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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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^  HOUSTON. £.,
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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 1518.
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: zl<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 logz + 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 logz + 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 singlevalued "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
multiplevaluedness 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 singlevalued 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) = logz + 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': wz<c, Im wsO) u fw€C', ,: wzj<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 logz + irr + 2nni = f(z). Thus, f
is continuous.
Thus , we have succeeded in defining a set S which
carries a singlevalued 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(nl)/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(nl)/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 m1.
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 11 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
multiplevalued "function" ,/(za) (zb) . 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(za) (zb) 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(za) (zb) 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 ,/(za) (zb) 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 ./(za) (zb) (zc) , 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<fea) (zb) (zc) (zd) , 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 '(za.j ) (za„) . . . (za ) , 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
m1 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
Cthe 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: jwz<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 ii is compatible with
G. Since G contains {cp } (since {cp } is itself
an analytic atlas), ii 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 cpitUi "* 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 ,
cpiocpo (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. 1315.
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
Injuu, + 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]) = wz. 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 onetoone correspondence with
c/n.
II
25
JJJ+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: ]zp<e}u{zeA: ] zp 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 pi 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 ((22t)cp(p) + (2tl)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 cppUj  W, in a neighbor
hood of p Q such that f(Uj)<=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 CauchyRiemann 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]) = wz,
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 = forr.
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
jwz < e and cp(p) = wz. 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. 34).
Then f is holomorphic on S. Likewise,
consider the Riemann surface T for z
and the corresponding function f (pp. 56).
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(Ui) contains a neighbor
hood of fCpjj). We can assume cp, : II,  W, is a chart
for S, and cp2 : U? " w ? a cnart f° r ^9 anc ^ f(Ui)cU 9 .
Then ep2 c f°CDj 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 cpi (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 icUj • 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 =
<Poir "(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
fok (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 if (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 . 1m . , f,
*°V (z) = a_ m z + ai _ m z + ..., a_ m t 0,
a,
m,, 1m , >.
= a z (1 + z + . . . ) •
m v a
m
In this case choose a such that a = a and h
m a lm
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. 2325 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^) = (pi».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
k1 k
VnG. Thus, N(q') = N(q) for q'^VflG. Thus, the
integervalued 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
("multiplevalued") 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
plevaluedness 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 (za) 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
(za) k = (zb + ba) k
 (b*) k (l + B^) k
n
= (ba) k E & &=&
n=0 ' n (ba) 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 , ^ (zb) 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 ki H f k in Vi n A k'
and such that there exist = t n <t,<...<t = 1 with
1 n
Y([t kl 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: lzlj < 1}
and f, is the principal determination of z 2 in A,
f (z) = (l+(zl))^ = A(zl)'
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 yi 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 axiomnamely , 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 onetoone 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 fc1 , 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(sr(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 xs .  < %r(s.). If yx] £ \ rZ } then
Ill 57
ys.  £ yx + xSj < %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
" k1' 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 k1 T k
U and U are [f ] and [f ] , respectively.
T k1 T k T k1 z T k z
In particular for z = y(t, _, ) we have
k1 k
so that f = f in a neighborhood of y(t, , ) , and by
T k1 T k • tc ~ i
analytic continuation f s f in A P, A Finally,
Ik1 T k T k1 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 yi 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) = [fjL, 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 (singlevalued) 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 kl s f k ° n A k1 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 kl' 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 up < , , 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. , u3 <€ , u6I,
i) ' k1 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 yi : I » T are paths in T having the same
end points , then Yq and yi 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
Yi are Thomotopic with fixed end points .
Ill 63
DEFINITION 7 . A connected topological space T is
simply connected if each pair of paths Yn an d yi 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 yi ar e Dhomotopic with
fixed end points , and by the covering homotopy theorem
the paths Yn an< 3 yi 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 socalled "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 singlevalued 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 ,(zz ) J (ww )
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., yi i s continuous) and obviously n yi = Y
Also define
VZW = [ ° ] Y(t) •
Then Yo is a path in M with ^=Yo = Y By hypothesis,
Yi (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. 6465).
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 multiplevalued
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 zplane 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 vv 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 wplane
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 wplane:
? 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 zplane, cut as described. Sup
pose we use three copies of the zplane, 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 wplane. This is in
dicated in the figure.
/
/ y>0
y<0
o / 2 v
y = Jv(u
T
D
wplane
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 onetoone
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 [gJu £ 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 seta 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 onetoone 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
onetoone .
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 (PpQi), 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 pcpi 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) = Qj^ (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 onetoone. This means that each element
of M is identified with an element of M in a one
toone 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 oneto
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 onetoone 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 , Qi = Q° o in a disk
Ai centered at 0. We choose Ai 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. 3741 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 m1.
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. 5253. 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 onetoone (small
t), then the mapping s (P(s ),Q(s )) is exactly ,
ntoone (small s). Likewise, the mapping s  (P, (uus ),
Qi (jus )) is exactly m toone. 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 onetoone. 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 onetoone for tfA, this
implies t = t ' . This shows that at least cp maps
U(PjQ>A) to A in a bijective fashion (onetoone
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 onetoone 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(Pi ,Qi , Ai ) 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 cpi°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 + (zt Q +t)) = P 1 (t 1 + p(zt Q +t))
= P 1 (t 1 + p(zt Q ) + [p(zt +t)  o(zt o )l).
The function
Pl (t) = p(zt Q +t)  p(zt )
satisfies pi(O) = and pf(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 onetoone 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 ftta \
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 onetoone
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
onetoone, 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. ,Pi). 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
so
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 multiplevalued 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)
ss„ ss_ ss„
= lim a(s)  a(s Q ) .
ss
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. 183. 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
multiplevalued 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  (za) (zb) ,
(zb)w  (za) ,
w  (zap (za 2 ) . . . (za 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 RiemannHurwitz 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 ve+f .
A theorem of topology is that this number is a topo
logical invariant of the surface and equals 22g:
ve+f = 22g.
Now S has certain branch points e,,...,e of orders
b,,...,b , respectively, b. s 1. Define
112
V = lb.
Jl 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. 4344. The RiemannHurwitz
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
ve+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,
(nvV)  ne + nf = 22g.
Since ve+f = 2 we can write this relation as
In  V = 22g,
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  (za, ) . . .(za),
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 + gl = 2 + g 1,
m2
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 3wz 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. 6873,
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 n1 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 CauchyRiemann 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 < ww  < 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 zzj <{
ww
o
116
Therefore, the residue theorem implies that for each
fixed z, zz 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 ww  < e And it is
clear that this number is a continuous function of
z for jzz  < 6, and is therefore constant. For
z = z Q we are counting the number of zeros of A(z ,w)
in ww 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 zz <c
Thus, j zz 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(si ) • 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 , ss 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 ai , • • • , 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 k1
R k1 = \ 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 , n1
n1 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 TD 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 onetoone (as it must), then the mapping
t  P(t) must be onetoone, 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 CD 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 (wf 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 CD) . Then for
some sufficiently large integer N we have near a
a (z) > zaj 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 < nCza 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 <, constza 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 za 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 cD 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 CD is finite,
sufficiently small that
1 the set CD is finite. Let a € C and choose e >
A = [z:  za  < 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, za 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(ml) = 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 ), (ml)^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 (k1)— <; arg t < k— and
v y m & m
(jl)~ 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 onetoone.
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 multiplevalued solutions
of A = by two singlevalued 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= (za) } gives
a series
1 a, (za)
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 RiemannHurwitz 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
1a 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[ (wV(e k ))
k=l
= IT (wV(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 , CD
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 \ (wV(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 CD. 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™ = za.
(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  Iza 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) <.  za  if e € S A n (z), z near a.
Thus
Mk
a k (z) < const za 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 ai, 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 QQ^ 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
n1
n1 .
f = I a. nV 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
n1 .
f(e, ) = L a.(z) V(e, ) J , 1 £ k < n.
k j0 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 socalled Vandermonde determinant and its
value is well known and easily seen to be
7T (V(e k )V(e^)),
lst<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
za = s .
Let
2ni/m
. = €
J
a 4  e J
Then for <. I <, m 1,
m/m. m.
za = (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 = za
so the Vandermonde has modulus bounded below by
N n(nl)
I i m "2
za
Thus, we have proved that each a is rational and
by definition
n1 .
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) = £ as 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 onetoone 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
ii 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 onetoone 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 :
goif (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, = \rz, 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 onetoone. 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
toone. 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((zt) 4 , (zt) 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 CauchyRiemann
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 realvalued 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 ii 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 = uf. 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 logz + Ref(z),
f (z) = — + ... (Laurent expansion
near 0) , a # 0.
Let ii be a chart near q , if(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 (CauchyRiemann equation) and thus
constant (since it is realvalued) . 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,
lzz o i< P * x *y zz o i= P v
where dS is the element of arc length on the circle
I zz 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) =
1az
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 1ae
10
+
io ie 
e  a e a
,  ie. i9 
1ae +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 socalled 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
ze ° 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
i9  16
i9v e ° ze
2rrP(z,e J ) = f— +
19 ,  11
e z 1ze
exhibiting P as a sum of two harmonic functions of z.
ie
To prove the fourth, assume j ze ° < 6/2. Then
ifl ie ie ie
e z s e e °e °z > 66/2 = 5/2, so that
2
I P(z,e l9 )d9 g j~ 1 I Z L • 2 n < 4^(11 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  ze ° < 5 ' ' , we obtain
i8
u(z)f(e °) < c
18
if I ze ° < 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 zz  .
u(z) = ^r f ■ ■** — 2 f( z + re lb )d6.
Z * J !re l6 (zz o ) Z °
This can be derived in the same manner,, or merely
by considering the change of variable zz = 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 zz  < jr , the Poisson integral
formula implies
where
l^(z)  5 c supu ,
VI 177
2 I I 2
. r  zz
I ' O ' I
C = SUP r— — r
i dx  13/ viz '
I I ^ 1 re  (zz )
zz < ?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 zz  <— 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 zz ia
u (z) = 4 f ' 2 u (z + re Ld )d9.
re  (zz )
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 (zz 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
lsrl * 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 zz' •
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: zz  < 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 lzz [ 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 SD 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 lz <,  e 1 z < 1 + zl so we obtain
1 z __ 1 z
1 ' (l+zj) e z (lz)
= i±Ui
lz
ij.
Therefore, since v(e ) > 0,
1lzl 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 16
Therefore, if jz < 6 and w < 5,
v<0) £ (±±V
15 16
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 ' , uu 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 n1
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 realvalued
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 u2 (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:  zw <r})
for sufficiently small r. Then u < u„, so in particular
UOep (z) < UpOcr, (z) .
Since u^or is continuous on j_w : zw 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 = uh 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 (SA)~. 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 realvalued 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 Mu
are subharmonic in © and nonpositive on 3S, so the
weak maximum principle implies M+u <; and Mu <,
in &. That is, M <, u <; M.
PROPOSITION 11. Let u be a continuous realvalued
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 complexvalued 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 SD (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 SD, harmonic in SD , and such that
u s f on SD . Moreover, we can assume
%%
?£>
inf u = inf f
SD 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 AiA,
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
SD by the formula
h 0
c
in D,D,
c in SD
Then v is continuous on SD, v s f on SD. and v is
c c c
harmonic in SD, and in D.D
If c <. inf f, then v
SD
is subharmonic on SD ; 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 SD. Therefore, criterion 3 of Propo
sition 10 is trivially satisfied at a point of 9D, .
Thus, v is subharmonic in SD Likewise, if c 2 sup f,
VI 199
then v is superharmonic in SD .
c
Let A = inf £, B = sup f . Then v A is subharmonic
3D 3D A
in SD , v R is superharmonic in SD , and v. <; v in
SD. This last inequality follows from the maximum
principle , since h_h. is continuous in A, Aj harmonic
in A.A , = BA 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 SD . We have of
course applied the corollary of Theorem 5 to the
connected Riemann surface SD , which is why u is
defined only on SD . But the inequalities (7) imply
that u can be extended to a continuous function on SD
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 logz + 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 logz .
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 logz + I 23 Re( n )
n^1
n
c z
= 2s(c . logz) + 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 CauchyRiemann
Inn J
equation
a(u  c logz  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 logz + 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«— nine
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 rlog r Z... q
a A = t ^ r rr [u(r'e i9 )log ru(re 18 )log r']d0
log rlog r 2ttj q e & J
1 1 (i p / iS N /~m //iBNmiimGj^
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 SD which is harmonic in SD 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 SD which
is harmonic in SD , then
sup u = sup u.
SD 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 SD which is sub
harmonic in SD . Choose a constant C such that sup u^C
SD
Let w be the unique function which is continuous in SD,
harmonic in D,D , such that
206 VI
w = u on 3D,
w  C in SD, .
Then w is superharmonic in SD and criterion 2 of
Proposition 10 implies u < w in D, D and therefore u ^ w
in SD. Let v = GHM of w (cf. p. 197). Then v is har
monic in SD and u s v <, w. Therefore, we can naturally
extend v to be continuous in SD by setting v = u = w on
dD . Condition 3 applies to v and thus
sup u < sup v = sup v = sup u
SD SD 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.
SD 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 SD,
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.
SD SD ?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 SD, harmonic in D D , such that
v. = u on ?D,
v. = A in SD, .
We define v the same way with A replaced by B. Then
B
v. is subharmonic. v„ is superharmonic in SD , and
A D
v. <, C, v. <, u, u < vui 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 SD 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 BA. Thus, condition 2 implies ^wiO.
Thus
B = sup u = sup v/, < C .
SD SD 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 SD, which is harmonic in SDT and which is
identically zero on SD, , and yet v ^ 0. The strong
maximum principle implies that v > in SD7. The
function cp is holomorphic on U, and therefore
Re(rp qj ) is harmonic on Ufp} 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 SD,,
1
t Re(cp" n cp n ) in B ±  {p]
f Cv in SD,
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
SD,
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
<; uRe( ¥ " n ^ n ) ^ A in D x  ip}.
Therefore, the function uRe(<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
SD such that u = Re(. £ ) on 3D and u is harmonic in
r r V4/ ' r r
SD . If u is constant on 3D,, then Proposition 15.3
implies u is constant in SD, (apply the criterion 3
both to u and u ) and thus u is constant in SD 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 SD, .
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 SD,/,. If v = lim v , then
r, ° J 3/2. , r,
k koo k
Proposition 7 implies v is harmonic in SDT .
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/~nin9\,„
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)(lr^) * £ 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 SD, ,
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 ) ,
qp
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)clog 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
+ 82 + 2^ .
l2r
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 12r
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. 168171 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 V9 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 (lr 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. 209211
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 logz + 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 + logcp 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  ii = loj
1
k
and we see that log ii  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 SD, , v > and v is harmonic in SD, , and v = on
3D, . As before, there exists a constant C > such
that
log  cd I ^ Cv in D^  Dj.
As before, the function
f Cv in SD, ,
{ log ml in D 1 ,
is superharmonic in S{pl. Much more trivially, the
function
220
VI
w
f in SD,,
■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
SD 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. 4664 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 = (zaib: 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 , taib
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 SD^; p. 214 showing that u = v in SD n ;
and p. 218, showing that our function u is a linear
combination of two functions bounded in SD, . 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 onetoone 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 onetoone 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 fa 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
SD 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 SD /0 . Let
k— n k 3/2
h = lim f in S  D. .
k„ n k l
VII 229
Then h is holomorphic on SD, 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 cpcp(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
<; 4f 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 Dw 2 , sa y
lim g = g, holomorphic in D^/o*
Now define
f h in SD" ,
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 SD0/9, 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 SD7, 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 onetoone . 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
wg
W0
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 onetoone. Since F is not constant and takes the
value 1, F takes every value z for zl < 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 onetoone 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
loghj = n log j co + loga + logl+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 51 ^ 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 onetoone . 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 onetoone
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(wa) 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 Jwa.
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: jzF(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 onetoone. 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 92
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 onetoone \
analytic equivalence of S onto C
f maps S onto C in a onetoone 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, ukHi^'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, pi+n09 : 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 ci 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
nfold composition of t~ , is just the mapping
z . z+n, ci+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 onetoone, 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 , adbc 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
lOist^)  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 zct
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,
moAcm (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
onetoone. 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 onetoone.
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(ze 1 )(ze 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
(zaj^) (za 2 ) (za 3 ) (za^),
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
fi.
where tt and V are the functions on M discussed in
Chapter IV, restricted to S. Thus,
2
V = 4(ne 1 )( T Te 2 )( T Te 3 );
fV = 4f(f(na) + (ae 1 )f)(f(na)+(ae 2 )f)(f(rra)
+ (ae 3 )f)
= 4f(l + (ae 1 )f)(l 4 ( a e 2 )f)(l + (ae 3 )f)
 4(ae 1 )(ae 2 )(ae 3 )f(f i ^)(f L.) (f  ^
Choose a complex number 3 = v , 4(ae,)(ae )(ae 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. 913 . Also,
it follows from the RiemannHurwitz 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 uuj , uuo € C have nonreal ratio and define as usual
n = [n, uu, + n.pUy: n,,n2 integers). Define
^ z Z C €Q (zC) C
We must first prove this series converges. If K is
a compact subset of CQ, then for z € K
258 VIII
3^4l \ z2+2 fi *ec 3 •
(zc) 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 CQ, 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 (2C) £
Replacing the "dummy" £ by ;, we therefore obtain
(p(z) = \+ Z (~ J — ^  K)
z z ren (zc) £
Next, we compute^); since the series f or ££> converges
uniformly locally, this can be done formally:
£/f>(z) = ^§ + £ " 2 3
z CGQ (zC)
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 0Kz
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 Tpj+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(ze 1 ) (ze~) (zeO •
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) = za+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 oneto
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(zC 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 onetoone.
If we assume that F:C/q  C/q is onetoone and
onto, then since F(0+q) = 0+n, we have
z+o = OKi « 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, jui+t^um 1 n, ,n2 integers] ,
then we have
aw, = n. , ij+n, 2^0
a'jj« — n« jjj +n^ ^ uuo
for certain integers n
., . Also ,
jk
Si = m, , aioj,+m 2 a ^2
Si 2 = m n\ axi+m22 aijj 2
VIII 267
for integers m., . Therefoie, 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 axo 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  (za 1 )(za 2 )...(za 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,
(ga(f)) 2 = a(f) 2  b(f) .
Let g x = ga(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 (fa k ) z n (fe)
2 _ 2 k=i k ., j=i J
§1 " a IP ~ x n 7 '
n (fa') z n (fB,0
ki K ji J
where the a's and g's are complex constants and the
numbers 9. and B .' are distinct. Let
J 3
m'
n (fap n ,
n (fa.) J_1
k=l *
Then
g z = n (f B .) n (fe/)
ji J ji 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 (fcu )
'?
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(ze, ) (ze~) (ze.,)
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 , 1,3k ,„
C Z + C Z = JZ
an impossible identity.
Critical points : By definition, z=°° is critical. Since
3
A(0,w) = w , z0 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, £Cz, 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  3QL + 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 22 ju . Thus, there is an element
e(2 2/ V k + t, 22 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+gl)).
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 it 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 tQ5t 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 — rr (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:^4Vj5 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 ■ 3t > <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 . a1 3ak , „ (al)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 = 2t = 2am,
so bm=0, and thus m=0, which is not allowed.
Case (a) . m+3£ = £<am
Here m=2t 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+3t = am<i
Here 2>l = (al)m and m+2t<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+(2a2)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<nH34
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 a1
■*— = 2aw  2az w
 = 2a(2al)w 2a " 2  2a (al)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(al) ,
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 41 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(bl) + 2(bl) , so
a(bl)+bl = b + gl,
g = a(bl) = 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  sZ +...)= — 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 + al = 2a + gl, 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
1B
V= f
so that f = tt— and g = y^ are meromorphic on S
Then
3 / \3 2
2+6g
im+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
P262 » 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(1TT ) ,
 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 (1tt" 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,
1TT
so also F takes every value 3 times . Then
F  Ftt = a + an ,
Fa
11 = F+£ '
Thus ,
v 3 + 11=5)1 a l
(F+a) J
((F+a)V) 3 + (Fa) 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, , ...zj,}. 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)  (wz ' z ) (w+z ' z ) (wz (JUZ '
> . 1/2 1/3 W 1/2 2 1/3 W . 1/2 2 1/3.
x (w+z  xz ) (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) = [(wz l/2 ) 3 z][(w+z 1/2 ) 3 z]
= [w 3 3z l/2 w 2 +3zwz 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 +3zwz)  (3z w +z )
 w 6 + (6z9z)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.
SD 3D
Thus , us 0.
b. Hyperbolic . Choose a nonconstant function f which
is continuous and realvalued 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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THE FONDREN LIBRARY
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