A subset Z of X is discrete if for any z ∈ Z, there exists an open neighborhood U of z so that U ∩ Z = {z}

1. Some Complex Analysis in Riemann surfaces

Definition 1.1. A topological space X is connected if the only nonempty open and closed subset of X is X itself.

Definition 1.2. Let X be a topological space. A subset Z of X is discrete if for any z ∈ Z, there exists an open neighborhood U of z so that U ∩ Z = {z}.

Theorem 1.1. (Coincidence Principle) Let f : D → C be a holomorphic function defined on an open connected subset D of C. Suppose that there exists a sequence of distinct points (zn) in D with a limit in D so that f (z_n) = 0 for all n ≥ 1. Then f = 0 on D.

Corollary 1.1. (Identity Theorem/Coincidence Principle) Let f, g : D → C be holomorphic func- tions defined on an open connected subset D of C. Suppose that f = g on a subset S of D having a limit point a ∈ D. Then f = g on D.

Proof. Let h : D → C be the function h = f − g. Then h : D → C is holomorphic. Since a is a limit point of S, we may choose distinct points a_n of S so that lim_n→∞a_n = a. Since f = g on S, h(an) = 0 for all n ≥ 1. By coincidence principle, h = 0 on D which implies that f = g on D.  Theorem 1.2. (Identity Theorem/Coincidence Principle for Riemann surfaces) Let X be a con- nected Riemann surface and f, g : X → C be holomorphic functions. Suppose there exists a subset S of X having a limit point a in X so that f = g on S. Then f = g on X.

Proof. Let h : X → C be the function h = f − g. Then h is holomorphic on X. To show that f = g on X is equivalent to show that h = 0 on X.

Let A be the subset of X consisting of points x so that there exists a neighborhood W of x, h|W = 0. Let us show that A is open. For p ∈ A, we can find an open neighborhood W of p such that h|W = 0. For any q ∈ W, W is an open neighborhood of q and h|W = 0. This implies that q ∈ A for any q ∈ W. Hence W ⊆ A. We see that q is an interior point of A. Since q is arbitrary in A, A is open.

Let us show that A is closed, i.e. let us show that A contains all of its limit points. Let p be a limit point of A. Let us choose a complex chart φ : U → C so that φ(p) = 0 and φ(U ) = B. Here B = {z ∈ C : |z| < 1}. Since h : X → C is holomorphic, h ◦ φ⁻¹ : B → C is holomorphic. Since p is a limit point of A, 0 is a limit point of φ(U ∩ A). We may choose a sequence of distinct points (zn) in φ(U ∩ A) so that limn→∞zn= 0. Since zn ∈ φ(U ∩ A), zn = φ(xn) for some xn ∈ U ∩ A. Since lim_n→∞z_n = 0, lim_n→∞x_n = p (φ is a homeomorphism). Since x_n ∈ A, h(xn) = 0. Furthermore, by continuity of h, h(p) = 0. Therefore

(h ◦ φ⁻¹)(zn) = h(xn) = 0, n ≥ 1.

Since B is connected and h ◦ φ⁻¹ : B → C is holomorphic, h ◦ φ⁻¹ = 0 on B by the coincidence principle. We find that h = 0 on U, i.e. h|_U = 0. Hence p ∈ A by assumption. This shows that A is closed.

Let us show that A is nonempty. In fact, we can show that a ∈ A. By assumption h = 0 on S.

We choose a complex chart φ : U → B so that φ(a) = 0. Then h ◦ φ⁻¹ is holomorphic on B. Since a is a limit point of S, 0 is a limit point of φ(S ∩ U ) ⊆ B. Since h = 0 on S, h ◦ φ⁻¹ = 0 on φ(S ∩ U ) having a limit point 0. By connectedness of B and the Corollary 1.1, h ◦ φ⁻¹ = 0 on B and hence h = 0 on U, i.e. h|U = 0. Since U is an open neighborhood of a, a ∈ A.

Since A is nonempty, open and closed in X, A = X by connectedness of X. Therefore h = 0 on X.

 Theorem 1.3. Let X be a connected Riemann surface and f : X → C be a holomorphic function, not identically zero. Then the zero set of f forms a discrete subset of X if Z is nonempty.

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Proof. Let Z be the zero set of f. By continuity of f, Z = f⁻¹({0}) is a closed subset of X. Let p be a point of Z. We want to show that there exists an open neighborhood U of p so that U ∩ Z = {p}.

Suppose not. Then p must be an accumulation point of p.¹

Since f : X → C is holomorphic, we may choose a complex chart φ : U^p → C around p so that φ(p) = 0 and f ◦ φ⁻¹ : φ(Up) → C is holomorphic. We may chose U^p so that φ(Up) = B is the open unit disk in C. Denote f ◦ φ⁻¹ by g. By definition, g : B → C is holomorphic. Let Z⁰ be the zero set of g in B, i.e. Z⁰= {z ∈ B : g(z) = 0}. Then g(0) = f ◦ φ⁻¹(φ(p)) = f (p) = 0 and hence 0 ∈ Z⁰. Since p is an accumulation point of Z, 0 is an accumulation point of Z⁰ (using the fact that φ is a homeomorphism). Hence we can find a sequence (z_n) in Z⁰ so that lim_n→∞z_n = 0 and g(z_n) = 0 for all n ≥ 1. Since B is open and connected, by coincidence principle (identity theorem), g must be the zero function, i.e. g(z) = 0 for all z ∈ B. For any q ∈ Up, φ(q) ∈ φ(U_p) = B. Hence g(φ(q)) = 0 for all q ∈ Up. Furthermore,

f (q) = (f ◦ φ⁻¹)(φ(q)) = g(φ(q)) = 0

for any q ∈ Up. We find that q ∈ Z for all q ∈ Up. This implies that Up ⊆ Z; hence p is an interior point of Z. Since p is arbitrary in Z, Z is open. We conclude that Z is a nonempty open and closed subset of X. Since X is connected, Z = X. This implies that f (x) = 0 for all x ∈ X which leads to the contradiction to the fact that f is not identically zero. Therefore p must not be an accumulation point of Z; hence we can find an open neighborhood U of p so that U ∩ Z contains exactly one point p. In other words, U ∩ Z = {p}.

 Theorem 1.4. (Maximum modulus Principle) Let f : D → C be a holomorphic function on an open connected set D ⊆ C. Suppose that there exists z⁰∈ D so that |f (z)| ≤ |f (z0)| for all z ∈ D.

Then f is a constant function.

Theorem 1.5. (Maximum modulus Principle for Riemann Surfaces) Let f : X → C be a holo- morphic function on a connected Riemann surface X. Suppose that there exists p₀ ∈ X so that

|f (p)| ≤ |f (p₀)| for all p ∈ D. Then f : X → C is a constant function.

Proof. Let A = {p ∈ X : f (p) = f (p0)}. Then A = f⁻¹({f (p0)}) is a closed subset of X by the continuity of f. Since p0∈ A, A is nonempty.

Let q ∈ A. We choose a complex chart φ : U → C so that φ(q) = 0 and φ(U ) = B. Then f ◦ φ⁻¹: B → C is holomorphic. Let g : B → C denote the holomorphic function g = f ◦ φ⁻¹. Since f (q) = f (p0), |f (p)| ≤ |f (p0)| = |f (q)| for any p ∈ U. Hence

|g(φ(p))| = |f (p)| ≤ |f (q)| = |g(0)| for any p ∈ U .

Since B is connected and g : B → C is holomorphic, by the maximum modulus principle, g is a constant function, i.e. g(z) = g(0) for all z ∈ B. This implies that

f (p) = g(φ(p)) = g(0) = f (q) = f (p0)

for all p ∈ U. Therefore p ∈ A for all p ∈ U. This implies that U ⊆ A which shows that q is an interior point of A. Since q is arbitrary in A, A is open.

By connectedness of X, A = X. Hence f (p) = f (p0) for all p ∈ A = X. We show that f is a

constant function. 

Corollary 1.2. Let X be a compact connected Riemann surface. If f : X → C is holomorphic, then f is a constant function.

Proof. The function F : X → R defined by p ∈ X 7→ |f (p)| is continuous. Since X is compact, F attains its maximum. We can find p0 so that F (p0) = max{F (p) : p ∈ X}. Thus

|f (p)| = F (p) ≤ F (p0) = |f (p₀)|

1If U ∩ Z contains at least one point other than p for any open neighborhood U of p, then U \ {p} ∩ Z is nonempty for any open neighborhood U of p. In other words, p is an accumulation point of Z.

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for all p ∈ X. By the connectedness of X and the maximum modulus principle for Riemann surfaces (Theorem 1.5), f is a constant function.