Are two complex manifolds ever homoeomorphic?

is biholomorphic, i.e. the coordinate change is bijective and a holomorphic mapping of open subsets of C, and the inverse mapping

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1 Chapter 1 Riemann surfaces 1.1 Riemann surfaces and holomorphic mappings Subject of the lecture are Riemann surfaces, i.e. 1-dimensional complex manifolds. We therefore begin by transferring the definition of the (real) manifold (perhaps known from the analysis lecture) to 1-dimensional complex manifolds. 1.1 definition. Let X be a topological space. (a) A (1-dimensional) complex (coordinate) chart of X is a pair (U, ϕ) from an open subset UX and a homeomorphism ϕ from U to an open subset U = ϕ [u] C (b) Two maps (U 1, ϕ 1) and (U 2, ϕ 2) of X are called holomorphically compatible if U 1 U 2 =, or if in the case of U 1 U 2 the change of coordinates [ transition function] ϕ 2 ϕ 1 1 ϕ 1 [U 1 U 2]: ϕ 1 [U 1 U 2] ϕ 2 [U 1 U 2] is biholomorphic, ie the coordinate change is bijective and a holomorphic mapping of open subsets of C. , and the inverse map is also holomorphic. ϕ 1 ϕ 1 2 ϕ 2 [U 1 U 2]: ϕ 2 [U 1 U 2] ϕ 1 [U 1 U 2] (c) A holomorphic atlas for X is a family A = {(U i, ϕ i) i I} of pairs of holomorphically compatible maps on X, so that (U i) i I is a coverage of X, ie i IU i = X. (d) A holomorphic atlas A is called maximal [maximal], if for every complex card (U, ϕ) of X that is compatible with all cards from A, (U, ϕ) A already holds. Holomorphic compatibility is obviously an equivalence relation on the set of complex maps of a topological space X. Therefore every holomorphic atlas A from X can be extended to a maximum atlas à by adding all maps from X to A that are compatible with the maps in A. 1

2 2 CHAPTER 1. RIEMANN'S AREAS 1.2 Definition. A topological space X has a countable basis of the topology if there is at most a countable family (U k) k IN of sets open in X such that for every further open set U of X and every x U a k IN with x U k U gives. Any such family (U k) is called the basis of the topology of X. 1.3 Definition. A Riemann surface is a path-connected, metrizable, Hausdorff topological space with a countable base of the topology, together with a maximum holomorphic atlas. When we talk about a map of a Riemann surface, we always mean a map from its maximum atlas. 1.4 Remarks. (a) Because every complex atlas can be expanded to a maximally complex atlas, a path-connected, metrizable, Hausdorff topological space X with a countable topology can also be made into a Riemann surface by defining a non-maximal complex atlas for X. (b) Our condition that Riemann surfaces should have a countable basis of the topology can be weakened too separable or too paracompact without changing the concept of the Riemann surface, i.e. for a path-connected, metrizable, Hausdorffian topological space X, for the one holomorphic atlas exists, the following applies: X has a countable base of the topology X is separable X is paracompact. In addition, the metrizability of X follows from the other properties. We demand these strong topological properties for Riemann surfaces here in order to have to fall back on as few results from the set-theoretical topology as possible. 1.5 Examples. (a) Every region (connected, open subset) GC is obviously a Riemann surface, because {(G, id G)} is a complex atlas for G. (b) The set Ĉ = C {} is called a Riemann sphere [Riemann sphere] when it is understood as a Riemann surface in the manner described below. The following task (a) shows that it is homeomorphic to the 2-dimensional unit sphere, which justifies the designation as a sphere. It is compact and therefore not biholomorphic to any region in C. It is also referred to as CP 1 because of the following task (b). For all future calculations we define for each z C: ± z = z ± =, and for z 0: z = z = z 0 =. The expressions ±, 0 and remain undefined. We first explain a topology on Ĉ by calling a subset U of Ĉ open if and only if UC is open in C in the usual sense and, furthermore, in case U there is an ε> 0 such that {z C z> 1 ε} U

3 1.1. RIEMANN'S SURFACES AND HOLOMORPHIC FIGURES 3 applies. Apparently, this turns Ĉ into a Hausdorff room. Because of the following task (a), like S 2, it is path-connected and metrizable and has a countable basis of the topology. Now we define: U 1 = C, ϕ 1: U 1 C, zz, U 2 = C {}, ϕ 2: U 2 C, z 1 z (where we remember our setting 1 = 0 for the last definition) . U 1 and U 2 are open subsets of Ĉ, U 1 U 2 = Ĉ, and the mappings ϕ 1 and ϕ 2 are homeomorphisms. The change of coordinates from ϕ 1 to ϕ 2 is given by U 1 U 2 = C, ϕ 2 ϕ 1 1 C: C C, z 1 z, which is obviously a biholomorphic map. Therefore {(U 1, ϕ 1), (U 2, ϕ 2)} is a holomorphic atlas for Ĉ, through which Ĉ becomes a Riemann surface. 1.6 Task. (a) Ĉ is homeomorphic to S2 via stereographic projection. (b) Ĉ is biholomorphic to the complex projective space CP1. A room always includes morphisms, i.e. images that are compatible with the structure of this room. 1.7 definition. Let X, Y be two Riemann surfaces and let f: X Y be a map. (a) f is called holomorphic or smooth if f is continuous and for every map (U, ϕ) of X and every map (V, ψ) of Y the function ψ f ϕ 1 ϕ [uf 1 [V]]: ϕ [uf 1 [V]] ψ [v] is holomorphic in the sense of function theory I or smooth in the sense of Analysis II (that is, as a function according to C = IR 2, real differentiable any number of times). In the case V = C one speaks of a holomorphic function. (b) f is called biholomorphic or conformal if f is a homeomorphism and both f and f 1 are holomorphic. (c) X and Y are called biholomorphically equivalent if there is a biholomorphic map f: X Y. The set of holomorphic functions f: X C is denoted by O (X). Obviously the constant functions lie in O (X), and with f, g O (X) we also have f ± g, f g O (X). In this sense, O (X) is a C-algebra, in particular a commutative ring with one. The biholomorphic mappings are the isomorphisms in the category of Riemann surfaces; two Riemann surfaces are isomorphic if they are biholomorphically equivalent. The following statement makes it easier to prove that an image is biholomorphic:

4 4 CHAPTER 1. RIEMANN'S SURFACES 1.8 Statement. (Osgood's theorem for Riemann surfaces.) Let X, Y be Riemann surfaces and f: X Y be a holomorphic map. If f is injective, then f [x] is open in Y, and f is a biholomorphic mapping onto the image f [x]. Proof. f is not constant, so from the openness theorem of function theory I it follows that f [x] is open in Y. Let x X and (U, φ) or (V, ψ) maps of X and Y around x and f (x), respectively. Then f: = ψ f φ 1 (on a suitable domain of definition in C) is holomorphic and injective. Therefore f has no zeros: If f had a zero in a z 0 C, for example of order k, then f would be a branched (k + 1) -leafed overlay on a neighborhood of z 0, i.e. certainly not injective. The theorem of the inverse function therefore shows that f 1 is holomorphic in f (φ (x 0)). It follows that f 1 is holomorphic in f (x 0). Many local statements of function theory I can easily be transferred to holomorphic functions in the new sense. The following task gives some examples. In the following, we will often use such statements without express proof. 1.9 Task. Let X, Y be Riemann surfaces. Show: (a) Identity Theorem. Let f, g: X Y be two holomorphic maps, and suppose that the set A = {x X f (x) = g (x)} has an accumulation point in X. Then f = g holds. [Hint. Use the identity theorem from Function Theory I to show that the set of accumulation points of A is contained in A o.] (B) Suppose that f: XY is a non-constant, holomorphic map and y Y. Then f 1 [{y }] is a discrete subset of X. (c) Maximum and Minimum Principle for holomorphic functions. Let f: X C be a holomorphic function. Then f does not attain a local maximum, and if it attains a local minimum at some x 0 X, then f (x 0) = 0 holds. [Hint. Use the analogous statement from Function Theory I.] (d) Suppose that X is compact. Then any holomorphic map f: X C is constant. (e) Suppose that X, Y are both compact. Either prove or give a counterexample for the following statement: Any holomorphic map f: X Y is constant. (f) Riemann's theorem on removable singularities. Let x 0 X and f: X \ {x 0} Y be a holomorphic map. Suppose that one of the following two statements holds: (i) f can be extended to a continuous function f: X Y. (ii) Y = C and there exists a neighborhood U of x 0 in X so that the function is bounded. f U \ {x0} Then f can be extended at x 0 to a holomorphic map f: X Y. The following statement shows that holomorphic maps correspond to I according to function theory. Ĉ the meromorphic functions 1.10 statement. Let X be a Riemann surface, x 0 X and f: X \ {x 0} C a holomorphic function. Then the following statements are equivalent:

5 1.2. DIFFERENTIAL FORMS 5 (a) f can be holomorphically (or continuously) continued in x 0 to a map f: X Ĉ. (b) There is a map (U, ϕ) of X with x 0 U such that (f U \ {x0}) ϕ 1: ϕ [u \ {x 0}] C in x 0 meromorphic in the sense of function theory I is. (c) There exists a number n Z, and a map (U, ϕ) of X with x 0 U such that the function (zz 0) n ((f U \ {x0}) ϕ 1): ϕ [u \ {x 0}] C in z = z 0: = ϕ (x 0) can be continued holomorphically (or continuously) by a value in C. If these statements hold, then the function f is identified with its continuation f from (a). In this case the number n Z from (c) is uniquely determined. It is called the order [order] of f in x 0 and is denoted by ord x0 (f). If n = ord x0 (f)> 0, then f (x 0) = 0 and one says that f in x 0 has a zero of the n-th order [zero, root of n-th order]. If n = ord x0 (f) <0, then f (x 0) = and we say that f has a pole of (n) -th order in x 0. If ord x0 (f) = 0, then f (x 0) C. We call holomorphic mappings f: X Ĉ of a Riemann surface X also meromorphic functions on X. We denote the space of meromorphic functions on X with M (X). For f, g M (X) we have f ± g, f g M (X) as well as for g 0 also f g M (X). Therefore M (X) is a field, namely the quotient field of the integrity ring O (X). In addition, for x 0 X ord x0 (f g) = ord x0 (f) + ord x0 (g) and ord x0 (f g) = ord x 0 (f) ord x0 (g). Proof of proposition The equivalence of holomorphism and continuity results in (a) from the version of Riemann's theorem of liftability from Exercise 1.9, in (c) from the classical Riemannian theorem. We now choose a map (U, ϕ) of X with x 0 U, and consider the function f: = (f U \ {x0}) ϕ 1: ϕ [u \ {x 0}] C, which is in the sense of Function theory I is holomorphic. This function can be holomorphically continued in ϕ (x 0) if and only if f can be holomorphically continued in x 0. The equivalence of statements (a) (b) (c) and also the other statements result from the corresponding results of function theory I. 1.2 Differential forms In this section we consider differential forms on Riemann surfaces, namely smooth or holomorphic 1-forms and 2 forms. We use differential forms on the one hand to define the derivatives of holomorphic or smooth functions on Riemann surfaces in a map-invariant manner, on the other hand we need them as objects for integration on Riemann surfaces. From now on, we often designate maps of Riemann surfaces X with (U, z) (instead of (U, ϕ)) in order to have more suggestive designations.

6 6 CHAPTER 1. RIEMANN'S SURFACES We want to define a derivative for smooth (especially holomorphic) functions f: X C. First we work on a map (U, z) of X. We write z = x + iy with x = Re (z), y = Im (z): U IR. We can then differentiate the function (f z 1): z (u) C C in the sense of analysis or function theory I. The partial derivatives fz: UC and fz: UC define, and thereby the x: = (fz 1) x differential operators z: = 1 2 y: = (fz 1) y () xiy and z: = 1 () 2 x + iy (Why this definition, especially why is the minus sign at this point? One reason is that the Cauchy-Riemann differential equations state that f U is holomorphic if and only if ffz = 0, and if so, then z agrees with the complex derivative of fz 1 in terms of function theory I.) The problem with these partial derivatives is that they are not invariant, that is, that they depend on the choice of map X. In more detail: If (z, u) and (z, ũ) f are two different maps of X with U Ũ, then z and f z on U Ũ generally do not match, the same is true for the z-derivatives. Instead, they have the transformation behavior described below: The map change mapping zz 1: z [u Ũ] z [u Ũ] is biholomorphic, therefore dzdzdz = 0, the derivative dz is holomorphic and zero-zero on U Ũ, and because of the chain rule, df dz applies = df dzdz df and dz dz = df () dzdz on U Ũ dz. () (We can) as an invariant derivative of f the totality of all pairs of derivatives df dz, df dz, i.e. the family {(df df = dz, df) (U, z) A} () dz together with the transformation formulas (). We now use these transformation formulas as a model for the definition of differential forms, more precisely: of 1-form definition. Let X be a Riemann surface with the maximum atlas A = {(U α, z α) α A} .. (a) A smooth or holomorphic or meromorphic differential form of degree 0 or 0 for short [differential form of degree 0 , 0-form] is a family f = (f α) α A of smooth or holomorphic or meromorphic functions f α on U α, so that for every pair α, β A with U α U β the transformation formula f α = f β on U α U β applies. We denote the space of the smooth 0-forms with C (X) = Ω 1 (X) = A 0,0 (X). There is another, more formal reason: If X is a region in C and z = id is the canonical global map, then dx and dy can be used as 1-forms (projections in the direction of the coordinate axes), and and x as vector fields (in the direction the coordinate axes). Then dx ((yx) = dy y) = 1 and dx ((y) = dy) x = 0. With dz = dx + idy and dz = dx idy, and the above definitions for, we therefore have zz dz (( z) = dz (z) = 1 and dz (z) = dz) z = 0. In other words, (,) is a dual basis of the vectors zz to the basis (dz, dz) of the 1-forms. We shall see later that this justification is carried over to differential forms on general Riemann surfaces.

7 1.2. DIFFERENTIAL FORMS 7 (b) A smooth or holomorphic or meromorphic differential form of degree 1 or 1-form for short [differential form of degree 1, 1-form] is a family ω = ((ω α, 1, ω α, 2) ) α A of pairs (ω α, 1, ω α, 2) of smooth or holomorphic or meromorphic functions on U α, so that for every pair α, β A with U α U β the transformation formulas ω α, 1 = ω β, 1 dz β dz α and ω α, 2 = ω β, 2 () dzβ dz α on U α U β apply. We denote the space of the smooth 1-forms by Ω 1 (X). Remember rule for the transformation formulas. ω α, 1 (z α) dz α and ω α, 2 (z α) d z α are invariant, i.e. independent of α A. We say that ω Ω 1 (X) is of type (1,0) or of type (0,1) [is of type (1,0), (0,1)] if A holds for all α : ω α, 2 = 0 or ω α, 1 = 0. We denote the space of smooth 1-forms of type (1,0) or (0,1) with A 1,0 (X) or with A 0.1 (X). This results in a decomposition as a direct sum Ω 1 (X) = A 1.0 (X) A 0.1 (X). (c) A smooth or holomorphic or meromorphic differential form of degree 2 or short 2-form [differential form of degree 2, 2-form] is a family ψ = (ψ α) α A of smooth or holomorphic or meromorphic Functions ψ α on U α, so that for every pair α, β A with U α U β the transformation formula ψ α = ψ β dz β dz α () dzβ dz α on U α U β applies. We denote the space of the smooth 2-forms with Ω 2 (X) = A 1,1 (X). Remember rule for the transformation formula.ω α (z α) dz α d z α is invariant, i.e. independent of α A. The maximum atlas of X is not required to define a differential form; it is sufficient to specify the functions or pairs of functions for the maps in some compatible atlas of X. Differential forms of degree 0 obviously correspond to global functions on X examples. Let X be a Riemann surface. (a) If ϕ 1, ϕ 2 differential forms of the same degree k {0,1,2} and f: XC are a smooth function, then we define ϕ 1 + ϕ 2 and f ϕ 1 component-wise, ie by acting on the individual functions or pairs of functions. In this way, differential forms of degree k result again. If, in addition, U X is an open set, a differential form ϕ U on U can be obtained from a differential form ϕ on X by restricting all individual functions to U. (b) If f: XC is a smooth function, then df defined by () is a smooth differential form of degree 1 to X. If f is holomorphic or meromorphic, then df is a holomorphic or meromorphic differential form of degree 1 and type ( 1.0) on X. For two smooth functions f, g: XC, the sum and product rule apply: d (f + g) = df + dg and d (fg) = f dg + g df.

8 8 CHAPTER 1. RIEMANN'S SURFACES (c) For regions G C and holomorphic functions f: G C, df = f (z) dz applies, if we denote the canonical global map for G with z = id G. (d) If ω = ((ω α, 1, ω α, 2)) α A Ω 1 (X), then for every α A: ω Uα = ω α, 1 dz α + ω α, 2 dz α . On the left-hand side of this equation, the restriction is naturally applied to the individual functions ω α, ν, and on the right-hand side are the differentials of the functions z α and z α in the sense of the definition (). In the future, we will often state 1-forms in this form, leaving out the index α and the restriction to U α if no misunderstandings are to be feared. Therefore a 1-form ω has the type (1,0) or (0,1) if and only if it is of the form ω = ω α dz or ω = ω α d z. In particular, for a smooth function f: X C in this sense df = f f dz + z z d z. If f is holomorphic or meromorphic, this representation simplifies to df = f z dz task. Prove the above rules, and all others that may be needed task. The wedge product. Let X be a Riemann surface, with atlas {(U α, z α) α A}. (a) Let two differential 1-forms ω, ω Ω 1 (X) be given, say ω = ((ω α, 1, ω α, 2)) α A and ω = ((ω α, 1, ω α , 2)) α A. For α A we define the functions ψ α = ω α, 1 ω α, 2 ω α, 1 ω α, 2. Show that ψ: = (ψ α) α A Ω 2 (X) holds. This 2-form ψ is called the wedge product of ω and ω, and is denoted by ω ω. Moreover, show that ω ω = ω ω holds. (b) Let a differential 2-form ψ = (ψ α) α A Ω 2 (X) be given. Show that for every α A we have the local representation ψ Uα = ψ α (dz α d z α). For smooth functions f: XC we also define the decomposition of df in the sense of the direct sum Ω 1 (X) = A 1,0 (X) A 0,1 (X), ie the partial differential operators df = f dz and zdf = fzd z. Hence, df A 1,0 (X), df A 0,1 (X) and df = d f + d f. The derivative df or df is called the holomorphic derivative or anti-holomorphic derivative of f. These terms are justified by the facts of the following task: 1.15 Task. Let f: XC be a smooth function on a Riemann surface X. Then f is holomorphic if and only if df = 0. If this is the case, then df corresponds to the usual complex derivative of f, more precisely: For any chart ( U, z) of X we have df U = (fz 1) dz, where on the right-hand side of the equation, denotes the usual complex derivative in the sense of Funktionentheorie I.

9 1.2. DIFFERENTIAL FORMS 9 We extend the differential operators d, d, d for 1-forms ω Ω 1 (X) by defining for ω = ω 1 dz + ω 2 dz with smooth functions ω 1, ω 2: d (ω 1 dz + ω 2 dz) = ω 2 z (dz dz), d (ω 1 dz + ω 2 dz) = ω 1 z (dz dz) and dω = d ω + d ω. For formal reasons we also set for every ψ Ω 2 (X): dψ = d ψ = d ψ = 0. Because the real partial derivatives interchange with one another, 2 therefore also holds for f C (X): () () d ( df) = dff zdz + zdz = fzzfzz (dz dz) = 0. So we remember: dd = 0. zz = 2 zz, and 1.16 exercise. Prove the following variant of the product rule: Let X be a Riemann surface, f C (X) and ω Ω 1 (X). Then d (f ω) = df ω + f dω holds. For the time being, we consider the so-called pullback as the last operation for differential forms. Let Y be another Riemann surface and let η: Y X be a non-constant, holomorphic map. We want to pull back differential forms on X to differential forms on Y by means of η. For 0-forms, i.e. global functions f C (X) = Ω 0 (X), this can be done simply by composition: η f: = f η Ω 0 (Y). For 1-forms ω Ω 1 (X), roughly represented locally as ω = f (z) dz + g (z) dz with respect to a map (U, z) of X, we define η ω Ω 1 (Y) by For two cards each (U, z) and η (f (z) dz + g (z) dz) = (f η) d (z η) + (g η) d (z η). (Ũ, z) of X we have d (z η) d (z η) = dz d z η, and therefore this definition actually defines a global differential form η ω Ω 1 (Y). For 2-forms we proceed in exactly the same way: For ψ Ω 2 (X) we define η ψ Ω 2 (Y) by the local definition η (f (z) (dz dz)) = (f η) (d (z η) d (z η)), which also describes a global 2-form because of () statement. Let X, Y be Riemann surfaces and η: Y X be a non-constant holomorphic map. (a) For f C (X) we have d (η f) = η df. (b) For ω Ω 1 (X) we have d (η ω) = η dω. Proof. For (a): We may assume, without loss of generality, that η is locally biholomorphic. If (U, z) is a map of X, then there exists an open subset V Y with η [y] X such that (V, z η V) is a map of Y. Then: d (η f) = d (f η) = (f η) (f η) () (z η) d (z η) + (z η) d (z η) = The proof of (b ) runs accordingly. (f z η) d (z η) + () (f z η) d (z η) = η df.

10 10 CHAPTER 1. RIEMANN'S SURFACES 1.3 Integration and Residual When defining differential forms, we were guided by the endeavor to describe the derivation of a smooth function on a Riemannian surface through a global object. Differential forms are also suitable objects for the opposite operation, integration, namely 1-forms are integrated along curves and 2-forms on surfaces (open or compact subsets of a Riemann surface). We define such integrals for differential forms whose carriers are contained in a map neighborhood U of the Riemann surface X. In this case it is sufficient to consider the case that the integration path or the integration area is also contained in U. So let (U, z) be a map of X, ω Ω 1 (X) with Tr (ω) U and γ: [a, b] U a smooth curve in U. In this case ω U = f (z) dz + g (z) dz and we define, almost like in function theory I, b (ω: = f (γ (t)) (z γ) (t) + g (γ (t)) (z γ) (t )) dt. γ a The nice thing about this definition is that it is independent of the choice of card. If (Ũ, z) is another map with Tr (ω) Ũ and γ [[a, b]] Ũ, say ω Ũ = f (z) dz + g (z) dz, then because of the transformation rule for 1 -Forms for all t [a, b] and because of the chain rule f (γ (t)) = f (γ (t)) dz dz (γ (t)) and g (γ (t)) = g (γ ( t)) (dz dz (γ (t))) (z γ) (t) = ((zz 1) (z γ)) (t) = (zz 1) (z (γ (t))) (z γ) (t) = dz dz (γ (t)) (z γ) (t), and thus ba (f (γ (t)) (z γ) (t) + g (γ (t)) (z γ) (t)) dt = ba (f (γ (t)) (z γ) (t) + g (γ (t)) (z γ) (t)) dt. This is the reason why 1- Shapes are the appropriate integrands for integration along curves. The same applies to the integration of 2-forms on surfaces: If ψ Ω 2 (X) with Tr (ψ) U, for example ψ U = f (z) dz dz, then we define ψ = (fz 1) (w) d 2 (w), U z [u] where on the right side there is a 2-dimensional (Lebesgue) integral over an area in C = IR 2. If h: G C C is holomorphic, then h 2 is equal to the determinant of the Jacobi matrix of h as a real differentiable function; from this and from the transformation theorem for multidimensional integrals it follows that this integral is also independent of coordinates. In order to be able to integrate globally defined differential forms in this way, we use so-called smooth decompositions of the one theorem. (Existence of a smooth decomposition of the one.) Let X be a Riemann surface, and U = (U α) α A an open cover of X. Then there exists a partition of the one [partition of unity] of X for U, that is a family (f α) α A of smooth functions f α: X IR with the following properties:

11 1.3. INTEGRATION AND RESIDUAL 11 (a) 0 f α 1 (b) Tr (f α) U α (c) Local finiteness: For every x X there is a neighborhood U (x) of x in X, so that all but finite many the f α on U (x) vanish. (d) α A f α = 1. The term decomposition of the one comes of course from property (d). Because of the local finiteness, the sum in property (d), when evaluated at one point, is only a sum of finitely many positive numbers. Question: Are there also holomorphic decompositions of one? In order to define integrals globally on Riemann surfaces X, we consider a holomorphic atlas A = {(U α, z α) α A} of X and, according to the above theorem, choose a suitable, Smooth decomposition of the one (f α) α A. If ω Ω 1 (X) and γ: [a, b] X are a smooth curve, we define ω: = f α ω. γ α γ In each case Tr (f α ω) is contained in the map environment U α, so that the integrals on the right were defined at the beginning of this section. The sum runs over all α A for which U α with γ [[a, b]] has a non-empty intersection; because γ [[a, b]] is compact, these are only finitely many, so that the sum results in a complex number. Finally, the definition according to the coordinate invariance calculation at the beginning of this section is independent of the choice of the atlas and the decomposition of the one. Correspondingly we define for ψ Ω 2 (X) and a (relatively) compact subset K of X ψ = f α ψ. K α K The integrals on the right are again of the type previously defined, and because K is compact, the sum can again be assumed to be finite. This definition is also independent of the atlas and the decomposition of the one. Proof sketch for theorem Because X has a countable topology, we can assume without restriction that A = N is countable, i.e. U = (U n) n N. One can show that possibly after a refinement of the open coverage (U n) the U n map environments to maps φ n of X are, so that the images φ n [U n] = B (0, rn) are balls of certain radii rn> 0, but that conversely already (φ 1 n [B (0, rn / 2)]) n N completely covers X. The refinement can also be chosen so that every U n only has a non-empty intersection with a finite number of other U m. In order to construct a decomposition of the one in this situation, we consider the function fa, b for real numbers a

12 12 CHAPTER 1. RIEMANN'S SURFACES One can show that f a, b is smooth. For r> 0 then the function gr (z) = fr / 2,2r / 3 (z) is a smooth function on C, which vanishes outside of B (0, 2r / 3) and on B (0, r / 2 ) is identical to 1. We continue the smooth function h n = g rn φ n on U n to a smooth function on X by setting it to zero outside of U n. Thus the two functions hn, 1 hn are a decomposition of the one to the two-element open cover {U n, mn U m} of X. With the help of the hn we can construct the desired decomposition of the one (fn) n IN: We set n 1 fn = hn (1 hl) for all n N. l = 1 Then obviously 0 fn 1 and Tr (fn) Tr (hn) U n. Because U n is only a non-empty intersection with finitely many other U m has, the condition of local finiteness applies. Finally, one shows by induction that for all n IN the following applies: nffn + (1 hl) = 1. l = 1 If x X is given, there is an n (x) N with φ n (x) (x) B ( 0, rn / 2) and thus hn (x) (x) = 1. For every nn (x) then f 1 (x) + ... + fn (x) = 1 and thus n IN fn (x) = Statement. Calculation rules for the curve integral. Let X be a Riemann surface and γ: [a, b] X be a smooth curve. (a) For ω 1, ω 2 Ω (X) and λ 1, λ 2 C, γ (λ 1ω 1 + λ 2 ω 2) = λ 1 γ ω 1 + λ 2 γ ω 2. (b) For f C (X) we have γ df = f (γ (b)) f (γ (a)). In particular, γ df = 0 if the curve γ is closed. (c) If ω Ω (X), Y is another Riemann surface, η: Y X is a non-constant, holomorphic mapping, and α: [a, b] Y is a smooth curve, then α η ω = η α ω. Proof. We first consider the case that both γ and Tr (ω), Tr (ω k) are contained in a map neighborhood U to a map (U, z) of X. In this case (a) is obvious. For (b) the following applies (where we again write z = x + iy with the real differentiable functions x = Re (z), y = Im (z): U IR): () df = ff zdz + zdz γ = = γ baba () fz (γ (t)) (z γ) (t) + fz (γ (t)) (z γ) (t) () fx (γ (t)) (x γ) (t) + ify (γ (t)) (y γ) (t) = f (γ (b)) f (γ (a)). For (c) we can again assume that z η is a map of Y. We write locally ω = ω 1 (z) dz + ω 2 (z) dz, then η (ω = (ω1 η) d (z η) + (ω 2 η) d (z η)) α = α ba (ω1 (η (α (t))) (z η α) (t) + ω 2 (η (α (t))) (z η α) (t)) dt = η α ω.

13 1.3. INTEGRATION AND RESIDUAL 13 For the general case, the statements result from this by using a decomposition of the one. For area integrals, Stokes' theorem, which we quote here without proof, also applies in this situation: 1.20 Theorem. Stokes' theorem. Let X be a Riemann surface, ω Ω 1 (X), K X a compact subset, the edge of which is smooth and bounded in a mathematically positive direction by the closed curve K. Then dω = ω. K Among the differential forms of degree 1, two classes play a special role, namely the closed and the exact 1-forms: 1.21 Definition. Let X be a Riemann surface and let ω Ω 1 (X). K (a) ω is called closed when dω = 0. (b) ω is called exactly [exact] if there is an f C (X) with df = ω. Such an f is called an antiderivative [primitive] to ω. The rule d d = 0 shows that every exact 1-form is also closed. The following exercise shows that the reverse of this statement is generally false. However, we will soon see that the inversion is correct on simply connected Riemann surfaces, i.e. every closed 1-form is also exact. Consider X = C, the smooth 1-form ω Ω 1 (X) that is given with respect to the global chart z = x + iy = id C of C by ω = yz 2 dx + xz 2 dy, and the closed smooth curve γ: [0,2π] C, t (cost, sint) that parameterises S 1. Show that ω is closed, compute γ ω and conclude that ω is not an exact task. Let X be a Riemann surface and ω A 1.0 (X). Show that ω is holomorphic if and only if it is closed Statement and definition. Let X be a Riemann surface, ω A 1,0 (X) a meromorphic 1-form of type (1,0) on X, and x X. For every map (U, z) of X with x U and cz (x ) = 0 there is exactly one c C, so that the 1-form ω z dz is exact on a neighborhood of x. The value of c does not depend on the choice of card z. This value is called the residue of ω in x and denoted by Res x (ω). Proof. Regarding a map (U, z) as in the statement, we can write ω = ω 1 (z) dz with a meromorphic function ω 1: U Ĉ. ω 1 z 1 is then on z [u] C a meromorphic function in the sense of function theory I. If c C is the coefficient for z 1 of the Laurent series

14 14 CHAPTER 1. RIEMANN SURFACES of this function at z = 0, then ω 1 z 1 c z has an antiderivative g on B (0, ε), and this is not the case for any other choice of c. Here ε> 0 is chosen so small that B (0, ε) z [u]. Then (ω c z dz) z 1 [B (0, ε)] has the antiderivative (g z) z 1 [B (0, ε)]. If (Ũ, z) is another map of X with x Ũ and z (x) = 0, then after the previous one there is exactly one c C such that ω c z d z has an antiderivative on a neighborhood of x. Then there is a neighborhood in U Ũ on which c c c zdz c z d z = z dz + c (1 z dz 1 z d z) has an antiderivative. The 1-form 1 z dz 1 z d z is holomorphic in x and therefore has an antiderivative in the vicinity of x. So c c z dz also has an antiderivative locally near x, and this is only possible for c c = 0. Let X, Y Riemann surfaces, ω A 1,0 (X) be a meromorphic 1-form of the type (1,0), y Y and f: Y X a holomorphic mapping that is locally injective at y.Then the transformation formula Res y (f ω) = Res f (y) (ω) applies. Proof. If z is a map function of X at f (y), then z f (if necessary, a restriction thereof) is a map function of Y at y. ω Res f (y) (ω) dz z has a holomorphic antiderivative g in the vicinity of f (y). Then d (gf) = f dg = f ω Res f (y) (ω) d (zf) zf, in other words: f ω Res f (y) (ω) d (zf) zf has a holomorphic locally at y Indefinite integral. Because of the uniqueness statement in Definition 1.24, the assertion statement follows from this. Let X be a Riemann surface, ω A 1,0 (X) a meromorphic 1-form of the type (1,0), x X, and (U, z) a map of X with x U and z (x) = 0 Then for every ε> 0 with B (0, ε) z [u]: Res x (ω) = 1 ω = 1 (z 1) ω. 2πi z 1 B (0, ε) 2πi B (0, ε) proof. This results from combining the corresponding classical statement of function theory I with the transformation formula of statement 1.25 for f = z 1. Meromorphic differentials of type (1,0) are also called abelian differentials. According to an old parlance, they are divided into three types: Abelian differentials of the first kind are holomorphic. In Abelian differentials of the second kind, their residual vanishes at every pole. And Abelian differentials of the third kind are all others. 1.4 Superpositions So far we do not have very many examples of Riemann surfaces: apart from the regions in C, essentially only the Riemann number sphere Ĉ. One possibility to construct further Riemann surfaces is provided by so-called overlays. The overlays that we learn about in this section are unbranched.