PMATH 950 (Riemann Surfaces) Notes

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1 PMATH 950 (Riemann Surfaces) Notes Patrick Naylor February 28, 2018 These are notes for PMATH 950, taught by Ruxandra Moraru in the Winter 2018 term at the University of Waterloo. These notes are not guaranteed to be complete, accurate, or even correct, so use at your own risk. Course Description (taken from the course syllabus): Riemann surfaces can be defined in several different, equivalent ways, for example as onedimensional complex manifolds, or as oriented two dimensional real manifolds. In addition, any compact Riemann surface can be embedded in projective space, thus giving it the structure of an algebraic curve. Riemann surfaces therefore appear in many areas of mathematics, from complex analysis, algebraic and differential geometry, to algebraic topology and number theory. This course will cover fundamentals of the theory of compact Riemann surfaces from an analytic and topological perspective. Sources include: Lectures by Dr. Ruxandra Moraru; Lectures by Dr. Benoit Charbonneau (Feb. 26 & 28); Notes by Ragini Singhal; O. Forster, Lectures on Riemann Surfaces. [Required text] 1

2 Contents 1 Preliminaries Defining a Riemann Surface Maps on Riemann Surfaces Branched and Unbranched Covers Sheaves and Things About Sheaves Analytic Continuation 19 3 Calculus on Riemann Surfaces and Differential Forms 24 2

3 1 Preliminaries 1.1 Defining a Riemann Surface We begin with some preliminary notions that give rise to our definitions of a Riemann surface. Recall that a topological space X is Hausdorff if its topology separates points. Examples R n with the metric topology is Hausdorff. 2. R n with the cofinite topology is not Hausdorff. Definition 1. A topological surface is a Hausdorff topological space that is locally homeomorphic to R 2 (or C with the usual identification). Equivalently, a topological surface is a real 2-manifold, or a complex 1-manifold. Examples C = R 2 ; 2. The upper half plane H = {z C : im z > 0}; 3. The graph of any continous map f : U C C with u open; 4. The Riemann sphere S 2 = C { } = P 1, equipped with stereographic projections at the poles. If we define S 2 := {(x, y, z) R 3 : x 2 + y 2 + z 2 = 1} then the projections φ 1 : S 2 \ N = R 2, φ 2 : S 2 \ S = R 2 are given by φ 1 (x, y, z) = (x/1 z, y/1 z) and φ 2 (x, y, z) = (x/1+z, y/1+z). That these are homeomorphisms is clear. In case it ever comes up, it s nice to have the descriptions of the maps on hand. Note that we re defining P 1 = C { } to be the one point compactification of C. The topology is defined by the condition that U P 1 is open if U C is open or U = (C\K) where K is some compact set. It s a fact that this topology is Hausdorff. The local charts describing P 1 as a surface are given by C P 1 by z z and C P 1 by z 1/z, with obvious conventions. In fact, another useful fact that might come up is the following: 3

4 Fact 1. Let X be a locally compact Hausdorff topological space and suppose f : X X is a homeomorphism. Then f extends to a homeomorphism of the one point compactification of X. In particular, homeomorphisms of R n always extend to homeomorphisms of S n. Definition 2. A homeomorphism φ : U X V C is called a complex chart. Two charts φ i : U i V i are called holomorphically compatible if the composition φ 2 φ 1 1 : φ 1 (U 1 U 2 ) φ 2 (U 1 U 2 ) is a biholomorphism. Definition 3. An atlas on X is a collection U = {(U α, φ α )} α of charts covering X. By a complex atlas we mean an atlas of charts which are pairwise holomorphically compatible. Definition 4. A Riemann surface is a pair (X, U) where X is a connected topological surface and U is a complex atlas for X. Examples Note that C, H (open subsets of C in general), and graphs of continuous functions all admit atlases with only one chart. They are thus trivally complex atlases. 2. The atlas for P 1 consists of the two charts given above. The appropriate compositions are easily checked to be biholomorphisms. Remarks In most cases we ll deal with atlases with only finitely many charts. 2. If the surface is compact, there are necessarily at least two charts. Definition 5. Two complex atlases U, U are called analytically equivalent if every chart in U is holomorphically equivalent with every chart in U. Note that this is an equivalence relation amongst complex atlases. Definition 6. By a complex structure Σ on a topological surface X, we mean an equivalence class of analytically equivalent complex atlases on X. Note that every complex structure Σ on X contains a representative U which is a maximal atlas. We can now (finally) give a formal definition of our objects of study! Not everyone assumes connectedness but we will. Definition 7. A Riemann surface is a pair (X, Σ) where X is a connected topological 4

5 surface and Σ is a complex structure on X. Examples The examples of open subsets of C and graphs of complex functions are all examples of Riemann surfaces. 2. The torus X := C/Γ where Γ is the lattice generated by integer combinations of two R-linearly independent vectors in C. We give X the quotient topology by the quotient map π : C X sends z [z], which by default makes π continuous. Thus X is connected, compact (the image of it s fundamental parallelogram) and Hausdorff (since π is an open quotient map). To show that X admits a complex structure, for each point p choose a small enough disk D p on which π gives a chart from C to D p. These are obviously compatible since the charts will compose to be the identity plus an integer if their domains intersect. Algebraic curves are another example of Riemann surfaces. Let p(z, w) be a non-constant polynomial in 2 complex variables. Then the the subset C = {(z, w) C : p(z, w) = 0} C 2 is called the algebraic curve determined by p. Definition 8. The algebraic curve C is called smooth at (z 0, w 0 ) if: ( p p = z, p ) 0 at (z 0, w 0 ) w Otherwise we call C singular at (z 0, w 0 ). Examples Let p(z, w) = w 2 z. Then p = ( 1, 2w) doesn t vanish, so p is smooth everywhere. 2. Let p(z, w) = w 2 z 3 (a cubic with a cusp). Then p = ( 3z 2, 2w) vanishes at (0, 0), where p has a cusp. Thus p is smooth everywhere except the origin. Proposition 1. Let C be an algebraic curve and let S = C \ {singular points of C}. Then S admits a natural complex structure that makes it into a Riemann surface. Recall the Implicit Function Theorem (complex version): Theorem (Implicit Function Theorem). Let C be an algebraic curve determined by p. Suppose that (z 0, w 0 ) is a point on C where p/ w(z 0, w 0 ) 0. Then there is a disk D 1 centered around z 0 and D 2 centered around w 0 and a holomorphic map φ : D 1 C D 2 C with φ(z 0 ) = w 0 such that 5

6 C (D 1 D 2 ) = {(z, φ(z)) : z D 1 } Basically, this lets us take a relation on two complex variables and convert it to the graph of an honest to goodness function. I ll include a proof when I have time to write it up. It s currently available as a handout. Proof of Proposition. Since points of S are smooth, at least one partial deriviative doesn t vanish, so p is locally the graph of a holomorphic function. By our remarks earlier, such objects are Riemann surfaces. Proof of the Implicit Function Theorem. [When I have time.] 1.2 Maps on Riemann Surfaces Having defined the objects of study, we ll now talk about functions on them. Definition 9. Let X be a Riemann surface and Y X an open subset. A function f : Y C is called holomorphic if its composition with any chart is a holomorphic map C C, i.e. for any chart ψ : U X C, the map f ψ 1 : ψ(u Y ) C is holomorphic. The set of all holomorphic functions on Y is denoted O(Y ). Remark 2. Since constant functions, sums and scalar multiples of holomorphic functions are again holomorphic, this makes O(Y ) into a C-algebra. Example 6. We find all the holomorphic functions on P 1. Let X = P 1 = C { }, and recall the stereographic projections given as φ 1 : U = C C by z z and φ 2 : V = C { } by z 1/z. Suppose that f : P 1 C is holomorhpic. Then the maps below are holomorphic maps on C: f φ 1 1 : φ 1 (C) C f φ 1 2 : φ 2 (C { }) C Moreover, f φ 1 (z) = f(z) and f φ 1 2 (z) = f(1/z). These have to agree on the intersection C. If we write out a Laurent series expansion for f and use uniqueness, we find that the only possible such expansion is a constant function. Hence O(P 1 ) = C. This isn t surprising because P 1 is compact, but more on that later. 6

7 Recall the following: Theorem (Riemann s Removable Singularity Theorem). Let U be an open subset of a Riemann surface X and let a U. Suppose that f O(U \ {a}) is bounded in some neighbourhood of a. Then f can be extended uniquely to some f O(U). We won t prove this, but it follows directly from the corresponding result over C. Definition 10. Suppose that X and Y are Riemann surfaces and f : X Y is continuous. Then f is called holomorphic if its composition with any pair of charts is a homolomorphic map from C C, i.e., for any charts ψ 1 : U 1 X V 1 C ψ 2 : U 2 Y V 2 C such that f(u 1 ) U 2, we have that the composition is holomorphic. ψ 2 f ψ 1 1 : V 1 V 2 If f and its inverse are holomorphic, we call f a biholomorphism. If f : X Y is a biholomorphism, we call X and Y isomorphic as Riemann surfaces. Examples Any holomorphic map f : X C is really a map between the Riemann surfaces X and C, so this definition subsumes the previous one. 2. This definition of holomorphic is obviously preserved by composition of maps (when it makes sense). 3. Linear fractional transformations (mobius transformations) are autormorphisms of P 1 in the precise sense given above [this might be an exercise at some point]. Recall that holomorphic functions are determined by their values on open sets, or more generally on sets with a limit point. We ll prove the analogous result for Riemann surfaces. Theorem 1 (Identity Theorem). Suppose that X and Y are Riemann surfaces and that f 1, f 2 : X Y are holomorphic. Suppose that there is a set A with a limit point in X. Then f = g on X 7

8 Corollary 1. If f and g agree on a nonempty open subset U X then f and g agree on X. We won t prove the identity theorem in its fullest generality, since it more or less follows in the same way that the version over C does, and with some fiddling with charts. We will prove the basic versions properly. Definition 11. Let X, Y be topological spaces. Let S X and x S. We say that x is an isolated point of S if there is an open neighbourhood U of x such that (U \ {x}) S =. Otherwise, x is called an limit point for S (some authors use accumulation point). Facts 2. Here are some facts about limit points. 1. S is closed if and only if it contains all of its limit points. 2. If x is a limit point of S and U is an open neighbourhood of x such that U S then x is a limit point of U S. 3. Let φ : X Y be injective and continuous. If x is a limit point of S then φ(x) is a limit point of φ(s). Theorem 2. Let D C be a domain and suppose that f, g are holomorphic on D. Suppose that f and g agree on a set A that has a limit point c in D. Then f and g agree on D. Proof. Here s a nice proof that s a little different from the one given in class. Wikipedia! Thanks Since D is open and connected, if we can show that f and g coincide on a set that is nonempty, open, and closed, we will be done. Write h = f g, so without loss of generality we suppose that h is zero on A, and we try to show that h is zero on D. Consider the set: S = {z D : h (k) (z) = 0 for all k} Note that S is closed because it s the intersection of the closed sets (h (k) ) 1 (0) (h is continuous). To see that S is open, let w S. Then h has a Taylor series expansion at w which is identically zero. By Taylor s theorem, h vanishes in a neighbourhood of w, and in so particular this neighbourhood is in S. Thus S is open. We now show that S is nonempty. Consider the Taylor expansion of h about c. Let m be the smallest integer for which h (m) (c) 0, so that we can write 8

9 h(z) = (z c) m h(z) where h is another holomorphic function that doesn t vanish at c. By continuity, we conclude that h(z) is nonzero in a neighbourhood B \ {c} of c. On the other hand, this contradicts the assumption that c is a limit point of A, since B \ {c} is an open set that isolates c. Thus c S, so S is nonempty. This completes the proof. Corollary 2. Suppose f, g are holomorphic on a domain D C. If f = g on a nonempty open subset W D then f = g on D. Proof. Again, let h = f g, and try to show h = 0 on D. Note that Z = h 1 (0) is closed by continuity, and so W Z. Hence h = 0 on W. In the previous theorem, set A = W, which is closed and hence contains a limit point. The conclusion follows. Corollary 3. Let f : X Y be holomorphic. If f is nonconstant, then f 1 (pt) is an isolated set of points. Proof. Suppose that f 1 (a) contains a point that isn t isolated, so that a is a limit point of A = f 1 (a). Then f(x) = a for all a A and A has a limit point, so by the identity theorem for Riemann surfaces we conclude that f(x) = a on X. Definition 12. Let X be a Riemann surface and let Y X. A meromorphic function on Y is a function f : Y Y C such that Y is open and f is holomorphic on Y, and: (i) Y \ Y contains only isolated points; (ii) for all points p Y \ Y we have lim x p f(x) =. We call the points of Y Y the poles of f. The set of all meromorphic functions on Y is denoted M(Y ). Examples Linear fractional transformations f(z) = (az + b)/(cz + d) are meromorphic functions on P 1. They have poles at d/c but are otherwise holomorphic. 2. Polynomials can be extended to a map P 1 P 1 by. Note that lim z p(z) = and is isolated. 3. The function f(z) = e 1/z is not meromorphic. It does have a singularity at z = 0 which is isolated, but is an essential singularity, in the sense that the limit condition above isn t met. 9

10 Remarks Condition (ii) above ensures we re getting poles in the usual sense. 2. Not every holomorphic function f : C C extends to a meromorphic function f : P 1 P 1. For instance, e z doesn t extend, since it s limit at infinity isn t really sensible. Theorem 3. Suppose X is a Riemann surface and f M(X). For each pole of f define f(p) =. Then the map f : X P 1 is holomorphic. Conversely, if f : X P 1 is holomorphic then either f is constant or else takes the value at an isolated set of points. Moreover, f X\f 1 ( ) is meromorphic. Remark 4. In other words, there is a one to one correspondence between holomorphic functions X P 1 and meromorphic functions X C. Proof. Suppose f M(X) and let P denote the set of poles of f. Define the extended function f : X P 1 by sending elements of P to. Note that by the limit condition, f is actually continuous. To check that the resulting map f : X P 1 is holomorphic, choose charts φ : U X V C ψ : U P 1 V C with f(u) U. Denote the composition g = ψ f φ 1 : V V. Away from the poles, g is holomorphic because f was. Note furthermore that for any p P, g(φ(p)) = ψ(f(p)) C which means we can find a neighbourhood W of φ(p) on which g is bounded. Then g is bounded and holomorphic on W \ φ(p), so by Riemann s removable singularity theorem, g can be extended to all of W while preserving holomorphy. Since this is a local construction and the poles are isolated, this completes this part of the proof. The converse statement follows from the identity theorem (we ve proved this already). The limit condition on the pole set follows from the continuity of f. Remark 5. The identity theorem also holds for meromorphic functions, since we ve proved it for any map between Riemann surfaces, and we just characterized meromorphic maps on X as holomorphic maps X P 1. 10

11 Theorem 4 (The Open Mapping Theorem). Let X and Y be Riemann surfaces and f : X Y be a non constant holomorphic map. Then f is an open mapping. This follows immediately from the corresponding result in C because being open can be checked locally. We won t prove this one. Theorem 5. Let D be a domain and f : D C be holomorphic. Then f is an open mapping. Theorem 6. Let X, Y be Riemann surfaces and X be compact. Suppose that f : X Y is holomorphic and non constant. Then f is surjective. In particular, Y is compact. Proof. Suppose that f is holomorphic and nonconstant. Then by the open mapping theorem, f is open, so f(x) is open. On the other hand, f is also a compact subset of Y so is closed. Since Y is connected, f(x) = Y. Corollary 4. Every holomorphic function on a compact Riemann surface is constant. In fact, this is true in general for complex manifolds: O(M) = C if M is complex. We ve recovered the results we proved earlier: O(P 1 ) = C, O(C 2 /Γ) = C. Note that on the torus, this says that the only holomorphic doubly periodic functions are constant. Moreover, if f : C C is a non constant doubly periodic meromorphic function, we can interpret this as a function f : T P 1. By the corollary, f takes every value in C, which is also an interesting fact. We can prove a bunch of things with the open mapping theorem. Theorem 7 (Louiville s Theorem). Every bounded holomorphic function f : C C is constant. Proof. Consider f : P 1 \ { } C which is bounded in a neighbourhood of. By Riemann s removable singularity theorem, f can be extended to P 1, but by the above corollary this means f is constant. In particular f C is constant. Theorem 8 (The Fundamental Theorem of Algebra). Every nonconstant complex polynomial has a root. Proof. View a polynomial p as a meromorphic function on P 1. Thus p is a holomorphic map p : P 1 P 1. Since P 1 is compact, p is surjective, so p 1 (0), so p has a root. [This has got to be one of the shortest proofs I ve every seen.] 11

12 Theorem 9 (Local Behavior of Holomorphic Mappings). Let X, Y be Riemann surfaces and f : X Y be holomorphic and non constant. Then there are local coordinate charts in which f looks like z z k for some k, and this k does not depend on the charts chosen. More precisely, let a X and write b = f(a). Then there is some k 1 and charts and f(u) U so that the following all hold: (i) a U, φ(a) = 0, b U, ψ(b) = 0; φ : U X V C ψ : U Y V C (ii) The composition f = ψ f φ 1 : V V is given by f(z) = z k. Proof. Pick coordinate charts around a and b. Then by translation we can certainly arrange for (i) to hold. To get (ii), write f locally as f(z) = z k h(z) where h(0) 0. Choose an analytic branch of the k th root function to locally write f(z) = (zh(z) 1/k ) k where the piece inside is holomorphic. Now just adjust the charts by setting φ = α φ, where α(z) = zh(z) 1/k. In terms of new local parameters, f has the required form. Remark 6. Note that for all points in a neighbourhood of b (except b), there are exactly k preimages. Locally, this means that f : X Y is a branched covering, with branch locus at b. There is a picture for this: x. We ll investigate this more in the next section. 1.3 Branched and Unbranched Covers We develop the minimum of covering space theory required to discuss its application to Riemann surfaces. Definition 13. Let p : X Y be a continuous map. For y Y, we call the set p 1 (y) the fiber over y and say that points in the fiber lie over y. If every fiber is discrete (finite), then we call p discrete (finite). Remark 7. Discrete maps of compact spaces are automatically finite, since the fiber is a discrete subset of a compact space. 12

13 Remark 8. Let f : X Y be a map of Riemann surfaces, and suppose X is compact. Then we ve seen that Y is compact, and so if f is discrete then it s automatically finite. Our motivating question for a bit will be the following: if X is not compact, then when can we ensure that f : X Y is still finite? Definition 14. A map f : X Y (usually of locally compact spaces) is called proper if f 1 (K) is compact for all K Y compact. Examples Any continuous map with X compact is necessarily proper. 2. Constant maps C C are not proper. Theorem 10. Let f : X Y be a non constant holomorphic proper map of Riemann surfaces. Then f is finite. Proof. Let y Y. Then f 1 (y) is discrete. By properness, f 1 (y) is compact and discrete, hence finite. We re going to work towards the statement that non constant, holomorphic, proper maps of Riemann surfaces are branched coverings. In particular, when we count with multiplicity, every point has the same number of preimages, which is a pretty remarkable fact. Definition 15. Let f : X Y be a non constant holomorphic map of Riemann surfaces. A point x X is called a branch point if f fails to be injective in any neighbourhood of x. The map f is called unbranched if f has no branch points. [This is equivalent to an etale cover in some sense.] Examples The map f : C C defined by z z k is branched at z = 0, but is unbranched on C. 2. The exponential map exp : C C is unbranched since its derivative never vanishes. 3. If f : X Y is holomorphic and non constant, then f is unbranched at points where the index in the local description can be taken as k = The quotient map π : C C/Γ is unbranched (exercise). Definition 16. Let f : X Y be continuous. We call f a local homeomorphism if for all x X, there is an open neighborhood U x of x such that f Ux : U x f(u x ) is a 13

14 homeomorphsm. We call f a covering map if for all y Y, there is a neighbourhood V y of y such that: p 1 (V y ) = j J U j where U j are open, disjoint, and f Uj is a homeomorphism. The typical picture of this definition looks like: ftp.ius a y Remark 9. Note that covering spaces are local homeomorphisms, but the converse is not true. Let X be the real line with two origins, and Y = R. Then the map X Y that identifies the origins in X is a local homeomorphism, but not a covering map. Any inclusion will also trivially provide an example if it isn t surjective. If both of these examples feel a bit like cheating, consider the partial covering map (0, 2) S 1 by x e 2πix. Then any neighbourhood around 1 S 1 C will fail the covering space requirement. In general, if we add (Hausdorff), properness, and locally compactness hypotheses then the converse does hold; i.e., proper maps between manifolds. Theorem 11. Suppose that p : X Y is a covering map and that Y is connected. Then all fibers of p have the same cardinality. In particular, p is necessarily surjective. Proof. Let y 0 Y and let y V Y such that: p 1 (V ) = j J U j where U j are all disjoint and p : U j V is a homeomorphism. Thus p 1 = J. We note that for any other y V, we must also have p 1 (y) = J since the same neighbourhoods work for y V and p Uj is a homeomorphism for each j. 14

15 Now let A = {y Y : p 1 (y) = J }. We ve just proved it s open, and note that it must be closed by the exact same claim above. Moreover, A is nonempty since y 0 A. Since Y is connected we conclude A = Y, and this completes the proof. We require two more facts: Lemma If X, Y are locally compact and f : X Y is proper then f is closed. In particular, proper maps of Riemann surfaces are both open (Open Mapping Theorem) and closed. 2. If f : X Y is closed then for all y Y and any open neighbourhood U X of f 1 (y), there is an open neighbourhood V of y such that f 1 (V ) U Proof of 1. There is a characterization of closed sets in locally compact spaces: A is closed if and only if A K is compact for all compact subsets K. It should follow from this, but I don t want to write it down. Proof of 2. Note that X \ U is closed, so that A := f(x \ U) is closed and y / A. Thus define V := Y \ A is an open neighbourhood of y such that f 1 (V ) U. After this digression into point set topology, we return to our study of maps between Riemann surfaces. Theorem 12. Let f : X Y be a non constant holomorphic map of Riemann surface. Then: 1. If f is unbranched then f is a local homeomorphism. The converse is also true but less useful. 2. If f is unbranched and proper then f is a covering map. Proof of 1. Suppose that f is unbranched. If x X, there is some neighbourhood U of x where f is injective. Since f is holomorphic and non constant, f is continuous and open; thus f U is a homeomorphism. The converse also holds since local homeomorphisms are locally injective. Proof of 2. By the first part, f is a local homeomorphism. Let y Y and let p 1 (y) = {x 1,..., x m } (it s finite by a previous lemma about proper maps of Riemann surfaces). Since p is a local homeomorphism, there are neighbourhoods W j of x j and V j of Y such that p : W j V j is a homeomorphism. Without generality, the W j are disjoint. Since W j is a neighbourhood of p 1 (y) by the previous lemma there is a neighbourhood U V 1 V m 15

16 such that p 1 (U) W 1 W m. Set U j = W j p 1 (U). Then the covering space requirement is met, and so p is a covering map. Remark 10. Note now that if we start with a non constant proper holomorphic map f : X Y of Riemann surfaces which is unbranced, then we get that f is automatically surjective (as a covering map), and so there is a well defined degree of f, given by the number of preimages in a fiber. This can be generalized to the branched case in an obvious way. Theorem 13. Suppose that f : X Y is a non constant holomorphic map of Riemann surfaces. Let A X be the set of branched points, and set B = f(a). Then: 1. A is closed and discrete in X; 2. If f is proper and D is any closed discrete subset then f(d) is also closed and discrete. In particular, if f is proper, then B is closed and discrete. Proof of 1. Set W = X \ A; then f is unbranched on W. Then W is open, and so A is closed. By the local description of holomorphic maps, A is also discrete. Proof of 2. Suppose D X is closed and discrete and that f is proper. Then f is closed, so f(d) is closed. To check that its discrete, supposed not. Then f(d) has a limit point, say some y 0. Then by definition, there is no open neighbourhood V of y 0 which separates y 0 from f(d). We now remark that if K Y is compact, we have f(d) K = f(d f 1 (K)). In particular, if K is finite, we see that f(d) K is finite. But now if V is any open neighbourhood of y 0 with compact closure, V f(d) is finite, which means we can easily separate y 0 from f(d). This is a contradiction so we conclude that f(d) is discrete. The message here is that proper maps are finite and so look like polynomials. Examples The exponential map is not proper since its not finite. 2. Polynomial maps p : C C are proper. The number of preimages in a generic fiber is the degree of the polynomial; this isn t really a coincidence. We now give the real definition of the degree of a map. Definition 17. Let X, Y be Riemann surfaces and let f : X Y be a non constant, proper, holomorphic map. Set A X to be the branch points of f, and let B = f(a). Set 16

17 X = X \ f 1 (B) X \ A (all points that map to a critical value), and Y = Y \ B. Then f : X Y is unbranched and has a well defined degree, say n. For any x X, set v(f, x) = k, where z z k is the unique local description of f at x. If y Y, set m := Note that m = n at the unbranched points. x f 1 (y) v(f, x) This is always a well defined integer by the below theorem: Theorem 14. Let X, Y be Riemann surfaces and f : X Y a non constant, proper, holomorphic map. Then there is an n N such that counting multiplicity, f takes every value n times. Proof. I will prove this eventually. It s posted online. This statement is actually quite strong. As a consequence, we ll see that the existence of maps between Riemann surfaces places strict requirements on the genus, degree etc. (Riemann-Roch). Corollary 5. Let X be a compact Riemann surface. Suppose there exists f M(X) with only one pole and that this pole has order one. Then X is isomorphic to P 1. Proof. We think of f as a holomorphic map f : X P 1, where poles of f are mapped to. Moreover, we have f 1 ( ) = 1. Since X is compact, f is proper, and so by the above theorem must take every value exactly once. Thus f is a degree one cover: X and P 1 are biholomorphic. 1.4 Sheaves and Things About Sheaves Definition 18. Let X, τ be a topological space. A presheaf (F, ρ) on X (of abelian groups) consists of: (i) A family F = {F(U)} U τ of abelian groups; (ii) A family ρ{p U V = res UV : F(U) F(V )} of group homomorphisms such that p U U = id F(U), and for which composition of restrictions satisfies res V W res UV = res UW. 17

18 Remarks We usually just write F and not (F, ρ). 2. Sheaves can take values in different categories; there are obvious definitions. 3. We call elements in F(U) sections over U: sections of a fiber bundle form a sheaf this may be the origin of this terminology. Elements of F(X) are called global sections. 4. The section over should be the zero object in the category. Examples Sheaves of functions (of specified type) are a natural example of a sheaf on a space. 2. The skyscraper sheaf (at p X) is defined by setting F(U) = G if p U, and F(U) = 0 otherwise. If we want F to be a sheaf, we ask for two additional properties: 1. [The locality/identity axiom] Suppose U is open and U = U i for some open sets U i. If f, g F(U) are such that f Ui = g Ui for all i, then f = g. 2. [The gluability axiom] Suppose U is open and U = U i for some open sets U i. If f i F(U i ) are such that f i Ui U j = f j Ui U j for all i, j, then there is some f F(U) such that f Ui = f i for all i. These are natural for sheaves of functions but not guaranteed in general. Note that the glued function in (2) is unique by (1) if F is a sheaf. Examples The skyscraper sheaf is a sheaf, hence the name. 2. [Non-example] Not every presheaf is a sheaf. (i) Consider X = {x, y} with the discrete topology. Define the section over x, y to be G, and the section over X to be G 3. Let restriction maps to x and y be proj 1 and proj 2, respectively. Then this presheaf clearly fails the identity axiom (but not gluability!). (ii) The presheaf of bounded functions on R fails the gluability axiom, but not the identity axiom. Another example of the above flavor is also possible. Sheaves also give us a way to talk about local behavior at points, through something called a stalk. 18

19 Definition 19. Let F be a presheaf on (X, τ) and let a X. The stalk at a is defined to be the quotient F(U)/ a U a where a is the equivalence relation that identifies f F(U) and g F(V ) if there is an open set W U V on which f and g are equal. This captures the notion of the germ of a function. Another way to describe this is as the direct limit of the diagram of F(U) where a U. When X is a Riemann surface and O is the sheaf of holomorphic functions, we recover O a as the germs of holomorphic functions there: this has a natural description as the analytic power series at a. In the example of the skyscraper sheaf, note that we get F a = G if a = p, and 0 otherwise. Remark 12. One thing that we can do is sheafify a presheaf: this fixes any problems that with gluing and restriction that we might have. We don t use it much since the sheaf of holomorphic is indeed a sheaf, but I ll add the material if it s needed later. The relevant proposition is below. Proposition 2. Let F be a presheaf on X. There is a sheaf F + on X called the sheafification of F whose stalks agree with F, and which is isomorphic to F if F is already a sheaf. Proof. Exercise (A2). 2 Analytic Continuation The motivation for this section is the following: given a holomorphic function f on a domain D C, what are the largest open set(s) on which we can extend f (it s unique by the identity theorem). The solution to this problem, roughly, will be to attach a topological space O to the sheaf of holomorphic functions, and consider its connected components. Somewhat miraculously, this turns out to be a covering of the original space X, and questions about holorphic functions on X can be answered in O. The problem in general is that extensions are multivalued: consider the complex square root function. This once we specify a branch, it s well defined on the complex plane minus a ray. It turns out that the largest domain will be unique and isomorphic to a Riemann surface. We ll formalize all of this now. 19

20 Definition 20. Let f O(U). An analytic continuation F of f is a tuple (Y, p, F ) such that: (i) Y is a Riemann surface; (ii) p : Y C is an unbranched holomorphic map; (iii) F : Y C is a holomorphic function such that F p 1 (U) = f p. We think of F as covering f. Our candidate for Y above will be the space O associated to O. Definition 21. Let X be a Riemann surface and F be a presheaf on X. We define F = x X F x to be the disjoint union of all the stalks. p : F X. Note that there is a natural projection map Proposition 3. The space F has a natural topology with respect to which p is a local homeomorphism. Proof. Denote the map which sends a section to a stalk by p x : F(U) F x. Note that this is well defined because the stalk is a local construction. We give F by declaring the sets [U, f] = {p x (f) : x U} to be a basis. Note the following (obvious) facts. 1. p([u, f]) = U; 2. [V, g] [U, f] implies V U; 3. If φ [U, f] then φ = p x (f) with x = p(φ). Let B be the set of such open sets. Note that B covers X, so we only need to check the intersection property of a basis. Suppose that φ [U, f] [V, g]. Then φ = p x (f) where x = p(φ). Similarly, φ = p x (g) where x = p(φ). We conclude that p x (f) = p x (g), i.e., f and g have the same germ at x. By the definition of stalk equivalence, there is an open set W U V such that f W = g W, and so φ [W, f W = g W ] [U, f] [V, g] 20

21 Now we check that p is a local homeomorphism. To check that p is continuous, let U τ and consider p 1 (U). We have p 1 (U) = {φ F : p(φ) U} = To check that the element on the right hand side is open, let φ p 1 (U). Then φ F x for some x U. Let φ = p x (f) for some f F(V ). Then if we denote W = U V, we see that φ [W, f W ] p 1 (U) Thus p is continuous. To see that p is a local homeomorphism, let φ F. As before let φ F x where x = p(φ). We now claim that p [U,f] is a homeomorphism on U. This is certainly a bijection, and we know that p is continuous, so we only need to show that p 1 [U,f] is continuous. But this is clear: if [V, g] [U, f], then x U F x This completes the proof. p 1 [U,f]([V, g]) = V U We specialize a bit to the case we re interested in. The following holds for any presheaf satisfying a precise version of the identity theorem, but we don t really care. Proposition 4. Let X be a Riemann surface and O be the sheaf of holomorphic functions on X. Then O is Hausdorff. Proof. Suppose that φ 1, φ 2 O are distinct. If p(φ 1 ) p(φ 2 ), then we re done since X is Hausdorff. If we have p(φ 1 ) = p(φ 2 ), then let f i F(U i ) where φ i = p x (f i ). Then φ i [U i, f i ]. Now let U U 1 U 2 be the connected component of U 1 U 2 containing x. Note that we still have φ i [U, f i ]. We claim that these are actually disjoint open sets. Indeed, suppose not. Then the germ of f 1 agrees with the germ of f 2 at some point in the domain U. By the identity theorem, we would conclude that f 1 = f 2 on all of U and so φ 1 = φ 2 (that s why we needed U to be connected). This is a contradiction, so we conclude that O is Hausdorff. Theorem 15. Let X be a Riemann surface and Y a Hausdorff topological space. Suppose that p : Y X is a local homeomorphism. Then there is a unique complex structure on Y which makes p a holomorphic map. Sketch of Proof. I won t prove this: it s posted as notes and it s a standard exercise in charts. This theorem also holds more generally. If you replace X with your favorite kind of manifold and replace complex with your favorite kind of category, the result is still true. 21

22 Essentially, pull back the manifold structure on X to Y via the local homeomorphism, and check that this is unique. We ll now work towards defining an analytic continuation. The idea will be to do this along paths, and then have a monodromy result that says this is well defined. Definition 22. Let X be a Riemann surface and let u : I X be a curve with endpoints a = u(0), b = u(1). Pick φ O a and ψ O b. Then we say that ψ is an analytic continuatino of φ along u if the following holds: 1. There is a partition 0 = t 0 < t 1 < t 2 < < t n = 1 of I and connected open sets U i with u([t i 1, t i ]) U i. 2. There are holomorphic functions f i O(U i ) such that φ = p a (f 1 ), ψ = p b (f n ), and such that f i agree in a precise sense: f Vi = f i+1 Vi, where V i is the connected component of U i U i+1 containing u(t i ). Lemma 2. Let X be a Riemann surface and let u : I X be a curve in X with endpoints a = u(0), b = u(1). Then the germ ψ O b is the analytic continutation of φ O a if and only if there exists a curve û : I O such that û(0) = φ, û(1) = ψ, and p û = u. In this case, we call û a lifting of u. Proof. Suppose first that ψ is an analytic continuation of φ along u. Then set û(t) = p u(t) (f i ) O u(t) if t [t i 1, t i ]. Then we have p û = u (note that û is well defined by the continuation data: f i have the same germs where they agree). To check that û is continuous, we only need to check the basis elements. Suppose that [U, f] O is open and let t û 1 ([U, f]) [0, 1]. Then û(t) = p u(t) (f) = p u(t) (f i ), so f and f i have the same germ at t (for some i). By definition of stalk equivalence, there is an open set W such that u(t) W and W U U i and for which f i W = f W. In particular, the germs of f and f i are the same in W. Then u 1 (W ) is open, contains t, and is contained in û 1 ([U, f]). Thus û is continuous and is a lifting of u. To prove the other direction, suppose that û is a lift of u. For each t I we have û(t) O u(t), where û(t) = p u(t) (f t ) for some f t O(U t ) and where U t X is an open set containing u(t). Thus û(t) [U t, f t ]. The collection of these sets covers û(i), and so by compacts admits a finite refinement. Denote this refinement by {[U i, f i ]} where i = 0,..., n. This gives a partition of I (throw in points if necessary to make the continutation requirement). Moreover, by stalk equivalence, the second condition will be met. This completes the proof. 22

23 What this says is that analytic continuations along a curve are in bijections with lifts of curves to O. They are essentially well defined by the following theorem. Theorem 16 (The Monodromy Theorem). Let X be a Riemann surface and u 0, u 1 be homotopic curves. Denote the homotopy by A : I I X, and the intermediate curves by u s. Suppose that φ O a admits an analytic continuation along each u s. Then the continuation along each u s well defined. One of the main consequences of this is that we will be able to extend functions on simply connected domains. Proof. Suppose u 0 and u 1 start at a X and end at b X. We can lift each u s to some û s : I O. Moreover, the homotopoy lifts to a homotopy  : I I O. Note that û s (0) = φ for all s, and that û s (1) p 1 (b) for each s. Note that these all have to be in the connected component Y of O that contains φ O a, i.e., that û s (t) Y for all s, t. Thus p Y : Y X is a holomorphic map of Riemann surfaces, where Y has a complex structure induced from X. Hence p Y is a holomorphic non constant map, so the fibers are discrete. In particular, {û s (1)} s I is a connected subset of the discrete space p 1 (b), so we conclude that û s (1) = ψ for all s and some ψ. We ll now discuss the happy case when X is simply connected. Definition 23. A topological space X is simply connected if π 1 (X) = 0, i.e., if any two curves with the same start and end points are homotopic. Corollary 6. Let X be a simply connected Riemann surface. Let a X and φ O a. Suppose that φ admits an analytic continuation along any curve starting at a. Then there exists a globally defined holomorphic function f with f a = φ. Proof. Define a function by f(x) = ψ x (x) where ψ x is the analytic continuation of φ from a to x along any curve. This is well defined by the above theorem, and holomorphic since f is locally the germ of a holomorphic function. Note that if X is not simply connected then not all germs can be analytically extended to all points along curves in a well defined way. Example 14. Consider f(z) = z. Note that f isn t analytic at 0, so our best hope would be to define a function f on C, which isn t simply connected. One can check, however, that paths on different sides of the origin produce different continutations. This is why we need to choose a branch of z on a domain which looks like C \ ray. 23

24 [This section needs finishing, but I m leaving it for now.] 3 Calculus on Riemann Surfaces and Differential Forms We now study differential forms on Riemann surfaces and the complexified tangent bundle. As a motivating example, note that we can define a sheaf E on C = R 2 in the following way: E(U) = {f : U C : f is infinitely differentiable wrt x, y} where by differentiable we mean with respect to the partials / x, / y. Moreover, we can define: and z := 1 ( 2 x i ) y z := 1 ( 2 x + i ) y Note that by the Cauchy-Riemann equations, the holomorphic functions are exactly the kernel of / z. Define a new sheaf by O = ker z. This is the sheaf of holomorphic functions on C. In a similar way, we can extend this idea to any Riemann surface, but we have to be careful. Let X be a Riemann surface and let Y X be open. Definition 24. A function f : Y C is differentiable (infinitely differentiable) if for all charts, φ : U Y V C on X, the composition f φ 1 : V C is differentiable. Remarks We only need to check differentiability on a set of charts that cover X, as usual. 2. For our purposes, differentiable will be synonomous with smooth. 3. Holomorphic implies smooth, but the converse is of course false. Now we define E(Y ) = {f : Y C : f diff on Y } to be the sheaf of smooth functions on X, with the obvious restriction maps. If we pick a coordinate chart φ : U C X at a, 24

25 with φ = z = x + iy, then we can also make sense of the partial derivatives with respect to x, y, z, z. They are defined concretely by f x = ( ) x (f φ 1 ) φ = φ x (f φ 1 ) Note that there s a lot implicit in this notation, but it s justified in the sense that we get the usual rules for addition, multiplication etc. Now we ll work on defining a cotangent space. If a X, we set m a to be the ideal of E a consisting of those germs of functions that vanish at a. Moreover, we set m 2 a to be the subset of m a consisting of those germs which vanish to second order, i.e., Remarks 14. m a = {η E a : η = [f] and f(a) = 0} { m 2 a = η m a : η = [f] and f x = f } y = 0 1. This is well defined, in that these ideals don t depend on the choice of chart or representatives. If we have any other chart around a, say (U, φ = z ) then is also zero. f x = f x x x 2. Note that m 2 is a vector subspace of m, so we can define the quotient space T (1) a := m a /m 2 a We call this the cotangent space of X at a. Note that this is complex 2-dimensional, even though X is complex one dimensional. This doesn t quite mesh with the usual tangent space, but there is good reason for this. 3. If f E(U) and a U then the differential d a f T a (1) d a f := [f f(a)] mod m 2 a. of f at a is the element Theorem 17. The space T a (1) has basis given by {d a x, d a y} or {d a z, d a z}. Moreover, if f E(U) for some open set U then and d a f = f x (a)d ax + f y (a)d ay d a f = f z (a)d az + f z (a)d a z 25

26 Proof. To see that these elements space T a (1), suppose that we have t = φ mod m 2 a, and φ = [f] for some f E(U) and f(a) = 0. Then using a Taylor expansion we can write but modulo m 2 a this is just φ = [f] = [c 1 (x x(a)) + c 2 (y y(a)) + higher order terms] c 1 [(x x(a))] + c 2 [y y(a)] and so we have t = c 1 d a x + c 2 d a y. To show that d a x and d a y are linearly independent, suppose that we have c 1, c 2 C such that c a d a x + c 2 d a y = 0 in T a (1). Then consider the function f = c 1 (x x(a)) + c 2 (y y(a)) E(U) We have [f] = c 1 d a x + c 2 d a y, but on the other hand we ve seen that c 1 = f/ x(a) = 0, and c 2 = f/ y(a) = 0. Thus these elements are linearly independent. A similar proof shows that {d a z, d a z} form a basis. We see that there are two natural vector subspaces of T (1) a. Definition 25. We define cotangent vectors of type (1, 0) and (0, 1). If (U, φ = z) is a coordinate chart with a U, then by above T a (1) = span C {d a z, d a z}. We set T 1,0 a T 0,1 a = Cd a z = Cd a z It turns out that span R {d a x, d a y} is exactly T a, but we want to keep information about the complex structure on X. That s why this is a bit bigger, even if it seems a bit morally wrong. Remark 15. This definition is independent of the coordinate chart chosen. another chart (U, z ) then note that d a z = z z (a)d az + z z (a)d a z The second term vanishes since φ and φ are holomorphic charts. Note: If we pick z ( ) z = φ z (φ φ 1 ) = 0 Hence while changing coordinate charts may scale the vector spaces Ta 1,0 and Ta 0,1, but the subspaces remain invariant. We thus have T (1) a = T 1,0 a T 0,1 a. 26