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1. Riemann Surfaces an N-Dimensional Topological Manifold

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1. Riemann Surfaces an N-Dimensional Topological Manifold RIEMANN SURFACES JIA-MING (FRANK) LIOU 1. Riemann surfaces An n-dimensional topological manifold is a Hausdorff space M such that every point n p 2 M has an open neighborhood homeomorphic to an open set of R : If X is a 2n-dimensional topological manifold, a complex chart on X is a pair (U; ') n such that ' is a homeomorphism from U onto an open subset of C : A complex atlas on a 2n-dimensional complex manifold is a family of complex charts f(Uα;'α)g such that fUαg is an open cover of X and on each nonempty intersection Uα \ Uβ 6= ;; the map −1 'β ◦ 'α : 'α(Uα \ Uβ) ! 'β(Uα \ Uβ) n is biholomorphic on C : Two complex altas are called holomorphically equivalent if their union is also a complex atlas. A holomorphic equivalence class of complex atlas is called a complex structure. Definition 1.1. An n-dimensional complex manifold is a connected 2n-dimensional topo- logical manifold together with a complex structure. An one dimensional complex manifold is called a Riemann surface. Let X be a Riemann and f(Uα;'α)g be a complex atlas on X: Let U ⊂ X be an open set. A continuous function f : U ! C is called holomorphic if for each nonempty intersection U \ Uα; the map −1 f ◦ 'α : 'α(U \ Uα) ! C is a holomorphic function. The set of holomorphic functions on U is denoted by O(U): It is a complex subalgebra of C(U); the algebra of complex-valued continuous functions on U: The ring O(U) depends only on the complex structure on X and is independent of the choice of complex atlas. Lemma 1.1. Two holomorphically equivalent complex atlas on X determine the same ring O(U) of holomorphic functions on U for every open set U of X: Proof. The proof is left to the reader. Lemma 1.2. (Maximum modulus principle) Let f be a holomorphic function on an open set U of C: If there exists x0 2 U so that jf(x0)j = maxfjf(x)j : x 2 Ug; then f is a constant function. Proof. See Ahlfors, Stein, e.t.c. Lemma 1.3. Let X be a connected topological space. The only nonempty closed and open subset of X is X itself. Proof. The proof is left to the reader. ∼ Proposition 1.1. If X is a compact Riemann surface, then O(X) = C: 1 2 JIA-MING (FRANK) LIOU Proof. The ring O(X) obviously contains constant functions. Now, we only need to show that O(X) consists of constant functions only. Let f : X ! C be a holomorphic function. Since X is compact, there exists x0 2 X such that jf(x0)j = M; where M = maxfjf(x)j : x 2 Xg: −1 Let E be the set of all points x so that f(x) = f(x0): In other words, E = f (f(x0)): Then E is a closed subset of X: Since x0 2 E; E is nonempty. To show that f is a constant function, it suffices to prove E = X: If we can show that E is also open, by Lemma 1.3, E = X: −1 Let x be any point of E: Then x belongs to Uα for some α: By definition, f ◦ 'α : 'α(Uα) ! C is holomorphic. Since 'α(Uα) is open in C;'α(x) is its interior point. More- −1 over, jf ◦ 'α j has maximum at 'α(x) on 'α(Uα): By the maximum modulus principle −1 (Lemma 1.2), f ◦ 'α must be a constant function on 'α(Uα): Since 'α is a homeomor- phism from Uα to 'α(Uα); f is a constant function on Uα: This shows that Uα is an open neighborhood of x contained in E: Then x is an interior point of E: Since x is arbitrary in E; E is open. This completes the proof of our assertion. Definition 1.2. Let f : X ! Y be a continuous function between Riemann surfaces. Assume f(Uα;'α)g and f(Vβ; β)g are complex atlas on X and on Y respectively. We say −1 that f is a holomorphic mapping if for any Vβ and any Uα with Uα \ f (Vβ) 6= ;; the map −1 −1 β ◦ f ◦ 'α : 'α(Uα \ f (Vβ)) ! C is holomorphic. Observe that if f : X ! Y is a continuous function, for every open set V of Y; the map f ∗ : C(V ) ! C(f −1(V )); h 7! h ◦ f ∗ is a C-algebra homomorphism. Moreover, the set f OY (V ) is a well-defined C-subalgebra of C(f −1(V )): Theorem 1.1. A continuous function f : X ! Y between Riemann surfaces is a holomor- ∗ −1 phic mapping if and only if f OY (V ) is contained in OX (f (V )) for every open set V of Y: Proof. Exercise. Definition 1.3. A homeomorphism f : X ! Y between two Riemann surfaces will be called an isomorphism if both f and f −1 are holomorphic mapping. Two Riemann surfaces are called isomorphic if there exists an isomorphism between them. It follows from the definition that isomorphisms between Riemann surfaces define an equivalence relation. Our goal in this topic is to classify all compact Riemann surfaces up to isomorphisms. RIEMANN SURFACES 3 2. Some examples of Riemann surfaces Any open subset of C is a Riemann surface; in general, any open subset of a Riemann surface is again a Riemann surface. 2 2 2 2 2 Example 2.1. The two dimensional unit sphere S = f(x1; x2; x3) 2 R : x1 +x2 +x3 = 1g has a structure of a compact Riemann surface. Let N = (0; 0; 1) denote the north pole of S2 and U = S2 n fNg: Denote S = (0; 0; −1) the south pole of S2 and V = S2 n fSg: Then fU; V g forms an open cover of S2: Define ' : U ! C and : V ! C by x1 x2 x1 x2 '(x1; x2; x3) = + i ; (x1; x2; x3) = − i : 1 − x3 1 − x3 1 + x3 1 + x3 Then ' : U ! C and : V ! C are homeomorphisms. 1 ◦ '−1 : n f0g ! n f0g; z 7! : C C z We leave to the reader to verify that f('; U); ( ; V )g is a complex atlas on S2 and hence determines a complex complex structure on S2: Our next example will be the one-dimensional complex projective space. 2 Let C n f(0; 0)g be equipped with the subspace topology induced from the standard 2 2 Euclidean topology on C : On C n f(0; 0)g; we define a relation (z0; z1) ∼ (w0; w0) () (z0; z1) = t(w0; w1) × 2 for some t 2 C = Cnf0g: The quotient space (C nf(0; 0)g)= ∼ modulo this relation is called 1 the one dimensional complex projective space and denoted by P : Denote the quotient map 1 2 by π: We equip P with the quotient topology. The equivalent class of (z0; z1) 2 C nf(0; 0)g 1 is denoted by (z0 : z1) called the homogeneuous coordinates on P : 3 2 2 2 3 Denote S = f(z1; z2) 2 C : jz1j + jz2j = 1g: Then S is closed and bounded. By Heine-Borel theorem, S3 is compact. Consider the map 3 1 f : S ! P ; (z0; z1) 7! (z0 : z1): 1 Then f is continuous and surjective. This implies that P is compact by the compactness of S3 and continuity of f: The continuity and the surjectivity of f can be proved as follows. 3 2 1 −1 The inclusion map ι : S ! C n f0g is continuous. For every open set U of P ; f (U) = −1 −1 −1 2 −1 ι (π (U)): Since π (U) is open in C n f(0; 0)g and ι is continuous, f (U) is open in S3 and hence f is continuous. Surjectivity of f is obvious. 1 Lemma 2.1. P is a Hausdorff space. Let U0 = f(z0 : z1): z0 6= 0g and U1 = f(z0 : z1): z1 6= 0g: Then fU0;U1g forms an open 1 cover of P : Define '0 : U0 ! C and '1 : U1 ! C by z1 z0 '0(z0 : z1) = ;'1(z0 : z1) = : z0 z1 Then 'i : Ui ! C are homeomorphisms for i = 0; 1: 1 1 Proposition 2.1. The family f(Ui;'i): i = 0; 1g is a complex atlas on P so that P is a compact Riemann surface. 4 JIA-MING (FRANK) LIOU 1 Proof. We have seen that P is a compact Hausdorff space. We only need to show that 1 f(Ui;'i): i = 0; 1g is a complex atlas on P : Observe that U0 \ U1 = f(z0 : z1): z0; z1 6= 0g and '0(U0 \ U1) = '1(U0 \ U1) = C n f0g: The transition function on U0 \ U1 is given by 1 ' ◦ '−1 : n f0g ! n f0g; z 7! : 1 0 C C z −1 −1 Thus '1 ◦ '0 is biholomorphic on C n f0g whose inverse map is '0 ◦ '1 : 1 2 Proposition 2.2. The Riemann surface P is isomorphic to S as Riemann surfaces. Proof. Exercise. 1 Definition 2.1. A holomorphic mapping f : X ! P is called a meromorphic function on X: The set of all meromorphic functions on X is denoted by M(X): Suppose π : X ! Y is a surjective continuous function between topological spaces. We say that π : X ! Y is a covering space of Y if for each y 2 Y; there exists an open −1 ` neighborhood V of Y and a family of disjoint open sets fUig so that π (V ) = i Ui and for each i; π : Ui ! V is a homeomorphism.
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