 

- 1 i0 
v°co (;x(p)+re J ) ^ M by definition of M. Thus, we con- 
clude that equality holds everywhere, so 

vocd" 1 (. p (p) + re 19 ) = M 

for < 9 ^ 2 n and small positive r. Thus, v = M near 
p, so A is open. 

It is now a simple topological argument to show 
that SA and S3 have a point in common. To see this, 
let Pq £ A and p, £ S-<§ be chosen arbitrarily, and 
use the connectedness of S to conclude that there 
exists a path y 1° S from p^ to p,. Since p.-. £ A 
and p, | A and the image of y is connected, there exists 
a point p.-, in the image of y such that P2 € dA . If a 
neighborhood of p., were disjoint from ©, it would also 
be disjoint from A, contradicting p~ € A . Therefore, 
p^, € (s . If p., € ©,• then since A is closed in &, 
p^ t A; since A is open, this contradicts the fact 
that p 2 g (S-A)~. Thus p 2 € &©• Since p 2 £ dA, 
v (p 2 ) = M by continuity of v. Since p.. £ a©, v(p 2 )<0 
by hypothesis. Therefore, M s 0, and the conclusion of 
2 follows. 

QED 

DEFINITION 4 . A continuous real-valued function u 
on a Riemann surface -S is subharmonic if it satisfies 
condition 4 of Proposition 10. 



190 VI 



Strong Maximum Principle . Let u be a subharmonic 

function on a connected Riemann surface S such that usO . Then 
either u <0onSoru=0onS. 

Proof : This is contained in the proof of 3 => 2 of 
Proposition 10. For the set A = [p€S: u(p) =0} is 
closed since u is continuous and is open by Condition 
4 of Proposition 10, and thus either A = S or A is 
empty . 

QED 

Weak Maximum Principle . Let S be a connected 
Riemann surface , and & a proper open subset of S such 
that & is compact . Let u be continuous on is and 
subharmonic on &. Assume u <, on 9&. Then u <, 0. 

Proof : This is again contained in the proof of 3=>2 
of Proposition 10. If M = max -u and if M > 0. let 
A = [p £ &: u(p) = M}. Then the argument proving 
3 => 2 shows that 3A and 33 have a point in common 
and thus M < 0, a contradiction. 

QED 

COROLLARY . If G is a proper open subset of a 
connected Riemann surface S such that & is compact , 
and if u is harmonic on & and continuous on & , then 

sup _ | u| = sup | u| . 



VI 191 

Proof : Let M = sup |u| . Then -M+u and -M-u 
are subharmonic in © and nonpositive on 3S, so the 
weak maximum principle implies -M+u <; and -M-u <, 
in &. That is, -M <, u <; M. 

PROPOSITION 11. Let u be a continuous real-valued 



function on a Riemann surface S . Then u is harmonic if 
and only if u and -u a re subha rmon ic . 

Proof : We only have to prove the "if" part of 
the assertion. Since the proposition deals with local 
properties, we can assume S is connected. By part 1 
of Proposition 10, if D is an analytic disk, then 
u <, u Q and -u <; (-u) D - But clearly (~u) n = -u^, so 
we have u <, u_ and -u <, -u D - Thus, u s u D . Therefore, 
u is harmonic in D. Since every point of S is contained 
in an analytic disk, u is harmonic in S. 

QED 

PROPOSITION 12. Let u ,u l3 ...,u be subharmonic 
====== — 1' n 

on a Riemann surface S, and let a-, ,a , . . . , a be non- 

^ 2' n 

negative real numbers . Then the functions 

a,u, + ... + a u , 
11 n n 

maxfu, , . . . ,u ) 
v 1 n' 

are subharmonic . Also , if D is an analytic disk, u n is 

subharmonic . 



192 VI 

Proof : This follows directly from the definition. 
Condition 4 of Proposition 10 asserts in that notation 

that for small positive r 

2 
V^O) < ^- J \o^ _1 (z+re i0 )de. 

Multiplying by a, and adding, the function a.,u, + . . .+a u 

is seen to satisfy condition 4. If u = max(u, ,...,u ), 

then we have 

":tt 





2,i 
Ui ecc (z) < ~— a uopxi (z+re )d9, 1 £ k s n 

K ZfTJ 



Therefore, 



-l l ^ n -l ie 

uo'j) (z) <; ■=— P uoco (z+re )dQ, 
2tt Jq 



proving that u is subharmonic. The last statement will 

be proved on p. 196. 

O^ED 

The basic theorem we need is the following. 
THEOREM 5. Let S be a connected Riemann surface 



and j? a nonempty family of subharmonic functions on S 
such that 

1. if u,v eg, then max(u,v) € n, 

2 . If. u - 3 and D is an analytic disk in 
S, then Up € a- 

Then supo is either harmonic in S or' sup g s ». 



VI 193 

Proof : Let U = sup 3. If D is an analytic disk 
in S, let IV be the functions on D defined by 

Hq - (u D : u e o)- 
Then ,-l is a family of harmonic functions on D and 

sup 3^ = U in D . 

For, u € implies Uj, € 3, so that any function in £j_ 

is the restriction to D of a function in 3 and thus 
sup 1. s U in D. On the other hand, u € implies 
u s u„ by Proposition 10.1. Therefore, U < sup ju 

in D. 

Now we apply Harnack's convergence theorem to 
the family x. on the Riemann surface D. We have to 
check that ru is directed upwards. So suppose u,v € r<. 
Let w = max(u,v), so that w € 3 by property 1. Then 
u £ w implies u~ <> w_ and v ^ w implies v-. «j w~, so 
that we have found w^ £ 3_. such that w n 2 tu, w^ 2 v_.. 
Thus 3 n is directed upwards. Thus, Harnack's conver- 
gence theorem implies that either sup jl. is harmonic or 
sup 3^ = =0. Therefore, either U is harmonic on D or 
U = co on D . 

Finally, we have the familiar connectivity argu- 
ment: if A = ip € S: U(p) = w} and B = {p € S: U(p) < »}. 
then A and B are disjoint open sets with union S. Since 



194 VI 

S is connected, either A is empty or A = S. Thus, 
either U < » on S or U = <a on S. If U < » on S, we 
have shown that U is harmonic in every analytic disk 
in S. Therefore, U is harmonic. 

QED 

Problem 8 . (The Dirichlet problem for an annulus) 

1. Prove the Weierstrass approximation theorem 
for a circle. That is, if f is a contin- 
uous complex-valued function on the circle 
|z| = 1 and e > 0, then there exists a 
finite sum 

g(z) = "a z (positive and 
negative n) 
such that |f(z)-g(z)| < e for |z| = 1. 

Hint : Use Proposition 5 and Proposition 1, 
with the obvious remark that the proof of 1 => 3 for 
the disk |z| < 1 gives two holomorphic functions defined 
on the entire disk. 

2. Consider the annulus r < |z| < 1, where 

< r < 1 is fixed. Let n be an integer. 
Exhibit the (unique) harmonic function 
which equals z for |z| =1 and equals 
for I zl = r. 



195 



3. Combine 1 and 2 to conclude that there 

exists a function u which is harmonic 
e 

for r < |z| < 1, continuous for r s |z| ^ 1 
(in fact, it will be harmonic for 
< |z| < oo) such that 

|u (z)-f(z)| < e for |z| = 1, 

u (z) = for |z| = r. 
e 

4. By a limiting argument, prove there exists 

a function u which is harmonic for r < |z| < 1. 
continuous for r <. |z| < 1, such that 

u(z) = f(z) for |z| = 1, 
u(z) = for |z| = r. 

5. Use this result and an appropriate conformal 
mapping to treat any continuous boundary 
values on |z| = r as well. 

Now we state a corollary, and we use the obvious 
terminology that a function w is superharmonic if -w is 
subharmonic . 

COROLLARY . Let w be a superharmonic function on a 
connected Riemann surface S . Let 

= [v: v subharmonic on S, v < w}. 

Then sup p is either harmonic in S o_r sup L T s -«. 



196 VI 

Proof : We have sup 3 = -« if and only if 3 is empty,, 
so we assume from now on that 3 is not empty. We verify 
the two properties required in Theorem 5. First, if v-^v^ 
€ 3, then clearly max(v-, ,v~) s w and Proposition 12 implies 
max(v, , v.-,) is subharmonic; thus, max(v,,vJ € 3. If D is 
an analytic disk in S, then for v € 3, 
v D < w D * w , 

the latter inequality being a consequence of criterion 1 
of Proposition 10 for super harmonic functions. So we 
need only check v~ is subharmonic. This amounts to 
checking the local criterion 3 of Proposition 10. This 
mean value criterion clearly holds at any point p e D (since 
v n is harmonic near p) and at any point p in S-D (since 
v„ = v is subharmonic near p) . So we consider p € §D and 
a chart cp in the complete analytic atlas for S, ^ defined 
near p. Since v is subharmonic and v < v~, we obtain for 

small positive r 

1 2tt -1 ifl 

V D (P) = V (P) « 27fJ v °tp (cp(p) + re )de 

< h r ^D^'^cpCp) + re i9 )de, 
establishing the criterion in this case as well. Thus, 
v~ is subharmonic on S. Now Theorem 5 implies sup 3 = » 

(which is impossible in this case) or sup 3 is harmonic. 



QED 



VI 197 

DEFINITION 5 . Let w be a superharmonic function on 
a connected Riemann surface. The function 

u = supt.v: v subharraonic on S, v^w} 

is called the greatest harmonic minorant of w. This 
terminology agrees 'with the obvious fact that if v is 
harmonic on S and v g w, then v < u. Moreover, if w 
has any harmonic minorant at all, then u is harmonic 
(not =-«=). Of course, our corollary shows that actually 
u is the greatest sub harmonic minorant of w, and is 
itself harmonic. We shall use the abbreviation 

u = GHM of w. 

As an application of these ideas, we show how to 
solve a certain kind of Dirichlet problem. 

PROPOSITION 13 . Let D be an analytic disk in a 
connected Riemann surface S and f : 3D - R a continuous 
function . Then there exists a function u which is 
continuous in S-D, harmonic in S-D , and such that 
u s f on SD . Moreover, we can assume 



%% 



?£> 



inf u = inf f 
S-D SD 



Remark. Nothing is claimed about the uniqueness of 
u. As we shall see, u is unique for certain S and not 
unique for other S. 



198 



V] 



Proof : By Definition 2 of analytic disk, there 
exists a chart cp: U - W in the complete analytic atlas 
for S such that x(D) is a disk A. Since A~ c W, there 
exists a concentric disk A, with A c A-, , A-, c W 




\ 



\ u 



V 



w 



Let D, = 



1' "1 

-1 



(A 1 ). If c c ? 



then by Problem 8 there ex- 
ists a unique function h 

which is continuous on A-i-A, 
harmonic in A, -A , and such 



that 



h s c on SA, j 

h s fojrj on SA. 
c 



Define a function v on 
c 

S-D by the formula 



h 0| 
c 



in D,-D, 



c in S-D- 



Then v is continuous on S-D, v s f on SD. and v is 
c c c 



harmonic in S-D, and in D-.-D 



If c <. inf f, then v 
SD 



is subharmonic on S-D ; for. the only points where we 

need to check the mean value criterion 3 of Proposition 

10 are on SD, , and there v takes the value c. But 
1 c 

the minimum principle implies h 2 c in A-, - A, and 
thus v 2 c in S-D. Therefore, criterion 3 of Propo- 
sition 10 is trivially satisfied at a point of 9D-, . 
Thus, v is subharmonic in S-D Likewise, if c 2 sup f, 



VI 199 



then v is superharmonic in S-D . 
c 

Let A = inf £, B = sup f . Then v A is subharmonic 
3D 3D A 

in S-D , v R is superharmonic in S-D , and v. <; v in 
S-D. This last inequality follows from the maximum 
principle , since h_-h. is continuous in A, -Aj harmonic 
in A-.-A , = B-A on 3A-, , = on 3lA, and thus h R - h A s 0. 

Let u be GHM of v_ . Then since v. is a subharmonic 

minorant of v^^ we have 
a 



(7) v A & u s v R 

and u is defined and harmonic on S-D . We have of 
course applied the corollary of Theorem 5 to the 
connected Riemann surface S-D , which is why u is 
defined only on S-D . But the inequalities (7) imply 
that u can be extended to a continuous function on S-D 
in exactly one way, namely by taking u = v. s v R = f on 
3D. 

Finally the last assertion of the proposition 
follows from 



A ^v A <. u <v R sB. 



QED 



Remark. The above analysis is typical in the 
sense that even when we wish to have boundary values for 
a certain harmonic function, the corollary to Theorem 5 
does not by itself give anything more than a harmonic 



200 VI 

function on an open set. Some other consideration, 
e.g. (7), is needed to obtain information about the 
function at the boundary. We shall see more instances 
of this phenomenon later. 

To complete the preliminary material, we need to 
obtain a representation for harmonic functions in an 
annuluSj analogous to the Laurent expansion of a holo- 
morphic function. 

PROPOSITION 14. Let u be a real harmonic function 



in an annulus a < j z | < b . whe re <, a < b < m . Then 
there exist unique complex numbers c, (a } , such that 
for a < I z I < b 



u(z) = c log|z| + Re(_E a z ) , 

and a is real . Furthermore , if a < a' < b' < b, then 
there exist constants K and {K } depending only on a ' 
and b ' ( and n ) such that 

|c| <; K supi |u(z)|: a' <- |z| <; b'}, 
|a n l ^ l^supiluCz)! : a' 5 |z| < b'j. 

Proof : The discussion on p. 162 implies 9u is holo- 
morphic for a < |z| < b. Therefore, the Laurent expan- 
sion of du exists, say 

Su= 2cz, a<|z|<b. 
- oo n 



VI 201 



By formula (3) of p. 165 , for n ^ -1 we have 

• i n+1 

c z = -3— c rr = 2 g Re ( n ) > 

n dz n n+I v n+1 

c_ 1 z" = c_ 1 ££ log z = c_ 1 2d log|z| . 

Now the Laurent expansion for >u converges uniformly 
on compact subsets of the annulus., and therefore the 
same is true for the integrated series, so we obtain 

c z 

Su = 2c_ 1 3 log|z| + I 23 Re( n ) 

n^-1 

n 
c z 

= 2s(c . log|z|) + a L 2Re( "^ ) 

_i nfO n 

= d(c log | z| + Re E a z n ) , 
n^O n 

2c 

where c = 2c n and a = . The Cauchy-Riemann 

-Inn J 



equation 



a(u - c log|z| - Re I a n z n ) = 



follows and shows there exists a function g holomorphic 
in a < | z | < b such that 

u = c log |z|+Re £ az +g(z). 
n^O n 

Taking imaginary parts, 

Im g = (Im c) log] z| . 

Since i log z = -arg z + i log | z | , this shows that 
g = i(Im c)log z, and thus g is defined only if Im c 
= 0. Then g is a real holomorphic function, and thus 



202 VI 

is constant. Thus, if g = a~, we have 

u(z) = c log|z| + Re r a z n , c, a~ real, 

a representation of the desired form. 

Nov; we obtain the uniqueness: if a < r < b, then 

, iBv . . 1 °° nin9.1«— n-ine 

u(re ) = c log r + -*• ^ a r e +77 rare 

v y & 2n 2 -as n 

- 03 

1 ' , 1 OT / n , -ri\ in9 

= c log r + t 1 (a r + a r )e 

Since this series converges uniformly for 0^6^ 'In, we 
obtain by integration 

1 2 ^ lfl 
-r — J 1 u(re )de = c log r + su , 

z ' u 

(8) 

I J "uCre^Je^^e - a^r™ + ^ r" m , m * 0. 

Here we have used the orthogonality relation 

1 r 2;T e ine e -im9 de f if m + n, 
~ I ^~' [ 1 if m = n. 

If we use the relations (8) for two different values 
r, r € (a,b), we can solve for all coefficients; 

L T '[u(re i6 )-u(r'e ie )]dq. 



log r-log r Z... q 

a A = -t ^ r -r-r [u(r'e i9 )log r-u(re 18 )log r']d0 

log r-log r 2ttj q e & J 

1 1 (i p / iS N /~m //iBN-mi-imGj^ 

a = ■ — — f Lu(re )r -u(r e )r je dP, 

m „ m . m 

Op-) -<f-) 



VI 203 

m ^ . This proves that the coefficients are uniquely 
determined by u, and at the same time shows easily how 
to obtain the estimates stated in the last half of the 
proposition . 

£ED 

COROLLARY . "Removable singularity theorem" Let u 
be harmonic and bounded in an annulus < | z | < b . Then 
there exists a harmonic function in the disk \z\ < b 
which agrees with u in the annulus < j z J < b . 

Proof : By Proposition 14 we have 



u(z) = c log |z| + Re(_| a z n ) , < |zj < b 

In formula (8) we let r - and we read off the relations 

c log r is bounded, 

a r + a r is bounded, m f . 
m -m 

Therefore, c = and m < implies a =0. Thus the 

r m 

expansion for u reads 

u(z) - Re( I a z n ) . < |z| < b. 
n 

The right side of this expression is harmonic in the 

disk | z | < b. 

QED 

COROLLARY . If u is harmonic and bounded for 
a < |z| < co and continuous for a <. |z| < », and if 



204 VI 



u(z) s for \z\ = a, then u = 0. 

Proof : In formula (8) let r - a to obtain c log a 

n m , — — -m ~ „ . , . 

+ Sq = 0, a a +a a =0. By the reasoning given 

in the previous corollary, a =0 for m > and c = 0. 

r J m 

Therefore, a _ = and a =0 for rn < . Thus, u = 0. 
m 



We are now almost ready to prove Theorem 4. But 
something rather strange will arise in the proof. 
Namely, we shall see that there is a certain dichotomy 
of Riemann surfaces which requires that the proof of 
Theorem 4 be quite dependent on this classification, 
although the statement of the theorem is the same in 
both cases. We present this phenomenon in the form 
of a proposition. 

PROPOSITION 15 . The following conditions on a 
connected Riemann surface S are equivalent. 

1. Every bounded subharmonic function on 
S is constant. 

2 . If D is any analytic disk and u is a 



bounded continuous nonnegative function 
in S-D which is harmonic in S-D and 
which vanishes identically on 3D, then 
u = 0. 



VI 205 

3. If D is any analytic disk and u is a 
bounded continuous function in S-D which 
is harmonic in S-D , then 

sup u = sup u. 
S-D SD 

4 . Same as 3 with "harmonic" replaced by 
1 ' subharmonic . ' ' 

2 ' . Condition 2 holds for some analytic disk 

D. 
3 ' . Condition 3 holds for some analytic disk 

D. 
4 ' . Condition 4 holds for some analytic disk 

D. 

Proof : We shall prove 1 => 2 =» 3 =» 4 and I' => 3 ' =» 4' 
=> 1. Since the assertions 2 => 2 ' , 3=3', and 4 =* 4 ' are 
trivial, the proposition will follow. The proof that 
2 =» 3 is identical to the proof that 2' => 3' and like- 
wise for 3 =» 4 and 3' => 4 ' . 

3 => 4 : As in the proof of Proposition 13, we choose 
a "concentric" analytic disk D-, with D c D. . Suppose u 
is a bounded continuous function in S-D which is sub- 
harmonic in S-D . Choose a constant C such that sup u^-C 

S-D 

Let w be the unique function which is continuous in S-D, 
harmonic in D,-D , such that 



206 VI 

w = u on 3D, 
w - C in S-D, . 

Then w is superharmonic in S-D and criterion 2 of 
Proposition 10 implies u < w in D-, -D and therefore u ^ w 
in S-D. Let v = GHM of w (cf. p. 197). Then v is har- 
monic in S-D and u s v <, w. Therefore, we can naturally 
extend v to be continuous in S-D by setting v = u = w on 
dD . Condition 3 applies to v and thus 



sup u < sup v = sup v = sup u 
S-D S-D 3D SD 



4 / =» 1 : Suppose u is a bounded subharmonic 
function on S. Let D be an analtyic disk on S for 
which condition 4 holds. Then 



sup u = sup u. 
S-D BD 



Therefore, 



sup u = sup u , 
S D" 



and since D is compact, we see that u assumes its 
maximum. By the strong maximum principle, u = constant 

1 => 2 : Let u be the function in the hypothesis of 
2. Define v on S by 

v = u in S-D, 
v = in D. 



VI 207 

Then v is subharmonic and bounded on S, so 1 implies 
v = constant. Thus, v = and it follows that u s 0. 

2^3 Let u be the function in the hypothesis 
of 3. Define 

A = inf u, B = sup u, C = sup u. 
S-D S-D ?D 

Then A <, C s B and we want to prove B = C. As in the 
proof of Proposition 13 and also the current proof that 
3 =» 4 we take a disk D.. and define v. to be continuous 
in S-D, harmonic in D -D , such that 

v. = u on ?D, 
v. = A in S-D, . 

We define v the same way with A replaced by B. Then 
B 

v. is subharmonic. v„ is superharmonic in S-D , and 

A D 

v. <, C, v. <, u, u < v-ui all of which follow from the 
A A d 

maximum principle. Let 

w, = GHM of min (u,C), 
w = GHM of v„ . 

Note that minCu^C) is superharmonic by Proposition 12 . 
Then the inequalities we have obtained for v. and v„ show 

V A £ W X £ U £ W 2 S V fi . 

Thus j, w, and w ? have continuous extensions to S-D with 

w, = w- = u on SD . Therefore,, w~-w, satisfies the hypothe- 



208 VI 

sis of 2, the required boundedness following from 

w,-w, < v R -v. s B-A. Thus, condition 2 implies ^-w-i-O. 

Thus 

B = sup u = sup v/-, < C . 
S-D S-D i 

^ED 

DEFINITION 6 . A noncompact connected Riemann 
surface satisfying the conditions of Proposition 15 
is a parabol ic Riemann surface. A noncompact connected 
Riemann surface not satisfying these condition is hyper - 
bolic . 

Examples . 

1. If S is compact and connected, S satisfies 
the conditions of Proposition 15. For 
suppose u is a bounded subharmonic function 
on S. Then u assumes its maximum, so the 
strong maximum principle implies u is 
constant . 

2. C is parabolic. We verify condition 2' for 
D - [z: |z| <1}. Suppose u is a function 
satisfying the hypothesis of Proposition 15. 
criterion 2'. By the second corollary on 

p. 203, u = 0. 

3. If S = {z: |z| < 1} has its usual complete 



VI 209 

analytic atlas, then S is hyperbolic. 
This is obvious, a nonconstant subhar- 
monic function which is bounded on S 
being, for example, z -> Rez. 

Finally , the stage is set for the proof of Theorem 
4. In the statement of the theorem, on p. 168, there is 
a given chart &: U - W in the complete analytic atlas 
for S, where U contains the given point p and ^(p) = 0. 
By a simple change of variable, we can assume that 

[z: |z| <; 2] c W. 
Let A = [z: |z| < r} and D = x~ (A ) for < r < 2. 

Proof of Theorem 4 in case S does not satisfy 
the conditions of Proposition 15 : By criterion 2', 
there exists a bounded continuous nonnegative function 
v in S-D, which is harmonic in S-DT and which is 
identically zero on SD, , and yet v ^ 0. The strong 
maximum principle implies that v > in S-D7. The 
function cp is holomorphic on U, and therefore 
Re(rp -qj ) is harmonic on U-fp} and in particular 
is harmonic on D 2 • On 6D-, , |cp| =1 so that cd =ip 
and thus cp -cp is purely imaginary and thus 
Re(cp n -ro ) = 0. Since v > on the compact set 3D 2 , it 
is bounded below by a. positive constant there. 
Therefore, there exists a constant C such that 



210 VI 



Re(rD~ n --c n ) I s Cv on SD. 



The same inequality holds trivially on 3D, , both sides 
vanishing, and therefore the weak maximum principle 
implies 

(9) |Re(co" n -co n )| <, Cv in Do - D, . 



Now we define 



•Cv in S-D,, 

1 

t Re(cp" n -cp n ) in B ± - {p] 



f Cv in S-D, 
1 1' 

." „ / -n n> 

^ Ke^cp ~rp ) ill li. 



I Re(.^ -rn ) in D, - Lp] 



Then w, and w~ are clearly continuous on S - [p] and (9) 

implies w, is superharmonic and w^ is subharmonic in 

S - {p}: it suffices to check the mean value property 

of Proposition 10.3 at points on SD, and at such a point 

w, = = Re(cp -co ) = the mean value of Re(co to ) on 

small circles centered at the point s the corresponding 

mean value of w., (since Re(r r n -co n ) £ - Cv on the part of 

the circle lying outside D-. ) . Thus, w, is superharmonic 

and a similar proof shows w- is subharmonic. 

Choose a constant A > 2C sup v. Let u = GHM of 

S-D, 



VI 211 

w + A. Note that trivially w^-w < A, and therefore 

Wo <; u <, w, + A . 

We have of course used here the existence of GHM on the 
Riemann surface S - {p } . Therefore, u is harmonic on 
S - [p] and our inequalities show 

<; u-Re( ¥ " n -^ n ) ^ A in D x - ip}. 

Therefore, the function u-Re(<c -cp ) is harmonic and 
bounded in D, - [p } , so the removable singularity theorem 
of p. 203 shows that there is a harmonic function h in 



D, such that 



u = Re(c2" n ) + h in D- 



1 

Since h = ReF for some holomorphic function F in D, 
(Proposition 1.4), we have proved Theorem 4 in this case, 
and we can even assert that no term log |tpj appears in 
the representation for u. 

Proof of Theorem 4 in case S does satisfy the con - 
ditions of Proposition 15 : By Proposition 13, there 

exists for < r < 1 a bounded continuous function u in 

r 

S-D such that u = Re(. £ ) on 3D and u is harmonic in 
r r V4/ ' r r 

S-D . If u is constant on 3D-,, then Proposition 15.3 
implies u is constant in S-D-, (apply the criterion 3 
both to u and -u ) and thus u is constant in S-D by 
Proposition 3, which is not true. Thus u is not constant 
on 3D-. , and therefore there exist unique coefficients a 
and 3 such that if v = a u + j3 • then 



212 VI 



max v = 1, min v = 
SD- L r SD 1 r 



Proposition 15.3 implies =s v s 1 in S-D, . 

By Proposition 8 there exists a sequence r, - such 

that v converges uniformly on compact subsets of D..-D, . 
r k * l 

Moreover, Proposition 15.3 applied to Do/o implies that 

v converses uniformly on S-D-,/,. If v = lim v , then 
r, ° J 3/2. , r, 

k k-oo k 

Proposition 7 implies v is harmonic in S-DT . 

We now -write down the Laurent expansion of Propo- 
sition 14 for v in the set D-,-D : 
r 2 r 

>^" 1 (z) = c(r)log | as | + Re I a.(r)z J , 



v o, 
r 



r < | z | < 2, 

where c(r) and a Q (r) are real, and formula (8) of p. 202 

shows 

c(r)iog s + a n (r) = f f " v T .o r / 1 ( se i9 )d9, 
(10) ° lv ' "0 r 

ai (r)s j + a (r) s" j = I j^ v o (p - 1 ( 8 e i9 )e" iJe d8, 

J n 

j ^ 0, 
for r < s < 2. 



VI 213 



-1/ i9\ n/~n-in9\,„ 
Now v o^ (re ) = a r Re(r e ) + 3 r 

-n e in9 -fe-^ 9 , 
= a r r j + p r . 



and therefore if we let s - r in the second part of (10) 
we obtain 

a (r)r- 3 + a (r) r" j = if j 2 1, j + n . 

Taking s = 1 in (10), 

|a (r) + a (r)| <. 2 if j s 1. 

Therefore, for j s 1 and j ^ n, 

2 ;> | 3j (r) + a (r)| = | aj (r) - aj(r)r 2j | 
2 



= | aj (r)|(l-r^) * £ l a j< r >l if ° < r < 7' 



thus, I a . (r) I <, 4 and therefore 
|a (r)| s 4r 2j 



Now the estimates in Proposition 14 imply that as 

r, - 0, the coefficients in the Laurent expansion for 

v converge to the coefficients in the Laurent expansion 
r k 

for v. Therefore, 



voq) (2) = c loglzl + Re(a z + £ a . z-* ) , 



^ ^7^ = n 1 ocr I r? I 4- Rofa Z~ n + £ a Z^ 

1 j 

1 < Izl < 2, 



214 VI 



since a (r, ) - for j s 1, j ^ n. Nov? we define u 



on S - [p] by 



u = v in S-D, , 



uocp" (z) = c log |z| + Re(a_ n z" n + I a.z^), 

< |z| < 2. 

It is clear that u is harmonic on S - {p}. The theorem 

will be proved once we establish that a ^ 0. This 
v -n 

involves a rather delicate argument. 

Suppose that a =0. Then the formula for u near 
rr -n 

p shows that there exists 

t = 1 im u ( q ) , 

q-p 

where -v> < I < «. By Proposition 15.3 applied to u and 

-u and small analytic disks containing p, we conclude 

that u = I on S - [p } ; therefore, -»<£<<». Now we 

shall prove that v - u uniformly on SD-, , and therefore 

k 
that max u = 1, min u = 0, contradicting the fact that 

u is constant. Again. Proposition 15.3 shows that it 

is sufficient to prove that v - u uniformly on SDi . 

r k $ 

For |z| = j and < r < ■*■ . 

|v r ocp" (z) - uo r£ " (z)| <|c(r)-c|log 2 + |a_ n (r)-a_ j 

N 
+ £ |a.(r)-a.| + £ (4 + 4)2~ J 
J J N+l 



VI 



215 



+ E 4r" 2j 2" j 



N 

C [|c(r)-c| + |a n (r)-a j + S |a (r)-a |] 
n ii 11 q j j 

+ 8-2 + -2^ . 



l-2r 



Therefore, if e > we can choose a fixed N such that 

8r^ 

8'2" N < -I and a k n such that z < t and r < 7 

3 U 1-2r 

1 " r k 

for k a k - Then we choose k^ s k^ such that 

N 

C [|c(r,)-c| + | a (r, )-a | + E |a.(r.)-a.|] <4 

n L| v W * ' -n v k' n 1 n ' j k j 3 



if k s k, . Therefore, if k s k, , 



j v -u| < e on 3Di , 
r -1 



QED 



Thus, we have completed the proof of Theorem 4. 
We have already indicated the use of this theorem in 
establishing the existence of meromorphic functions, 
shown on pp. 168-171 in the proof of Theorem 3. In the 
next section we shall give further applications. 

Remark. In the above proof in the second case, 
the second part of (10) in the case j = n was not used. 
But we get additional information by using this formula 



216 VI 



/ v n , 7 — r- -n 1 ,-. -1/ i9v -inS ,, 

a(r)s +a (r)s = — ; vpm (se )e d 

n v/ -n v/ rr 1q y * 



Letting s = r, we obtain 

2 

/ \ n _l — ~7 — v ~ n „-n 1 [» TT /„in8. -in8\ -in6, r 

a (r)r + a (r) r = a r -> — (e -fe )e d8 

n -n r ^ n j ^ 

+ * r I r 2 V-9 d9 

r w .' 

- a r r 



We already know from s = 1 that 



Therefore, 



a (r) + a (r) £ 2 
n v y -n v ' ' 



a - a (r) + a (r) r I < 2r 
r -n v ' -n ' 



Taking imaginary parts, we conclude 

| In a_ n (r)l (l-r 2n ) < 2r 2n , 
showing that Im a_ (r) - as r - 0. Therefore, letting 



r = r, , we have 
k 



a = lim a (r, ) is real. 



We obtain from this remark the following result: 

COROLLARY TO THEOREM 4 . Let S be any connected 
Riemann surface and let p € S. Let ©: U - W be a chart 
in the complete analytic atlas for S with p 6 U and rp(p)=0 



VI 217 

Let n be a positive integer . Then there exists a 
harmonic function u £n S - {p} such that for z near 

uo v " 1 (z) = ReC^g-) + Ref(z), 

where a is a nonzero complex number and f is holomorphic 
in a neighborhood of 0. Moreover , u can be taken to be 
bounded outside a neighborhood of p . 

Proof : If S is hyperbolic, our proof on pp. 209-211 
already contained this result with a = 1. The boundedness 
of u away from p has also been shown in this case. 

If S is compact or parabolic, the assertion about 
the boundedness of u is automatic. What we must do is 
eliminate the term involving log |z| . By the previous 
remark, we have obtained a harmonic function v on 
S - [p] such that near 

v°af (z) ■ c log |zj + Re (-—) + Re g(z), 

z 

where g is holomorphic near 0, c is real, and a ^ is 

real. If c = 0, we are through. Otherwise, we replace 

m by the chart ourx., where uu is a fixed complex number 

with uu = i. Applying our result in this case, we 

obtain a harmonic function w on S - {pi such that near 



w°«f 1 (z) = d log|z| + Re(-\) + Reh(z), 

iz 

where h is holomorphic near 0, d is real, and b 4- is 

real. We have left out a trivial intermediate calculation 



lib vi 



here. Now define 



d 

U = W - — V 

c 



Then u is harmonic on S - L pj. and near 

uoqf 1 ^) = Re (4-) + Re(h(z) -"£ g(z)), 



n' x v ' c 
z 

whe re 

a -5 - ^ * 0. 

u 1 c 

.QED 

Also in the next section we shall require the 
existence of a Green's function on a parabolic Riemann 
surface . 

DEFINITION 7. Let S be a connected Riemann sur- 



face and p € S. A function g defined on S - {p} is a 
Green ' s function if 

1. g is positive and harmonic in S-{pj; 

2. if -j is an analtyic chart near p with 
c(p) = 0, then g + log|cp| is harmonic 
in a neighborhood of p; 

3. if h has properties 1 and 2, then g < h. 

We first remark that condition 2 is independent of 
the particular chart x- since any other analtyic chart 
V can be expressed as 



VI 



219 



sji = a-(l + E a k cp ), 



and so 

log | i|i| = loj 



1 



k 



and we see that log |i|i| - log |cp| is harmonic in a 
neighborhood of p. 

PROPOSITION 16 . Let S be a connected hyperbolic 
Riemann surface and p € S. Then there exists a unique 
Green's function on S - {p}. 

Proof : Uniqueness is clear by property 3 of a 
Green's function. The proof of existence is like the 
proof of Theorem 4 in the hyperbolic case. We set the 
problem in the framework of all the notation on the top 
of p. 209 . Thus,, v is a bounded continuous function 
on S-D, , v > and v is harmonic in S-D, , and v = on 
3D, . As before, there exists a constant C > such 
that 

log | cd I ^ Cv in D^ - D-j. 



As before, the function 

f -Cv in S-D, , 
{ -log| ml in D 1 , 



is superharmonic in S-{pl. Much more trivially, the 
function 



220 



VI 



w- 



f in S-D,, 
■log |cp| in D 1 



is subharmonic in S-(p}. Let g be the least harmonic 

ma jo rant of w., • As in Definition 5, g is harmonic on 

S - [p] and if A «> C sup v, then w ? <, w, + A. so that 

S-D 1 z L 

w 2 ^ g < ^ + A. 

By the removable singularity theorem, g + log | cp I is 
harmonic in D, , so property 2 follows. Also since 
Wo > 0, also g s and since g is not constant, the 
strong maximum principle implies property 1. To check 
property 3. suppose h has properties 1 and 2. Then 
h + log | c| is harmonic in D, and is positive on dD, , 
so the minimum principle for harmonic functions implies 
h + log | £ 1 > in D, . Therefore, h > w on S - ip}, 
and the definition of g therefore implies g <, h. 

QED 

Rema rk . We can prove even more. Namely, if h is 
positive and superharmonic on S - {p} and h + log 1^1 is 
superharmonic near p, then g s h. It's exactly the 
same proof. 

Problem 9 . Find the Green's function for the unit 
disk [z: |z| < 1}. 



VII 221 

Chapter VII 

CLASSIFICATION OF SIMPLY CONNECTED RIEMANN SURFACES 

As an application of the results of the previous 
chapter, we are going to prove that every simply connected 
Riemann surface is analytically equivalent to the sphere 
C, the complex plane c, or the unit disk z: |z| < 1} c c. 
These cases are exclusive, of course, since the compact- 
ness of the sphere shows it is not even homeomorphic to 
the plane or disk; and the plane and disk, though homeo- 
morphic, are not analytically equivalent (Liouville's 
theorem) (see p. 42 ) . 

We shall require some slight generalizations of 
some of the basic results of Chapter III. Namely, we 
shall require a permanence of func tional relations 
generalizing that of p. 66, and a monodromy theorem 
generalizing that of p. 64. in addition, we shall 
require a generalization of Lemma 2 on p . 117 which 
deals with unrestricted analytic continuation. 

The framework for this discussion has just been 
mentioned - the analytic continuation of meromorphic 
functions defined on arbitrary Riemann surfaces, rather 
than C. Given a Riemann surface S, we can form defini- 
tions as at the beginning of Chapter III and speak of M_. 
the sheaf of germs of . meromorphic functions on S. All 
the material of pp. 46-64 can be discussed with 
very little change. The applications we have in mind 
are given in the next two lemmas. 



222 VII 



LEMMA 1 . Let p € S, a simply connnected Riemann 
surface . Let u be harmonic on S - {p} such that if $ 
is an analytic chart near p with cp(p) = 0, then 

-1 °° k 

uoco (z) = Re( Z a, z ), z near 0, 

k=N k 
where a ^ . Here -co < N < « . Then there exists 
a meromorphic function f on S such that 

(1) Re(f) S u. 

Proof : It is obvious that we may define f near 
p by setting 

-1 °° k 
fore (z) = I a,z , z near 0. 

k=N K 

It is now a question of continuing f analytically to 
all of S. The generalized principle of the permanence 
of functional relations implies that the analytic 
continuation will always satisfy the identity (1). 
Briefly, the reason is that if f is meromorphic in an 
analytic disk D and Re(f) = u in a neighborhood of 
some point of D, then Re(f) = u holds throughout D 
(see Proposition 3 of p. 167). 

The second point is that analytic continuation is 
possible along every path in S with initial point p. 
The reason is that Proposition 1.4 of p. 165 shows that 
(1) holds locally, this and the permanence of functional 



VII 223 

relations combine as in the proof of Lemma 2 on p . 117 
to show that the process of analytic continuation never 
"stops . "' 

Now we have the hypothesis needed to apply the 
monodromy theorem, and the lemma is proved. 

QED 

LEMMA 2 . Let p € S., a simply connected Riemann 
surface . Let u be harmonic on S - L pj such that if cp 
is an analytic chart near p with co(p) = 0> then 



1 °° \e 

Uoco (z) = log |z| + Re( E a, z ), 

k=0 K 



z near . 

Then there exists a holomorphic function f on S 
such that 



(2) |f| = e U . 



Proof ; The outline of the proof is the same as 
in the previous lemma. First, we prove that f exists 
near p. Using e °' ' = | z | , we naturally choose 

foco (z) - z exp( I a,z ) , z near 0. 
k=0 k 

Then (2) obviously holds near p. Second, we apply the 
permanence of functional relations to show that (2) 
remains valid under analytic continuation of f . The 



224 VII 

point to be checked is that if f is holomorphic in an 
analytic disk and (2) holds in a neighborhood of some point, 
then (2) holds throughout the disk. This follows as 
before since log |fj is harmonic. One might think 
there is trouble here at zeros of f; by (2), however, 
if f has a zero along some path of analytic continuation, 
then (2) will have been violated before the zero is 
reached , 

Third, (2) holds locally at least. For locally 
we can write u = ReF, F holomorphic (we are not new 
treating neighborhoods of the exceptional point p) . 
Then we set f - e , implying (2) . Therefore, as 
before, analytic continuation is possible along every 
path from p. We are also using in this step the fact 
that (2) determines f locally essentially uniquely. 
That is, any other choice of f is just f multiplied 
by a constant of modulus 1, since holomorphic functions 
with constant modulus must be constant. (A similar 
fact about (1) was used implicitly in the proof of 
Lemma 1; in that case functions satisfying (1) have 
constant differences ■ ) 

Fourth, the monodromy theorem finishes the 
proof. 

QED 

Next, a technicality. 



VII 225 

LEMMA 3 . Let E be any bounded nonempty set in 
C . Then there exist complex numbers a and 3 , a ^ , 
such that if 

E = {az + p: z £ E) , 

then 

sup [ I w I : w 6 E } - 1 , 

inf { I w I : w € E} = \. 

Proof : Let a = inf j.Rez: z € E} and choose b such 
that a 4- ib € E (using the boundedness of E) . Let 
E l = (z-a-ib: z ? E } . so that inf (Rez: z 6 E, } = 0, 
e E~. Define for t 2 

m(t) = inf {j z+t| : z 6 E, }, 
M(t) = sup [j z+t| : z f E }. 

Then m and M are continuous increasing functions, 
m(0) = 0, and the boundedness of E, implies tj - 1 
as t - 90. Choose t such that !! = \. Let c = M(t) 
and 

E = [5±£ : z € E 1 }. 

Then E satisfies the conditions of the lemma, and 

1 , t-a-ib 

a = c ' 3 : c ' 

jQED 
Classification Theorem . Any connected , simply 



226 ., VII 

connected Riemann surface is analytically equivalent to 
the Riemann sphere , the complex plane , or the unit disk . 

Proof : Let S be the connected, simply connected 
Riemann surface. We have three cases to consider. 

S is compact : By the corollary to Theorem 4 on 
p. 216 , if p f s, then there exists a harmonic function 
u on S - L p] such that in terms of a given analytic 
chart c near p with cp(p) = 0, 

u n -1 (z) = Re(-) + ReF(z), z near 0, 

a + 0, 

where F is holomorphic near 0. By Lemma 1 there exists 
a meromorphic function f on S such that 

Re(f) = u. 

Now the only pole of f is the point p, and this is a 
pole of order 1. Thus, f takes the value co exactly one 
time. By Proposition 9.1 of p. 44 , f takes every value 
in C exactly one time. That is, f: S - £ is an analytic 
equivalence between S and C, proving the result in this 
case . 

S is parabolic : If p 6 S , and qj is an analytic 
chart in a neighborhood of p, then by the corollary to 
Theorem 4 on p . 216, there exists a harmonic function 
u on S - {p} such that 

uo-'^z) = Re(-|) + ReF(z), z near 0, 

a * 0, 



VII 227 

where F is holomorphic near 0. We are assuming cp(p) = 0. 
The construction of u shows that u is bounded outside 
any neighborhood of p. For this see p. 212, where 
< v < 1 in S-D^; p. 214 showing that u = v in S-D n ; 
and p. 218, showing that our function u is a linear 
combination of two functions bounded in S-D-, . Note that 
we are tacitly assuming that cp is rescaled if necessary 
to guarantee the existence of Dp, the analytic disk given 
by | cp| < 2. 

By Lemma 1 there exists a meromorphic function f 
on S such that 

Re(f) = u. 

Note that f has a pole only at p, that p is a simple 
pole, and that Ref is bounded outside any neighborhood 
of p. We wish to obtain another function with the 
stronger property that |f| is bounded outside any neighbor- 
hood of p . 

To do this let a-.< a^ < a-. <• • • be a sequence of 
positive integers. Since these are real numbers tending 
to oo and the real part of f is bounded outside any 
neighborhood of p, and since f is one-to-one in a neigh- 
borhood of p, it follows that for sufficiently large n 
there exists a unique p £ S such that f (p ) = a . By 
eliminating the first few terms in the sequence, we can 

assume this holds for all n. Also, it is clear that p -p. 
Since Ref is bounded outside any neighborhood U of p, we 



228 VII 



obtain for sufficiently large n 

I f -a I > a -Ref s \a outside U, 
1 n' n z n 

and therefore since we can assume f is one-to-one in 

Uj p - has a simple pole exactly at p and is bounded 

outside any neighborhood of p • By Lemma 3 there exist 

constants a and 8 such that if 
n ' n 



then 



f = -z- — + 3 > 
n f-a H n 



sup I f I = 1 , 

1 i n i 

V D i 

inf |f | = \ 
D 2- D l 



Since S is parabolic. Proposition 15.4 on p . 205 implies 

sup |f| - 1 
S-D 1 n 

(|f | is subharmonic) . 

By Proposition 8 on p . 178, there exists a subsequence 

n-, < n , < • • - such that lim f exists uniformly on compact 

k-»3o k 

subsets of D 2 -Dp and then Proposition 15.4 on p. 205 again 

implies lim f exists uniformly on S-D /0 . Let 
k— n k 3/2 

h = lim f in S - D. . 
k-„ n k l 



VII 229 



Then h is holomorphic on S-D, and |hj < 1. By renaming 
all the sequences, we can assume n, = k . 

Now we consider f in D^, where f has a simple 
n 2' n v 

pole at p and no other pole. Thus, we may write 

g n 
f = LJ in D„ 

n cp-cp(P ) 2 

where g is holomorphic in D 2 • Note that in D 2 ~DT 

|g n -g m l - lWP n )Jf n - [«r«p<P n >]fJ 

* l«r«p(P n )Mf n -fJ + lv(P n ) - q»Cp m ) 1 1 f m l 

<; 4|f -f | + |a>(pj " cp(p )\, 

' n m ' ^ r n ^ r m ' 

and therefore the sequence g converges uniformly on, 
say, SD^, 2 (since p -» p, ep(p ) - 0) . By the maximum 
principle, g converges uniformly in D-w 2 , sa y 



lim g = g, holomorphic in D^/o* 



Now define 

f h in S-D" , 

fp = J f in D 3/2 • 

Then we see that f is well defined, is meromorphic on 

S } and has at most a pole of first order at p and no other 

poles. Since g - g uniformly in D~ /^ and f - h uniformly 

in S-D0/9, we obtain the result that f - f uniformly in 
3/2 n p 3 

D^-D,. Therefore, 



230 VII 



sup | f | = 1 , 
P 



D 2- D l 



D 2 -D 1 



proving that f is not constant. Since S is parabolic, 

the nonconstant function If I cannot be bounded, and 

since If I < 1 in S-D7, it follows that f really does 
P l P 

have a pole at p. 



Summarizing the construction thus far, we have 

shown that for every p £ S there exists a meromorphic 

function f on S such that f has a pole of order 1 at 
P P 

p and f is bounded outside every neighborhood of p. 

These conditions essentially uniquely determine 

f . For if f has the same properties, then there is 
p p _ 

a unique complex number a 4- such that f - af has 

no pole at p, and is therefore a bounded meromorphic 

function on all of S. Since S is parabolic, f - af 

v P P 

is constant, and thus 

f p - af p + 3- 
Conversely, for any constants a and 3, a ^ 0, the 

function af + 3 has properties similar to those of f . 
P P 

Also, as we have previously discussed at the top 

of p. 228 , for a given fixed p, the function g _ r — 7— -r 

p" P 
has a simple pole at q if q is in a sufficiently small 



VII 231 

neighborhood of p, and a _ r — j — x has no other poles. 

P" P q 
Futhermore, this function is bounded outside any neigh- 
borhood of q, if q is sufficiently near p. Thus, by 
the remark above, 

f -f (q) q 

P P H 

where a and 3 are constants depending only on q. Thus, 
for q sufficiently near p there exists a Mobius trans- 
formation T such that 

q 

f = T of . 

q q p 

Now let p be a fixed point in S and let 
r o 

A = [p € S: 3 Mobius transformation T 

such that f = Tof } . 

P P 

r r o 

Then p £ A, and the argument just given shows that A is 

open . The same argument shows that A is closed . In both 

cases we rely on the fact that the Mobius transformations 

form a group under composition. Since S is connected, 

A = S. 

Now we prove that f is one-to-one . Suppose 
Po 

f P <p> - f P <i> • 

*o K o 

Then there exists a Mo'bius transformation T such that 

f = Tof . 
P P^ 



232 VII 



Therefore, 

- = f (p) = T(f (p)) = T(f (q)) = f (q). 
" *o *o v 

Since f has pole at p only, q = p . 

Now we prove that C - f (S) cannot have more than 

^o 
one point. Otherwise, there are two complex numbers 

<z,p 4 f (S) - note that definitely co € f (S) . Since 
^o ^o 

f (S) is simply connected, the monodromy theorem implies 

^o 
there exists a holomorphic determination of 



w-g 
W-0 




for w € f (S); choose that determination which is 1 at 

^o 
w = co . Define 

F = 



Then F is holomorphic on S and F(p ) = 1. Furthermore, 

F never takes the value zero and it is impossible for 

F(p) = -F(p'). For if this holds, then F(p) 2 = F(p') 2 , 

which implies f (p) = f (p') and p = p', since f 

^o - o ^o 

is one-to-one. Since F is not constant and takes the 

value 1, F takes every value z for |z-l| < e, some e > 

Therefore, 

|F(p) + 1| a e for all p € S. 

Thus, tttt is a bounded holomorphic function on S and is 
therefore constant since S is parabolic. This is a 



VII 233 



contradiction . 



Now we cannot have f (S) = f by Proposition 9.2 

^o 

on p. 44, since S is not compact. Therefore, there 

is a unique complex y such that f (S) = C - iy) • 

^o 
Therefore, 






is a one-to-one analytic mapping of S onto C, and thus 
forms the desired analytic equivalence between S and C, 

S is hyperbolic : Here we use Proposition 16 and 

let g be the unique Green's function on S - [p}. By 

Lemma 2 there exists a holomorphic function f on S 

P 

such that 

If I =- e" 8 P. 
i p 1 



Then 



1. f p (p) = 0, 

2. |f | < 1. 
1 P 1 

3. f is holomorphic on S, 

4. f does not vanish on S - ip}, 

5. if h is a function on S satisfying 1,2,3 
then |h| <_ |f p |. 



234 VII 

We have to prove the last statement. By 1, if cp is an 
analytic chart with tp(p) = 0, then near p 

h = arp n (1 + 3 cp + . . . ) , 

where we can assume a ^ and n s 1. Thus, near p we 
have 

log|hj = n log j co| + log|a| + log|l+p<xH- ...|, 

showing that 

-lojdJiU log u 

n 
is harmonic near p. Also, — °J — L > by 2 and is 

harmonic away from zeros of h. Let 
h = min ( g , zi2lM). 

Then h is superharmonic on S - [p], h > 0, and near p 

h + log | cp| = min (g + log | cp| , n + log ' ^ ^ 

is superharmonic, being the minimum of two harmonic 
functions near p. By the minimal property of the Green's 
function, 

g p * h. 
Thus f 

6 p n ' 

so iloglhl I 

~g n n 

|f p l = e P s e = |h| , 



VII 235 

and thus 

ih * if r < if i. 
ii i pi i p i 

This proves property 5. 

Now let p,q € S and set 

f - f (q) 
h --£ £— . 

P H P 
Then since all numbers involved have modulus less 
than 1, we see that h(q) = 0, jhj < 1, h is holo- 
morphic on S. Therefore, property 5 above implies 



M * if q i • 

Since h(p) = -f (q) , we obtain in particular 
|f p (q)l « |£ (P)|. 



By symmetry we conclude 

|f p (q)l = If q (p)l for all p,q € S. 

(In terms of the Green's functions ^ this relation 
states 

g p (q) = s q (p)- ) 

Thus, we conclude that the holomorphic function 

Jl_ satisfies 
f q 



236 VII 



1 5-1 ^ 1 on S, 

q 



h(p) 

f q ( P ) 



= l 



By the strong maximum principle., it follows that J2_ 

q 

is constant, and in particular 



(3) |-|-| S 1 on S. 

x q 

Now we prove that f is one-to-one . Suppose f (q) = f (q') 

Then h(q') = and by (3) f (q') = 0. By property 4 
of f , we conclude that q' = q. 

q h h 

Therefore, for any p € S, f is a one-to-one 

P 

analytic mapping of S into the unit disk A = iz: | z | <1 } . 
We now prove that f (S) = L. If this is not the case, 
then a simple topological argument shows that there 
exists 

a € af (S), |a| < 1. 

Since f (S) is open, a | f (S) . Choose p, ,p^,p„ , . . . in 

S such that 

f (p ) -a. 
p vt n 7 

Since f (S) is simply connected and a & f (S) , the 
p p 

monodromy theorem implies there exists an analytic 
determination of log(w-a) for w € f (S) . Note that 



VII ^37 

Re log(f - a) = log |f - a| < log 2 . 

Let T be a Mdbius transformation mapping 

{z: Rez < log 2} 

onto & and such that T(log(-a)) = 0. Consider the func- 
tion 

F = T°log(f -a) . 

Then F is holomorphic on S and |F| < 1, F(p) = T(log(-<x))=0 
By property 5 of f , 



F S f . 

ii I p i 

We therefore conclude successively that 
f p (P n ) - a - , 
log (f (p n )-a) - », 
Tolog(f (p n )-a) - a£, 

i.e., 

|F(p n )| - 1, 

and thus 

|f (p )| - 1. 

Thus, | ex j =1, a contradiction. 

.QED 
COROLLARY . " THE RIEMANN MAPPING THEOREM " Let S be 

a connected , simply connected open subset of C with S j> C. 



238 VII 

Then there exists an analytic equivalence of S and the 
unit disk . 

Proof ; We have only to show that the Riemann surface 
S is hyperbolic. We proceed as on p. 232. If a € C - S, 
then there exists a holomorphic determination of Jw-a. 
for w € S. Define F(w) = v w ~ ex • Then one shows that 
F(xaj) = -F(w') implies w = w' by squaring both sides, so 
that it is impossible that F(w) = -F(w'). Suppose w € S. 
Since F is an open mapping, there exists e > such that 
F(S) includes the set [z: jz-F(w ) | <e } . Therefore, F(S) 




O' 

bounded, nonconstant holomorphic function on S, proving 
that S is hyperbolic. 

QED 

Now we want to indicate some applications of the 
classification theorem. The first of these is a trivial 
application, but answers the question of which Riemann 
surfaces are homeomorphic to a sphere. Cf. p. 42. 

THEOREM 1 . Let S be a connected compact Riemann 
surface . Then the following conditions are equivalent . 

A 

1 . S is analytically equivalent to C . 

2. S is homeomorphic to C. 

3 . S is simply connected . 



VII 239 

4 . There exists a meromorphic function f 

on S such that every meromorphic function 
on S is a rational function of f . 

5 . There exists a meromorphic functio n f on 
S having a simple pole at some point and 
no other pole . 

Proof : 1 => 2 : Trivial. 

2 => 3 : Trivial, since a sphere is simply- 

connected . 

3 => 1 : Follows from the classification theorem, 

A 

1 =» 4: We can assume S = C and we then take 



f(z) = z. The result is immediate. 
4^5 : We prove that tne function f of 4 
must be one-to-one. Suppose p,q 6 S, p 4- q- By Theorem 
3 of Chapter VI, there exists a meromorphic function g on 
S such that g(p) ^ g(q) • By condition 4, there exists 
a rational function A such that g = A°f. Thus, A(f(p)) 
^ A(f(q)), which implies f(p) ^ f (q) • By Proposition 9-2 
of Chapter II, f takes every value the same number (one) 
of times, so f takes the value » one time. 



5^1 : By Proposition 9.2 of Chapter II 

A . 

:o C m a one-to-one \ 
analytic equivalence of S onto C 



f maps S onto C in a one-to-one fashion. Thus, f is an 



QED 



240 VII 

We are now going to discuss the next easiest case. 
Theorem 1 is concerned with a compact surface of genus 0. 
We shall next discuss the compact surfaces of genus 1. 
This case is already so involved that we shall devote a 
separate chapter to it. 



VIII 241 

Chapter VIII 
THE TORUS 

Our use of the classification theorem in proving 
Theorem 1 of the previous chapter is rather disappointing. 
For we have applied the classification theorem in the 
compact case only, and the proof of this case occupies 
only half of p. 226,, whereas the proof of the other 
two cases requires eleven more pages. Essentially all that 
has been used is Theorem 4 of Chapter VI and its corol- 
lary. In this chapter we shall get to use the full 
force of the classification theorem in discovering what 
all the "analytic" tori are. I.e., we shall "classify" 
the analytic tori. 

At first glance, it perhaps seems that the classi- 
fication theorem, which is addressed to simply connected 
surfaces, could not be used on tori, which are mani- 
festly not simply connected. In any case, the utility 
of the classification theorem would be minute if it 
had no application to anything but simply connected 
surfaces. Indeed, the theorem states essentially that 
simply connected Riemann surfaces are trivial in a 
certain sense. 

One of the primary applications of the classification 
theorem is to the universal covering surface of an 



242 VIII 

arbitrary connected Riemann surface. The universal 
covering surface is a connected Hausdorff space, and 
can be made into a Riemann surface in a natural way, 
as we shall see in Lemma 1. Also, it is simply con- 
nected, so the classification theorem applies. Once 
we know that the universal covering surface is analyt- 
ically equivalent to the sphere, plane, or disk, then 
standard topological methods can be invoked to obtain 
analytic information about the original surface. Ac- 
tually, in the case of a torus the universal covering 
surface is obviously the plane, topologically ; the 
"covering map" is also rather obvious; and as a result 
in this chapter not even the definitions of the concepts 
mentioned in this paragraph will be given. But the 
topologically alert reader will know the general 
setting of what follows. 

DEFINITION 1 . If T and S are topological spaces and 
f: T - S , then f is a local homeomorphism if for every 
point p € T there exist a neighborhood U of p and a 
neighborhood V of f(p) such that f is a homeomorphism 
of U onto V. 

LEMMA 1 . Let T be a Hausdorff topological space and 
S a Riemann surface . Let f : T - S be a local homeomor - 
phism . Then there exists a unique complete analytic 
atlas on T such that f is an analytic function from the 
Riemann surface T to S . 



VIII 243 

Proof : First we prove uniqueness , so we suppose 
first that T is a Riemann surface. If p £ T, there 
exist a neighborhood U of p and a neighborhood V of f(p) 
such that f : U -» V is a homeomorphism and V is the do- 
main of an analytic chart cp: V - cp(V) on S. Let f-, be 
the restriction of f to U. Then since f is analytic, 
f, is an analytic equivalence of U onto V, so cpo f , must 
be an analytic chart on U. Knowing an analytic chart in 
a neighborhood of each point of T implies that we know 
the complete analytic atlas for T, so the uniqueness 
follows . 

Conversely, we use the above procedure to define 
charts cpof, on T. We now show these charts form an 
analytic atlas. If we have another choice, U, V, cp, 
and f, (the restriction of f to U) , then where the 
composition is defined we have 

~~ _x ~~ -l -l *** -1 

Cpof o(cpof ) = Cpof of ocp = Cpocp 

since f,oL = identity (we might have to decrease the 
sizes of everything to achieve this) . Since S is a 
Riemann surface, cpocp" is holomorphic , and thus we have 
an analytic atlas for T. We have to show that f is now 
analytic, but this is clear. For, on U we have 



f = f, = co o (cpo f .. ) 



'44 VIII 

and this is a composition of two analytic functions. 
Thus , f is analytic in a neighborhood of any point of T. 

QED 

LEMMA 2. Let T and S be Riemann surfaces and 



f : T - S an analytic local homeomorphism . Let T-, be a 
Riemann surface and g: T-, -. T a continuous function 
such that fcg is analytic . Then g is analytic . 

T 1 - g ^ T 



fog 



\ ' 



S 



Proof : Given p € T, , there exist neighborhoods U-, 
of p, U of g(p) , and V of f(g(p)) such that g : IL -» U 
and f : U -» V is an analytic equivalence. Then on U-, we 
have 

g = f^o(fog) , 

where f-, is the restriction of f to U. Thus, g is ana- 
lytic. 

QED 

Now that the preliminaries are finished, we are 
ready to discuss tori. The situation is this: S is a 
Riemann surface which is homeomorphic to a torus . 



VIII 245 

The problem is to discover what kind of 
analytic atlas S can have. What we shall do is prove 
that S is analytically equivalent to one of the c/g 
discussed in Problem 1 of Chapter II, p. 24- The 
problem is essentially to find the complex numbers 
<jj, and m« such that Q = [n-, uk-Hi^'ju^ : n, ,n« integers}. 

To start with it is convenient to choose a topolog - 
ical representation of S as c/fi for some q which we can 
pick arbitrarily. Thus, choose arbitrary complex r 
and on whose ratio is not real. Then we suppose that 
S is the set of all cosets [z] = [z+n, p-i+n-09 : n 1 ,n~ 
integers } , and S is made into a Hausdorff space in the 
way described in Problem 1. We thus have a concrete 
representation of S as a topological space, but the 
analytic atlas for S is unknown. In particular, it is 
probably not the analytic atlas described in Problem 1, 
unless we happened to choose p, and p 9 correctly. We 
reiterate that we are going to prove it is_ such an 
analytic atlas with the proper choice of c-i and o„. 

Now we are ready to apply Lemmas 1 and 2. First, 
let C be the complex plane as a topological space with- 
out the usual complete analytic atlas and let 

tt: C - S 

be the natural mapping defined by tt(z) - [z]. Clearly, 
n is a local homeomorphism, so Lemma 1 shows there is a 



246 VIII 

unique way to make C a Riemann surface such that rr is 
analytic. Let C denote this Riemann surface; C is 
homeomorphic but not necessarily analytically equivalent 
to C as we usually consider it as a Riemann surface. 

There are obvious translations on c". For example, 
let 

* * 

t,: c - c 



be defined by 



t, (z) ■ Z+c 



Then since n(z+p-,) = [z+p,] = [z] = tt(z), we have 

not-, = n . 

Or, we have a commutative diagram. 

tl 
C* >- C* 

\ i 

\ TT ; IT 

i 

X f 

s 

By Lemma 2, t, is analytic. Likewise, t2 is analytic, 
where t ? (z) - z+p ? . Two obvious facts about these trans- 
lations are that t, and t ? commute: 

t l° t 2 = t 2° t l ' 
and that t-. and t 2 generate an Abelian group: if n-^ and 



VIII 247 



x\j are any integers , the mapping 



n l n 2 

tot 

which stands for n-,-fold composition of t, composed with 
n--fold composition of t~ , is just the mapping 

z -. z+n-, c-i+n2P2 • 

Now we apply the classification theorem to C , which 
is simply connected, connected and not compact. Thus, 
c" is analytically equivalent to C or the disk 
A = [z: |z|<lj. In spite of appearances, it does not 
seem obvious that c" is equivalent to C and not A, which 
is indeed the case. It is clear that this question must 
be faced; cf. p. 42. Let ? = C or A as the case may be, 
and let f be the analytic equivalence: 



Using t, and t„ , we now define corresponding mappings 
of ? to itself: 

A 1 = f" 1 t x of , 

A 2 = f_1 t 2° f * 

Then the properties of t, and t« are obviously reflected 
in A, and A« : A, and A« are analytic maps of ? onto ?, 
they are both one-to-one, they commute, for 



248 VIII 



A 1 °A 2 - (f" 1 ot 1 =f)o(f" 1 ot 2 of) 

= f" 1 °t 1 ot 2 cf 

= f" 1 °t 2 °t 1 of 

= A oA 

and they generate an Abelian group, with the formula 



Now we remark that the only analytic equivalences of A 
onto A or of C onto C are Mobius transformations. Thus, 
A, and A~ are both Mobius transformations. 

It turns out that the thing relevant to our discus- 
sion is the fixed point structure of A, and A~. Suppose 
now that A is a Mobius transformation of the form 



A(z) = ^~k , ad-bc 4 . 
v ' cz+d ' 

A point z € C is a fixed point of A if A(z) = z. That is, 



az+b _ 
cz+d 



Observe that ^ is a fixed point if and only if c = 0. 
If c i 0, the above equation can be written 

? 
az+b = cz +dz , 

a quadratic equation for z, which has either two roots 
or one root. Thus, every Mobius A has one or two fixed 



VIII 249 

points in C (note that if c=0 , we can write 

A(z) = az+b . and then A has two fixed points if and 

only if a ^ 1) . 

Now if A is Mobius and an equivalence of A 
onto A, and if z is a fixed point of A, then the 
con/jugate of z with respect to SA is also a fixed 
point of A. For suppose w is the conjugate of z 
(that is, w = l/z) . Then a property of Mobius transforma- 
tion is that they preserve conjugacy - thus, A(z) and 
A(w) must be conjugate with respect to A (a A) . But 
A(z) = z and A(BA) = SA, so we see that z and A(w) 
are conjugate with respect to gA. Thus, A(w) = w. 

It is obvious that t-, has no fixed points in 
C*. Thus, A-, has no fixed points in ?. If ? = C, 
we must have therefore 

A 1 (z) = z+w 1 , 

and likewise 

A2(z) = Z+UO2 

Here u;^ and [*)„ are nonzero complex numbers. 

If ?=A, then if A. has only one fixed point, 
it must be on SA. This follows since A, has no 
fixed points in A and since the conjugate with respect 

to SA of a fixed point of A-, is also a fixed point 

of A-, . Likewise, if A-. has two fixed points, they 
both lie on 3 A. 



250 VIII 

Proof that ?=C . Suppose the contrary, that ?=A. 
There are two cases to consider. Suppose first that A, 
has two fixed points, a and 3. Then 

A ] _(A 2 (a)) =A 2 (A 1 (a)) = A 2 (a) , 

so A 2 (a) is a fixed point of A, . Likewise, A 2 (s) 
is a fixed point of A, . So either Ao(a) = a and 
A 2 (s) = S, or A 2 (a) = 3 and A 2 (b) = a. Now define 
the Mobius transformation 

Then the transformation 

moA, om 

maps to and °= to °° , and thus is multiplication 
by a complex number a-, . Since m(sA) is a straight 
line through 0, it follows that m(A) is a half plane 
bounded by a straight line through 0. And the mapping 
z -> a,z maps this half plane onto itself. Thus, a, 
is a positive real number. That is, 

m°A o m " (z) = a, z , 0<a,<» 
If we have A ? (cc) = a and A ? (s) = B, then also 
m°A 2 °m~ (z) = a 2 z, 0<a 2 <=° 
On the other hand, if A 2 (a) = 3 and A 2 (8) = a, then 



VIII 251 



mcA^om maps to °= and °° to 0. Thus, for some 



nonzero complex b, 



a -1/ \ b 

moA °m (z) = — 

z N ' z 



Then it follows that 



(m°A 2 °nf )o(moA 2 °m" ) (z) = ^-^ = z 



so that also 

A 2 °A 2 = identity . 

But then also t^tj = identity, a contradiction since 
tpot^ is translation by 2p ? . Therefore, we conclude 



that 



moA.om (z) = a.z, j=l,2 



Now we need a lemma. 



LEMMA 3 . Let x,y€R. Then there exist integers 
m, , n, such that for each k , m, and n, are not 
both zero , and 



lim (m, x + n, y) = 



Proof : We can obviously assume x and y are not 
both zero and that — = e is irrational. Let N be an 

y 

positive integer. For l^j^N+1 there exists a unique 



252 VIII 



integer I. such that 

< j? - I. < 1 . 

Among the N intervals (0,i), (^, |) , . .., (^1, 1) 
there must be one which contains two of the numbers 
2% - -t . , say for j, and j 9 . Then 

lOis-t^) - o 2 ;- 4 J2 )i <s • 

Now we apply this lemma to the real numbers 

log a, and log a 9 to obtain m, log a, -f n, log a 9 -> 0. 

m, n. 
Exponentiating 3 a, <; a 9 - 1. Thus, for each z we 

have 

m, n, , 
moA 1 K oA 2 K °m" 1 (z) - z . 

Therefore, since m and m are continuous, 

m, n, 
A 1 °A 2 K (z) - z 

for each z , and thus 

m, n, 
ti t 2 k (z) - z 

for each z. This says z + m, c-, + n, p« - z , and thus 

m,p n +n,^ -0. This contradicts the fact that n, 
k K l k. 2 1 

and 2 have a nonreal ratio; cf. the discussion under 
Problem 1 . 

The only other case is the case in which A-, and 



VIII 253 

A 9 each have only one fixed point. Suppose A, (a) = a. 
As we saw on p. 250 , A (a) is a fixed point of A, , 
so also A 9 (a) = a. Let m be the Mobius transformation 

iG 

/ \ e 
m(z) = 

v J z-ct 

The m(o) = <= and m(^£>) is a straight line. We choose 
9 to force this straight line to be parallel to the 
real axis. Then moA, °m "" maps °° to °° and has no 
other fixed point , so 

m r -A, om (z) = z4a. 

Since m^A c rn maps the associated half plane onto itself, 

a -.(EH. Likewise, 

moA-cm (z) = z + a„ , a^6R 

By Lemma 3, there exist integers m, and n, such that 
m,a, + n,a„ - 0. Therefore, as in the argument above 
we obtain m, c -■ + a Co -0, a contradiction. 

Thus , we have now completely contradicted the 
assumption that ?=A. The only other possibility must 
hold. Thus, ?=C . 

Now from p. 249 we know that A.(z) ■ z+,x. , j=l,2. 
Exactly as in the above discussion, it follows that 
ju-, and <\jry have a nonreal ratio. Thus, if we define 
Q = [n,uj, + n ? (ju ? : n-, ,'n^ integers}, we have a Riemann 
surface C/n as defined in Problem 1. 



254 VIII 

Consider the diagram 



C* 



| tt jn, 

y y 

F 

s -*- — C/q 

where the map n-. is z - z + Q. What we want to do is 
obtain an analytic function F from c/fi to S. First, 
we can define a function F by 

F(z + n) = TT°f(z). 

This makes sense, for if z + = z' + Q, then 

z = z ' + n, u, + n., jj„ for some integers n, and n„, 

and thus 

TT°f(z) = nof(z' + n,'ju. + n.pWn) 

n l n 2 
= nofCA^oA^Cz')) 

n n 2 

= not 1 i ot 2 Z (f(z')) 

= nof(z'). 

Thus, F is uniquely defined such that F°tt-, = rrof. Since 
n-, is locally an analytic equivalence, we have F = rrofon-, 

locally and thus F is analytic. Since n and f are 
surjec tions, so is Forr-, and thus so is F. Finally, F is 
one-to-one. For, suppose F(z+Q) = F(z'+q). Then rr°f(z) = 
nof(z'), so that there exist integers n-, and n- such that 

f(z) = f(z') + n 1 P 1 + n 2 P 2 

n l - n 2 
= t 1 i ot 2 Z °f(z'). 



VIII 255 



Thus, 



z = f i ot 1 1 =t 2 2 of(z / ) 
n n. ? 

= a/oa/Cz') 



= z ' + ri-, a, + n^ jj« 



Thus, z + a = z' + n, proving F is one-to-one. 

We shall now formally state a theorem which 
includes the above discussion. We need to recall 
Definition 2 of Chapter V. 

THEOREM 1 . Let S be a compact , connected Riemann 
surface . Then the following conditions are equivalent . 

1 . S is analytically equivalent to the 

Riemann surface of a polynomial 

2 
w - 4(z-e 1 )(z-e 2 ) ( z " e 3 ); 

where e, ,e ? ,e, are distinct complex numbers 

2 . S is analytically equivalent to the 

Riemann surface of a polynomial 

2 



whe 



(z-a-j^) (z-a 2 ) (z-a 3 ) (z-a^), 
re a, , a ^ a-, a, are distinct complt 



numbers . 

3 . S is homeomorphic to a torus . 

4. S is analytically equivalent to a torus 
of the form c/Q of Problem 1. 



256 VIII 

Proof : 1 => 2 ; This is simple algebra. We can 
assume S c m is the Riemann surface of the polynomial 
1. Let a € C, a 4- e, or e 9 or e . Define 



f--i. 



where tt and V are the functions on M discussed in 
Chapter IV, restricted to S. Thus, 

2 
V = 4(n-e 1 )( T T-e 2 )( T T-e 3 ); 

fV = 4f(f(n-a) + (a-e 1 )f)(f(n-a)+(a-e 2 )f)(f(rr-a) 
+ (a-e 3 )f) 

= 4f(l + (a-e 1 )f)(l 4- ( a -e 2 )f)(l + (a-e 3 )f) 

- 4(a-e 1 )(a-e 2 )(a-e 3 )f(f- i -^)(f- -L.) (f - ^ 

Choose a complex number 3 = v , 4(a-e,)(a-e )(a-e 3 ) and let 

f 2 v 



Thus, 



s 2 = f ( f -i^>< f -i^M f -^>- 



Nov? f and g are meromorphic on S and f takes every value 
2 times, since the same is true of tt. Nov? consider 
the diagram of p. 156 



VIII 257 



f,g\ A,v 

6x6 



The polynomial equation satisfied by f and g is easily 
seen to be irreducible and it is of degree 2 in g. 
Therefore, Theorem 2 of Chapter VI implies $ is an 
analytic equivalence. This proves 1 =» 2. 

2 => 3 : This follows trivially from the cutting 
and gluing process described on pp. 9-13 . Also, 

it follows from the Riemann-Hurwitz formula of p. 112. 
In this case V = 4 and n = 2, so that g = 1. 

3 => 4 : This is the content of the discussion 
preceding this theorem. 

4 =» 1 ' Now we have to do some work. In fact, we 
need to introduce some rather classical and famous 
concepts of the theory of elliptic functions. Suppose 
that uu-j , uuo € C have nonreal ratio and define as usual 
n = [n-, uu, + n.pUy: n,,n2 integers). Define 

^ z Z C €Q (z-C) C 

We must first prove this series converges. If K is 
a compact subset of C-Q, then for z € K 



258 VIII 

|-3^-4l- \- z2+2 fi *e|c|- 3 • 
(z-c) 2 c 2 l<*-oVl 

We now check that the series of constants 

s Id" 3 
cVo 

converges. Since uk and ^ have nonreal ratio, it 
follows that for any 8€[0,2rr], id-, cos 9 + Wo sin 9 ^ 0. 
Since this is a continuous function of 9, there exists 
a positive constant 5 such that 

|uj, cos + u)« sin fi| s *, , 0^9 ^2rf 

Multiplying both sides of this inequality by a positive 
number, we obtain 



/ 9 9 

|xuu, + y^2 1 s ^ x + y ' x anc * y rea i 



2 

Therefore, summing on squares in H , we obtain 



Z kl" 3 « 6" 3 E (n^)" 3 ' 2 

C^n n, ,n 9 

^ r not both 

zero 

- 5 IE (n 1 +n 2 ) 

k=l max(| n;L | , jn 2 |)=k 

£ *" 3 S 4(2k+l)k' 3 < » . 
k=l 

Therefore, the series def ining /"^converges uniformly on 
any compact subset of C-Q, and one sees likewise that 



VIII 259 

for any lattice point ' , the series for/"5(z) - ^ 

(2- Co) 

converges uniformly on any compact subset of C-(fi-[£ ]). 
Therefore , (■- is meromorphic on C and has poles of order 
2 at each C6Q. The function/^) is called the Weierstrass 
pe- function . 

The first remark to be made is that/^ is an even 
function. For, 

(Pi-*) = V £ (- 1 — 2 - \) ■ 
z Z £€fl (-2-C) £ 

Replacing the "dummy" £ by -;, we therefore obtain 
(p(-z) = \+ Z (~ J — ^ - K) 

z z ren (z-c) £ 

Next, we compute^); since the series f or ££> converges 
uniformly locally, this can be done formally: 

£/f>(z) = ^§ + £ " 2 3 
z CGQ (z-C) 

c*o 



= -2 



7 — 775 

(z- ) 



Thus, if £ Q c 



/£'(z+0 = -2 



cen (z+r -:) 3 



260 VIII 

- -2 I. 1 

where we have replaced the "dummy" r by Z+C . This 
implies that 

J5(z+x-,) -(fc(z.) s constant. 

e tl 
^ is even, 



"1 
We evaluate this constant by setting z = -y-. Since 



<£Cjl) -^(-^) - o , 



and thus the constant is zero. Therefore, using the 
same argument for ju 2 , 

( ^/(z+,ju 1 ) =<£(z) , 
/^6(z+uj 2 ) =^6(z). 

These relations imply that we can regard^? as a 
meromorphic function on C/q, whose value at z+Q is just 
/fc)( z ) • To keep track of the notation, lety/) be this 
function: 

Then since^) has double poles at the points in Q and i.. 
other points , we see that^ has a double pole at 0-Kz 
and no other pole. Since c/fi is a compact Riemann 



VIII 261 

surface , /£> takes every value 2 times . 
Likewise, if we define 

then <^> takes every value 3 times , since^> has triple 



poles at the points of Q. We shall be interested in 



particular in the zeros of /p . If ; ?o but j> 

then since/p is odd , being the derivative of an even 

function , 

Since ^Co^^^Cq)- 00 ' and thus^ (jCj = 0. Thus, 
<£ <lV^ = ° ' 

Now define 

jb(^ 2 +Q) = e 2 , 
/£? (-juu ^+2X2+0) = e 3 

Since X-< takes every value 3 times , we have found 



/ 

all of its zeros. And these must therefore also be 



simple zeros of L . Also, we see that /t> takes the value 
e-^ 2 times at Tp-j+fi (a zero of t> ) and likewise for e„ 



262 VIII 

and e~. In particular, e, ,e„,e- are distinct. 
Now consider the meromorphic function 

on C/o. The numerator and denominator have poles only 
at 0+0, and near there we have the Laurent development 

( 2 + ^ 2 4 + 

-=4+... . 



f 1 4- ^ 3 l + 

(~2+...) -g+--- 
z z 



Thus, the function has no pole at 0+Q , and in fact is 

equal to 4 at O+o. The only other possibilities for 

X]_ liJ 2 ,JJ 1 ' ,} 2 
poles are at y- + Q, j- + a, and j h -~ 1- Q. But 

at these points the numerator has zeros of order 2 and 

the denominator has zeros of order 2. Thus, the function 

has no pole at all, and is thus constant. Therefore, 



ck K - 4 <J5- e i>Ur e 2Kir e 3> • 

This is the classical differential equation for^> 
As on p. 257, consider the diagram 

C/q -- ^ T 



6x6 



VIII 263 

Since A-> takes every value 2 times and the algebraic 
equation relating A3 and Ap has degree 2 in Ad and is 
irreducible, Theorem 2 of Chapter VI implies § is an 
analytic equivalence. And T is the Riemann surface of 
w z "-4(z-e- 1 ) (z-e~) (z-eO • 

QED 

Even among the tori C/o there are lots of equivalences 
We now treat this problem. 

THEOREM 2 . Two tori C/o and C/fl are analytically 
equivalent if and only if there exists a nonzero complex 
a such that 

Q = aQ . 

Proof : If a = afi, then we can define a mapping 
C/ .. -> t'Vi by the formula z+Q - az+Q , and this is easily 
seen to be an analytic equivalence. 

Conversely, suppose F:C/q - C/fi is an analytic 
equivalence. If F(0+Q) = a+o , then let 

t: c/n -» c/n 

be defined by 

t(z+Q) = z-a+n . 

Then t is an analytic equivalence and 

toF(0+fi) = t(a+Q) - O+o . 



264 VIII 

8y considering t°F instead of F, we see that there is 
no loss of generality in assuming 

F(0+Q) = O+o . 



Given z €C choose arbitrarily w €C such that 
o J o 

F(z Q +fi) = w q +Q . 

Let n : C -* C/fi and rr:C -. C/q be the canonical mappings and 
choose a neighborhood U of w such that ff is one-to- 
one on U. Then for z near z consider the mapping 

g^ (z) = ff^CFCz+n)) . 

w o 

This is a holomorphic function in a neighborhood of z, 

and if we choose a different w' such that F(z +q) = w '+Q , 

then w' = w + C , where C~€Q, and the associated ft 
o o b o o ' 

is thus equal to the original ff + ? . Thus , 

g^oo - g„ («) + : • 

o o 
It follows that 



Hz" §w < z > 
o 

is well defined near z in the sense that it is independent 

of the choice w . Thus , we can define a function h on 
o ' 

C by the formula 

h(z) = Iz"^ < z) ' z near z o • 
o 



VIII 265 



Then h is holomorphic on C and since for z' = z + " , 
r o o o 

r €fi, we can take w to be the same and thus for z near 
o ' o 

z' 

o 

g^ (z) = n _1 (F(z+n)) 

o 

= n" 1 (F(z-C o +Q)) 

o 

it follows that h(z) = h(z-'" ) for z near z '. Hence, 
N o o 

h(z ') = h(z'-f ) = h(z ). so h represents a meromorphic 
o o "o o 

function on C/q. But h is holomorphic and thus constant, 
say h=a . 

Therefore, following the above notation we have 

g^ (z) = az 4 b 
o 

for z near z . Applying rf to both sides we obtain 

F(z+n) = az+b+Q for z near z . 

o 

Here b is a constant which can depend on z . By a 
connectivity argument it is easy to see that the constants 
b which can appear here must differ from each other 
only by elements of Q. Since F(0+Q) = 0+Q, the b assoc- 
iated with z = must itself belong to Q. Therefore, 
we have proved 

F(z+q) = az+o 



266 VIII 

This much has been done assuming only that F is analytic 
and not necessarily one-to-one. 

If we assume that F:C/q - C/q is one-to-one and 
onto, then since F(0+q) = 0+n, we have 

z+o = O-Ki « F(z+Q) = F(0+n) 
» az+n = 0+Q 

That is , 

z€q « az€Q 

But this means that Q = aQ. 

QED 
Of course, we can describe the relation Q = aQ 
algebraically rather than geometrically. If 

Q = [n, lo, +n ? !jLv ? : n, ,n~ integers] , 

Q = [n, ju-i+t^um 1 n, ,n2 integers] , 

then we have 



aw, = n-. -, ij-|+n, 2^0 



a'jj« — n«-| jj-j +n^ ^ uuo 



for certain integers n 



., . Also , 
jk 



S-i = m-, -, aioj,+m-| 2 a -^2 



Si 2 = m n\ ax-i+m22 aijj 2 



VIII 267 



for integers m., . Therefoi-e, we have the product of 
matrices 



n ll n 12\ f m ll m 12\ A ° 



n 21 n 22/ V m 21 m 22/ \° l 



Therefore, the product of the determinants is 1, or 

n ll n 22 " n 21 n 12 = ±X * 

Conversely, if this equation holds, then the relations 
expressing ajj-, and a-xo in terms of £, and 1 ? can be 
inverted, and thus fi = a . 

Almost as an afterthought we mention that if S is 
a compact, connected Riemann surface, then a necessary 
and sufficient condition that there exist a meromorphic 
function on S which takes every value 2 times is that 
S be analytically equivalent to the Riemann surface of 
a polynomial 

2 
w - (z-a 1 )(z-a 2 )...(z-a t ) 

where a.. ,a„ , . . . ,a are distinct. 

The proof is almost trivial. If S is the Riemann 
surface of the above polynomial , then the function rr 
takes every value 2 times since the polynomial has degree 
2 in w. Conversely, suppose f is a meromorphic function 
on S which takes every value 2 times. By the proof of 
the corollary on p. 160, there exists a meromorphic 



268 VIII 

function g on S such that f and g satisfy a polynomial 
equation which is irreducible and has degree 2 in g. 
Thus, for certain rational functions a and b, 

g 2 - 2a(f)g + b(f) = . 

Completing the square, 

(g-a(f)) 2 = a(f) 2 - b(f) . 

Let g x = g-a(f) and f x = a(f) 2 - b(f). Then 

Si " f l • 

Since f, is a rational function of f , we can write 
m .-, n 

n (f-a k ) z n (f-e-) 

2 _ 2 k=i k ., j=i J 

§1 " a IP ~ x n 7 ' 

n (f-a') z n (f-B,0 
k-i K j-i J 

where the a's and g's are complex constants and the 

numbers 9. and B .' are distinct. Let 
J 3 

m' 

n (f-ap n , 
n (f-a.) J_1 

k=l * 

Then 



g z = n (f- B .) n (f-e/) 
j-i J j-i J 



Thus, we have produced distinct complex numbers a-, , . . . ,a 
(>t=n+n ') and a meromorphic function g„ such that 



VIII 269 



2 l 
gi - n (f-cu ) 



'? 

k=l 



This type of Riemann surface is called hyperelliptic . 



270 



Appendix 
FINAL EXAMINATION 

1. Let a and b be relatively prime positive integers. 
Analyze the Rieraann surface of the polynomial 

A(z,w) = w 2a - 2z b w a + 1 . 
Do the same for the polynomial 

B(z,w) = z 2a - 2w b z a + 1 . 

Be sure to compute the genus in each case and check 
that they are equal. 

2. Let A(z,w) be an irreducible polynomial of degree 
at least 2 in w. Prove that there does not exist 
a rational function f such that 

A(z,f(z)) = . 

3. If A(z,w) is an irreducible polynomial and S is the 
Riemann surface of A, prove that S cannot have 

exactly one branch point (of possibly high order) . 

3 3 

4. The Riemann surface of the polynomial w + z - 1 

is easily seen to have genus 1. Thus, it is homeo- 
morphic to a torus and by our general theorem is 
analytically equivalent to the Riemann surface of 
a polynomial of the form 

w - 4(z-e, ) (z-e~) (z-e.,) 

Find such a polynomial explicitly. 



271 



Hint . Use algebra only. 

5. Prove that the sum of two algebraic functions is 
algebraic. Compute explicitly a polynomial A(z,w) 
such that 

ACz.z^ + z 1 / 3 ) , . 

6. Are the following (noncompact) Riemann surfaces 
parabolic or hyperbolic? 

a. A compact Riemann surface minus a point. 

b. A Riemann surface minus the closure of an 
analytic disk. 

c. A Riemann surface on which a Green's function 
exists . 

SOLUTIONS TO PROBLEMS 6 AND 7 



3 3 

Problem 6 (p. 139)- Analysis of A(z,w) = w - 3zw + z 

Irreducible : If not, A must have a linear factor, so 

o 
A = (w+a) (w +3w+y) , 



where a,3,y are polynomials in z which must satisfy 

a + B = , 
a 8 + Y = - 3z s 

ay = z 

The third relation shows that a = cz , where c ^ and 
k€ {0 ,1 ,2 ,3 j ; solving for a and s and using the second 
relation shows that 



272 



2 2k , -1,3-k ,„ 

C Z + C Z = -JZ 



an impossible identity. 

Critical points : By definition, z=°° is critical. Since 

3 
A(0,w) = w , z-0 is critical. For other z, we look for 

solutions of the pair of equations A(z,w) = and 



f£ - 3w 2 - 3z - . 



That is, 

( 3 3 

! w - 3zw + z J = , 

2 

w = z 

Thus, w - 3w + w =0, sow =2w and since w^O we 

have w = 2 1 ' 3 u> k , where 2 1/3 > and w = e 2nl/3 , k = 0,1,2 
Thus , 

z = 2 2/3 jj 2k , k=0,l,2. 



Puiseaux expansions : 

z=0 . First, we argue heuristically . If w-. ,w„ ,w. are 

the zeros of A , then 



w. + w~ + w„ = , 

(*) J w i w ? + W 2 W 3 + w 3 w i = ~3z , 

1 3 

W 1 W 2 W 3 = " z * 

If there is no branching, these are all holomorphic near 
z=0 and |w, |£C|z|, contradicting the second line of (*) . 
If the branching is of order 2, then each |w, | is 



273 



1/3 
asymptotic to const jzl for some integer I. The 

third line of (*) shows 1=3, again contradicting the 

second line. The only other possibility is a branch 

point of order 1 and a holomorphic solution e(t,tQ). 

For this solution we have 

t 3 Q 3 - 3t 2 Q + t 3 - . 

Thus , 

tQ 3 - 3Q + t = . 
Thus, Q(0) = 0, so we let Q - tQ 1 and find 

t 4 Q 3 - 3tQ 1 + t = . 



Thus , 



t 3 Q 3 - 3Q-L + 1 = . 



The derivative of this polynomial with respect to Q-, 
equals -3 at t=0, so the implicit function theorem 
implies Q-. exists with Q-,(0) =1/3. Thus, the Riemann 
surface has an element 

e(t,t 2 /3 + . . .) . 

o 
The branched element we represent as e(t ,tQ) and 

find 

t Q - 3t J Q + t = . 
Thus , 



274 



Q 3 - 3Q + t 3 = 



At t=0 there is a solution Q(0) = ,J3 and the implicit 
function theorem again can be applied to provide an 
element 

e(t 2 , v /3t + ...) . 

z= m . The heuristic analysis is similar. Now we try 
e(-r,~*) and obtain 

- 3 3 -2 - 3 
t J Q J -3t Q + t J = 0; 

Q 3 -3tQ +1=0. 

At t=0 we obtain 3 distinct solutions Q(0) = -ur , so 
we find 3 unbranched solutions 



•<i 


=l + .. 


.) 




2 


■ 



z=2 ju (k=0,l,2) Again we omit the heuristics, 

1 /3 k 
except to note that w=2 u, is exactly a double root, 

so there is at least one unbranched solution ,' and the 

1 /3 k 
corresponding root is -2-2 ju . Thus, there is an element 

e(2 2/ V k + t, -2-2 1/3 ,, k + ...) . 



Now we see whether the other two solutions are branched, 
If not , then there is an element 



A 275 



,,,2/3 2k . . l/3 k , .<, . . 
e(2 a + t, 2 (u + ct + . . .) 



where c^O and £sl. Then 

<2 1/3 u, k + ct 4 + ...) 3 -3(2 2/ V k + t)(2 l/3 «, k + ct* + 

+ (2 2/3 a 2k + t) 3 = . 



Expanding and simplifying, the coefficient of t on the 
left side is 

o.ol/3 k , ,. 4/3 4k _ ,. l/3 k / n 

(this holds even if <t=l) . Thus , the other two solutions 
are branched and we obtain the element 

e(2 2 /V k + t 2 , 2 l 'h k + ...) . 

Observation : The total branching order is V=4 (first 

order branch points at 0, 2 m ) and n=3, so the genus 

is (recall V = 2(n+g-l)). 

Another example of an algebraic function . 

Let 

A(z,w) = 2zw 5 - 5w 2 + 3z 2 . 



Then 



■as = lOzw^ - lOw . 

Sw 



Now the critical points are z=0 , z=°° , and for the others 
we obtain 



275 



|£ = = 10w(zw 3 -l) . 
dW N 

If w=0, then A=0 => z=0, which we are not now considering. 
Thus , 

zw 3 =l and A = = 2w 2 - 5w 2 + 3z 2 , 
so 

z 2 =w 2 . Therefore, z 2 w 6 =l=w 8 . 

Thus, if uu = e 2ni ^ 8 , w=uj k , ^ k * 7 , and z = uu~ 3k . We 
still must check z = w : uu = uj , which is valid. So we have 

found all the critical points, and we now analyze them. 

-3k k 

z - uu . The only possible multiple value for w is uu 

and at these points 

A 3 

^-j = 40zw - 10 = 30 d 0. 

aw 
We guess a branch point occurs , so we try for an element 
e(oT 3k -i-t 2 , uu k -rtQ) . Then 

2(uf 3k -H: 2 )(aj k +tQ) 5 -5(uu k +tQ) 2 +3('if 3k +t 2 ) 2 ^ . 
Expanding , 

2(uf 3k +t 2 ) (u) 5k +5u) 4k tQ+10« J u 3k t 2 Q 2 +. . .) 

-5(w 2k +2,Ao+t 2 Q 2 ) +3(uf 6k +2uf 3k t 2 +t 4 ) s o; 

10 J k tQ+20t 2 Q 2 +2, JJ 5k t 2 +10x 4k t 3 Q+. . . 

-10<ju k tQ-5t 2 Q 2 +6uf 3k t 2 +3t 4 ^ 0; 



2 
dividing by t , 



15Q + 8uj + IOju K tQ + . . . + 3t z = , 



277 



where the omitted terms vanish at t=0. At t=0 we can 
let 



Q(0) 



/ g, 5k 
= V — r-r- (either determination) 



and note that at t=0 and for this value of Q(0) the 
above expression has its derivative with respect to Q 
equal to 

30Q * . 

Thus , the implicit function theorem is in force and we 
obtain branch points 



-3k. Jl k^/jJuj; 



5k 
e(au Jrv +t^, d:^4V--j5- t+. ..)> 0^k^7 



In order to treat the critical points and °° we 
look for meromorphic elements of the form 

e(t ,t Q) , 

where m and I are integers (m^O) and Q is holomorphic 
near 0, Q(0)^0. Then 

2t m+5i Q 5 _ 5t 2^ Q 2 + 3t 2m g Q _ 

We now try to juggle m and i, to obtain some definite 
information as t->0. Thus, we would like to have at 
least two exponents of t in this equation coincide and 
to correspond to the dominant terms near t=0. Obviously 
it is impossible to have all three exponents coincide. 
Thus, the various possibilities in this case are 



278 



(a) m + 51 = 21 < 2m , 

(b) 21 - 2m <_m + 51 , 

(c) m + 51 = 2m < 21 . 

In case (a) we have m ■ -3-t > <t, so i < 0. Thus, we must 
have I = -1, m = 3, and the equation for Q becomes 

20 5 - 5Q 2 + 3t 8 = . 

3 
Thus, 2Q(0) = 5 and the derivative with respect to Q 

is 10Q 4 - 10Q = 15Q ^ for Q(0) . Thus, the implicit 

function theorem shows we obtain the branch point 

3 5 1/3 1 
e(f, (|) ^+ ...) . 

In case (b) we have m=l<3lsol>0. Choosing m = I = 1 
gives 

2t 4 Q 5 - 5Q 2 + 3 i . 

Again we obtain solutions corresponding to either choice 



of Q(0) and we get two regular elements 

1/2 1/2 

e(t,(|) ' t+...), e(t,-(|) ' t+...) 



In case (c) we have m = 51 < K so I < 0. Thus, we must 
l--\ , m=-5, and we obtain 

2Q 5 - 5t 8 Q 2 +3=0. 



Again we obtain the branch point 

-5 3 1/5 -1 

e(t \ -(|) t l 



279 



This completes the analysis of this example except 
for the observation that the branch point corresponding 
to z=°° is of order 4 and thus all five "sheets" of the 
Riemann surface are branched at e in a single cycle. 
This proves that A is irreducible . 

Notice the total branching order here is V = 8 + 2 
+ 4 = 14 , so the genus g satisfies 

7 = n + g - 1 = 4 + g, 

or g=3. 

3 a 

Problem 7 (p. 139). Analysis of A(z,w) = zw - 3w + 2z , 

cA 2 
a any integer. Now ■¥— = 3zw - 3, so critical points 

oW r 

other than z^O, 00 , come from solving 

2 

zw z = 1, 

-2w + 2z a = . 

a ?a+l 

So w=z and thus z =1. Let b=2a+l and 

2ni/b 

uu = e . 

Then we have the critical points z = uj , £ k s 1 2a+l | - 1; 

ak 
the corresponding double value of w is «j . Here is a 

good place to present a criterion for branching: suppose 

z o ^«= is a critical point for a polynomial A and that 

A(z .w o ) = M( z w ) = o but ^~(z ,w )*0. Then any 
N O / BW x O' o' ?»z v o o' ^ 

element e(z +t ,Q(t)) in the Riemann surface for A 
must be a branch point if Q(0) = w . That is, m>2. 



For suppose m=l . Then for t near 



280 



A(z Q +t,Q(t)) = 0. 

Differentiate this identity with respect to t and set 
t=0 to obtain 

= 4^(z ,w ) + |4(z 5 w )Q'(0) = |~(z ,w ) ^ 0, 

hz K o' o y dw v o' o' x v ' Bz v o' o y ' 

a contradiction. 

k ak 
In the present case we have z = uo , w = oj , and 

dA, N 3 . a-1 3ak , „ (a-l)k n 
^(z o ,w o ) = w Q + 2az Q = oj + 2a «A 0. 

Here we also have more information. Namely, 

2 

- (z ,w ) = 6z w / 0, so w is exactly a double root. 
2 s - o> o' o o 7 ' o L 

c*W 

Thus, the meromorphic element in this case has m=2 , and 
we can express it as 

e(o k + t 2 , w ak + ...), o £ k * |2a+l| - 1. 

To examine the critical points z=0 and » consider 
elements 

e<t m ,t 4 Q), Q(0) ^ 0. 



Then 



t m+34 Q 3 _ 3t ^ Q + 2t am m Q 



For these exponents to be equal we require m = -2-t = -2am, 
so bm=0, and thus m=0, which is not allowed. 
Case (a) . m+3£ = £<am 

Here m=-2-t so we must have £=1 , m=-2, and 
l+2a<0, or -t=-l, m=2 , and l+2a>0. We obtain 



281 



in either case 

Q 3 - 3Q + 2t' b ' = , 
and we have the branch point 

e(t~ 2 , ,/3 t* 1 *...) ( top signS if a>0 

[bottom signs if a<0 . 

Case (b) . m+3-t = am<i 

Here 2>l = (a-l)m and m+2-t<0. The equation is 

Q 3 - 3t" m_2 ^Q +2=0, 

so 

Q(0) 3 = -2 . 

We note that 0>3m+6<t = 3m+(2a-2)m ■ brn. We 
can take m=+l if and only if asl(mod 3), and 
we thus obtain smooth solutions only: 

e(t +i j _ 2 1 /V a - ; r + ...) J top signs if a2 ° 

[bottom signs if a<0 , 

1/3 
where we use all three determinations of 2 ' . 

If a^l (mod 3) , we must choose m=+3 and we have 

a branch point of order 2. 

Case (c) . £=arn<nH-34 

Thus , am<m+3am , or bm>0 . We can take m=+l , 

<t=+a and obtain the smooth solution 

„/v±l 2.±a , , 



182 



Summary . 

If a > 0, z = corresponds to a first order branch point, 
z = oo corresponds to a second order branch 

point if a | l(mod 3), to no branch point 
if a s l(rnod 3) '. 
If a < 0, z - °° corresponds to a first order branch point, 
z = corresponds to a second order branch 

point if a 4 l(mod 3), to no branch point 
if a s l(mod 3) . 

Thus, V = |2a+l| + 1 + { 2 if a ^ 1 ( mod 3 > 

[0 if a = l(mod 3) , 



a if a ^ l(mod 3) , a > 0, 

r. J a - 1 if a = l(mod 3) , a => 0, 

- tj - z. - <^ 

|a| - 1 if a 4 l(mod 3), a < 0, 
|a| - 2 if a = l(rnod 3) , a < 0. 



283 



SOLUTIONS TO FINAL EXAM PROBLEMS 



1 . A(z ,w) = w 



2a 



2z w + 1 



^ dA „ 2a- 1 b a-1 
■*— = 2aw - 2az w 



| = 2a(2a-l)w 2a " 2 - 2a (a-l)zV 1 " 2 



V 



;W 



f§ = - 2bz b -V 
Sz 



Critical points : z = » by definition is critical 

rA 3 b 

Suppose — = and A = 0. Then w = z so 

a 2b 2b , , 2b , . 
A=z - 2z +l=-z +1. 



Let uu = e 



TTl 

b 



Then z = 



s k s 2b - 1, and 



a b bk / i \k 
w = z = uu = (-1) . 

Thus, for each z = uu there are a distinct solutions 

2 

of A(z,w) = 0. Since -^-4 = 2a 2 w 2a " 2 d , we have 

Sw ?A 2b- 1 

no more than double roots. And since ~~ = - 2bz t 0, 

we have a branch point of order 1 (cf. p. 279 ) 

associated with each solution w of A(uu ,w) = 0. 

Thus , there are a branch points of order 1 lying over 

each z = u) , so the branching associated with these 

critical points is 2ab . 

Now consider z = ». Look for elements of M 

of the form 



e(t- m ,t*Q) 



284 



where m > and Q(0) i 0. Substituting, 



t 2a^ Q 2a . 2t a^-bm Q 2 + ± m Q 



As on p. 280 , we have 3 cases: 
Case (a) . 2a£ = al - bm < 

Here I < and aK, = - bm. If we choose m = a 
and K> = - b , we obtain 

Q 2a - 2Q a + t 2ab , 
and thus 

Q(0) a = 2. 

Then we obtain 

e(t- a , 2 l/a t- b + ...)• 

Since a and b are relatively prime, this is 
an element of M. 
Case (b) . al - bm = < 2a<t, 

Here K> > and aK, = bm. Choose m = a and I = b , 
obtaining 

t 2ab Q 2a _ 2 Q a + 1 s , 

a 1 
so that Q(0) = ■£. Then we obtain 

_1 

e(t~ a ,2 a t b + ...). 

Case (c) . 2a£ = < aj, - bm 

Here K, * and bm < , which is impossible. 
Summarizing, at z = « we have two branch points, 
each of order a - 1. Thus, V = 2ab + 2(a-l) , 



285 



so 



or 



ab + a - 1 = 2a + 



g = ab - a . 

2a b a 
Second part . B = z - 2w z +1 

The equation B = is 

, 2a,, a, -a 
b z 4-1 z +z 

W = = 7y 

„ a l 
2z 



Thus, we simply obtain branch points at z = 0, z = <=, 

1/b 



2a 
and where w = 0, which is z +1=0. Since 



"&*) 



the branch points at finite z are of order b - 1 
and at z = or °° are of order b - 1 as well, since 
a and b are relatively prime. Thus, 
V = 2a(b-l) + 2(b-l) , so 

a(b-l)+b-l = b + g-l, 
g = a(b-l) = ab - a. 

Alternate solution : Solve for w : 

w = z + Vz -1 (either determination). 

By inspection there are branch points over the roots 
of z =1, each .of order 1. This gives 2ba to the 
total branching. Then near z = *> we have w ^ z ± z , 
or w - 2z on half the sheets and 



286 



a b b n 1 -2b , v 1 

W =Z - Z ( 1 - -s-Z +...)= — C + • • • 

z 2z D 

on the other half. Thus, w a 2 z ' gives a 
branch point of order a - 1 and w a: 2 -l/ a z _ b/a of 
order a - 1. So 

V = 2ba + 2 (a - 1) , 
and 

ab + a-l = 2a + g-l, or g = ab - a. 

2. Let n be the degree of A with respect to w. Let 
S. be the Riemann surface of A. Let S be the com- 
ponent of M which contains all the germs 

[f ] =e(a+t ,f (a+t)) , assuming f is rational. By 
hypothesis, ScS*. Clearly, n:S - t is an analytic 
equivalence, so S is compact. As S. is connected, 
S = Sa. Thus, n:S. - t is an analytic equivalence. 
But tt restricted to S. takes every value n times. 
Thus , n = 1 . 
Alternate solution : Suppose f is rational: f(z) = Q )"( 
in lowest terms. Let 

B(z,w) = Q(z)w - P(z) . 

Then A(z,f(z)) - B(z,f(z)) = V z. Thus, Lemma 
3 on p . 121 implies A and B have a common factor. 
Since A is irreducible and B is linear in w, this 
implies that A = const B, which shows A has degree 1 

3. We prove something more general. We have tt:S - U, 



287 



taking every value n times. Here n > 1, since 

otherwise tt is an analytic equivalence and there 

are no branch points. Suppose S has branch points 

e, , . . . ,e and that rr( e i ) = • • • = "(e ) = z , Since 

C - fz } is simply connected, the corollary on 

p. 119 implies there exists a meromorphic function 

f on t - [z } such that A(z,f(z)) = 0. (Actually, 

the corollary is stated for regions in C and 

holomorphic f, but the generalization to this case 

is easy.) By familiar estimates, f grows at most 

like a power of z - z as z - z . Thus , f is 
v o o ' 

rational and the previous problem implies n = 1, 

a contradiction. 

3 3 
Let S be the Riemann surface of w + z - 1. Then 

tt and V restricted to S satisfy V + tt = 1. 

Let 

1-B 



V= f 

so that f = tt— and g = y^- are meromorphic on S 

Then 

3 / \3 2 

2+6g 



i-m+M 



so that 



Now 



- g 2 = ^(f 3 -2). 
let g x = 2V<5"g 5 obtaining 



2S8 



g 2 = 4(f 3 -2). 

Now apply the argument which appears on p. 256 and 

P-262 » concluding that S is analytically equivalent 

2 3 
to the Riemann surface of the polynomial w - 4(z -2) 

A little care seems to be needed at this step. 

Namely, we need to know that f takes every value 

2 times in order to be able to apply Theorem 2.4 

on p. 158 . We do this by checking that f takes the 

value 2 times , or that V + tt takes the value °° 

2 times. This is easy. The surface S has 3 smooth 

■ / ^ 
sheets over °°, and if uu = e 171 ' (cube root of -1) , 

then on these three sheets we have respectively 

„ /. -3.1/3 

V = lutt(1 - tt ) , 

„ 2 n -3x1/3 

V = UU TT(1 - TT ) , 

,, _ 3,, -3.1/3 

V = (JUTT(1-TT ) , 

- 3 1/3 
where (1 - tt ) ' is the principal determination 

near tt = °°. Thus, in the first two cases we have 

V + tt = tt[1 + <ju+...] 

and 

V + tt = tt[1 + uu 2 + . . .] 

and thus V + tt takes the value « one time on each 
sheet. On the third sheet 



V + tt = n[l -(1-tt" 3 ) 1/3 ] 



289 



= ^ + ... , 
3tt 

and thus V + rr takes the value on this sheet (at 

the point lying over z = ») . Thus , V + n takes 

the value <*> exactly 2 times. 

Alternate solution : Define 

F - a y±^ on S, 

1-TT 

so also F takes every value 3 times . Then 

F - Ftt = a + an , 

F-a 

11 = F+£ ' 



Thus , 



v 3 + 11=5)1 a l 
(F+a) J 



((F+a)V) 3 + (F-a) 3 = (F+a) 3 
((F+a)V) 3 = 2(3F 2 a + a 3 ) . 



Let 



Then 



G = (F+a)V 
(24a) 1/3 



24aG 3 = 6aF 2 + 2a 3 ; 
F 2 = 4G 3 - *i . 
5. We write the hypothesis in the following way. 

On a disk AcrC are given meromorphic functions f 
and g such that for certain polynomials A(z,w) 
and B(z ,w) , 



290 



A(z,f(z)) = B(z,g(z)) = 0, z<EA. 

We can assume A and B are irreducible. Let 

z, ,...,z N be the critical points of either A or 

B. Then define for z *' z. 

J 



C(z,w) = | | (w - a - 3) 



A(z,a)=0 
B(z,3)=0 

By the usual symmetry argument, C is a polynomial 
in w with coefficients which are holomorphic functions 
of z€t - [z, , ...z-j,}. By the usual estimates, these 
coefficients have polynomial growth at these exceptional 
points, and thus are rational functions of z. Ob- 
viously, C(z,f(z) + g(z)) = for z€A. 

To do the second part we use the above formula. 
The required polynomial is therefore 

r( \ t 1/2 1/3 W , 1/2 1/3 W 1/2 ,1/3) 
C(z,w) - (w-z ' -z ) (w+z ' -z ) (w-z -(JUZ ' 

> . 1/2 1/3 W 1/2 2 1/3 W . 1/2 2 1/3. 
x (w+z - x-z ) (w- z - x z ) (w+z - x z ) 

o*/q i / o t'q 

where x = e and z and z are any values 
of the roots. After multiplying all these terms 
together we are bound to get a polynomial. Here is 
the arithmetic: take the terms #1,3,5 together and 
likewise #2,4,6 to obtain 

C(z,w) = [(w-z l/2 ) 3 -z][(w+z 1/2 ) 3 -z] 
= [w 3 -3z l/2 w 2 +3zw-z 3/2 -z] 

x [w 3 +3 z l/2w 2 +3zw+z 3 / 2 -z] 



291 



, 3,, ,2 /Q 1/2 2^ 3/2.2 
= (w +3zw-z) - (3z w +z ) 

- w 6 + (6z-9z)w 4 + (-2z)w 3 + (9z 2 -6z 2 )w 2 

- 6z^w + z - z J 

= w 6 - 3zw 4 - 2zw 3 + 3z 2 w 2 - 6z 2 w + z 2 - z 3 . 

.a. Parabolic . We check Proposition 15.2 of p. 204. 

If S is the compact Riemann surface and pGS , 

let D be an analytic disk in S - {p}, and let u 

be a bounded continuous nonnegative function in 

S - fp] - D which is harmonic in S - {p} - D and 

= on 3D. Since u is bounded near p, u has a 

unique extension to a harmonic function in S - D . 

As S - D is compact, the maximum principle holds 

and implies that sup u= sup u= 0, so u ^ 0. 

S-D 3D 
Thus , us 0. 

b. Hyperbolic . Choose a nonconstant function f which 
is continuous and real-valued on the boundary of 
the analytic disk in question. By Proposition 13 
on p. 197 , there exists a harmonic function u in 

the Riemann surface minus the closed disk, continuous 
up to the boundary, where it equals f. Moreover, 
u is bounded. Since f is not constant, u is not 
constant. Thus, u is a bounded nonconstant subharmonic 
function, and we apply Proposition 15.1 of p. 204. 

c. Hyperbolic . By definition (p. 218 ) there is a point 
pGS and a function g on S - {p} satisfying the 
conditions of Definition 7 of Chapter VI. Since 



292 



g - =° as one approaches p, if A is a sufficiently 
large constant the function u = min (g,A) is super- 
harmonic and not constant. In fact, u is super- 
harmonic on S since u = A near p. As < u ^ A , 
u is also bounded. Apply Proposition 15.1 of Chapter 
VI. 



293 



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Mackey, G. , Lectures on the Theory of Functions of a Complex 
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Narasimhan, M. , et al , Riemann Surfaces , Tata Inst, of 
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294 



Riemann, B., Collected Works , Dover, N. Y. , 1953. 

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Stoilow, S. , Lecons sur les Principes Topologiques de la 
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Weyl , H . , The Concept of a Riemann Surface , 3rd ed . , 
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