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Math 213a - Complex Analysis

Taught by Wilfried SchmidNotes by Dongryul Kim

Fall 2016

This course was taught by Wilfried Schmid. We met on Tuesdays and Thurs-days from 2:30pm to 4:00pm in Science Center 216. We did not use any text-book, and there were 13 students enrolled. There was a take-home final andalso an oral exam. The course assistance was Yusheng Luo, a graduate student.

Contents

1 September 1, 2016 41.1 Holomorphic functions . . . . . . . . . . . . . . . . . . . . . . . . 4

2 September 9, 2016 72.1 Continuing the proof . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 September 11, 2016 103.1 Finishing the proof . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 A bunch of corollaries . . . . . . . . . . . . . . . . . . . . . . . . 11

4 September 13, 2016 144.1 More and more corollaries . . . . . . . . . . . . . . . . . . . . . . 144.2 Laurent series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 September 15, 2016 185.1 Isolated singularities . . . . . . . . . . . . . . . . . . . . . . . . . 185.2 Residue theorem and its applications . . . . . . . . . . . . . . . . 19

6 September 20, 2016 216.1 The argument principle and total order . . . . . . . . . . . . . . 216.2 Projective 1-space . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7 September 22, 2016 257.1 Automorphisms of P . . . . . . . . . . . . . . . . . . . . . . . . . 257.2 Conjugacy classes of matrices . . . . . . . . . . . . . . . . . . . . 267.3 Automorphisms of ∆ . . . . . . . . . . . . . . . . . . . . . . . . . 27

1 Last Update: August 27, 2018

8 September 27, 2016 298.1 Automorphisms and conjugacy classes . . . . . . . . . . . . . . . 29

9 September 29, 2016 329.1 Riemann surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 329.2 Mittag-Leffler theorem . . . . . . . . . . . . . . . . . . . . . . . . 33

10 October 4, 2016 3510.1 Mittag-Leffler on an open set . . . . . . . . . . . . . . . . . . . . 3510.2 Infinite products . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

11 October 6, 2016 3811.1 Weierstrass problem and product theorem . . . . . . . . . . . . . 39

12 October 11, 2016 4212.1 The Gamma function . . . . . . . . . . . . . . . . . . . . . . . . 4212.2 Gamma as an infinite product . . . . . . . . . . . . . . . . . . . . 44

13 October 13, 2016 4613.1 Legendre duplication formula . . . . . . . . . . . . . . . . . . . . 4613.2 Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

14 October 18, 2016 5014.1 Functional equation for ζ . . . . . . . . . . . . . . . . . . . . . . 50

15 October 20, 2016 5315.1 Hadamard’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 54

16 October 25, 2016 5616.1 Poisson-Jensen formula . . . . . . . . . . . . . . . . . . . . . . . . 56

17 October 27, 2016 5817.1 Proof of Hadamard’s theorem . . . . . . . . . . . . . . . . . . . . 58

18 November 1, 2016 6018.1 Poisson’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 6018.2 Normal family . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

19 November 3, 2016 6319.1 Riemann mapping theorem . . . . . . . . . . . . . . . . . . . . . 63

20 November 8, 2016 6520.1 Caratheodory’s theorem . . . . . . . . . . . . . . . . . . . . . . . 65

21 November 10, 2016 6821.1 Riemann surfaces again . . . . . . . . . . . . . . . . . . . . . . . 69

2

22 November 15, 2016 7122.1 A fundamental domain for Γ(2) . . . . . . . . . . . . . . . . . . . 71

23 November 17, 2016 7423.1 Universal covering space for the thrice-punctured sphere . . . . . 7423.2 Invariant form of Schwartz lemma . . . . . . . . . . . . . . . . . 75

24 November 29, 2016 7624.1 Big Picard theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 76

25 December 1, 2016 7925.1 Compact Riemann surfaces coming from C . . . . . . . . . . . . 7925.2 Elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3

Math 213a Notes 4

1 September 1, 2016

There will be either a take-home and an oral exam, or a sit-down final. Thereare no textbooks, but Ahlfors and Greene-Krantz is will be a reference.

We are going to do a rapid review of complex analysis. Then we are going totalk about Mittag-Leffler and Weierstrass theorems and refinements, and thennormal families and Riemann mapping theorem with refinements. Next we aregoing to do learn the Picard theorems, and then special functions like the Γfuncitons.

We will typically look at an open set U ⊆ C and a function f : U → C.Some conventions we will use is if z ∈ U then we write z = x+ iy and for suchfunctions f(z), we write f(z) = u(x, y) + iv(x, y) with u, v : U → R real.

1.1 Holomorphic functions

Theorem 1.1. The following conditions of f(z) = u+ iv are equivalent:

(1) for each z ∈ U , f is differentiable in the complex sense, i.e.,

f ′(z) = limh→0

f(z + h)− f(z)

h

exists.

(2) for each z ∈ U , f ′(z) exists and z 7→ f ′(z) is continuous.

(3) the map (x, y) 7→ (u(x, y), v(x, y)) is a C1 map which satisfies the Cauchy-Riemann equations ∂u/∂x = ∂y/∂v and ∂u/∂y = −∂v/∂x.

(4) the function f is continuous and∫∂D

f(z)dz = 0 for every Green’s domainD ⊆M , i.e., D that is relatively compact in U and ∂D is a disjoint unionof piecewise C1 simple closed curves.

(5) the function f is continuous and∫∂Rf(z)dz = 0 for every rectangle R

whose closure is contained in U .

(6) the function f is continuous, and for every z0 ∈ U , there exsits an openneighborhood V ⊆ U and F : V → C such that f |V = F ′ with F having aderivative at every z ∈ V .

(7) for every z0 ∈ U , there exists an open neighborhood V in U and a powerseries

∑∞n=0 an(z − z0)n that converges absolutely on X with value equal

to f .

One remark is that if f, g : U → C are differentiable in the complex sense atz ∈ U , then f+g, fg, cg for any c ∈ C, and f/g if g(z) 6= 0, are all differentiablein the complex sense and the usual rules of differentiation apply. Also, (2)is Cauchy’s definition of a “holomorphic function” where (7) is Weierstrass’definition of a holomorphic function. One funny anecdote is that when JohnTate taught this course, he used only (7) to develop the whole theory. Thereason was because if you are working for instance in Qp, then you have to doeverything in terms of power series.

Math 213a Notes 5

The implication (2) ⇒ (4) is Cauchy’s theorem, and (1) ⇒ (4) is Goursat’stheorem. Statement (3) can be formulated as the Jacobian matrix of the C1

map (x, y) 7→ (u(x, y), v(x, y))(∂u∂x

∂v∂x

∂u∂y

∂v∂y

)=

( ∂v∂y −∂u∂y− ∂v∂x

∂u∂x

)is either 0 or a positive multiple of a special orthogonal matrix, i.e., the C1 mapis conformal.

Proof of (2) ⇒ (3). If f ′(z) exists at z, then so do

limh→0h∈R

f(z + h)− f(z)

h, lim

k→0k∈R

f(z + ik)− f(z)

ik

and both equal f ′(z). The expression for the first one is

limh→0h∈R

u(x+ h, y)− u(x, y)

h+ i lim

h→0h∈R

v(x+ h, y)− v(x, y)

h

and the second one is

limk→0k∈R

u(x, y + k)− u(x, y)

ik+ i lim

k→0k∈R

v(x, y + k)− v(x, y)

ik.

Comparing the real and imaginary parts, you get the result.

Proof of (3) ⇒ (4). We apply Green’s theorem to

f(z)dz = (u+ iv)(dx+ idy) = (udx− vdy) + i(vdx+ udy).

Then ∫∂D

f(z)dz =

∫∂D

(udx− vdy) + i

∫∂D

(vdx+ udy)

=

∫D

(−∂v∂x− ∂u

∂y

)dxdy + i

∫D

(∂u∂x− ∂u

∂y

)dxdy = 0.

Proof of (1) ⇒ (5). Suppose f satisfies (1), but∫∂Rf(z)dz 6= 0 for some R

whose closure is in U . We may suppose that∫∂Rf(z)dz = 1. Divide R into 4

equal sized rectangles. Then there must exists at least one of the four, call itR1 such that |

∫∂R1

f(z)dz| ≥ 1/4. Continue inductively and we get a sequence

of subrectangles R1, R2, . . . with |∫∂Rn

f(z)dz| ≥ 4−n, diamRn = 2−n diamR,

and lengthRn = 2−n lengthR.Now the intersections of all the rectangles Rn will be a single point z0 ∈ U

because it is a nested sequence of compact sets. Becasue f is differentiable atz0, given any ε > 0 there is a N such that n ≥ N implies∣∣∣∣f(z)− f(z0)

z − z0− f ′(z0)

∣∣∣∣ < ε

Math 213a Notes 6

for z ∈ ∂Rn. Then

4−n ≤∣∣∣∣∫∂R

f(z)dz

∣∣∣∣ =

∣∣∣∣∫∂Rn

(f(z)− f(z0))dz

∣∣∣∣≤∣∣∣∣∫∂Rn

f ′(z0)(z − z0)dz

∣∣∣∣+ ε

∫∂Rn

|z − z0|dz

= ε

∫∂Rn

|z − z0|dz ≤ ε lengthRn diamRn.

Because this is true for every ε, we get a contradiction.

Math 213a Notes 7

2 September 9, 2016

2.1 Continuing the proof

Let us recall where we ended last time. We are working in the settting off : U → C with f(z) = u(x, y) + iv(x, y) where z = x+ iy. So far, we have (2)⇒ (3) ⇒ (4) ⇒ (5) and (2) ⇒ (1) ⇒ (5).

Proof of (5) ⇒ (6). May as we suppose that U = ∆(z0, r), wher ∆(z0, r) de-notes the open disc centered at z0 with radius r.

For z ∈ ∆(z0, r) define the two paths in ∆(z0, r) from z0 to z, given as inthe following diagram.

R

z

z0

γ′zγ′′z

Figure 1: The paths γ′z and γ′′z

By (5), we see that ∫γ′z

f(z)dz =

∫γ′′z

f(z)dz

since γ′′z − γ′z = ∂R. Let us define

F (z) =

∫γ′z

f(z)dz =

∫γ′′z

f(z)dz.

On horizontal lines, dz = dx and on vertical lines, dz = idy. This implies

∂F

∂xf,

∂F

∂y= if

by the fundamental theorem of calculus. Write F (z) = u(x, y) + iv(x, y). Sincef is continuous, the functions u and v are both C1. Also

∂u

∂x= <f, ∂v

∂x= =f, ∂u

∂y= −=f, ∂v

∂y= <f

and so u and v satisfies the Cauchy-Riemann equations, which is condition (3).But we need that F satisfies (2), i.e., is differentiable. But why is (3) ⇒ (2)?

Recall that the Cauchy-Riemann equations are equivalent to the Jacobian ofthe map (x, y) 7→ (u, v) is multiplication by a complex number when we makethe identification R2 ∼= C. Then the derivative of the map, thought of as a mapfrom R2 ∼= C to itself, is multiplication by a complex number. So F ′ exists atevery point. So F ′ = f .

The proof of (2) ⇒ (7) depends on the following lemma:

Math 213a Notes 8

Lemma 2.1. Suppose f satisfies (2), and let D ⊆ U be a Green’s domain. Forany z ∈ D,

f(z) =1

2πi

∫∂D

f(ζ)

ζ − zdζ.

Proof. Let us choose η > 0 that less less than the distance of z and ∂D. (Byconvention, the distance between z and ∅ is +∞.) Define Dη = D−clos ∆(z, η).This is a Green’s domain in U − {z}, and its boundary is

∂Dη = ∂D − {|ζ − z| = η}.

Since (2) ⇒ (4), and since ζ 7→ f(ζ)/(ζ − z) satisfies (2) on U − {z}, we have∫∂D

f(ζ)

ζ − zdζ =

∫|ζ−z|=η

f(ζ)

ζ − zdζ.

By (2), given ε > 0, if η is sufficiently small and |ζ − z| = η then

|f(ζ)− f(z)− (ζ − z)f ′(z)| ≤ εη.

So

ε

∫|ζ−z|=η

|dζ| ≥∣∣∣∣∫|ζ−z|=y

f(ζ)

ζ − zdζ − f(z)

∫|ζ−z|=η

dζ

ζ − z− f ′(z)

∫|ζ−z|=η

dζ

∣∣∣∣=

∣∣∣∣∫|ζ−z|=η

f(ζ)dζ

ζ − z− 2πif(z)− 0

∣∣∣∣.As ε→ 0, we get the lemma.

Proof of (2) ⇒ (7). Choose a z0 ∈ U . Let R between distance between z0 and∂D. Pick a 0 < r < R.

Suppose that |z − z0| < r. Then

1

ζ − z=

1

(ζ − z0)− (z − z0)=

1

ζ − z0

1

1− z−z0ζ−z0

.

Let D = ∆(z0, r). For z ∈ ∆(z0, r) and ζ ∈ ∂D, we have the geometric series

1

ζ − z=

∞∑k=0

(z − z0)k

(ζ − z0)k+1

which converges absolutely and uniformly1 in ζ and z proved |z − z0| ≤ r − δfor some small δ > 0.

1This, by convention, means that it absolutely converges and the absolute convergence isuniform. It is literally not what it says, but people usually mean this.

Math 213a Notes 9

Then the lemma tells us that

f(z) =1

2πi

∫∂D

f(ζ)

∞∑k=0

(z − z0)k

(ζ − z0)k+1

=

∞∑k=0

(z − z0)k1

2πi

∫∂D

f(ζ)dζ

(ζ − z)k+1=

∞∑k=0

ak(z − z0)k,

where

ak =1

2πi

∫∂D

f(ζ)dζ

(ζ − z0)k+1.

This convergence is uniform and absolute in z.

We note that (7) is a local statement, so U can be replaced, for example,by a disc centered at z0, of radius at most the distance between z0 and ∂U .Also, what we have actually proved is stronger than (7). We have proved thatif (2) is true, then for any z0 ∈ U , the function f(z) can be expressed asf(z) =

∑∞k=0 ak(z− z0)k on ∆(z0, r) given that r is less than the distance of z0

and ∂U , and that the convergence is absolute and uniform.Consider a formal power sereis

∑∞k=0 ak(z − z0)k for z0, ak ∈ C so that z0 is

the center of the power series.

Lemma 2.2. There exists a uniquely determined R, 0 ≤ R ≤ ∞, such that

(1) for any r with 0 < r < Rand |z − z0| ≤ r, the series converges uniformlyand absolutely.

(2) if |z − z0| > R then the series diverge.

This R is called the radius of convergence. Suppose for |z − z0| < R,

s(z) =

∞∑k=0

ak(z − z0)k.

Then s(z) is differentiable at z and

s′(z) =

∞∑k=0

kak(z − z0)k

absolutely and uniformly. This new function s′ has the same radius of conver-gence with s(z).

If we have this lemma, we are done.

Proof of (7) ⇒ (2). Assume the lemma. Then on ∆(z0, r) with 0 < r < R, s(z)is differentiable. Its derivative s′(z) is also differentiable and hence continuous.

Proof of (6) ⇒ (7). Locally, we can say f = F ′ and hence F satisfies (2). ThenF can be expressed locally as a power series with positive radius of convergence,and by the lemma, same is true for f = F ′.

Math 213a Notes 10

3 September 11, 2016

3.1 Finishing the proof

So we need to prove this lemma to finish the proof of the theorem. Let∑∞k=0 ak(z−

z0)k be a formal power series, with z0, ak ∈ C.

Lemma 3.1. There exists a uniquely determined 0 ≤ R ≤ ∞, such that

(1) for z ∈ C, |z−z0| < R, the series converges absolutely and the convergenceis uniform and absolute on any compact subset of ∆(z0, R)

(2) the series diverges for z ∈ C, |z − z0| > R.

This R is called the radius of convergence of the power series. For z ∈ ∆(z0, R),let s(z) denote the limit. Then s(z) is differentiable in the complex sense ands′(z) =

∑∞k=1 kak(z − z0)k−1 with this series having the same radius of conver-

gence.

Proof. We define

R = sup{|z − z0| : the series converges at z}.

Now suppose 0 < r < R. Then there exists z1 with r1 = |z1 − z0| such thatthe series converges absolutely at z1 and r < r1 ≤ R. Also convergence impliesboundedness of the terms: there exists an M such that |akrk1 | ≤M , i.e., |ak| ≤Mr−k1 . For |z − z0| ≤ r, we have

|ak(z − z0)k| ≤ |ak|rk ≤Mrk

rk1.

Therefore the series converges absolutely and uniformly.Now we want to say the same thing for convergence of the series

∑kak(z−

z0)k−1. Choose r2 with r < r2 < r1. Because the series converges also for |z −z0| = 0, we can easily check that the series converges absolutely and uniformlyon ∆(z0, r).

Let us define for z ∈ ∆(z0, r), and let

s(z) =

∞∑k=0

ak(z − z0)k.

Consider h ∈ C where 0 < |h| < r − |z − z0|. The difference quotient can be

Math 213a Notes 11

computed as∣∣∣∣s(z + h)− s(z)h

−∞∑k=1

kak(z − z0)k−1

∣∣∣∣ =

∣∣∣∣ ∞∑k=2

ak

k∑l=2

(k

l

)(z − z0)k−lhl−1

∣∣∣∣≤∞∑k=2

|ak|k∑l=2

(k

l

)|z − z0|k−l|h|l ≤M

∞∑k=2

k∑l=2

(k

l

)rk−l

rk1|h|l

= M

∞∑k=2

((r + |h|)k − rk

|h|rk1− k r

k−1

rk1

)≤M

∞∑k=2

k(r + |h|)k−1 − krk−1

rk1

= Mr−11

(1

(1− (r + |h|)/r1)2− 1

(1− r/r1)2

).

Letting h → 0, this sum goes to zero. This shows that∑ak(z − z0)k is differ-

entiable and its derivative is∑kak(z − z0)k−1.

As I proved last time, this lemma implies the theorem.

Definition 3.2. A function f : U → C is said to be holomorphic or complexanalytic if it satisfies any one (equivalently, all) of the statements (1)–(7) inthe theorem.

3.2 A bunch of corollaries

There are some consequences.

Corollary 3.3 (Cauchy integral formula). Suppose f is holomorphic on U withz ∈ U . If D ⊆ U is a Green’s domain with z ∈ D, then

f(z) =1

2πi

∫∂D

f(ζ)

ζ − zdζ.

Corollary 3.4. Let f : U → C be holomorphic and z0 ∈ U . Let R be thedistance of z0 and ∂U (R = +∞ if U = C). Then on ∆(z0, R),

f(z) =

∞∑k=0

ak(z − z0)k

with appropriately chose ak, convergence is absolute, and uniform and absoluteon a compact subset of ∆(z0, R). This power series has radius of convergenceat least R. Moreover, formally differentiated series has the same radius of con-vergence as the original series, and converges to f ′(z).

The argument for (2) ⇒ (7) contains the formula for ak:

ak =1

2πi

∫|ζ−z0|=r

f(ζ)

(ζ − z0)k+1dζ

where 0 < r < R.

Math 213a Notes 12

Corollary 3.5. If f : U → C is holomorphic, then f has derivatives of allorders (in the complex sense). Moreover, if z ∈ U and D ⊆ U is a Green’sdomain containing z, then

f (k)(z) =k!

2πi

∫∂D

f(ζ)

(ζ − z)k+1dζ.

Proof. You can do this by differentiating the Cauchy integral formaula (butneeds to be justified). Alternatively, use the previous corollary.

Corollary 3.6 (Morera’s theorem). Suppose f : U → C is continuous andfor any rectifiable closed curve γ in U the integral

∫γf(z)dz = 0. Then f is

holomorphic.

The inverse is not always true. For instance, the integral of 1/z over theunit circle is 2πi. If the curve is cohomologous to zero, then the integral shouldbe zero.

We define the differential operators

∂

∂z=

1

2

( ∂∂x− i ∂

∂y

),

∂

∂z=

1

2

( ∂∂x

+ i∂

∂y

).

Note that with this notation, ∂/∂z sends z 7→ 1, z 7→ 0, and likewise ∂z/∂z = 0and ∂z/∂z = 1.

Corollary 3.7. Suppose f : U → C is C1. Then f is holomorphic if and onlyif ∂f/∂z = 0.

Corollary 3.8. Suppose f : U → C is holomorphic and let f(z) = u(x, y) +iv(x, y). Then u and v are harmonic.

Proof. We have

0 =∂2f

∂z∂z=

1

4

( ∂2

∂x2+

∂2

∂y2

)f.

Corollary 3.9. Suppose f : U → C is holomorphic and z0 ∈ U , U is holomor-phic and f 6≡ 0. Then there exists a uniquely determined integer n ≥ 0 and aholomorphic function g : U → C such that f(z) = (z − z0)ng(z) with g(z0) 6= 0.

Proof. First suppose f 6≡ 0 on any neighborhood of z0. Then for z near z0, thefunction f can be presented as

f(z) =

∞∑k=0

ak(z − z0)k.

The ak cannot all be zero.Let n be the least integer such that an 6= 0. Then

f(z) = (z − z0)n∞∑k=n

ak(z − z0)k−n

Math 213a Notes 13

and we can let the sum on the right hand size be g(z).This proves that {z ∈ U : f 6≡ 0 in any neighborhood } is open in U . Like-

wise the complement is open in U . Because U is connected and f 6≡ 0 on U , itmust be that f 6≡ 0 on some neighborhood of z0.

Math 213a Notes 14


4.1 More and more corollaries

We proved this corollary last time.

Corollary 4.1. Suppose f : U → C is holomorphic and z ∈ U , with f 6≡ 0.Then there exists an integer n ≥ 0 and a holomorphic g : U → C such thatg(z0) 6= 0 and f(z) = (z − z0)ng(z).

If n ≥ 1 says that f has a zero at z0 (or “vanishes at z0”), and n is theorder of zero.

Corollary 4.2. Under the same hypotheses, S = {z ∈ U : f(z) = 0} is discretein U .

Corollary 4.3. Suppose fn : U → C is holomorphic with n = 1, 2, . . .. Iffn → f uniformly on a compact sets (or equivalently, locally uniformly) then fis holomorphic.

Proof. Let R be a closed rectangle with R ⊆ U . Then int(R) is a Green’sdomain in U , so

∫∂Rfndz = 0. Then

∫∂Rfdz = 0. This implies that f is

holomorphic.

Corollary 4.4 (“Open mapping theorem”). Suppose f : U → C is holomorphic,and U is connected, with f non-constant. Then f(U) is open.

Proof. Suppose z0 ∈ U , and let f(z)−f(z0) = (z− z0)ng(z) with g(z0) 6= 0 andn ≥ 1. Since g(z0) 6= 0, the function g(z)1/n has a locally defined holomorphicfunction, because log g(z) is locally well-defined. Define h(z) = (z − z0)g(z)1/n.Then h(z0) = 0 and h′(z0) = g(z0)1/n 6= 0 and satisfies

f(z) = f(z0) + h(z)k.

Since h(z0) = 0 and h′(z0) 6= 0, the map h, viewed as a map from R2 → R2, hasa nonzero differential at z0 = x0 + iy0, namely multiplication by h′(z), hencethe differential is invertible. By the inverse function theorem, h(z) has a localinverse from an open neighborhood of h(z0) = 0 onto an open neighborhood ofz0. That is, the inverse of z0 under h covers a full neighborhood of 0. Then mapw 7→ wn is surjective on the neighborhood of zero. Because f(z) = f(z0)+h(z)n,the image of f includes a neighborhood of f(z0).

Corollary 4.5 (“Inverse function theorem”). Suppose f : U → C is holo-morphic and 1-to-1. Then by the open mapping theorem f(U) is open, and bythe hypothesis, f−1 : f(U) → U is well-defined set-theoretically. Then f−1 isholomorphic.

Proof. As in the proof of the previous corollary, for z0 ∈ U I can write f(z) =f(z0) + h(z)n where h(z) = (z − z0)g(z)1/n. Also note that we may as wellsuppose that U is connected. Then f is not constant.

Math 213a Notes 15

So f(z) = f(z0) + h(z)n. Arguing as in the previous corollary, z 7→ f(z) islocally n-to-1 near z0. So we must have n = 1. Then f(z) = f(z0)+(z−z0)g(z)with g(z0) 6= 0. Viewed as a map from R2 to R2, this f has a invertibledifferential at z0. So by the inverse function theorem of multivariable calculus,f has a C1 inverse locally near z0. Since f : U → C is globally 1-to-1, this provesthat f−1 : f(U) → U is at least C1. At any f(z0) ∈ f(U), the differential off−1 (thought of as R2 → R2) is the inverse of the differential of f at z0, i.e.,the inverse of multiplication by f ′(z0) (thought of as R2 → R2). But thatis multiplication by 1/f ′(z0) and so it satisfies to Cauchy-Riemann equations.That is f−1 is holomorphic.

Corollary 4.6 (“Maximum principle”). Suppose f : U → C is holomorphicand U connected, with f non-constant. Then the function z 7→ |f(z)| cannotassume a maximum at any point of U .

Proof. This follows from the open mapping theorem.

Let ∆ = ∆(0, 1).

Corollary 4.7 (Schwarz lemma). Suppose f : ∆→ C is holomorphic, f(0) = 0,|f(z)| ≤ 1 on ∆. Then for z ∈ ∆, |f(z)| ≤ |z|, and |f ′(0)| ≤ 1. Both of theseinequalities are strict (except for |f(0)| = 0) unless f(z) = cz for some c ∈ Cwith |c| = 1.

Proof. The function f(z)/z is holomorphic on ∆ since f(z) has a zero at 0.Define gr(z) = f(rz)/rz for 0 < r < 1. This is holomorphic on an openneighborhood of closure of ∆. Also |gr(z)| ≤ 1/r on ∂∆. This immplies that|gr(z)| ≤ 1/r for all z ∈ ∆ by the maximum principle. By continuity, |f(z)/z| ≤1 for all z ∈ ∆. Either f(z)/z is constant, in which case the statement is OK, orf(z)/z is not. Then by the maximal principle, |f(z)/z| < 1 and f ′(0) = limf(z)/z

and hence |f ′(0)| < 1 again by maximal principle.

This is a very important statement, especially in the context of complexmanifolds.

4.2 Laurent series

Suppose 0 ≤ r1 < r2 ≤ ∞ and z0 ∈ C. Let

A = {z ∈ C : r1 < |z − z0| < r2}

Theorem 4.8 (Laurent). Suppose f : A→ C is holomorphic. Then for z ∈ A,

f(z) =

∞∑k=−∞

ak(z − z0)k

converges locally uniformly and absolutely on A. More precisely,∑∞k=0 ak(z −

z0)k converges uniformly and absolutely on compact subsets of ∆(z0, r2), and

Math 213a Notes 16

∑−1k=−∞ ak(z − z0)k converges uniformly and absolutely on compact subsets of

C− clos ∆(z0, r1). Finally,

ak =1

2πi

∫|ζ−z0|=r

f(ζ)

(ζ − z0)k+1dζ

for any r with r1 < r < r2.

Proof. Suppose ρ1, ρ2 are chosen so that r1 < ρ1 < ρ2 < r2. Let A = {z ∈C : ρ1 < |z − z0| < ρ2}. This is a Green’s domain in A. The map ζ 7→f(ζ)/(ζ − z0)k+1 is holomorphic on A, so

0 =

∫∂A

f(ζ)

(ζ − z)k+1dζ =

∫|ζ−z0|=ρ2

f(ζ)

(ζ − z2)k+1dζ −

∫|ζ−z0|=ρ1

f(ζ)

(ζ − z0)k+1dζ.

This shows that the formula that was stated is independent of r.For z ∈ A, by Cauchy

f(z) =1

2πi

∫∂A

f(ζ)

ζ − zdζ =

1

2πi

∫|ζ−z0|=ρ2

f(ζ)

ζ − z0dζ − 1

2πi

∫|ζ−z0|=ρ1

f(ζ)

ζ − z0dζ.

Now

1

ζ − z=

1

(ζ − z0)− (z − z0)=

∑∞k=0

(z−z0)k

(ζ−z0)k+1 for |ζ − z0| = ρ2 and |z − z0| < ρ2,

−∑−1k=−∞

(z−z0)k

(ζ−z0)k+1 for |ζ − z0| = ρ1 and |z − z0| > ρ1.

So

f(z) =1

2πi

∞∑k=0

(∫|ζ−z0|=ρ2

f(ζ)

(ζ − z0)k+1dζ

)(z − z0)k +

1

2πi

−1∑k=−∞

(∫|ζ−z0|=ρ1

f(ζ)

(ζ − z0)k+1dζ

)(z − z0)k.

This converges as stated in the theorem.

As a special case, r1 = 0 and r2 = r. I will write ∆∗(z0, r) = {z ∈ ∆(z0, r) :z 6= z0} to denote the punctured disc. Suppose f(z) is holomorphic on ∆∗(z0, r).Then there exists ak for −∞ < k <∞ such that

f(z) =

∞∑k=−∞

ak(z − z0)k

converges uniformly and absolutely on compact subsets of ∆∗(z0, r).We call

a−1 =1

2πi

∫|ζ−z0|=ρ

f(ζ)dζ

is the residue of f at z0, and is denoted by Res f |z0 . Also, the series∑∞k=−∞ ak(z−

z0)k is called the principal part of f at z0.

Math 213a Notes 17

The function f has a removable singularity at z0 if ak = 0 for all k ≤ −1.We say that f has a pole of order n if a=0 for k < −n and a−n 6= 0. We saythat f has a essential singularity at z0 if ak 6= 0 for infinitely many k < −1.

Suppose U ⊆ C is open, and S ⊂ U is a discrete subset with f : U − S → Cholomorphic. Then for any z0 ∈ S, the function f : ∆∗(z0, r) → C is alsoholomorphic for some small r = r(z0) > 0. So one has the notion of removablesingularity, pole, essential singularity, residue, in the context. One says that “fis holomorphic on U except for isolated singularities” at the points of S.

Math 213a Notes 18


Suppose U ⊆ C is open and S ⊆ U be a discrete subset. One calls a holomorphicfunction f : U − S → C holomorphic on U with isolated singularities atS.

5.1 Isolated singularities

If z0 ∈ S then there exists a η > 0 such that ∆(z0, η) ∩ S = {z0}. So f :∆∗(z0, η) → C is holomorphic, and hence we have notions of residue, pole,isolated singularities. We say that f is meromophic on U if all the singularitiesare poles.

Theorem 5.1. For z0 ∈ S,

(1) f has a removable singularity at z0 if and only if |f(z)| is bounded nearz0.

(2) f has a pole at z0 if and only if limz→z0 |f(z)| = +∞.

(3) f has an essential singularity at z0 if and only if for any δ with 0 < δ � 1,f(∆∗(z0, δ)) is dense in C.

The (3) is called the Casorati-Weierstrass theorem, and in fact, f(∆∗(z0, δ))is all of C possibly except for one point. This stronger result is called BigPicard’s theorem, and is much much harder to prove.

Proof. (1) The forward direction is obvious. Now suppose |f(z)| is boundednear z0. As before, we know that f : ∆∗(z0, η)→ C is holomorphic. For k ≤ −1and 0 < r < η, we have

|ak| =∣∣∣∣ 1

2πi

∫|ζ−z0|=r

f(ζ)

(ζ − z0)k+1dζ

∣∣∣∣ ≤ max|ζ−z0|=r

|f(ζ)|r−k.

Because the right hand side goes to zero as r → 0 if k ≤ −1, we see that ak = 0for all k ≤ −1.

(2) For the forward direction, we can write f(z) = (z − z0)−ng(z) for somen ≥ 1 and g holomorphic with g(z0) 6= 0. So |f(z)| → ∞ as z → z0. For theother direction, suppose |f(z)| → +∞ as z → z0. Then after making η smallerif necessary, f(z) 6= 0 on ∆∗(z0, η). Then z 7→ 1/f(z) has an isolated singularityat z0, and it is locally bounded near z0. So 1/f(z) has a removable singularityat z0. Then 1/f(z) = (z − z0)ng(z) for some holomorphic g with g(z0) 6= 0.Then f has a pole of order n at z0.

(3) The backwards direction follows from (1) and (2). Now suppose that fhas an essential singularity. Suppose 0 < δ < η and f(∆∗(z0, δ)) is not dense inC. Then there exists ω0 ∈ C and ε > 0 such that f(∆∗(z0, δ)) ∩∆(w0, ε) = ∅.Then |f(z) − ω0| ≥ ε for z ∈ ∆∗(z0, δ) and so 1/(f(z) − ω0) is holomorphic on∆∗(z0, δ), bounded by ε−1. Then g(z) = 1/(f(z) − ω0) is holomorphic even atz0, because it is a removable singularity. Then f at z0 is either a removablesingularity or a pole.

Math 213a Notes 19

5.2 Residue theorem and its applications

Let U ⊆ C be and open set, and S ⊆ U discrete. A function f is holomorphicon U excpt for isolated singularities at points of S. Let D ⊆ U be a Green’sdomain and ∂D ∩ S 6= ∅.

Theorem 5.2 (Residue theorem). In this situation,∫∂D

f(z)dz = 2πi∑

z∈D∩SRes f(ζ)|z.

(Note that D ∩ S is finite.)

Although it is an important statement, the proof is a triviality.

Proof. Enumerate S ∩ D = {z1, . . . , zN}. For each j with 1 ≤ j ≤ N , we canchoose rj > 0 such that clos ∆(zj , rj) ⊆ D, and clos(∆(zj , rj))∩clos(∆(zi, ri)) =

∅ for i 6= j. Consider D = D −⋃

clos(∆(zj , rj)). This is a Green’s domain inU − S and then

0 =

∫∂D

f(z)dz =

∫∂D

f(z)dz −N∑j=1

∫|z−zj |=rj

f(z)dz =

∫∂D

f(z)dz − 2πi

N∑j=1

Res f |zj .

The residue theorem is useful in computing certain definite integrals. I’llgive you three examples.

Example 5.3. Let R(x, y) be a rational function without poles on the unitcircle in R2. Then∫ 2π

0

R(cos θ, sin θ)dθ =1

i

∫|z|=1

R(z + z−1

2,z − z−1

2i

)dzz

= 2π∑|ζ|<1

Res(R(z + z−1

2,z − z−1

2

)z−1)∣∣∣ζ.

Example 5.4. Let R(x) be a rational function with no poles on R, and as-sume that |R(x)| = O(x−2) as x → ±∞. For a ∈ R, we want to com-pute

∫∞−∞ eiaxR(x)dx. We may assume a ≥ 0 because otherwise we can take

the complex conjugate. Then eiaz is bounded on =z > 0. For r � 0, letDr = {z ∈ C : |z| < r,=z > 0}. This is a Green’s domain on C. If r is largeenough, R(z) has no poles outside ∆(0, r). Then by the Residue Theorem,

2πi∑=z>0

Res(eiaζR(ζ))|z =

∫∂Dn

f(z)eiazdz

=

∫ r

−reiaxR(x)dx+

∫|z|=r,=z≥0

eiazR(z)dz.

Math 213a Notes 20

Because∣∣∣∣∫|z|=r,=z≥0

eiazR(z)dz

∣∣∣∣ ≤ πr sup|z|=r|zR(z)| = O(rR(r)) ≤ const · r−1.

So the second integral goes to zero as r →∞, and therefore∫ ∞−∞

eiaxR(x)dx = −2πi∑=z>0

Res(eiaζR(ζ))|z.

Example 5.5. Let R(x) be a rational function without poles on R≥0. Lets /∈ N, s > 0 be a real number and suppose that |R(x)| = O(x−s) as x→ +∞.Because s is not an integer, the integral∫ ∞

0

xs−1R(x)dx

converges absolutely. We want to evaluate this domain. Choose 0 < δ � 1,1� r <∞, and 0 < η � 1 so that all poles of R lie in D(r, δ, η) = {z : δ < |z| <r, 0 < arg z < 2π − η}. This is a Green’s domain in C− {re−ηi/2 : 0 ≤ r <∞}and contains all poles of R(z). Because log z has a wel-defined branch aroundD, the function z1−s also has a well-defined holomorphic determination withagrees with the normal definition on R>0. Then

2πi∑z

Res(ζs−1R(ζ))|z =

∫∂D

zs−1R(z)dz

=

∫ r

δ

xs−1R(x)dx− ei(2π−η)s

∫ r

δ

xs−1R(ei(2π−η)x)dx

+

∫ 2π−η

0

reiθ(s−1)R(reiθ)rdeiθ −∫ 2π−η

0

δeiθ(s−1)R(δeiθ)δdeiθ.

First letting η → 0, we get

2πi∑z

Res(ζs−1R(ζ))|z = (1− e2πis)

∫ r

δ

xs−1R(x)dx

+

∫ 2π

0

rs−1eiθ(s−1)R(reiθ)rdeiθ −∫ 2π

0

δs−1eiθ(s−1)R(δeiθ)δdeiθ.

Letting r →∞ and δ → 0, we get

2πi∑z

Res(ζs−1R(ζ))|z = (1− e2πis)

∫ ∞0

xs−1R(x)dx.

Although this method does not work for s complex, if we get an holomorphicanswer, then the answer for s complex should also be the same.

Math 213a Notes 21


Recall that for an open set U ⊆ C, a function f being meromorphic on U meanshaving isolated singularities, all of which are poles (or removable).

Definition 6.1. For a z0 ∈ U , the order of f at z0 is

(i) n ≥ 1 if f has a zero of order n,

(ii) 0 if f is holomorphic on some neighborhood of z0 with f(z0) 6= 0,

(iii) −n with n ≥ 1 if f has a pole of order n at z0.

Locally this means that f(z) = (z − z0)ng(z) with g(z) holomorphic on aneighborhood of z0 and g(z0) 6= 0. Then the logarithmic derivative is

f ′

f(z) =

n

z − z0+g′

g(z).

Because g′/g is holomorphic near z0, we see that Res(f ′/f)|z0 is the order of fat z0.

6.1 The argument principle and total order

Theorem 6.2 (Argument principle). Let D ⊂ U be a Green’s domain such thatf has no zeros or poles on ∂D. Then

1

2πi

∫∂D

f ′

fdz

is the total order of f in D, i.e., the total number of zeros in D minus the totalnumber of poles, both counted with appropriate multiplicity.

Proof. This follows from the Residue theorem.

Why is this called the argument principle? We can write f(z) = r(z)eiθ(z)

for some C∞ functions r and θ, except at zeros and poles. On ∂D this iswell-defined with r > 0, and (f ′/f)dz = dr/r + idθ so∫

∂D

f ′

fdz = 0 + i

∫∂D

dθ = i(total variation of θ along ∂D).

Corollary 6.3 (Rouche’s theorem). Let U ⊆ C be open and connected, withf, g meromorphic functions on U . Let D ⊂ U be a Green’s domain such that fand g has no zeros nor poles on ∂D. Suppose |f(z)− g(z)| < |g(z)| for z ∈ ∂D.Then the total order of f and g in D coincide.

Proof. On ∂D, |(f/g)(z) − 1| < 1. This means that the image of ∂D underf/g in C lies in ∆(1, 1), which is in C − R≥0. This means that log(f/g) has awell-defined branch on ∂D. Then

1

2πi

∫∂D

(f ′f− g′

g

)dz =

∫∂D

d logf

g= 0,

which is the difference between the total order of f and total order of g.

Math 213a Notes 22

Corollary 6.4 (Fundamental theorem of calculus). Let P (z) = anzn + · · · +

a1z + a0 with an 6= 0. Then P (z) has exactly n zeros in C, counted withmultiplicity.

Proof. Suppose R > 0 and very big. Then on ∂∆(0, R),

|P (z)− anzn| ≤n−1∑k=0

|ak|Rk < |ak|Rn

if R is very large. Apply Rouche’s theorem on ∆(0, R) and find that P (z) hasas many zeros in ∆(0, R) as zn.

Corollary 6.5 (Open mapping theorem). Suppose U ⊆ C is open, connected,and f : U → C is holomorphic. Let z0 ∈ U and w0 = f(z0). For everysufficiently small ε > 0, there exists δ > 0 such that for w ∈ ∆(w0, δ), f(z)−whas exactly many zeros (counted with multiplicity) in ∆(z0, ε) as the order ofzero of f(z)− w0 at z0.

Proof. We know that there exists η > 0 such that ∆(z0, η) is contained in U ,and f(z)−w0 6= 0 on ∆∗(z0, η). For 0 < ε < η, the disc ∆(z0, ε) ⊂ U is a Green’sdomain, and f(z) − w0 6= 0 on ∂∆(z0, ε). Let δ = min|z−z0|=ε|f(z) − w0| > 0.Now suppose w ∈ ∆(w0, δ). Then for z ∈ ∂∆(z0, ε), w ∈ ∆(w0, δ), we have|w − w0| < δ ≤ |f(z)− w0|. Then

|(f(z)− w)− (f(z)− w0)| < |f(z)− w0|.

Now apply Rouche, with f(z)− w playing the role of f and f(z)− w0 the roleof g. This implies that the total number of f(z)−w0 on ∆(z0, ε) is equal to theorder of zero of f(z)− w0 at z0.

6.2 Projective 1-space

I should say something about what is coming up. We are going to define theprojective 1-space and define holomorphic/meromorphic functions on this space.Then we are going to talk about automorphisms of the unit disc. Although thismay seem unmotivated, this is important because if you have a Riemanniansurface, its universal cover is either the projective 1-space, C, or the unit disc.In particular, the proof of the Picard theorems will depend on this.

The group C∗ = C− {0} acts on C2 − {0} by scalar multiplication.

Definition 6.6. We define P1 (or simply P1 as C∗ \ (C∗−{0}) which is the setof lines through the origin in C2.

The group SL(2,C) acts on C2−{0}, and this commutes with the C∗ action,and hence induces an action on P. Topologize P by giving it the quotienttopology induced by C2 − {0} → P. Define ϕ0 : C → P, ϕ∞ : C → P byϕ0(z) = line through

(z1

)and ϕ∞(z) = line through

(1z

). Then

(i) ϕ0, ϕ∞ are continuous and injective,

Math 213a Notes 23

(ii) imϕ0 ∪ imϕ∞ = P,

(iii) ϕ−1∞ ϕ0(C) = C∗, ϕ−1

0 ϕ∞(C) = C∗,(iv) ϕ−1

0 ϕ∞(C∗) = C∗ = ϕ−1∞ ϕ0(C∗),

(v) for any z ∈ C∗, ϕ−10 ϕ∞(z) = 1/z, ϕ−1

∞ ϕ0(z) = 1/z,

(vi) via ϕ0, we can identify C ∪ {∞} → P1.

The proof is an exercise.Suppose U ⊆ P is open. Then define the notion of a holomorphic/meromorphic/function

with isolated singularities as those f for which f ◦ ϕ0 and f ◦ ϕ∞ have thoseproperty on C. Equivalently, this is when

(i) f has these properties on U ∩ C and

(ii) z 7→ f(1/z) has these properties on some neighborhood of 0, when∞ ∈ U .

Recall the notion of residue of a function at isolated singularity z0:

1

2πi

∫|z−z0|=δ

f(z)dz.

So I can, and should, think the residue as associated to the holomorphic 1-formf(z) with isolated singularities.

Then a special case of the residue theorem states:

Theorem 6.7. Suppose fdz is a holomorphic 1-form with isolated singularitieson P. Then ∑

z0∈PRes(fdz)|z0 = 0.

Proof. Choose r > 0 so big that fdz has no singularity outside clos ∆(0, r)except possibly at ∞. Then

1

2πi

∫∂∆(0,r)

f(z)dz =∑

z0∈∆(0,r)

Res(fdz)|z0 .

So we need to show that this integral also equals −Res fdz|∞. This is becausea small circle around infinity is a large circle, with orientation reserved by z 7→1/z.

Corollary 6.8. Suppose fdz is a holomorphic 1-form with isolated singularitieson P. Then the total number of zeros minus the total number of poles of f ′/fdzis equal to zero.

Suppose f is a meromorphic function on P. Then we can choose a polynomialp(z) which as exactly the same zeros as f on C. Likewise, we can choose apolynomial q(z) that has exactly the same poles as f on C. Then the function

h = fq

p

Math 213a Notes 24

has no zeros or poles on C, but it may have zero or pole at ∞. So either h or1/h has no poles on P. So h : P→ C is holomorphic, possibly zero at ∞. Thush is constant by the maximal principle. Thus we have proved:

Theorem 6.9. Any meromorphic function defined on P is rational, i.e., is aquotient of polynomials.

Essentially the same argument proves

Theorem 6.10 (Liouville’s theorem). Any bounded entire function (i.e., holo-morphic function on C) is constant.

Proof. Think of it as a function on P with possibly a singularity at ∞. Thatsingularity is removable, and so we can use maximal principle.

Math 213a Notes 25


Theorem 7.1. If f(z)dz is a meromorphic 1-form on P, then∑ζ∈P Res(f(z)dz)|ζ =

0.

Now suppose f is a meromorphic function on P. Then df/f is a meromorphic1-form and thus has sum of residues zero. What about Res(df/f)|∞? Letw = 1/z. Then w = 0 corresponds to z =∞, and so

Resdf

f

∣∣∣∞

= Res∂f/∂w

fdw∣∣∣w=0

= order of f at ∞.

So we conclude that∑ζ∈P ord f |ζ = 0. Indeed, df/f is invariant of local coor-

dinates and its residue is the order of f .

7.1 Automorphisms of PWe have define the notion of meromorphic/holomorphic functions on open U ⊆P. What about the notion of a holomorphic map F : U → P? First of all, werequire continuity. Suppose z0 ∈ U . There are two cases:

(i) limz→z0 F (z) ∈ C - then F holomorphic near z0 as map into P is equivalentto F holomorphic near z0 as a map into C.

(ii) limz→z0 F (z) = ∞ - then F holomorphic near z0 is equivalent to 1/Fbeing holomorphic near z0 as a map to C.

So we have a notion of an automorphism F : P → P, which means that it iscontinuous, bijective, and holomorphic in both directions.

Recall that GL(2,C) acts on C2, and the actions commutes with scalarmultiplication. So GL(2,C) acts as a group of automorphisms on P. The centerof this action is C∗1.

We have identified P ∼= C∪{∞}, with z corresponding to the line(z1

). Then

a matrix g act as

gz =

(a bc d

)(z1

)=

(az + bcz + d

)=az + b

cz + d.

These are call the linear fractional transformations. This makes sense evenwhen cz + d = 0, because we can interpret it as ∞.

We denote

SL(2,C) = {g ∈ GL(2,C) : deg g = 1},PGL(2,C) = GL(2,C)/center = GL(2,C)/C∗ = SL(2,C)/{±1}.

Theorem 7.2. The action of GL(2,C) on P induces an isomorphism SL(2,C)/{±1} ∼=Aut(P). Furthermore, SL(2,C) acts on P1 in a triply transitive manner.

Math 213a Notes 26

Proof. Suppose g ∈ SL(2,C) acts trivially: (az + b)/(cz + d) = z. Then settingz = 0 and z =∞, we see that b = c = 0. So the map SL(2,C)/{±1} → Aut(P)is injective.

Now suppose P → P is an automorphism. Then z 7→ F (z), z 7→ F (1/z),z 7→ 1/F (z), and z 7→ 1/F (1/z) are holomorphic where defined. That is, F canbe interpreted as a meromorphic function. Since F : P1 → P1 is bijective, itmust have a single zero and a single pole. That is, F (z) = az + b/cz + d. Thisshows that SL(2,C)/{±1} → Aut(P) is surjective.

Why is this action triply transitive? It suffices to show that if z0, z1, z∞ ∈ Pare distinct, then there exists a g ∈ SL(2,C) such that gz0 = 0, gz1 = 1, gz∞ =∞. Define the linear fractional transformation g as

z 7→ z − z0

z − z∞z1 − z∞z1 − z0

.

By construction, this sends z0 → 0, z∞ →∞, and z1 7→ 1.

7.2 Conjugacy classes of matrices

To compute the conjugacy classes of SL(2,C), we can use the Jordan canonicalform. Any g ∈ SL(2,C) is conjugate to one of the following:

±1,

(λ 0

0, λ−1

), ±

(1 10 1

).

The first one is trivial, and the second one which is z 7→ λ2z has two fixed points0 and ∞. The third one has only one fixed point ∞, and it is the unipotentone.

Define

SU(2) =

{(α β−β α

): α, β ∈ C, |α|2 + |β|2 = 1

},

SU(1, 1) =

{(α ββ α

): α, β ∈ C, |α|2 − |β|2 = 1

}.

These are all contained in SL(2,C). Also SU(2) is the group of matrices whichpreserves the standard inner product on C2, of determinant 1, and SU(1, 1) isthe group of matrices which preserves the standard indefinite hermitian formon C2, of determinant 1.

Theorem 7.3. The action of SU(2) on P is transitive and induces SU(2)/{±1} ↪→Aut(P). Any finite subgroup Γ ⊂ SL(2,C) is SL(2,C) conjugate to a subgroupof SU(2).

Proof. Suppose Γ ⊂ SL(2,C) is a finite subgroup. Define an inner product (, )Γ

in terms of the standard one on C2 as

(v1, v2)Γ =1

#Γ

∑γ∈Γ

(γv1, γv2).

Math 213a Notes 27

Then (γv1, γv2)Γ = (v1, v2)Γ, so this new (, )Γ is Γ-invariant. Any two innerproducts on C2 are GL(2,C) conjugate since there is an orthonormal basis, uptoGL(2,C) conjugacy, the two are the same. So under someGL(2,C) conjugate,and hence SL(2,C) conjugate, Γ preserves (, ) i.e. is a subgroup of SU(2).

Constructing this new invariant from an old one is called Weyl’s unitarytrick.

Proposition 7.4. Any g ∈ SU(2) is conjugate to(eiθ 00 e−θ

)for some θ ∈ R/2πZ.

Proof. g will have an eigenvector v1 with eigenvalue absolute value 1. Thentake a unit vector that is perpendicular to v1. Then v2 will be an eigenvector,corresponding to the reciprocal of the previous eigenvalue.

7.3 Automorphisms of ∆

From now one, we will write

∆ = ∆(0, 1), H = {z ∈ C : =z > 0}, c =1√2

(1 −i−i 1

)∈ SU(2).

This c is called the Cayley transform.

Lemma 7.5. The map c is an isomorphism from H to ∆.

Proof. For x ∈ R ∪ {∞},

c · x =x− i−ix+ 1

= ix− ix+ i

∈ ∂∆.

So c maps R ∪ {\} to ∂∆ isomorphically. So c must map H onto one of theconnected components of P− ∂∆. But because c · i = 0, c maps H to ∆.

Theorem 7.6. (a) The action of SU(1, 1) on P preserves ∆, is transitive on∆, and induces an isomorphism SU(1, 1)/{±1} → Aut(∆).

There are more statements, but let me just state and prove this one.

Proof. SU(1, 1) contains the two subgroups(eiθ 00 e−iθ

),

(cosh t sinh tsinh t cosh t

).

The former acts by rotation, and the latter maps 0 7→ tanh t. So SU(1, 1) actstransitively on ∆.

Math 213a Notes 28

Suppose F : ∆→ ∆ is an automorphism. By Schwartz’s lemma, we see that|F (z)| ≤ 1 for all z ∈ ∆ implies |F ′(0)| ≤ 1. By symmetry, between F and F−1,we see that |1/F ′(0)| ≤ 1 and so |F ′(0)| = 1. This implies that F (z) = e2iθz.Therefore

F (z) = e2iθz =

(eiθ 00 e−θ

).

Math 213a Notes 29


8.1 Automorphisms and conjugacy classes

There is a bijection

c =1√2

(1 −i−i 1

): H → ∆.

Theorem 8.1. (a) The action of SU(1, 1) on P preserves ∆ ⊆ P. This actionis transitive on ∆, and induces SU(1, 1)/{±1} ∼= Aut(∆).

(b) Similarly, SL(2,R)/{±1} ∼= Aut(H).

(c) The isotropy groups of SU(1, 1) acting on ∆ and those of SL(2,R) actingon H are connected and compact.

(d) c−1 SU(1, 1)c = SL(2,R).

(e) Every g ∈ SU(1, 1) is conjugate in SU(1, 1) to exactly one of the following:

±(

1 00 1

)(center),

(eiθ 00 e−iθ

)for − π < θ < 0 or 0 < θ < π (elliptic),

±(

cosh t sinh tsinh t cosh t

)for 0 < t <∞ (hyperbolic), ±

(1± i

2 ± 12

± 12 1∓ i

2

)(unipotent).

(f) Every g ∈ SL(2,R) is conjugate, in SL(2,R), to exactly one of:

±(

1 00 1

),

(cos θ sin θ− sin θ cos θ

)for − π < θ < 0 or 0 < θ < π,

±(et 00 e−t

)for 0 < t <∞, ±

(1 ±10 1

).

Proof. (a) Suppose

g =

(α ββ α

)∈ SU(1, 1),

and let eiθ ∈ ∂∆. Then

g · eiθ =αeiθ + β

βeiθ + α= e−iθ

αeiθ + β

αe−θ + β∈ ∂∆.

So g : ∂∆→ ∂∆. So either g maps ∆ to ∆ or P− clos(∆). But g · 0 = β/α with|α|2 − |β|2 = 1, and so g · 0 ∈ ∆. This shows that g : ∆ → ∆, and the rest of(a) was proved last time.

(b) The action of SL(2,R) fixes R ∪ {∞}. Hence any g ∈ SL(2,R) maps Hto either H or P− clos(H). But

g =

(a bc d

): i 7→ ai+ b

ci+ d=

(ad− bc)ic2 + d2

+bd+ ac

c2 + d2

Math 213a Notes 30

which is in H. So g : H → H is a homeomorphism, and we get SL(2,R)/±1 ↪→Aut(H). Since H and ∆ are isomorphic, every automorphism of H comes fromsome g ∈ SL(2,C). Suppose g ∈ SL(2,C) maps H → H. Then g : R ∪ {∞} →R ∪ {∞}. SL(2,R) acts on R ∪ {∞} in a triply transitive manner. So thereexists g1 ∈ SL(2,R) such that g · 0 = g1 · 0, g · 1 = g1 · 1, and g · ∞ = g1 · ∞.Then g−1

1 ◦g fixes 0, 1,∞ individually. Therefore g−11 ◦g = ±1, so g ∈ SL(2,R).

Hence SL(2,R)/{±1} ∼= Aut(H). The fact that SL(2,R) acts transitively on Hfollows from the corresponding fact about SU(1, 1) action on ∆, since H ∼= ∆.

(d) This follows from (a), (b), and the existence of c : H ∼= ∆.(c) It suffices to prove this for the SU(1, 1)-action on ∆, by (d). Since

SU(1, 1) acts transitively on ∆, it suffices to show that the isotropy subgroupof SU(1, 1) at 0 is {(

eiθ 00 e−θ

): θ ∈ R/2πZ

}.

This is clear because

g =

(α ββ α

)maps 0 to 0, then β/α = 0 or β = 0. Then |β|2 = 1 because |β|2 − |α|2 = 1.

(e) + (f) The Jordan canonical form applies to real 2 × 2 matrices whichhave real eigenvalues. So any g ∈ SL(2,R) which as real eigenvalues λ, λ−1

is GL(2,R)/center ∼= SL±(2,R)/{±1}. Suppose g ∈ SL(2,R) has eigenvaluesλ, λ−1 ∈ R. Then g is SL±(2,R)-conjugate to exactly one of

±(

1 00 1

), ±

(1 10 1

), ±

(et 00 e−t

)for 0 < t <∞.

Now there is a small problem of whether the elements of SL(2,R) will be, notSL±(2,R)-conjugate but SL(2,R)-conjugate to one of these. Because ( 1 0

0 −1 )

fixes both ( 1 00 1 ) and ( e

t 00 e−t

), things that are SL±(2,R)-conjugate to them are

also SL(2,R)-conjugate to them. However, conjugation by ( 1 00 −1 ) sends ( 1 1

0 1 ) to( 1 −1

0 1 ). So the question is whether they are SL(2,R)-conjugate or not. It turnsout that they are not.

Now let us consider the case when g ∈ SL(2,R) has eigenvalues λ, µ, not bothreal. Then {λ, µ} = {λ, µ} and λµ = 1, so λ = eiθ for some θ /∈ πZ. Since theeigenvalues are distinct, there are two linearly independent eigenvectors in C2.The must be each other’s complex conjugate. Since SL(2,R) acts transitivelyon pairs of linearly independent vectors in R2, up to SL(2,R)-conjugacy we mayassume the eigenvectors are

(±i1

)(as points in P). Since c : H → ∆ maps i to

0, asking which g ∈ SL(2,R) fix i is the same as asking which g ∈ SU(1, 1) fix

0. They are precisely those of the form ( eiθ 00 e−iθ

), and so the isotropy subgroup

of SL(2,R) at i is {(cos θ sin θ− sin θ cos θ

): θ ∈ R/2πZ

}.

Math 213a Notes 31

The question remains whether ( eiθ 00 e−iθ

) SU(1, 1)-conjugate to ( e−iθ 00 eiθ

). It

turns out that they are not SU(1, 1)-conjugate to each other, and so ( cos θ sin θ− sin θ cos θ )

is not SL(2,R)-conjugate to its inverse. This proves (f), and an easy computa-tion translates (f) to (e).

Note that(eiθ 00 e−θ

)z = e2iθz,

(1 10 1

)z = z + 1,

(et 00 e−t

)z = e2tz.

Elliptic elements of SU(1, 1), respectively SL(2,R) have exactly one fixedpoint in ∆, respectively H, and the other fixed point in P−clos(∆), respectivelyP−clos(H). Hyperbolic elements elements of SU(2,R), respectively SL(2,R) hastwo fixed points, located at the boundary R ∪ {∞}, respectivly ∂∆. Unipotentelements have one fixed point on the boundary.

Theorem 8.2.

Aut(C) ∼={(

a b0 1

): a ∈ C∗, b ∈ C

}with the latter acting by z 7→ az + b.

Proof. I leave it as an exercise. What needs to be proved is that any continuousfunction on C to P.

Math 213a Notes 32


SupposeX is a Hausdorff space, and Γ a group acting onX by homeomorphisms.We say that Γ acts without fixed points if for any x0 ∈ X and γ ∈ Γ,γx0 = x0 implies that γ acts on X as the identity map. (It is more commonto require γ = e in this case, but we are using this definition because we wantto allow centers.) We also say that Γ acts properly discontinuously, if forevery x0 ∈ X there exists an open neighborhood U0 of x0 such that {γ ∈ Γ :γU ∩ U 6= ∅} is finite. It is easy to see that Γ acting property discontinuouslyis equivalent to (i) {γ ∈ Γ : γx0 = x0} is finite and (ii) there exists an opensubset U0 such that γU0 ∩ U0 = ∅ if γx0 6= x0, and γ ∈ Γ and γx0 = x0 impliesγU0 = U0.

As an exercise, prove the following fact, which is tricky. Suppose X islocally compact and Hausdorff, and Γ acts properly continuously. Then themap X → Γ \X induces a Hausdorff topology on Γ \X. If Γ acts also withoutfixed points, then X → Γ \X is a covering map.

9.1 Riemann surfaces

Now suppose that X is a connected Riemann surface, meaning that it is a1-dimensional complex manifold. Let X be its universal cover. Then X clearlyis a Riemann surface as well, and Γ = π1(X) acts on X by biholomorphismswith a natural identification X = Γ \ X.

Theorem 9.1 (Kobe, 1907). Up to isomorphism, the only connected, simplyconnected Riemann surfaces are P, C, and H ∼= ∆.

The notion of the Riemann surface actually was established after 1907, butonce it was established, it was clear that Kobe had proved this theorem.

The only Riemann surface with covering space P is P, because any automor-phism of P has a fixed point.

To understand those connected Riemann surfaces whose universal coveringspace is C, we need to know the group of automorphisms of C which act properlydiscontinuously without fixed points. Aut(C) = {z 7→ az + b}, and so the onlysuch subgroups are the additive group C which are free of rank 1 or 2. Thereforethese Riemann surfaces are, up to isomorphism,

C, C/Z ∼= C∗, C/(Z + τZ) with τ ∈ H.

The third one is an elliptic curve, and they contain a lot of arithmetic informa-tion.

Now let us look at the Riemann surfaces whose universal covering is H.They are all of the form Γ \ H where Γ ⊆ SL(2,R) is a discrete subgroup,containing −1, and not containing elliptic elements (this implies Γ acts properlydiscontinuously, without fixed points).

Math 213a Notes 33

9.2 Mittag-Leffler theorem

Let U ⊆ C be open.

Lemma 9.2. Let {fm} be a sequence of holomorphic functions on U whichconverges locally uniformly on U to a function f . Then f is holomorphic, andf ′n → f ′ also locally uniformly.

Proof. We already saw that f is holomorphic, by using the characteristic ofintegrating over an rectangle. Now suppose z0 ∈ U , and choose r > 0 so thatclos(∆(z0, r)) ⊆ U . Then ∆(z0, r) is a Green’s domain, and for z ∈ ∆(z0, r),

f ′n(z) =1

2πi

∫|ζ−z0|=r

fn(ζ)dζ

(ζ − z)2

and similarly for f . Because fn → f uniformly on ∂∆(z0, r), and 1/(ζ − z)2 isbounded for |z − z0| < r/2, we see that f ′n → f ′ uniformly.

Let {fn} be a sequence of meromorphic functions on U .

Definition 9.3. The sequence {fn} is locally unipolar if each z0 ∈ U hasan open neighborhood V0 such that, for an appropriately chosen integer n0 =n0(z0, V0), n,m ≥ n0 implies that fm − fn is holomorphic on V0.

Definition 9.4. The sequence {fn} is unipolar on U , if for each z0 ∈ U thereexists n0, V0 as in the previous definition, and in addition fn − fn0

convergeslocally uniformly on V0. In this situation, define limn→∞ fn = f so that withV0, n0 as abouve, limn→∞(fn − fn0

) = f − fn0on V0.

This definition of the limit makes sense, because (i) if n1 > n0 then (fn −fn0

) = (fn − fn1) + (fn1

− fn0) and (ii) restricting the domain does not affect

the limit. As a direct consequence of the definition, fn → f locally uniformlyon U − {poles of f}. Hence f ′n → f ′ locally uniformly on this open set.

Suppose {zk}k≥0 is a set of distinct points in C without points of accumu-lation. Equivalently, zk → ∞ in P. Suppose further, for each k we are given apotential principal part at zk, Pk(z) =

∑Nk`=1 ak,l(z− z0)−l. Then Pk(z) is holo-

morphic on C − {zk}, and hence on ∆(0, |zk|). Now choose rk ≥ 0 and εk > 0such that 0 < rk < |zk| unless zk = 0 in which case rk = 0, and rk → +∞ ask → +∞, and

∑εk < ∞. Then we can choose Qk for each k, a polynomial

obtained by taking enough terms of the Taylor expansion of Pk at 0 so that|Pk(z)−Qk(z)| < εk on ∆(0, |zk|).

Let

sn =

n∑k=0

(Pk(z)−Qk(z)).

This function has poles at z1, . . . , zn with principal part Pj at zj . Also

|sm(z)− sn(z)| ≤n∑

k=m

|Pk(z)−Qk(z)| ≤n∑

k=m

εk

Math 213a Notes 34

for m > n. So we can conclude that {sn(z)} is unipolar, and converges uniformlylocally on C. This proves the following theorem.

Theorem 9.5 (Mittag-Leffler). Let {zk} be a sequence of distinct points in C,with zk →∞ and Pk(z) a sequence of potential principal parts at zk. Then thereexists a meromorphic function f on C which has principal part Pk at zk, k ≥ 0,and no other poles. This f can be expressed as

f(z) =

∞∑k=0

(Pk(z)−Qk(z)) + g(z),

with g(z) entire.

For example, f(z) = π2/ sin2 πz is meromorphic on C and has poles atk ∈ Z with principal part (z − k)−2 at z = k. Note that for |z| < |k|/2,|(z − k)−2| < 4k−2, and we can choose Qk = 0 and εk = 4k−2, rk = |k|/2. Sothe conclusion is

π2

sin2 πz=

∞∑k=∞

1

(z − k)2+ g(z).

What is g(z)? Note that |sinπz| = 12 |e

iπx − e−iπx+πy| and so |1/ sinπz| → 0 as|y| → ∞ uniformly on |x| ≤ 1/2. Also on |x| ≤ 1/2 and k > 0, |(z − k)−2| ≤4k−2, and so ∣∣∣∣ ∞∑

k=−∞

1

(z − k)2

∣∣∣∣ ≤ ∣∣∣∣ 1

z2

∣∣∣∣+

∞∑k=1

∞k2,

and hence for all x because it is periodic of period 1. The conclusion is thatg is periodic of period 1, and bounded on {|<(z)| ≤ 1/2} and hence globallybounded. Therefore g is a constant, and thus

π2

sin2 πz=

∞∑k=−∞

1

(k − z)2+ c

for some c constant.

Math 213a Notes 35

10 October 4, 2016

I had shown that

π2

sin2 πz=

∞∑n=−∞

1

(z − n)2+ c

for some constant c. We had also shown that π2/ sin2 πz → 0 as |y| →+∞ uniformly in x. We now claim that c = 0, and it suffices to show that∑∞n=−∞

1(iy−n)2 → 0 as |y| → ∞. This is because

∣∣∣ ∞∑n=−∞

1

(iy − n)2

∣∣∣ ≤ ∞∑n=−∞

1

n2 + y2=

1

y2+ 2

∞∑n=1

1

n2 + y2

≤ 1

y2+ 2

∞∑n=1

∫ n

n−1

dx

x2 + y2=

1

y2+ 2

∫ ∞0

1

n2 + y2

=1

y2+ 2|y|−1

∫ ∞0

dx

x2 + 1= O(|y|−1).

This shows that c = 0.Recall that (d/dx) cotx = −1/ sin2 x on the real axis, so it is true also for

complex z. Hence

d

dzπ cotπz = − π2

sin2 πz.

Likewise 1/(z − n)2 = −(d/dz)(1/(z − n)). Then

∞∑n=−∞

1

(z − n)2= − d

dz

(1

z+

∞∑n=1

( 1

z − n+

1

z + n

))because the sum in the right hand side converges, locally uniformly. This impliesthat

π cotπz =1

z+ 2

∞∑n=1

z

z2 − n2+ c

and since both cotπz and the right hand size are odd functions of z, we havec = 0.

10.1 Mittag-Leffler on an open set

Suppose U ⊆ C is open and {zk} a sequence in U without points of accumulationin U . Let zk 6= zl if k 6= l, and for each k consider a potential principal partPk(z) =

∑Nkl=1 ak,l(z − zk)−l at zk.

Theorem 10.1 (Mittag-Leffler). There exists a meromorphic function f on Uwith principal part Pk at zk and no other poles. It is determined up to additionof a holomorphic function on U .

Math 213a Notes 36

Proof. We may assume U 6= C. Initially assume that ∂U is compact (but Uneed not compact) and {zk} is bounded. Let δk = dist({zk}, ∂U) > 0. I wantto claim that δk → 0 as k → ∞. If not, we can go to a subsequence of {zk}such that δk ≥ δ > 0 for all k. Since {zk} is bounded, there must be a point ofaccumulation z∞, which can be either on U or ∂U . But the latter is precludedby dist({zk, ∂U}) ≥ δ > 0. This proves the claim.

Now for each k, choose wk ∈ ∂U such that |zk − wk| = δk. Consider theLaurent series of Pk(z) centered at wk. Consider the Laurent series of Pk(z)centered at wk:

1

z − zk=

1

(z − wk)− (zk − wk)=

1

z − wk

∞∑n=0

(zn − wkz − wk

)n=

∞∑n=1

(zk − wk)n−1

(z − wk)n.

Given ε > 0, we can choose Mk large enough so that∣∣∣ 1

z − zk−

Mk∑n=1

(zk − wk)n−1

(z − wk)n

∣∣∣ < ε for |z − wk| ≥ 2δk.

Because we have estimated 1/(z − zk) by a polynomial in 1/(z − wk), we canget the same type of estimate for∣∣∣ 1

(z − zk)l−(Mk∑n=1

(zk − wk)n−1

(z − wk)n

)l∣∣∣uniformly on {|z −wk| ≥ 2δk}. So for each k there exists a polynomial withoutconstant terms Qk in 1/(z − wk) such that |z − wk| ≥ 2δk implies |Pk(z) −Qk(z)| < 2−k.2

Qk has no poles in U and Pk has a pole at zk only, and |zk −wk| = δk → 0,with wk ∈ ∂U . This implies that the partial sums of

∑∞k=1(Pk(z) − Qk(z)) is

locally equipolar on U , and locally uniformly convergent. This prove that

f(z) =

∞∑k=1

(Pk(z)−Qk(z))

has the require property.Note that in this construction, both Pk(z) and Qk(z) are holomorphic on

a neighborhood of ∞, vanishing at ∞, and∑∞n=1(Pk(z) − Qn(z)) converges

uniformly on some neighborhood of ∞, where it has the value 0.Now let us look at the general case. By translating U if necessary, we can

arrange 0 ∈ U and zk 6= 0 for all k. Since {zk} has no point of accumulationin U , {0} /∈ clos({zk}). Now define U = {1/z : z ∈ U}, and define zk = 1/zk.We know that {zk} are a bounded sequence and zk 6= zl for k 6= l, and also∂U is compact because 0 ∈ U . So the previous argument applies to U , {zk}:there exists a meromorphic function f(z) on U with poles at zk, with specifiableprincipal part Pk at zk, no other poles, holomorphic at ∞.

2This is known as the “pole moving lemma”.

Math 213a Notes 37

Now define f(z) = f(1/z). Then f will be meromorphic on U and holomor-phic at a neighborhood of 0, with poles at zk and principal part Pk(1/z) at zk.How do I have to choose Pk so that f has principal part Pk at zk? We need

Pk

(1

z

)=

Nk∑k=1

ak,l

(1

z− 1

zk

)−l≡ Pk(z) =

Nk∑k=1

ak,l(z − zk)−l

where ≡ shall mean equivality modulo a function which is holomorphic on U .We have

Pk

(1

z

)=

Nk∑l=1

(−1)lak,lzlzlk(z − zk)−l =

Nk∑l=1

(−1)lzlk((z − zk) + zk)l(z − zk)−l

=

Nk∑l=1

(−1)lak,l

l∑r=0

(l

r

)z2l−rk (z − zk)r−l

≡Nk∑l=1

l−1∑r=0

(−1)lak,l

(l

r

)z2l−rk (z − zk)r−l

where r− l < 0 in all cases. We want this to be equal to Pk. We can make it soby downward induction.

10.2 Infinite products

We want to construct holomorphic functions, with zeros at specified locations, ofspecified order, with no other zeros. To do this directly, I need infinite products.

Definition 10.2. Let {cn} be a sequence of complex numbers. The infiniteproduct

∏∞k=0 ck exists if for N � 0 the limit limn→∞

∏nk=N ck exists and is

nonzero. In that case, limn→∞∏nk=1 ck also exists.

The definition implies that ck 6= 0 for all but finitely many k. One canobserve that if

∏∞k=1 ck converges then ck → 1 as k → ∞. This is because for

n > N

n+1∏k=N

ck −n∏

k=N

ck = (cn+1 − 1)

n∏k=N

ck

and as n→∞ we get limn→∞ cn = 1.

Math 213a Notes 38

11 October 6, 2016

Last time I introduced the convergence of a infinite product. This time I willuse a different notation.

Definition 11.1. The infinite product∏∞k=1(1 + ak) converges if for some N ,

limn→∞∏∞k=N (1 + ak) exists and is nonzero.

If so,

N−1∏k=1

(1 + ak) limn→∞

n∏k=N

(1 + ak)

is independent of N . Also ak 6= −1 for all but finitely many k. In the following,define a “branch” of log z, z 6= 0 requiring −π < = log z ≤ π.

Lemma 11.2. The product∏∞k=1(1 + ak) converges if and only if

(i) ak 6= −1 for all but finitely many k, and

(ii)∑∞k=N log(1 + ak) exists for N � 0.

Proof. We may as well suppose ak 6= −1. If∑∞k=1 log(1 + ak) converges, then

limn→∞

∞∏k=1

= limn→∞

exp

n∑k=1

log(1 + ak).

Because exp is continuous, we get ⇐.For ⇒, assume ak 6= −1 for all k, and

∏∞k=1(1 + ak) = P exists. We can

assume that P /∈ iR≥0. Then if we exponentiate

n∑k=1

log(1 + ak)− logP

and take the limit as n→∞, we get 1. Then∣∣∣ n∑k=1

log(1 + ak)− logP∣∣∣

becomes arbitrarily small in R/2πiZ as n → ∞. We know that∏∞k=1(1 + ak)

converges and so 1 + ak → 1 as k → ∞. So we cannot have jumps by integralmultiples of 2πi for an large. The conclusion is that there exists m ∈ Z suchthat ∣∣∣ n∑

k=1

log(1 + ak)− logP − 2πim∣∣∣→ 0

as n→∞. Then∑

log(1 + ak) converges.

Math 213a Notes 39

11.1 Weierstrass problem and product theorem

Definition 11.3. The infinite product∏∞k=1(1 + ak) converges absolutely if∑∞

k=N |log(1 + ak)| converges for N � 0.

We have

log(1 + z) = z − z2

2+z3

3− · · · = z

(1− z

2+z2

3− · · ·

)and that 1−z/2+· · · is a holomorphic function on ∆(0, 1) with value 1 at z = 0.So |z|/2 < |log(1 + z)| < 2|z| for |z| < δ for δ > 0. Therefore

∏∞k=1(1 + ak)

converges absolutely if and only if∑∞k=1 ak converges absolutely.

Let U ⊆ C be an open set, and let {fn} be a sequence of holomorphicfunctions on U such that

∑∞n=1 fn converges locally uniformly and absolutely

on U . (Again, this means that the convergence is absolute and the absoluteconvergence is locally uniform.) In that situation,

∏∞k=1(1+fk) converges locally

uniformly and absolutely on U . Let F (z) be the limit. Then for z0 ∈ U ,F (z0) = 0 if and only if 1 + fk(z0) = 0 for some k. If U is connected andfk 6≡ −1 for all k, then

ordF |z0 =

∞∑k=1

ord(1 + fk)|z0

and the latter sum is finite.Let {zk} be a sequence in C, and zk 6= zl for k 6= l, without points of

accumulation in C (or equivalently, |zk| → ∞ as k →∞). Suppose we are givennk ∈ Z>0 for each k.

Weierstrass problem. Construct an entire function W (z) which has a zeroof order exactly nk at zk, with no other zeros.

Of course you can ask this for any open subset of C, but then the problemhas a different problem.

What should we multiply to get this? We would like to multiply stuff like(1 − z/zk)nk to get the right orders, but this gets big as z → ∞. So we alsomultiply ePk(z) for some polynomial Pk(z). And there is another problem ofzk = 0.

As a slight modification of notation, always include z0 = 0 and zk 6= 0 for allk > 0, but allow n0 = 0. We can choose rk, εk for k > 0 such that 0 < rk < |zk|,rk →∞, and

∑∞k=0 εk <∞.

Now fix k and consider any |z| ≤ rk. Then |z/zk| ≤ rk/|zk| < 1. The series

−∞∑l=1

1

l

( zzk

)l= log

(1− z

zk

)converges absolutely and uniformly on |z| ≤ rk.

Math 213a Notes 40

Lemma 11.4. We can choose mk > 0 so that, for |z| ≤ rk,∣∣∣1− (1− z

zk

)nkenk

∑∞l=1

1l ( zzk

)l∣∣∣ < εk.

Proof. We have∣∣∣log(

1− z

zk

)+

mk∑l=1

1

l

( zzk

)l∣∣∣ =∣∣∣ ∞∑k=mk+1

1

l

( zzk

)l∣∣∣ ≤ 1

mk

∑l=mk+1

∣∣∣ zzk

∣∣∣l≤ (rk/|zk|)mk+1

mk

1

1− rk/|zk|.

Multiply by mk and exponentiate, and conclude that we can make that thingin the statement as close to 1 as I please for |z| ≤ rk by choosing mk largeenough.

Now using this lemma, we can choose the polynomials Pk(z) nicely so thatthe function converges locally uniformly and absolutely. If n0 > 0 then we cansimply multiply zn0 . This proves:

Theorem 11.5 (Weierstrass product theorem). There exists an entire functionF (z) which has a zero of order exactly nk at zk, and no other zeros. The mostgeneral such function can be expressed as F (z) = eg(z)W (z) with g(z) entireand

W (z) = zn0

∞∏k=1

(1− z

zk

)nkenk

∑mkl=1

1l ( zzn

)l

with an appropriate choice of mk, which makes the product locally uniformlyand absolutely convergent.

Suppose it is possible to choose mk uniformly. Then the least possible choiceof m = mk is called the genus of the Weierstrass product. In this situation,W (z) is uniquely determined. If in addition, g(z) is a polynomial (necessarilyunique modulo a multiple of 2πi) then the larger of m and deg g is called thegenus of F (z). It is not obvious from what I have said, but if there are morezeros, this forces the function to grow more rapidly.

Example 11.6. Consider F (z) = sinπz, which has first order zeros at z =k ∈ Z. Does the product z

∏k∈Z\{0}(1 − z/k) converge locally uniformly and

absolutely? This would be equivalent to saying that∑z/k converges uniformly

absolutely. The answer is not, because∑

1/k =∞. What about

z∏

k∈Z\{0}

(1− z/k)ez/k?

Because log(1− z/k) + z/k is O(k−2), this converges absolutely. So

sinπz = eg(z)z∏

k∈Z\{0}

(1− z

k

)ez/k = eg(z)z

∞∏k=1

(1− z2

k2

).

Math 213a Notes 41

This needs to be justified, but for now I am going to work formally. Take thelogarithmic derivative. Then

π cosπz

sinπz= g′(z) +

1

z+

∞∑k=1

(−1

k

1

1− z/k+

1

k

1

k

1

1 + zk

)= g′(z) +

1

z+

∞∑k=1

2z

z2 − k2= g′(z) + π cotπz.

So g′(z) = 0, and so eg(z) = c 6= 0. That is,

sinπz = cz

∞∏k=1

(1− z2

k2

).

Differentiating and setting z = 0 yields π = c. In conclusion,

sinπz = πz

∞∏k=1

(1− z2

k2

).

What about the logarithmic derivative? Assume that∑fk converges locally

uniformly and absolutely. Then( ∞∏k=1

(1 + fk))−1 d

dz

∞∏k=1

(1 + fk) = limn→∞

( n∏k=1

(1 + fk))−1 d

dzlimn→∞

∞∏k=1

(1 + fk)

= limn→∞

( n∏k=1

(1 + fk)−1 d

dz

n∏k=1

(1 + fk))

= limn→∞

n∑k=1

fk1 + fk

.

So logarithmic derivatives of Weierstrass product can be computed formallyterm by term.

Math 213a Notes 42

12 October 11, 2016

12.1 The Gamma function

The Gamma function is defined as

Γ(z) =

∫ ∞0

e−ttz−1dz.

Note that tz−1 = t<z−1 for t ≥ 0 and so this converges absolutely for <z > 0.So the above formula defines a holomorphic function on {<z > 0}, because theintegral of it over any rectangle will be zero.

Doing integration by parts, we get

Γ(z) =1

z

∫ ∞0

e−td(tz) =1

z

((e−ttz)|∞t=0 +

∫ ∞0

e−ttzdt)

=1

zΓ(z + 1).

We can use this identity to analytically continue Γ(z) to {<z > −1}, but witha first order pole at z = 0, with Res Γ(z)|z=0 = Γ(1). In a similar way, we have

Γ(z) =1

zΓ(z + 1) =

1

z(z + 1)Γ(z + 2) = · · · = 1

z(z + 1) · · · (z + k)Γ(z + k + 1).

(∗)

So inductively we can continue Γ(z) to all of C with first order poles at z = −k,k ∈ Z ≥ 0.

Proposition 12.1. For k ≥ 1, Γ(k) = (k − 1)! and for k ≥ 0 the residueRes Γ(z)|z=−k = (−1)k/k!.

Proof. Evaluate (∗) at z = 1 and find Γ(k+z) = (k+1)!Γ(z) = (k+1)!. Becausewe have

Γ(z) =1

(z + k)− k1

(z + k)− (k − 1)· · · 1

(z + k)− 0Γ(z + k + 1)

and evaluating near z = −k, we get

Γ(z) =(−1)k

k!

1

z + k(1 + (z + k)(holomorphic function))

and so the residue at z = −k is Res Γ(z)|z=−k = (−1)k/k!.

Why are people interested in this function? One of the big tools in numbertheory is L-functions, and these satisfy functional equations that are expressedin terms of the Gamma function.

Recall that

sinπz = πz∏

k∈Z\{0}

(1− z

k

)ez/k

Math 213a Notes 43

and this converges locally uniformly and absolutely. Define

G(z) =

∞∏k=1

(1 +

z

k

)e−z/k,

which is an entire function, with first order zeros at z = −k for k ∈ Z≥1. ThenG(z − 1) is entire and has the same zeros as G(z) plus one other at z = 0. So

G(z − 1) = zeg(z)G(z)

for some entire function g(z).

Lemma 12.2. g(z) = γ is a constant, called Euler’s constant, and

γ = limn→∞

(1 +

1

2+ · · ·+ 1

n− log n

)∼ 0.577 · · · .

Note that

γ =

∫ ∞1

( 1

bxc− 1

x

)dx.

Proof. We know that G(z − 1) = zeg(z)G(z) and so

∞∏k=1

(1 +

z − 1

k

)e−(z−1)/k = zeg(z)

∞∏k=1

(1 +

z

k

)e−z/k.

Now let us take logarithmic derivatives. We get

∞∑k=1

( 1

z + k − 1− 1

k

)=

1

z+ g′(z) +

∞∑k=1

( 1

z + k− 1

k

).

Then

g′(z) = −1

z+

∞∑k=1

( 1

z + k − 1− 1

z + k

)= 0.

This shows that g = γ is a constant.Now let us evaluate at z = 1 and take the logarithm. Then

0 = γ +

∞∑k=1

(log(

1 +1

k

)− 1

k

)= γ − lim

n→∞

( n∑k=1

1

k− log(n+ 1) + log 1

)= γ − lim

n→∞

(1 +

1

2+ · · ·+ 1

n− log n

).

So we have

G(z) =

∞∏k=1

(1 +

z

k

)e−z/k, G(z − 1) = zeγG(z).

Math 213a Notes 44

Define H(z) = eγzG(z) so that H(z) is entire and has zeros at z = k for k ∈ Z≥1

with no other zeros. We have

H(z − 1) = eγz−γG(z − 1) = zeγzG(z) = zH(z).

What do we know about 1/(zH(z))? It is meromorphic, has first order poles atz = −k for k ∈ Z≥0, and no other poles. Then

1

(z + 1)H(z + 1)=

1

H(z)=

z

zH(z).

So 1/(zH(z)) has exactly the same polar locations as Γ(z) and behaves the sameway under z 7→ z + 1.

12.2 Gamma as an infinite product

Theorem 12.3.

Γ(z) =1

zH(z)=

1

ze−γz

∞∏k=1

(1 +

z

k

)−1

ez/k.

This requires a lot of work to prove, so we first state some corollaries.

Corollary 12.4. Γ(z) has no zeros.

Corollary 12.5. Γ(z)Γ(1− z) = π/ sinπz.

Proof. We have

Γ(z)Γ(1− z) = Γ(z)(−z)Γ(−z)

=1

ze−γz

( ∞∏k=1

(1 +

z

k

)−1

ez/k)

(−z) 1

−zeγz( ∞∏k=1

(1− z

k

)−1

e−z/k)

=1

z

∏k∈Z\{0}

(1− z

k

)−1

ez/k =π

sinπz

Define

fn(t) =

{(1− t/n)n 0 ≤ t ≤ n,0 t > n.

Lemma 12.6. The functions fn satisfy the following:

(1) 0 ≤ fn(t) ≤ e−t for all t and n > 0.

(2) limn→∞ fn(t) = e−t.

Proof. For (1), we may assume 0 < t < n. We need to show log fn(t) ≤ −t, i.e.,n log(1− t/n) + t ≤ 0. This is true becuase

n log(

1− t

n

)= −n

∞∑k=1

tk

knk+ t = −

∞∑k=2

tk

knk−1= O(n−1)

This shows both (1) and (2).

Math 213a Notes 45

Proof of Theorem 12.3. We have

Γ(z) =

∫ ∞0

e−ttz−1 = limn→∞

∫ n

0

(1− t

n

)ktz−1dt,

by the dominated convergence theorem. This is the crucial step, and I willcontinue next time.

Math 213a Notes 46

13 October 13, 2016

Let us recall that we have defined

Γ(z) =

∫ ∞0

e−ttz−1dt

for <z > 0 and Γ(z + 1) = zΓ(z). We had shown

Γ(x) = limn→∞

∫ n

0

(1− t

n

)ntx−1dt

for x > 0. We have∫ n

0

(1− t

n

)ntx−1dt =

1

x

∫ n

0

(1− t

n

)nd(tn) =

n

n

1

x

∫ n

0

(1− t

n

)n−1

txdt

=n

n

1

x

n− 1

n

1

x+ 1

∫ n

0

(1− t

n

)n−2

tx+1dt

= · · · = n

n

1

x

n− 1

n

1

x+ 1· · · 1

n

1

x+ n− 1

∫ n

0

tx+n−1dt

=n

n

1

x

n− 1

n

1

x+ 1· · · 1

n

1

x+ n− 1

1

x+ nnx+n =

n!nx

x(x+ 1) · · · (x+ n)

=1

xnxn!

n∏k=1

1

x+ k=

1

xex logn

n∏k=1

(1 +

x

k

)−1

=1

xex(logn−1−1/2−···−1/n)

n∏k=1

(1 +

x

k

)−1ex/k.

Taking limn→∞, we get

Γ(x) =1

xe−γx

∞∏k=1

(1 +

x

k

)−1

ex/k.

So I have shown this formula for <x > 0 (and hence for all z for which bothsides make sense).

Suppose f is a continuous function on R>0 satisfying appropriate growthconditions at 0 and ∞. Then the function

M(f, s) =

∫ ∞0

f(t)ts−1dt

is called the Mellin transform of f(t) for s ∈ C where this makes sense. ThenΓ is just the Mellin transform of e−z.

13.1 Legendre duplication formula

Theorem 13.1 (Legenedre duplication formula). For any z ∈ C,

√πΓ(2z) = 22z−1Γ(z)Γ

(z +

1

2

).

Math 213a Notes 47

Proof. The derivative of the logarithmic derivative of Γ is

d

dz

Γ′(z)

Γ(z)=

d

dz

(−1

z− γ −

∞∑k=1

( 1

z + k− 1

k

))

=1

z2+k = 1

∞1

(z + k)2=

∞∑k=0

1

(z + k)2.

Then the derivative of the logarithmic derivative of Γ(2z) is

2d

dz

Γ′(2z)

Γ(2z)= 4

∞∑k=0

1

(2z + k)2=

∞∑k=0

1

(z + k/2)2.

Likewise we have

d

dz

( ddzΓ(z)Γ(z + 1/2)

Γ(z)Γ(z + 1/2)

)=

∞∑k=0

1

(z + k)2+

∞∑k=0

1

(z + k + 1/2)2=

∞∑k=0

1

(z + k/2)2.

This implies that

Γ(z)Γ(z +

1

2

)= eaz+bΓ(2z).

Now we know Res Γ(z)|z=0 = 1 and Γ(1) = 1. Also from Γ(z)Γ(1 − z) =π/ sinπz we see that Γ(1/2) =

√π. Taking the residues of both sides at z = 0,

we get

Γ(1

2

)= Res

(Γ(z)Γ

(z +

1

2

))∣∣∣z=0

=1

2eb.

The value at z = 1/2 give us

Γ(1

2

)= ea/2+b.

From this we get ea = 1/4 and eb = 2√π.

13.2 Dirichlet series

Suppose {an}n≥1 is a sequence of complex numbers such that |an| = O(nc) forsome c ∈ R.

Definition 13.2. The series L(s) =∑∞n=1 ann

−s is the Dirichlet series withcoefficients an, where s = σ + it for σ, t ∈ R. The series converges locallyuniformly and absolutely for σ > c+ 1.

Let us define

σ0 = inf{σ ∈ R :∑∞n=1|an|n

−σ <∞}.

Math 213a Notes 48

This is the abscissa of the absolute convergence of the Dirichlet series–it con-verges uniformly and absolutaly on {s ∈ C : <s ≥ σ0 + ε} for any ε > 0.

The convergence behavior of the Dirichlet series is much more complicatedthan that of the power series. That is, it can have uniform convergence to theleft of σ0.

Now suppose that (i) a1 = 1 and (ii) amn = aman whenever (m,n) = 1.Then formally at least,

∑n

ann−s =

∏p

( ∞∑k=0

apkp−ks)

=∏p

(1 +

∞∑k=1

apkp−ks).

Because ∑p

∞∑k=1

|ak|p−kσ ≤∞∑n=1

|an|n−σ <∞,

we conclude that

L(s) =

∞∑n=1

ann−s =

∏p

(1 +

∞∑k=1

apkp−ks)

and both the sum and the product converges uniformly and absolutely for <s >σ0 + ε.

Example 13.3. The Riemann zeta function

ζ(s) =

∞∑n=1

n−s

has abscissa of absolute convergece σ0 = 1, and so it is holomorphic (at least)on {s ∈ C : <s > 1}. This can also be written as

ζ(s) =∏p

(1 +

∞∑k=1

p−ks)

=∏p

(1− p−s)−1.

In particular, ζ(s) 6= 0 if <s > 1.

Theorem 13.4 (Riemann). ζ(s) extends to all of C as a meromorphic functionwhich has only one pole of order 1 at s = 1 with residue Res ζ(s)|s=1 = 1.Moreover,

ζ(1− s) = 2(2π)−sΓ(s) cosπs

2ζ(s).

Proof. We have

n−sΓ(s) =

∫ ∞0

e−ntts−1dt

Math 213a Notes 49

and so for t ≥ 0 and <s > 1,

ζ(s)Γ(s) =

∞∑k=1

∫ ∞0

e−ktts−1dt = limn→∞

∫ ∞0

e−t1− e−nt

1− e−tts−1dt.

By dominated convergence theorem,

limn→∞

∫ ∞0

e−te−nt

1− e−tts−1dt = 0

locally uniformly in s for <s > 1. This shows that

ζ(s)Γ(s) =

∫ ∞0

ts−1

et − 1dt.

Math 213a Notes 50

14 October 18, 2016

We had seen that

ζ(s)Γ(s) =

∫ ∞0

ts−1

et − 1dt

for <s > 1.

14.1 Functional equation for ζ

To compute this integral, for 0 < r1 < r2 consider a curve Cr1,r2 that startsfrom r2, goes around the circle of radius r2 counterclockwise to r2, goes to r1

along the real axis, goes around the circle of radius r1 clockwise, and then goesto r2 along the real axis. The two integrals along the real axis does not cancelbecause ts−1 changes by a factor of e2πi(s−1) = e2πis. Thus if r1, r2 /∈ 2πZ,∫

Cr1,r2

ts−1dt

et − 1= (1− e2πis)

∫ r2

r1

ts−1dt

et − 1+

∫t=r2eiθ

ts−1dt

et − 1−∫t=r1eiθ

ts−1dt

et − 1.

On the other hand, by the residue theorem,∫Cr1,r2

ts−1dt

et − 1= 2πi

∑r1≤2π|k|≤r2

Res( ts−1

et − 1

)∣∣∣s=2πik

.

First suppose 0 < r1 < r2 < 2π, <s > 1, and then let r1 → 0. Then we get

0 = (1− e2πis)

∫ r2

0

ts−1dt

et − 1+

∫t=r2eiθ

ts−1dt

et − 1.

For 0 < r < 2π, define

Ir(s) = (e2πis − 1)

∫ ∞r

ts−1dt

et − 1+

∫t=reiθ

ts−1dt

et − 1.

This is an entire function of s! But on <s > 1, we can write

Ir(s) = (e2πis − 1)

∫ ∞0

ts−1dt

et − 1+ (1− e2πis)

∫ r

0

ts−1dt

et − 1+

∫t=reiθ

ts−1dt

et − 1

= (e2πis − 1)

∫ ∞0

ts−1dt

et − 1.

This is independent of r provided 0 < r < 2π and <s > 1. So let us defineI(s) = Ir(s). The conclusion is that

(1) I(s) is an entire function.

(2) for <s > 1, (e2πis − 1)ζ(s)Γ(s) = I(s).

Math 213a Notes 51

From this we immediately see that ζ(s) extends to a meromorphic function onC.

Where are the poles of ζ(s)? Potentially they are at s ∈ Z, but we knowthat it has no poles for <s > 1. Also Γ(s) has first order poles at s ∈ Z≤0,which cancels the zeros of e2πis − 1. So the only possible poles occur at s = 1.We can quickly calculate the residue. We have

Res ζ(s)|s=1 =I(1)

Γ(1)Res

1

e2πis − 1

∣∣∣s=1

=1

2πiI(1)

=1

2πi

∫t=reiθ

dt

et − 1=

2πi

2πi= 1.

Now the identity

(e2πi2 − 1)ζ(s)Γ(s) = (e2πis − 1)

∫ ∞r

ts−1dt

et − 1+

∫t=reiθ

ts−1dt

et − 1

holds for any s. Now suppose 0 < r < 2π as before and r2 > r, r2 /∈ 2πZ. Then∫Cr,r2

ts−1dt

et − 1= 2πi

∑0<2π|k|<r2

Rests−1

et − 1

∣∣∣s=2πik

.

For 0 < 2πk < r2,

Rests−1

et − 1

∣∣∣t=2πik

= (ts−1|t=2πik) Res( 1

et − 1

)∣∣∣t=2pik

= eiπ(s−1)/2(2πk)s−1,

Rests−1

et − 1

∣∣∣t=−2πik

= (ts−1|t=−2πik) Res( 1

et − 1

)∣∣∣t=−2pik

= e3iπ(s−1)/2(2πk)s−1.

So ∫Cr,r2

ts−1dt

et − 1= 2πi

∑0<2πk<r2

(2π)s−1ks−1(eiπ(s−1)/2 + e3iπ(s−1)/2).

As r2 →∞, we get

limr2→∞

∫Cr,r2

ts−1dt

et − 1= 2πi(2π)s−1(eiπ(s−1)/2 + e3iπ(s−1)/2)ζ(1− s)

provided that 1−<s > 1, i.e., <s < 0. So for <s < 0,

limr→∞

ts−1dt

et − 1+ limn→∞

∫t=(2n+1)πeiθ

ts−1dt

et − 1−∫t=reiθ

ts−1dt

et − 1

= 2πi(2π)s−1(eiπ(s−1)/2 + e3iπ(s−1)/2)ζ(1− s).

Lemma 14.1. For |t| = (2n− 1)π for n ∈ Z>0, |et − 1| is bounded away from0 uniformly in t.

Math 213a Notes 52

If this is true, then

limn→∞

∫t=(2n+1)πeiθ

ts−1dt

et − 1= 0

given that <s < 0.

Proof. Because et lies on the circle of radius e<t centered at the origin, |et−1| ≥|e<t − 1|. So for |<t| ≥ δ with 0 < δ � 1, we have

|et − 1| ≥ |e<t − 1| ≥ min{eδ − 1, 1− e−δ}.

So for |<t| ≥ δ, the claim is good. For |<t| < δ, t = <t+ i=t, so =t ∼ (2n−1)π.So et − 1 ∼ −2.

Math 213a Notes 53

15 October 20, 2016

We have an entire function, defined for but independent of 0 < r < 2π,

I(s) = (e2πis − 1)

∫ ∞r

ts−1dt

et − 1+

∫t=reiθ

ts−1dt

et − 1

and I(s) = (e2πis− 1)ζ(s)Γ(s) for <s > 1. This showed that ζ(s) is a meromor-phic function with a single pole of order 1 at s = 1.

We also have

2πi(2π)s−1(eiπ(s−1)/2 + e3iπ(s−1)/2)ζ(1− s) = limn→∞

∫Cr,(2n−1)π

ts−1dt

et − 1

= −I(s) = −(e2πis − 1)ζ(s)Γ(s).

Therefore

(2π)s(eiπs/2 − e3iπs/2)ζ(1− s) = −(e2πis − 1)ζ(s)Γ(s),

and therefore

(2π)s sinπs

2ζ(1− s) = sinπsζ(s)Γ(s).

So

ζ(1− s) = 2(2π)−s cosπs

2Γ(s)ζ(s).

This finishes the proof of Theorem 13.4.Suppose <s < 0. Then <(1−s) > 1 and we have ζ(1−s) as a Euler product.

This shows that ζ(1 − s) 6= 0 for <s < 0. Still for <s < 0, Γ(s) cos(πs/2) hasa pole if and only if s is an integer and cos(πs/2) 6= 0. This shows that ζ(s)first order zeros at the strictly negative even integers, and has no other zerosfor <s < 0. These are the trivial zeros of ζ(s).

Γ has a first order pole of residue 1 at s = 1, and cos 0 = 1, and ζ(1− s) hasa first order pole at s = 0 with residue −1. This shows that

ζ(0) = −1/2.

Recall the Legendre duplication formula√πΓ(2s) = 22s−1Γ(s)Γ(s + 1/2).

Then√πΓ(s) = 2s−1Γ(s/2)Γ((s+ 1)/2) and so we get

√πζ(1− s) = 2(2π)−s cos(πs/2)2s−1Γ(s/2)Γ((s+ 1)/2)ζ(s).

Because Γ((s+ 1)/2)Γ((1− s)/2) = π/ cos(πs/2), we get

√πΓ(1− s

2

)ζ(1− s) = π1−sΓ

(s2

)ζ(s).

Math 213a Notes 54

To make it more symmetric, we can write it as

π−(1−s)/2Γ(1− s

2

)ζ(1− s) = π−s/2Γ

(s2

)ζ(s).

Define

ξ(s) = π−s/2Γ(s/2)ζ(s).

Then

ξ(s) = ξ(1− s).

Is this an accident? In the view of algebraic number theory, this ζ(s) encodesthe information of all primes, and the other part π−s/2Γ(s/2) can be thought asattached to the completion R of Q. Then ξ(s) is the completed zeta function.

15.1 Hadamard’s theorem

Let f(z) be an entire function, written as

f(z) = zn0eg(z)∏k

(1− z

zk

)e∑mkl=1

1l ( zzk

)l,

setting nk = 1 for k > 0 but allowing repetitions of zk. Recall if it is possibleto choose mk = m independent of k and if g(z) is a polynomial, then the genusof f is the larger of the minimal choice of m and the degree of g(z). Otherwise,the genus of f is ∞.

For r > 0, define M(r) = max{|f(z)| : |z| = r}. We know that if f is notconstant, M(r) is strictly increasing (by the maximal principle) and limn→∞ =+∞ by Liouville’s theorem. Then

λ(f) = lim supr→∞

log logM(r)

log r

is well-defined unless f is a constant—from now on we suppose f is non-constant.This λ(f) is called the order (of growth) of f

Theorem 15.1 (Hadamard). If g is a non-constant entire function then eitherboth or neither of “the genus of f” and λ(f) are finite. In the finite case,genus of f ≤ λ(f) ≤ genus of f + 1.

What this means is that if a function has many zeros, this forces the functionto grow rapidly. This theorem will take some time to prove.

Observe that λ = λ(f), in the finite case, means that M(r) ≤ erλ+ε

forevery ε > 0, and for no ε < 0. This implies, in particular, if f1, f2, f1f2 arenon-constant, then λ(f1f2) ≤ max(λ(f1), λ(f2)). Also if g(z) is a non-constantpolynomial, then λ(eg(z)) = deg g.

Math 213a Notes 55

Proof of λ ≤ genus + 1. Define

Em(w) = (1− w)e∑ml=1

1lw

l

.

Then if suffices to show that

log|Em(w)| ≤ C(m)|w|m+1

for an absolute constant C(m), because g will be a product of functions of theform Em(w) and eg(z).

We prove this claim by induction on m. For m = 0, we have

log|E0(w)| = log|1− w| ≤ log(1 + |w|) ≤ |w|.

For m ≥ 1,

log|Em(w)| ≤ log|Em−1(w)|+ 1

m<wm ≤ C(m− 1)|w|m +

1

m|w|m

≤ C(m− 1) + 1/m

|w||w|m+1.

So we are done when w is bounded away from 0. If |w| < 1, then

logEm(w) = −∞∑

l=m+1

1

lml =

1

m+ 1wm+1ϕ(w)

for some holomorphic ϕ(w). For, say |w| ≤ 1/2, we have log|Em(w)| ≤ c/(m+1)|w|m+1.

Math 213a Notes 56

16 October 25, 2016

Recall Hadamard’s theorem: if f is an entire function, then

genus of f ≤ order of f ≤ genus of f + 1.

We have proved the second inequality. The proof of the first inequality dependson the Poisson-Jensen formula.

Before stating and proving the formula, we remark that if θj ∈ R/2π, then∫ 2π

0log|eiθ − eiθj |dθ converges absolutely. To see this, we may suppose θj = 0.

Then log|eiθ − 1| ≈ log|θ| as θ → 0. This is integrable.Suppose that f is holomorphic on a neighborhood of ∂∆. Then

∫log|f(eiθ)|dθ

converges absolutely. To see this, write f(z) = g(z)∏nj=1(z − eiθj ) whre g is

holomorphic and nonzero on a neighborhood of ∂∆. Then∫ 2π

0

log|f(eiθ)|dθ =

∫ 2π

0

log|g(eiθ)|dθ +

n∑j=1

∫ 2π

0

log|eiθ − eiθj |dθ.

This converges absolutely.

16.1 Poisson-Jensen formula

Theorem 16.1 (Poisson-Jensen formula). Suppose f(z) is holomorphic on aneighborhood of cl(∆). Let z1, . . . , zn be the zeros of f on cl(∆), enumeratedwith multiplicity. Then for z ∈ ∆, z 6= z1, . . . , zn,

log|f(z)| =N∑j=1

log∣∣∣ z − zj1− zjz

∣∣∣+1

2π

∫ 2π

0

<(eiθ + z

eiθ − z

)log|f(eiθ)|dθ.

Corollary 16.2 (Jensen’s formula). Under the same hypotheses,

log|f(0)| =∑

log|zj |+1

2π

∫ 2π

0

log|f(eiθ)|dθ.

We are first going to prove Jensen’s formula.

Proof. Step 1. Suppose f(z) 6= 0 for z ∈ cl(∆). Then f(z) = eg(z) for someholomorphic g(z) defined on a (possibly smaller) neighborhood of cl(∆). Then

log|f(0)| = <g(0) = < 1

2π

g(eiθ)

dθ =

1

2π

∫ 2π

0

log|f(eiθ)|dθ.

Step 2. Suppose f has zeros z1, . . . , zn in ∆ and no zeros on ∂∆. Thenf(z) =

∏Nj=1(z − zj)/(1− zjz)h(z), where h is holomorphic on a neighborhood

Math 213a Notes 58

17 October 27, 2016

17.1 Proof of Hadamard’s theorem

Apply Jensen’s formula in the following way. Let f be an entire function. Weneed to show that g ≤ λ, where g is the genus of f and λ is the order of f .Multiplying f by a polynomial does not affect either λ or γ. So I may supposef(0) 6= 1. Also, I can assume f has infinite many zeros, since it is easy if f hasfinitely many zeros. Enumerate them as

0 < |z1| ≤ |z2| ≤ |z3| ≤ · · ·

with nk = 1, and define N(r) = #{zk : |zk| ≤ r} and M(r) = max{|f(z)| : |z| =r}. For r > 1, define fr(z) = f(rz). Using Jensen’s formula on fr, we get

0 =∑|zk|≤r

log∣∣∣zkr

∣∣∣+1

2π

∫ 2π

0

log|f(reiθ)|dθ,

and thus

−∑|zk|≤r

log∣∣∣zkr

∣∣∣ ≤ logM(r).

Applying this inequality to 2r, we get

−∑|zk|≤r

log∣∣∣zk2r

∣∣∣ ≤ − ∑|zk|≤2r

log∣∣∣zk2r

∣∣∣ ≤ logM(2r).

Recall that for any ε > 0, if r � 0 then logM(r) ≤ rλ+ε. So log 2N(r) ≤(2r)λ+ε. Because k ≤ N(|zk|+ 1), we get

log 2k ≤ (2(|zk|+ 1))λ+ε,

for r � 0. Then for any ε > 0, there exists k � 0 such that k ≤ |zk|λ+ε becausethe choice of ε allows to absorb 2, log 2, and 1.

Define m = bλc. Then λ < m+1. Choose ε > 0 so that λ+ ε < m+1. Thenfor k �, we have |zk|−m−1 ≤ k−(m+1)/(λ+ε) for k � 1. Thus

∑∞k=1|zk|−m−1 <

∞. It follows that

P (z) =

∞∏k=1

(1− z

zk

)e∑mk=1( z

zk)l

converges uniformly and absolutely on compact sets. Therefore the genus ofP (z) ≤ m ≤ λ.

We can now write f = ehP (z) for some entire function h. It suffices to showthat h is a polynomial of degree at most m. Let us apply Poisson-Jensen to

Math 213a Notes 59

fr(z) = f(rz). For a fixed z and |z| < r,

log|f(z)| =N∑k=1

log∣∣∣r−1z − r−1zk

1− r−2zzk

∣∣∣+1

2π

∫ 2π

0

<eiθ + r−1z

eiθ − r−1zlog|f(reiθ)|dθ

=

N∑k=1

log∣∣∣rz − rzkr2 − zzk

∣∣∣+1

2π

∫ 2π

0

<reiθ + z

reiθ − zlog|f(reiθ)|dθ.

Note that if F is a holomorphic function, then

F ′

F=

∂

∂z(logF ) =

∂

∂z(logF + log F ) = 2

∂

∂zlog|F (z)|.

Thus

f ′

f= 2

∂

∂zlog|f(z)|

= 2∂

∂z

N∑k=1

log∣∣∣rz − rzkr2 − zzk

∣∣∣+ 2∂

∂z

∫ 2π

0

<reiθ + z

reiθ − zlog|f(reiθ)|dθ

=

N∑k=1

( 1

z − zk+

zkr2 − zzk

)+

1

2π

∫ 2π

0

2reiθ

(reiθ − z)2log|f(reiθ)|dθ.

Because f = ehP (z),

f ′

f= h′ +

∞∑k=1

( 1

z − zk+ polynomial of degree ≤ m− 1

),

with the sum converging uniformly and absolutely. Then

∂m

∂zmf ′

f= h(m+1) + (−1)mm!

∞∑k=1

1

(z − zk)m+1.

Let me finish next time.

Math 213a Notes 60

18 November 1, 2016

Because

f ′

f(z) =

∑|zk|≤r

{ 1

z − zk+

zkr2 − zzk

}+

(m+ 1)!

2π

∫ 2π

0

2r

(reiθ − z)2log|f(reiθ)|dθ,

differentiating m times we get

dm

dzmf ′

f(z) = (−1)mm!

∑|zk|≤r

1

(z − zk)m+1+m!

∑|zk|≤r

zm+1k

(r2 − zzk)m+1

+(m+ 1)!

2π

∫ 2π

0

2r

(reiθ − z)m+2log|f(reiθ)|dθ.

If suffices to show that the two last terms tned to 0 as r → ∞, since thenh(m+1)(z) ≡ 0.

Recall that m = bλc. We have∣∣∣∣ ∑|zk|≤r

zm+1k

(r2 − zzk)m+1

∣∣∣∣ ≤ ∑|zk|≤r

∣∣∣ 1

( r2

zk− z)m+1

∣∣∣ ≤ N(r)1

(r − |z|)m+1.

Since N(r) ≤ rλ+ε and we can set λ+ ε < m+ 1, we get that∑|zk|≤r(r

2/zk −z)−m−1 tends to 0 as r →∞. For the next term,∣∣∣∣∫ 2π

0

2r

(reiθ − z)m+2log|f(reiθ)|dθ

∣∣∣∣ ≤ Cr−m−1 logM(r)

for r � 1 and again choosing ε > 0 such that λ+ ε < m+ 1 we see that it tendsto 0 as r →∞. Therefore h(m+1)(z) ≡ 0 and so h is a polynomial of degree atmost M .

We have defined m = bλc and from this had deduced the uniform andabsolute convergence on compact sets of

∞∏k=1

(1− z

zk

)e∑ml=1( z

zk)l.

This shows that the genus of the canonical Weierstrass product for f is at mostm. If so, in the previous argument replace m in the definition of the Weierstrassproduct by m1 ≤ m with m1 the genus of the Weierstrass product for f . Thisfinishes the proof of Hadamard’s theorem.

18.1 Poisson’s formula

Suppose u(x, y) is a harmonic function defined on some neighborhood of clos(∆).Recall that this means that u is C2 and( ∂2

∂x2+

∂2

∂y2

)u(x, y) = 0.

Math 213a Notes 61

For 0 < r < 1 define

Pr(θ) = <{1 + reiθ

1− reiθ}

=1

2

{1 + reiθ

1− reiθ+

1 + re−iθ

1− re−iθ}

=1− r2

1 + r2 − 2r cos θ.

Then Poisson’s formula states that for z = x+ iy = reiθ ∈ ∆,

u(x, y) =1

2π

∫ 2π

0

Pr(θ − ϕ)u(eiϕ)dφ.

Lemma 18.1. There exists a holomorphic function f(z) defined on a neighbor-hood of clos(∆) such that <f(x+ iy) = u(x, y).

Proof. Consider the 1-form

∂u

∂xdy − ∂u

∂ydx.

Therefore there exists a real valued C2 function v(x, y) on a neighborhood ofclos(∆) such that dv = (∂u/∂x)dy− (∂u/∂y)dx. Then by the Cauchy-Riemannequations f(z) = u(x, y) + iv(x, y) is holomorphic.

Corollary 18.2. Real valued harmonic functions on an open set in R2 do nothave a maximum unless they are constant.

Proof. If not, by scaling and translation we may suppose u is defined on clos(∆)and the maximum occurs in ∆. Write u = <f as in the lemma and defineF (z) = eu(z). Then |F (z)| = eu(x,y) cannot have a maximum. This shows thatu cannot have a maximum, unless u is a constant.

As before, suppose u is harmonic on a neighborhood of clos(∆), and u = <fwith f holomorphic. Define F (z) = ef(z). Write z = x + iy = reiθ ∈ ∆. ByPoisson-Jensen applied to F and r = 1,

u(x, y) = log|F (z)| = 1

2π

∫ 2π

0

<{eiϕ + reiθ

eiϕ − reiθ}u(eiϕ)dϕ

=1

2π

∫ 2π

0

Pr(θ − ϕ)u(eiϕ)dϕ.

This argument, applied to ur(x, y) = u(rx, ry), implies Poisson’s formula forharmonic functions u(x, y) defined on ∆, provided it extends continuously toclos(∆) by letting r ↘ 1.

More generally, this applies if u is harmonic on ∆ which have Lp boundaryvalues on ∂∆ (p ≥ 1) or even distribution boundary values, or even hyperfunc-tion boundary values.

Math 213a Notes 62

18.2 Normal family

Theorem 18.3 (Riemann mapping theorem). Suppose U ⊆ C is open, con-nected, and simply connected with U 6= C. Then U is biholomorphic to ∆.

Suppose U ⊆ C is open connected, and let F be a family (i.e., set) ofholomorphic functions on U .

Definition 18.4. F is a normal family if every sequence {fn} in F has asubsequence which converges locally uniformly. (Note that in that case thelimit is holomorphic.)

Lemma 18.5. The following two conditions are equivalent:

(a) F is locally uniformly bounded.

(b) F is a normal family.

This is a special case of Ascoli’s theorem.

Proof. Let us first prove (b) ⇒ (a). Suppose F satisfies (b) but not (a). Thenthere exists z0 ∈ U such that F is not uniformly bounded on any neighborhoodof z0. Hence for n ∈ N there exist zn ∈ U and fn ∈ F such that |fn(zn)| ≥ nand zn → z0. Then {fn} cannot have a subsequence which converges uniformlyon any neighborhood of z0.

For (a) ⇒ (b), suppose F satisfies (a). We claim that F is locally uniformlyequicontinuous. Suppose z0 ∈ U . Choose r > 0 and M > 0 such that z ∈∆(z0, r) implies z ∈ U and |f(z)| ≤ F for any f ∈ F , with clos ∆(z0, r) ⊆ U .Now suppose 0 < ρ < r and z1, z2 ∈ ∆(z0, ρ). Then

|f(z1)− f(z2)| ≤ 1

2π

∫∂∆(z0,r)

∣∣∣ 1

ζ − z1− 1

ζ − z2

∣∣∣Mrdθ ≤ Constr,ρ2π

M |z2 − z1|,

by Cauchy’s integral formula. Then F is uniformly equicontinuous on ∆(z0, ρ).

Math 213a Notes 63

19 November 3, 2016

Let U ⊆ C be open, and F be a family of holomorphic functions on U . We haveshown that if F is normal then F is locally uniformly bounded, and that if Fis locally uniformly bounded, then F is locally uniformly equicontinuous. Weneed to show that if F is locally uniformly bounded, then it is normal.

Choose a countable dense subset {sk} ⊆ U . Given a sequence of functionsin F , choose a subsequence that converge at s1 (which exists since {|f(sk)| :f ∈ F} is bounded). Then choose a subsequence of the first subsequence whichconverges also at s2, and continue. Take the diagonal subsequence and call it{fn}. By construction, for each k, fn(sk) converges. Let some arbitrary z0 ∈ Ube given, and choose r > 0 such that clos(∆(z0, r)) ⊆ U . We shall show that{fn} converges uniformly on ∆(z0, r), since z0 was arbitrary, this will implythat {fn} converges locally uniformly.

Let ε > 0 be given. Then by local uniform equicontinuity, there exists a δ > 0such that z1, z2 ∈ clos(∆(z0, r)) and |z1 − z2| < δ implies |f(z1)− f(z0)| < ε/3for all f ∈ F . The disks ∆(sk, δ) cover clos(∆(z0, r)). So by compactness,finitely many ∆1≤j≤m∆(skj , δ) cover clos(∆(z0, r)). We know that {fn(skj )}converges. So for these finitely many skj , we can choose N such that m,n ≥ Nand 1 ≤ j ≤ m implies |fm(skj ) − fn(skj )| < ε/3. Now suppose z ∈ ∆(z0, r).Then there exists j such that |z − skj | < δ. Then for m,n ≥ N ,

|fm(z)− fn(z)| ≤ |fm(z)− fm(skj )|+ |fm(skj )− fn(z)|+ |fn(skj )− fn(z)| < ε.

So I have proved Lemma 18.5.

19.1 Riemann mapping theorem

Riemann mapping theorem. Let U ⊆ C be open, connected, simply con-nected, and U 6= C. Then U ∼= ∆.

Proof. After translating U if necessary, we may suppose 0 /∈ U . Then log z hasa well-defined holomorphic determination on U because U is simply connected.Therefore z 7→

√z = elog z/2 also has a well-defined holomorphic determination

on U . By the open mapping theorem, U1 = image of U under z 7→ z1/2 is openand U ∼= U1 since w 7→ w2 is the inverse. If w ∈ U1 then −w /∈ U1 becausew2 ∈ U has a “unique” square root. Becuase U1 is open, there exist a, δ > 0such that U1 ∩ ∆(a, δ) = ∅. Let U1 be the image of U1 under z 7→ (z − a)−1.Then U ∼= U1

∼= U2, and U2 is a bounded set. Translating and scaling U2 allowsus to find 0 ∈ U3 ⊆ ∆ such that U3

∼= U2. So after all this, I can assume withoutloss of generality that 0 ∈ U and U ⊆ ∆. Of course, U is open, connected, andsimply connected.

Define

F = {f : U → ∆ : f(0) = 0, f is one-to-one , f ′(0) ∈ R>0}.

Clearly F 6= ∅ because the identity is an element, and by the lemma, F is anormal family. Let M = sup{f ′(0) : f ∈ F} > 0. Choose a sequence in F

Math 213a Notes 64

such that the derivative at 0 converges to M , and go to the subsequence thatconverges locally uniformly. The conclusion is that there exists a {fn} ⊆ Fsuch that f ′n(0) → M and fn → f locally uniformly. Clearly f(0) = 0 andf ′(0) = limn→∞ f ′n(0) = M . So we need to show that f is one-to-one andf : U → ∆.

Suppose that z0 ∈ U and w0 = f(z0). Because f is non-constant, thereexists a δ > 0 such that f(z)−w0 6= 0 on clos(∆(z0, δ))\{z0}, because the zerosare isolated. Let ε = minζ∈∂∆(z0,δ)|f(z) − w0|, and let fn → f be uniform on∂∆(z0, δ). Choose N such that n ≥ N implies |fn(z)− f(z)| < ε on ∂∆(z0, δ).Then

|(fn(z)− w0)− (f(z)− w0)| < ε ≤ |f(z)− w0|

for z ∈ ∂∆(z0, δ) and n ≥ N . By Rouche, there exists a zn ∈ ∆(z0, δ) such thatfn(zn) = w0 for n ≥ N . Suppose there exists z0 ∈ U such that z0 6= z0 andf(z0) = w0 = f(z0). Then apply the previous argument with small enough δ forboth z0 and z0, and conclude that there exist zn ∈ ∆(z0, δ), zn ∈ ∆(z0, δ) suchthat fn(zn) = wn = fn(zn). Since δ is small enough, ∆(z0, δ)∩∆(z0, δ) 6= 0 andso zn 6= zn. This contradicts the injectivity of fn. Therefore f ∈ F .

So we have constructed f ∈ F such that f ′(0) = M ≥ h′(0) for everyh ∈ F . If U = ∆, then there is nothing to prove. Suppose a ∈ ∆ \ U . DefineS(z) = (z − a)/(1 − az). Then S(a) = 0 and |S(eiθ)| = 1 and so Σ : ∆ ∼= ∆.Then S ◦ f : U → ∆ is one-to-one and S ◦ f(z) 6= 0 for all z ∈ u. Arguing asbefore, we conclude that S ◦ f has a well-defined holomorphic square root, i.e.,h : U → ∆ such that S ◦ f = h2. Write σ for the squaring map: σ(z) = z2.Then σ ◦ h = S ◦ f and so f = S−1 ◦ σ ◦ h. Since f and S are one-to-one, so ish.

Define T : ∆ ∼= ∆, T (z) = α(z− b)/(1− bz), where b = h(0) and α is chosenso that d

dzT ◦ h(0) ∈ R>0. (Note that the derivative cnanot be 0 because h andS are one-to-one.) We have

f = S−1 ◦ σ ◦ T−1(T ◦ h).

We know that T ◦ h ∈ F because T ◦ h(0) = 0. So ddzT ◦ h|z=0 ≤ f ′(0) = M .

On the other hand, S−1 ◦ σ ◦ T−1 : ∆ → ∆ and 0 7→ 0. Also it is clearly nota multiple of z. So by Schwartz’s lemma, |d/dz(S−1 ◦ σ ◦ T−1)|0| < 1. Thiscannot be possible, and thus f is a biholomorphism between U and ∆.

Math 213a Notes 65

20 November 8, 2016

Let C ⊆ R2 be a Jordan curve, i.e., the image of ∂∆ = R/2πZ under an injectivecontinuous map. Then the inverse map is continuous because images of closedsets are images of compact sets and hence closed. Therefore the original mapsis a homeomorphism. The Jordan curve theorem then states that if C ⊆ R2

is a Jordan curve, then R2 \ C has two connected components, one of which isunbounded, and C is the boundary of each of these. Assume this.

20.1 Caratheodory’s theorem

Let U ⊆ C be open, connected, simply connected, and bounded. Suppose alsothat ∂U is a Jordan curve. Let F : ∆ ∼= U be the essential unique biholomor-phism guaranteed by the Riemann Mapping Theorem.

Theorem 20.1 (Caratheodory). In this construction, F extends to a contin-uous map F : clos ∆ → closU , whose restriction to ∂∆ is a homeomorphismonto the Jordan curve ∂U .

For 0 < r < 1, let Cr = ∆ ∩ ∂∆(1, r). Let γr = F (Cr). Now let usparameterize Cr as z = 1− reiθ for −θr < θ < θr. Explicitly, θr = cos−1(r/2).Composing this parametrization with F , we get a parametrization of γr whichis real analytic. Therefore γr has a well-defined length `(γr) which may be +∞.

Lemma 20.2. The function r 7→ `(γr) is Lebesgue integrable and∫ 1

0`(γr)

2/rdr <∞.

Proof. The function r 7→ `(γr) is the pointwise limit of the length of the F -image of the portion of Cr parametrized by −(1− 1/2n)θr < θ < (1− 1/2n)θrwhich is continuous as a function of r. So r 7→ `(γr) is measurable.

F is conformal with scaling factor |F ′(z)| at z. So

`(γr) =

∫ θr

−θr|F ′(1− reiθ)|rdθ

where +∞ is allowed as value. Then∫ 1

0

`(γr)2

rdr =

∫ 1

0

r

(∫ θr

−θr|F ′(1− reiθ)|dθ

)2

dr

≤∫ 1

0

(∫ θr

−θr|F ′(1− reiθ)|2

∫ θr

−θr1dθ

)dr

≤ π∫ 1

0

∫ θr

−θr|F ′(1− reiθ)|2rdθdr ≤ π · (area of U) <∞.

Note that∫ δ

0dr/r = +∞ for every δ > 0. Thus `(γr) cannot be bounded

away from 0 on any interval (0, δ) with δ > 0.

Math 213a Notes 66

Corollary 20.3. There exists a sequence 1 > r1 > r2 > · · · > 0 such that∞ > `(γr1) > `(γr2) > · · · with rn → 0 and `(γrn)→ 0.

Lemma 20.4. If `(γr) < ∞ for some r, then F extends continuously to Cr =clos(Cr).

Proof. If θn ↗ θr then F (1− reiθn) converges in C and the limit is independentof a particular choice of θn. The same is true for θn ↘ −θr.

With rn chosen as before, let

an = limθ↘−θr

F (1− rneiθ), bn = limθ↗θr

F (1− rneiθ)

with γn for γrn . Observe that F : ∆ → U is a homeomorphism, and thus Fextends to a homeomorphism of 1-point compactification. Therefore an ∈ bn ∈∂U . If we know the statement of the theorem, an would be different from bn.But we don’t know this yet.

Lemma 20.5. |an− bn| → 0 as n→∞. For n� 1, either an = bn or an 6= bnand ∂U − {an, bn} has two connected components, exactly one of which, call itδn, satisfies diam(δn) < diam(∂U).

Proof. Let us parametrize ∂U as R/2πZ 3 θ 7→ w(θ). Then both θ 7→ w(θ) andits inverse are continuous, hence uniformly continuous. Note that |an − bn| ≤`(γn) → 0. Thus as n → ∞, the points an, bn ∈ ∂U get infinitely close. Hencew−1(an) − w−1(bn) ∈ R/2πZ tends to 0 as n → ∞. By uniform continuity ofω, δn, which is the image under w of “w−1(an) − w−1(bn)” will have diametertending to zero as n→∞.

There are two possibilities: an = bn or an 6= bn. Define δn = {an} = {an, bn}if an = bn and define δn = δ ∪ {an, bn} if an 6= bn. Also let γn = γn ∪ {an, bn}.

We know that γn and δn are one-to-one images of open intervals of undercontinuous maps F,w. Also γn ⊆ U and δn ⊆ ∂U with {an, bn} the intersectionof the closures of γn and δn. The conclusion is that

γn ∪ δn =

{γn if an = bn

γn ∪ δn

is a Jordan curve. Let Un denote the interior o γn ∪ δn. Note that diamUn =diam ∂Un ≤ diam δn + `(γn). This goes to 0 as has been shown already.

Lemma 20.6. For n� 1, Un = F (∆ ∩∆(1, rn)).

Proof. δn ⊆ ∂U and γn ⊆ U . Thus γn ∪ δn is idsjoint from C \ clos(U). ThusC\clos(U) is in the exterior of γn∪ δn. Now suppose w0 ∈ ∂U \ δn. Now supposew0 ∈ ∂U \ δn. Then w0 has an open neighborhood in C which is disjoint fromδn and γn. Thus the exterior of γn ∪ δn contains both points in U and points inC\clos(U). That means the exterior of γn∪ δn contains C\clos(U) and ∂U \ δn,some points in U .

Math 213a Notes 67

∆ \ Crn has two connected components. Thus its F -image also has twoconnected components, both of which are disjoint from γn ∪ δn because γn =F (Crn) and δn ⊆ ∂U . So each of these can be contained either in the interiorof exterior of γn ∪ δn. Then by exclusion one of the components must be theinterior of γn ∪ δn and the other contained in the exterior.

As n→∞, we have⋃

(∆\clos ∆(1, rn)) = ∆ and so⋃F (∆\clos ∆(1, rn)) =

U . But diam(γn ∪ δn) = diamUn → 0. Therefore we conclude that F (∆ ∩∆(1, rn)) = int(γ ∪ δn) for n� 1.

Note that closUn = Un ∪ γn ∪ δn is compact, and decreases as n increases.Thus

⋂n�1 closUn 6= ∅ by the finite intersection property. We know that

diam closUn → 0.

Corollary 20.7. The set⋂r>0 clos(F (∆ ∩∆(1, r))) consists of a single point.

Math 213a Notes 68

21 November 10, 2016

Let us recall where we are. U ⊆ C is an open, connected, simply connected,bounded domain with ∂U a Jordan curve. Let F : ∆→ U be a biholomorphism.We have shown that there exists a sequence rn with 0 < rn < 1 such thatrn → 0 as n→∞ and diamF (∆ ∩∆(1, rn)) = diam(clos(F (∆ ∩ (1, rn))))→ 0as n→∞. Then

⋂n clos(F (∆ ∩∆(1, rn))) consists of a single point whith lies

on ∂U .In this argument, we can replace 1 by any α ∈ ∂∆. Let ∩r>0 closF (∆ ∩

∆(α, r)) = {a(α)} so that a(α) ∈ U . Define F : clos ∆→ closU such that

F (z) =

{F (z) if z ∈ ∆,

a(z) if z ∈ ∂∆.

Lemma 21.1. F is continuous.

Proof. For z ∈ ∆, this is clear. Suppose α ∈ ∂∆. We know that (clos δ)∩∆(α, r)is a neighborhood for the topology of clos ∆ at α. We also know that

diam closF (∆ ∩∆(α, r))→ 0.

So it suffices to show that F (clos ∆ ∩ ∆(α, r/2)) ⊆ closF (∆ ∩ ∆(α, r)). It isobvious that F (∆ ∩∆(α, r/2)) ⊆ closF (∆ ∩∆(α, r)). Thus it suffices to show:

F (∂∆ ∩∆(α, r/2)) ⊆ closF (∆ ∩∆(α, r)).

Suppose β ∈ ∂∆(α, r/2). Then ∆ ∩∆(β, r/2) ⊆ ∆(α, r). We know that a(β) ∈closF (∆ ∩∆(β, r/2)) ⊆ closF (∆ ∩∆(α, 1)). This shows that F is continuous.

Corollary 21.2. F : clos ∆→ closU is surjective.

Proof. By continuity, F (clos ∆) is compact, and is contained in closU but con-tains U . So F (clos ∆) = closU .

By construction, F (∂U) ⊆ ∂U and F (∆) = F (∆) = U .

Corollary 21.3. F (∂∆) = ∂U .

To complete the proof of Caratheodory’s theorem, it suffices to show:

Lemma 21.4. F : ∂∆→ ∂U is one-to-one.

Proof. Suppose for some α, β ∈ ∂∆, α 6= β but F (α) = F (β). Define

`α = F ({tα : 0 ≤ t < 1}), `β = F ({tβ : 0 ≤ t < 1}).

Then `α = clos(`α) = `α ∪ {a(α)} and `β = clos(`β) = `β ∪ {a(β)}. Defineγ = `α ∪ β = `α ∪ `β . This is a Jordan curve in closU whose intersection with

Math 213a Notes 69

∂U consists of a single point {a(α)} = {a(β)}. Let Uγ denote the interior ofthis Jordan curve.

We know that C \ closU is open and disjoint from γ. Thus C \ closU ⊆exterior of γ. Any point p in ∂U \{a(α)} is disjoint from γ, and every neighbor-hood of p contains points from C \ closU as well as points in U . In particular,∂U \ {a(α)} ⊆ exterior of γ. We know that ∆ \ ({tα : 0 ≤ t < 1} ∪ {tβ : 0 ≤t < 1}) has two connected components. By exclusion, the F -imeage of one ofthese must be Uγ and the F -image of the other must lie in the exterior of γ. Sothe F -iamge of the one of the two components of ∂ \ {α, β} must get mappedto a(α) = a(β) because: suppose δ is a point in this component of ∂∆ \ {α, β}that goes in the Uγ direction. Then {tδ : 0 ≤ t < 1} is a curve in ∆ which denseto a(δ) as t→ 1. But a(δ) ∈ closUγ ∩ ∂U = {a(α)}.

This shows that if F : ∂∆ → ∂U fails to be one-to-one, then it is constanton some open nonempty /θ-interval I of {eiθ}. We are free to tranlate U andmay suppose without loss of generality that F -image of this open interval is 0.Then F (z) for z ∈ ∆ cannot be zero. By Jensen’s formula, for 0 < r < 1,

log|F (0)| = 1

2π

∫ 2π

0

log|F (reiθ)|dθ.

By uniform continuity of F , for θ ∈ I, F (reiθ) → 0 uniformly in θ. Thenit follows that the right hand side of the interval is −∞ as r → 1. ThenF (0) = 0.

This finishes the proof of Caratheorody.

21.1 Riemann surfaces again

Recall that the only connected, simply connected Riemann surfaces are P, C, H.The surface P doesn’t cover any other surface, because every automorphism hasa fixed point. The next one C is properly covers only C/Z ∼= C∗ or C/L whichis topologically a 2-torus. Aside from the “few” cases covered by C, and apartfrom P, connected Riemanns surfaces arise as covered by H: the surface Γ Hwhere Γ = π1(Γ H is determined as a subgroup of Aut(H) ∼= SL(2,R)/{±1}up to inner automorphism. Then Γ ⊆ SL(2,R) with {±1} ⊆ Γ arises as afundamental group of a Riemann surface if and only if

(1) Γ ⊆ SL(2,R) is discrete,

(2) Γ contains no elliptic elements.

For N ∈ N, define the principal congruence subgroup of level N as

Γ(N) = {γ ∈ SL(2,Z) : γ ∼= 12×2 mod N}.

Note that SL(2,Z/nZ) is a well-defined finite group (but GL(2,Z/nZ) is notunless N is prime). Then Γ(N) can be thought as a kernel of SL(2,Z) →SL(2,Z/nZ). So Γ(N) ⊆ SL(2,Z) is a normal subgroup of finite index.

Math 213a Notes 70

Lemma 21.5. For N ≥ 2, Γ(N) contains no elliptic elements. (But for N = 2only, −1 ∈ Γ(N).)

Proof. Suppose γ ∈ Γ(N) is elliptic. Then γ must have eigenvalues eiθ forθ /∈ πZ. So tr γ = 2 cos θ for θ ∈ πZ, and in particular, tr γ ∈ (−2, 2). Buttr γ ∼= 2 (mod N). So the only possibilities are N ≤ 3 and tr γ ∈ {−1, 0, 1}.

For N = 3, tr γ = −1 so γ = ( a bc d ). Write a = 1 + 3k and d = −2 − 3k.Then

1 = ad− bc = −2− 9k − 9k2 − bc ≡ −2 (mod 9).

This is impossible.The remaining case is N = 2. In this case, write a = 1+2k and d = −1−2k.

Then

1 = ad− bc = −1− 4k − 4k2 − bc ≡ −1 (mod 4)

and we again get a contradiction.

Next time I will prove the following theorem:

Theorem 21.6. The set C\{0, 1} has H as a universal cover with fundamentalgroup Γ(2)/{±1}.

Corollary 21.7 (Little Picard theorem). Any non-constant entire functionomits at most one value.

Proof. Assume that f : C → C \ {0, 1}. (We can assume this because this isgeneral up to translation, rotation, and scaling.) We have a universal coveringof C \ {0, 1} by H. Then by topological monodromy theorem, we can lift thisto get f .

H

C C \ {0, 1}

hol.

f

hol.

f

Compose f with H ∼= ∆. Then we get a bounded entire function, which mustbe constant. Then f is constant.

Math 213a Notes 71


Let Γ ⊆ SL(2,R) be a subgroup.

Definition 22.1. Γ acts property discontinuously on H if every z0 ∈ H hasa neighborhood U such that U ∩ γU 6= ∅ for only finitely many γ ∈ Γ.

You will show in the homework that:

Proposition 22.2. Γ ⊆ SL(2,R) is discrete if and only if Γ acts properlydiscontinuously on H. Also H → Γ\H is a covering map if and only if Γ actsproperly discontinuously without fixed points.

For simplicity, say “Γ acts without fixed points” if γz = z for γ ∈ Γ impliesγ = ±1.

Theorem 22.3. Γ(2)\H ∼= C\{0, 1}, and this is the universal cover of C\{0, 1}with fundamental group Γ(2).

Definition 22.4. A fundamental domain for the action of Γ on H is a closedsubset F ⊆ H such that

(a) ∂F consists of a finite number of smooth curves,

(b) {γ ∈ Γ : γF ∩ F 6= ∅} is finite,

(c)⋃γ∈Γ γF = H.

(d) z1, z2 ∈ F , z1 6= z2, z2 = γz1 for γ ∈ Γ implies z1, z2 ∈ ∂F .

22.1 A fundamental domain for Γ(2)

Let us write

F = {z ∈ C : |<z| ≤ 1, |z − 1/2| ≥ 1/2, |z + 1/2| ≥ 1/2}.

Proposition 22.5. F is a fundamental domain for the action of Γ(2) on H.If z1, z2 ∈ ∂F are distinct but Γ(2)-related, then z2 = −z1.

F

0 1−1

Figure 2: The fundamental domain F

Math 213a Notes 72

Proof. Define

t =

(1 20 1

), s =

(1 0−2 1

), w =

(0 1−1 0

).

Then s, t ∈ Γ(2) and although w ∈ Γ(2), it normalizes Γ(2), i.e., wΓ(2)w−1 =Γ(2). Also w ≡ w−1 mod 12×2 and so w,w−1 have the same action on H.

Suppose z, tz ∈ F . Since tz = z + 2, it follows that <(z) = −1 and <(tz) =+1. Then tz = −z. Because w{|z − 1/2| = 1/2} is the w-image of the circleof radius 1/2, centered at 1/2, so must be a circle or straight line. Because itw(0) =∞ and w(1) = −1, it is a straight line through −1 and perpendicular toR. So

w{|z − 1/2| = 1/2} = {<z = −1} and similarly w{|z + 1/2| = 1/2} = {<z = 1}.

Now

s{z ∈ H : |z − 1/2| = 1/2} = wtw−1{z ∈ H : |z − 1/2| = 1/2}= wt{z ∈ H : <z = −1} = w{z ∈ H : <z = 1}= {z ∈ H : |z + 1/2| = 1/2}.

Also

s{1/2 = 1/2eiθ : 0 < θ < π} = {−1/2− 1/2e−iθ : 0 < θ < π}

and so if s maps z ∈ ∂∆(1/2, 1/2) ∩H to ∂∆(−1/2, 1/2) ∩H, then sz = −z.To complete the proof, we proceed in the following steps:Step 1. Suppose γ ∈ SL(2,Z) with z ∈ H. Then =γz ≤ max{=z, 1/=z}.

To see this, write γ = ( a bc d ) with a, b, c, d ∈ Z and ad− bc = 1. Then

=γz = =az + b

cz + d= =adz + bcz

|cz + d|2=

=z|cz + d|2

.

If c = 0, then d = ±1 and we may assume d = 1 without loss of generality.Then im γz = =z. If c 6= 0, then

=γz ==z

|cz + d|2≤ =z|c|2(=z)2

≤ 1

=z.

Step 2. For z ∈ H, the supremum sup{=γz : γ ∈ Γ(2)} is assumed, i.e.,there is a maximal value.Suppose {γn} ⊆ Γ(2) is a sequence such that =(γnz) → sup{=γz : γ ∈ Γ(2)}.Because =(tkγnz) = =γnz, we may assume that |<γnz| ≤ 1. By Step 1 anda compactness argument, we may suppose—by going to a subsequence—thatγnz → z∞ in H with |<z∞| ≤ 1. By the homework problem, Γ(2) acts propertydiscontinuously without fixed points, so H → Γ(2)\H is a covering map. So z∞has a neighborhood U in H such that γU ∩U = ∅. But then because γnz → z∞,there exist γnz, γmz ∈ U and then (γmγ

−1n )U ∩ U 6= ∅.

Math 213a Notes 73

Step 3.⋃γ∈Γ(2) γF = H.

We know that any Γ(2)-orbit in H contains some point z such that =γz ≤ =zfor all γ ∈ Γ(2), and also |<z| ≤ 1. Then =(sz) ≤ =z and so |2z − 1| ≥ 1.Likewise, applying s gives |2z + 1| ≥ 1 and so z ∈ F .

Step 4. Suppose z, γz ∈ G, with γ 6= ±1 and γ = ( a bc d ) ∈ Γ(2). Then Γis one of the following modulo multiplication by ±12×2: s, s−1, t, t−1, and theidentification by γ is one of those we have described above.Let us first look at the case c = 0. Then a = d = ±1 and we may assumewithout loss of generality that a = d = 1. Hence γ = ( 1 2k

0 1 ) for some k ∈ Z.Because γz = z+ 2k, we have either <z = −1 and k = 1 or <z = 1 and k = −1.Suppose now that c 6= 0 and c = ±2d. Then we may suppose d = 1 and c = ±2and a = 1± 2b.

γz =az + b

cz + d=

az + b

d(1± 2z)=

(1± 2b)z + b

±2z + 1=b(2z ± 1) + z

2z ± 1=

z

1± 2z+ 2k

for k = b/2 ∈ Z. So γz = s∓1z + 2k. By assumption, z, γz ∈ F , and z ∈ Fimplies s±z ∈ {z ∈ H : |z − 1/2| ≤ 1/2} ∪ {z ∈ H : |z + 1/2| ≤ 1/2}. Theconclusion is that z and γz must both be in the boundary. Finally, if c 6= 0 andc 6= 2d, then we claim that |cz + d| > 1. Assume this for now. Then

=γz ==z

|cz + d|2< =z.

If γ−1 lands in one of the previous two conditions, then we are done by symmetry.So we may assume that both γ and γ−1 is in this case. Then =z < =γz and weget a contradiction. Thus it suffices to prove our claim |cz + d| > 1. Supposethat |cz + d| ≤ 1. Then |z ± d/|c|| ≤ 1/|c|. We know that c is a nonzero eveninteger. If c = ±2, then |z ± d/2| ≤ 1/2. Since c 6= ±2d, we have |d| > 1, butthis gives a contradiction. If c = 2k with |k| > 1, in which case |z±d/|c|| ≤ 1/|c|is a semicircle that is disjoint from F ∈ C.

This completes the proof of the proposition, and in particular, {γ ∈ Γ(2) :γF ∩ F 6= ∅} is finite.

Math 213a Notes 74


Last time, we have proved:

Proposition 23.1. F is a fundamental domain for the action of Γ(2) on H.If z, γz ∈ F and z 6= γz, then z, γz ∈ ∂F and γz = −z.

23.1 Universal covering space for the thrice-punctured sphere

Theorem 23.2. There exists a holomorphic covering map Φ : H → C \ {0, 1}with covering group Γ(2). Moreover, Φ can be chosen so that limz→0 Φ(z) = 0,limz→1 Φ(z) = 1, and limz→∞Φ(z) =∞.

We have already seen that this implies the “Little Picard theorem”.

Proof. Define F+ = F ∩{<z ≥ 0} and F− = F ∩{<z ≤ 0}. Then F = F+∪F−and F+ ∩ F− = iR>0. Also F− is the image of F+ under z 7→ −z.

Define ϕ+ : (clos. of F+ ⊆ P)→ (clos. of H ⊆ P) as a homeomorphism andholomorphic on interiors—such a map exists by Riemann–Caratheodory—suchthat ϕ+(0) = 0, ϕ+(1) = 1, and ϕ+(∞) = ∞, which can be done becauseAut(H) acts on ∂H = R ∪ {∞} in a triply transitive manner. Then the onlyway they can fit with the right orientation is

ϕ+(iR>0) = R<0, ϕ+(1 + iR>0) = R>1,

ϕ+(F+ ∩ |z − 1/2| = 1/2) = {x ∈ R : 0 < x < 1}.

Now define ϕ : (clos. of F ⊆ P)→ (clos. of H ⊆ P) as

ϕ(z) =

{ϕ+(z) z ∈ clos. of F+ ⊆ Pϕ+(−z) z ∈ clos. of F− ⊆ P.

We know that ϕ is well-defined and holomorphic on (the interior of) F possibleon iR>0.

We claim that ϕ is holomorphic even on points of iR>0. The easiest way todo it is to integrate over rectangles. It suffices to show that

∫∂Rϕdz = 0 for

every rectangle R in the interior of F . Suppose R passes thorough the imaginaryaxis. Define R+

n = R ∩ {<z > 1/n} and R−n = R ∩ {<z < −1/n}. Since ϕ iscontinuous, it is uniformly continuous on a compact set. Thus∫

∂R

ϕdz =

∫∂R+

ϕdz +

∫∂R−

ϕdz = limn→∞

∫∂R+

n

ϕdz + limn→∞

∫∂R−n

ϕdz = 0.

So ϕ is holomorphic on the F .By construction, ϕ : F → C \ {0, 1} is surjective because we already have

things not on the real axis, and also we have R<0, (0, 1), and R>1. Moreover itis injective except possible on the boundary of F in H, where it is 2-to-1, butthe two inverse images of any point are related by the action z 7→ −z, i.e., by

Math 213a Notes 75

Γ(2). So ϕ : F → C \ {0, 1} is surjective, one-to-one module the identificationson ∂F effected by the action of Γ(2).

Define Φ : H → C \ {0, 1} by

Φ(z) = {ϕ(γ−1z) : γ ∈ Γ(2), z ∈ F}.

This is well-defined, because if γ1z = γ2z with γ2 6= ±γ1 for z ∈ F , then γ−12 γ1

must preserve z, which must lie on ∂F , and γ−12 γ1z = −z1. By construction, Φ

is continuous (as composition of continuous function which agree on overlaps),Γ(2)-invariant, and 1-to-1 modulo the action on Γ(2). Also Φ is holomorphicexcept possibly on points in H which are either Γ(2)-conjugate to points in{iR>0 ± 1} and points in w{iR>0 ± 1}. These two cases can be treated exactlylike the earlier case on iR>0.

This “showing holomorphy using continuity” is called the reflection prin-ciple. If f is continuous and holomorphic except for on a line, then it is holo-morphic on the line.

23.2 Invariant form of Schwartz lemma

Theorem 23.3 (Big Picard theorem). In any neighborhood of an essentialsingularity, a holomorphic function assumes every value, with at most one ex-ception, infinitely often.

For example, f(z) = e1/z has an essential singularity at z = 0. Big Picardimplies little Picard. This is because any entire function that is not a polynomialcan be thought of having an essential singularity at z =∞.

Let H and ∆ have invariant metrics

dx2 + dy2

(1− |z|2)2on ∆,

dx2 + dy2

y2on H.

Let d∆ and dH be the corresponding distance functions. They are Aut(∆)(resp.Aut(H))-invariant.

Theorem 23.4 (Invariant form of Schwarz lemma). Suppose F : H → H isholomorphic and z1, z2 ∈ H with z2 6= z1. Then dH(F (z1), F (z2)) ≤ dH(z1, z2)and the inequality is strict unless F is a biholomorphic. Equivalently, the samething holds for F : ∆→ ∆.

Proof. Because dh is Aut(H)-invariant, it suffices to prove that if F : H → H isholomorphic and F (0) = 0 and z ∈ ∆, then d∆(F (z), 0) ≤ d∆(z, 0). By radialsymmetry, we have d∆(z, 0) = d∆(|z|, 0) and d∆(F (z), 0) = d∆(|F (z)|, 0).

We know that d(||z||, 0), for z ∈ ∆, is strictly increasing in terms of |z|.Then what we must prove is |F (z)| ≤ |z| and that this is inequality is strictunless F is a biholomorphism. This is precisely the Schwartz lemma.

Theorem 23.5 (Ahlfors). Suppose M is a hermitian manifold, whose holo-morphic sectional curvatures are negative and uniformly bounded away from 0.Let F : ∆ → M be holomorphic. Then there exists a constant ccurv such thatdM (F (z1), F (z2)) ≤ ccurvd∆(z1, z2).

Math 213a Notes 76


It’s been two weeks since last class, so let remind you where we stood. Weproved the theorem:

Theorem 24.1. There exists a holomorphic covering H → C \ {0, 1} withcovering group Γ(2).

We further constructed a surjective holomorphic map H → C \ {0, 1} whichdrops to a one-to-one map Γ(2)\H → C \ {0, 1}. We did not exactly show thatH → C \ {0, 1} is a covering, but a homework problem implies this. We alsoproved the invariant form of Schwartz lemma.

24.1 Big Picard theorem

The next business is to prove the big Picard theorem.

Theorem 24.2 (Big Picard theorem). In any neighborhood of an essentialsingularity a holomorphic function takes on any value infinitely often, with atmost one exception.

It suffices to show that if f : ∆∗ → C \ {0, 1} is holomorphic then f has aremovable singularity or a pole at the origin. The reason this suffices is becausewe can shrink the disc as much as we like.

Consider the map p : H → ∆∗ with p(z) = e2πiz. This is a covering withcovering group generated by z 7→ z + 1. Suppose f : ∆∗ → C \ {0, 1}. Then weget a commutative diagram

H H

∆∗ C \ {0, 1}

p

f

π

f

by lifting the map f to f . Suppose z0 ∈ H. Then p(z0 + 1) = p(z0) sof(z0 + 1) = γf(z0) for some γ ∈ Γ(2). But γ can be chosen independently ofz0, at least locally, because Γ(2) is discrete, and hence globally by holomorphy.Thus there exists a γ ∈ Γ(2) such that f(z + 1) = γf(z) for all z ∈ H.

Before finding out what γ is, let us think how unique γ is. We may replacef by z 7→ γ1f(z) for any fixed γ1 ∈ Γ(2). This has the effect of replacing γ byγ1γγ

−11 . More than that, Γ(2) is normalized by SL(2,Z). So the finite group

SL(2,Z)/Γ(2) acts on C \ {0, 1}.3 Replacing f by z 7→ γ1f(z) has the effect ofreplacing f by A ◦ f where A : C \ {0, 1} → C \ {0, 1} is induced by γ1. Sothe conclusion is that in proving the theorem, we may replace γ by γ1γγ

−11 for

γ1 ∈ SL(2,Z).

3We don’t need to know that this action is, but this finite group is S3 and it acts onP \ {0, 1,∞} by permuting these three points in a fractional linear manner.

Math 213a Notes 77

Recall that H ∼= G/K with G = SL(2,R) and K = SO(2), which is theisotropy subgroup of SL(2,R) at i ∈ H. So we can choose gn ∈ SL(2,R) forn ∈ Z>0 such that f(in) = gn · i. Then by the Schwartz lemma,

dH(f(in+ 1), f(in)) ≤ dH(in+ 1, in) =

∫ 1

0

dt

n=

1

n

while on the other hand,

d(f(in+ 1), f(in)) = dH(γf(in), f(in)) = dH(γgn · i, gn · i) = dH(g−1n γgni, i).

The conclusion is hat dH(g−1n γgni, i) ≤ 1/n. So as n → ∞, the element

g−1n γgn ∈ SL(2,R) lies in any open neighborhood of SO(2). The coefficients

of the characteristic polynomial of any g ∈ SL(2,R) are continuous functions ofg, and constant on conjugacy classes. This shows that γ has eigenvalues e±iθ.But Γ(2) contains no elliptic elements. Therefore γ has eigenvalues eiθ withθ ∈ πZ. So γ has eigenvalues both 1 or both −1. Because we are allowed tochange γ by the center, we may suppose without loss of generality that γ isunipotent or the identity.

Case (a) γ = 1.Then f(z + 1) = γf(z) = f(z). That is, we can think f has a holomorphicmap from ∆∗ to H, or equivalently, can lift f to a map π−1 ◦ f : ∆∗ → H.Comparing this with H ∼= ∆, we see that π−1 ◦ f has a removable singularity.

Case (b) γ 6= 1, γ is unipotent.We now that γ has a unique fixed point. Since Γ(2) ⊆ GL(2,Q), the fixed pointmust lie in Q ∪ {∞}. The group SL(2,Z) acts transitively on Q ∪ {∞}. Thisis because γ = ( a bc d ) sends γ(∞) = a

c . Then by the previous remark, we maysuppose γ∞ =∞. Then

γ =

(1 2k0 1

)for k ∈ Z \ {0}.

Let us define a new map h : H → C as h(z) = f(z)− 2kz. Then

h(z + 1) = f(z + 1)− 2k(z + 1) = γf(z)− 2k(z + 1)

= f(z) + 2k − 2k(z + 1) = h(z).

The fact that h is invariant under the fundamental group implies that there isa map h : ∆∗ → C that makes

H C

∆∗ C

h

h

commute.

Math 213a Notes 78

Proposition 24.3. For y � 1 and 0 ≤ x ≤ 1, the function |h(z)| is boundedby a polynomial in y where z = x+ iy.

Assume this for now. We have h(z) = h(t) where t = e2πiz. Then |t| = e−2πy

and so y = (1/2π) log|t|−1. Because e2πiz for 0 ≤ x ≤ 1 covers all of ∆∗,the function |h(t)| is bounded by a polynomial in log|t| for |t| � 1, by theproposition. Then h has a removable singularity at 0. This gives

f(z) = h(z) + 2kz = h(e2πiz) + 2kz

= (h(0) + a1e2πiz + a2e

4πiz + · · · ) + 2kz ∈ H.

So 0 < =f(z) = =h(0) + 2ky +O(e−2πy) for y � 1. It follows that k > 0. Alsoas f(z) = h(0) + 2kz +O(e−2πy). It follows that π−1 ◦ f(e2πiz) = h(0) + 2kz +O(e−2πy) for 0 ≤ x ≤ 1. Then f has a pole of order k at 0.

Proof of proposition 24.3. Because h(z) = f(z) − 2kz, it suffices to show for0 ≤ x ≤ 1, |f(z)| is bounded by a polynomial in y. We have

dH(f(z), i) ≤ dH(f(z), f(i)) + const ≤ dH(z, i) + const

≤ dH(x+ iy, iy) + dH(iy, i) + cosnt ≤ 1

y+ log y + const ≤ log y2

for y � 1. So dH(f(z), i) ≤ dH(iy2, i). Recall that c : H → ∆ with c(z) =(z − i)/(z + i) relates dH to d∆. Then d∆(c(f(z)), ci) ≤ d∆(c(iy2), c(i)). Writef(z) = w ∈ ∆. Then

d∆

(w − iw + i

, 0)≤ d∆

( iy2 − iiy2 + i

, 0)

for y � 1 and |t| � 1. It follows that |(w − i)/(w + i)| ≤ |(y2 − 1)/(y2 + 1)|.Because y � 1, we have∣∣∣y2 − 1

y2 + 1

∣∣∣ =∣∣∣1− y−2

1 + y−2

∣∣∣ = |1− 2y2 + · · ·| ≤ 1− y−2.

Likewise ∣∣∣w − iw + i

∣∣∣ =∣∣∣1− i/w1 + i/w

∣∣∣ =∣∣∣1− 2i

w+ · · ·

∣∣∣ ≥ 1− 4

|w|

for |w| � 1. It follows that |w| ≤ 4y2. Therefore |f(z)| ≤ 4y2 for 0 ≤ x ≤ 1 andy � 1.

Math 213a Notes 79

25 December 1, 2016

25.1 Compact Riemann surfaces coming from CRecall that connected Riemann surfaces with universal cover C are, up to iso-morphism, C, C∗, and C/L where L ⊆ C is a lattice, i.e., a Z-module spannedin an R-basis of C. The first two spaces are not very interesting.

Let L ⊆ C be a lattice. What can be said about holomorphic functions onC/L? Because it is compact, it takes a maximal value at some point. Then wesee that it is constant. But consider meromorphic functions on C/L. These arecalled elliptic functions. Suppose f is an elliptic function, and L = Zω1⊕Zω2

where {ω1, ω2} is an R-basis of C.

Theorem 25.1. Let ζ1, . . . , ζr ∈ C/L be an enumeration of the poles of f .Then ∑

Res f |ξi = 0.

Proof. For a ∈ C, define Fa = {a + t1ω1 + t2ω2 : 0 ≤ t1, t2 ≤ 1}. This is afundamental domain for the action of L on C. We can choose a so that none ofthe poles lie on ∂Fa. Then∑

ζ∈C\L

Res f |ζ =∑η∈Fa

Res f |η =1

2πi

∫∂Fa

f(z)dz.

But because f is periodic, the integral along the opposite sides cancel out.

Let f be an elliptic function. Let ζ1, . . . , ζr be an enumeration of zeros andpoles of f in C/L. Let ni be the order of f at ζi (so n > 0 if f has a zero andn < 0 if f has a pole).

Corollary 25.2.∑ni = 0.

Proof. Apply the theorem to f ′/f .

Corollary 25.3. Suppose f is a non-constant elliptic function. Then the totalnumber of poles in C/L, counted with multiplicity is at least 2.

Proof. Suppose f has a first order pole at ζ ∈ C/L and no other poles. ThenRes f |ζ = 0 and so there are no poles. It follows that f is constant.

Theorem 25.4. C/L1 ∼= C/L2 if and only if L2 = αL1 with α ∈ C∗.

Proof. Suppose C/L1∼= C/L2. Consider the following diagram.

C C

C/L1 C/L2

f

∼=

Math 213a Notes 80

Then f(z + w1) = f(z) + ϕ1 for ω1 ∈ L1 and ϕ1 ∈ L2. That means that f ′ isL1-periodic and so f ′ is constant. This shows that f(z) = αz + b with α ∈ C∗.It follows that L1 = αL2.

Let L = Zω1 ⊕ Zω2 be a lattice. After rescaling, we may suppose that theordered R-basis is positively oriented, i.e., going from ω1 to ω2 involves rotationby an angle θ with 0 < θ < π. So we can write C/L ∼= C/Lτ where τ = ω2/ω1

and Lτ = Z⊕ Z/τ . By construction 1, τ is a positively oriented Z-basis and soτ ∈ H. Thus C/L ∼= C/Lτ for some τ ∈ H.

A natural question is: with τ1, τ2 ∈ H, when are C/Lτ1 ∼= C/Lτ2? By theprevious argument, they are isomorphic if and only if there is an α ∈ C∗ suchthat Z ⊕ Zτ2 = α(Z ⊕ Zτ1). This is when 1, τ2 and α, ατ1 are related by somematrix in SL2(Z). Then

ατ1 = aτ2 + b, α = cτ2 + d.

This is possible if and only if τ1 and τ2 are related by the action of someγ ∈ SL(2,Z) on H. This proves:

Theorem 25.5. The set of connected, compact Riemann surfaces with universalcover C modulo isomorphism is naturally isomorphic to SL(2,Z)\H, with τcorresponding to C/Lτ .

Here, this SL(2,Z) ∼= C when unramified. Historically, this is the first answerto what we now call a “moduli problem”.

25.2 Elliptic curves

Suppose now L is a lattice, where L = Zω1 ⊕ Zω2 with ω2/ω1 ∈ H. Theanswer was given by Weierstrass. Fix L as above, and define the Weierstrass℘-function as

℘(z) =∑

ω∈L\{0}

( 1

(z − ω)2− 1

ω2

)+

1

z2.

This function is a function depending on L.

Theorem 25.6. The series converges locally uniformly on C/L to a meromor-phic function on C, which is L-periodic, hence an elliptic function on C/L withsecond order pole at 0 ∈ C/L and no other poles in C/L.

Note that C/L has a natural structure of an abelian group. Then we cansay that ℘(z) is an even function. ℘′ is also an elliptic function, with pole oforder 3 at 0 ∈ C/L, no other poles, and as a function on C, odd.

Theorem 25.7. The field of elliptic functions on C/L is generated over C by℘ and ℘′. Moreover,

(℘′)2 = 4℘3 − g2℘− g3

with some g2, g3 ∈ C.

Math 213a Notes 81

Define F : C/L→ P2 where F (z) = [1, ℘(z), ℘′(z)].

Theorem 25.8. This gives an isomorphism between C/L and a smooth curvein P2—smooth closed 1-dimensional complex submanifold.

So what is the image? Because (℘′)2 = 4℘3 − g2℘ − g3, the image is theclosure in P2 of the curve y2 = 4x3 − g2x− g3. Let us look at what happens atz = 0. We have

F (z) =[1,

1

z2+ · · · ,− 2

z3+ · · ·

]= [z3, z + · · · ,−2 + · · · ].

So z = 0 corresponds to [0, 0, 1], which is the point at infinity. Therefore F (C/L)intersects the “line at ∞” in P2 at the point [0, 0, 1] and the order of the inter-section is 3. By general nonsense in algebraic geometry, F (C/L) intersects anyline (i.e., copy of P1) in P2 with multiplicity 3.

Recall that C/L has a structure of an abelian group. Can you see that interms of the curve F (C/L) ⊆ P2? The additional law, in terms of this curve, isthat a+ b+ c = 0 if there exists a line π such that π ∩ F (C/L) = {a, b, c} withmultiplicity.

If I had another lecture, I could have proved all of this. But the conclusionis that all elliptic curves (i.e., compact Riemann surfaces, covered by C) are ofthe form: completion in P2 of the affine curve y2 = 4x3−g2x−g3 with additionlaw algebraic.

Index

abscissa, 48absolute convergence, 39argument principle, 21

big Picard theorem, 76

Caratheodory’s theorem, 65Casorati-Weierstrass theorem, 18Cauchy integral formula, 11Cayley transform, 27

Dirichlet series, 47

elliptic function, 79Euler’s constant, 43

fundamental domain, 71fundamental theorem of calculus,

22

Gamma function, 42genus, 40

holomorphic, 11

inverse function theorem, 14

Legendre duplication formula, 46linear fractional transformation, 25Liouville’s theorem, 24

maximum principle, 15Mellin transform, 46meromorphic function, 18Mittag-Leffler theorem, 35

normal family, 62

open mapping theorem, 14, 22order, 21order of growth, 54

Picard theorembig, 75little, 70

Poisson-Jensen formula, 56pole, 17principal congruence subgroup, 69principal part, 16projective space, 22properly discontinuous action, 71

radius of convergence, 9reflection principle, 75residue, 16residue theorem, 19Riemann mapping theorem, 62Riemann surface, 32Riemann zeta function, 48Rouche’s theorem, 21

Schwarz lemma, 15, 75singularity, 17special unitary matrices, 26

uniformization theorem, 32unipolar, 33

Weierstrass ℘-function, 80Weierstrass problem, 39Weierstrass product theorem, 40

82

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