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by: Jayde Lang


Marketplace > Rice University > Mathematics (M) > MATH 428 > TOPICS IN COMPLEX ANALYSIS
Jayde Lang
Rice University
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This 36 page Class Notes was uploaded by Jayde Lang on Monday October 19, 2015. The Class Notes belongs to MATH 428 at Rice University taught by Staff in Fall. Since its upload, it has received 5 views. For similar materials see /class/224932/math-428-rice-university in Mathematics (M) at Rice University.




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Date Created: 10/19/15
VECTOR BUNDLES ON RIEMANN SURFACES SABIN CAUTIS CONTENTS 1 Differentiable Manifolds 2 Complex Manifolds 21 Riemann Surfaces of Genus One 22 Constructing Riemann Surfaces as Curves in P2 23 Constructing Riemann Surfaces as Covers 24 Constructing Riemann Surfaces by Glueing 3 Topological Vector Bundles 31 The Tangent and Cotangent Bundles 32 lnterlude Categories7 Complexes and Exact Sequences 33 Metrics on Vector Bundles 34 The Degree of a Line Bundle 35 The Determinantal Line Bundle 36 Classi cation of Topological Vector Bundles on Riemann Surfaces 37 Holomorphic Vector Bundles 38 Sections of Holomorphic Vector Bundles 4 Sheaves 4 1 Cech Cohomology 4 2 Line Bundles and Cech Cohomology 43 Riemann Roch and Serre Duality 4 4 Vector bundles7 locally free sheaves and divisors 4 5 A proof of Riemann Roch for curves 5 Classifying vector bundles on Riemann surfaces 51 Grothendieck7s classi cation of vector bundles on ll 5 2 Atiyah7s classi cation of vector bundles on elliptic curves References 1 DIFFERENTIABLE MANIFOLDS Two topological spaces X and Y are horneornorphic if there exist continuous maps f X7Y andg Y7X suchthat gofz39dX and fogz39dy This is denotedXEY Exercise 1 Show that 1 and the open unit interval 071 are not homeomorphic Exercise 2 Show that 071 and the real line R are homeomorphic Warning to show X Y it is not enough to nd a continuous map f X 7 Y which is 1 1 and onto because the inverse map f 1 may not be continuous For example7 the natural inclusion f 071 7 1 is continuous and bijective but the inverse f 1 is not continuous at W A manifold of dimension n is a Hausdorff topological space M such that every point p E M has a neighbourhood p E U C M which is homeomorphic to the open unit ball in R In other words7 M locally looks like R Example The real line R7 the unit circle 1 and the open interval 071 are one dimensional manifolds The semiopen interval 071 is not a manifold since there is no open neighbourhood of 1 homeomorphic to an open interval in R Example The sphere 827 the torus Sl gtlt 817 the open cylinder 071 gtlt 81 as well as any open subset of R2 are two dimensional manifolds An open cover of a topological space X is a collection of open subspaces Ua such that UaUa X A manifold M is compact if every open cover of M has a nite subcover or equivalently if every sequence of points C M contains a subsequence which converges to a point poo E M By de nition7 an n dimensional manifold M is covered by open sets Ua which are home omorphic via maps fa Ua 7 R to an open ball in R Denote U04 Ua U and fa f o f1 faUa 7 f Ua One should think of M as being built from these balls by glueing Ua to U along U04 as dictated by fa ie p is identi ed with fa p The open sets faUa C R are called charts and the functions fa between them are called transitioafuactz39oas By construction7 the transition functions satisfy o fa f5 on U04 Ua m U f vofa few on Ua w U04 U Uw Conversely7 one can reconstruct M given the open cover Ua together with transition func tions fa satisfying these two conditions Example Consider Sl 2 y2 1 C R2 We can use as an open cover the sets U1 1 7 071 and U2 1 7 0771 Then we have homeomorphisms fl Ul 7 R given by f17y and f27y One can check that ffl R 7 U1 is given by tgt gt 7 Then fgl f2 0 ffl R 7 R is given by t gt gt Thus 1 is obtained by glueing R to itself along R R 7 0 using the transition function f21t A manifold has the extra structure of a dz crcatz39ablc manifold if all the maps all the transition functions are not only continuous but also 0 in nitely differentiable ln dimensions at most three7 these two notions coincide 7 in other words7 a topological manifold of dimension one7 two or three has a unique differentiable structure This is not true in higher dimensions For example7 by a result of Milnor7 the seven sphere S7 has exactly 28 different differentiable structures 2 The following is a short summary of classi cation results for compact manifolds o The only compact 0 dimensional manifold is the point 0 The only compact 1 dimensional manifold is the unit circle 81 0 There are two types of surfaces orientable and non orientable The orientable sur faces are classi ed by their genus g In other words7 for each 9 E Z20 there is exactly one surface of that genus The sphere has genus zero7 the torus T2 Sl gtlt Sl has genus one7 the double torus containing two holes has genus two and so on 3 dimensional manifolds are classi ed by the geometrization conjecture of Thurston Recently7 a proof was proposed by Perelman and it is being veri ed at the moment The classi cation of 4 manifolds was essentially established by Freedman although the classi cation of di erentiable 4 manifolds remains open Manifolds of dimension at least 5 both topological as well as differentiable have in principle been classi ed 2 COMPLEX MANIFOLDS A differentiable manifold of dimension n is modelled on R and 0 functions Analogously7 we can use C together with holomorphic transition functions to construct complem manifolds of complex dimension n real dimension 2n Locally7 a complex manifold will have charts Ua with maps fa Ua a C such that the transition functions fa from faUa U C C to f Ua U C C are biholomorphic instead of just homeomorphisms Example The 2 sphere 52 has the structure of a complex manifold since 52 can be gotten by glueing two copies of C along C C 7 0 using the transition function 2 gt gt This construction is analogous to the way we constructed the circle 1 by glueing together two copies of R along R R 7 0 Given two complex manifolds M1 and M2 a morphism f M1 a M2 is a continuous map such that its restriction to charts is holomorphic We say f is an isomorphism if it has a holomorphic inverse Before continuing we mention a few useful results about holomorphic maps 0 Open Mapping Theorem A nonconstant holomorphic map C a C maps open sets to open sets 0 A holomorphic map which is injective and surjective is biholomorphic ie it has an inverse which is also holomorphic A compact complex manifold of complex dimension one is known as a Riemann surface In a way7 these are the simplest compact7 complex manifolds and the source of some incredibly beautiful mathematics It turns out a complex manifold is automatically orientable see section 35 Thus Rie mann surfaces are topologically classi ed by their genus The sphere genus g 0 has only one complex structure In other words7 two Riemann surfaces which are homeomorphic to the 2 sphere then are actually isomorphic as Riemann surfaces ie there is a biholomorphic map between them This result follows from the following more general theorem Theorem 21 Uniformization Theorem Any simply connected complecc manifold of di mension one is isomorphic to precisely one of the following spaces 3 o the Riemann sphere C o the complep plane C o the hyperbolic plane H z E C 17712 gt 0 lntuitively7 we think of C as C with a point added at in nity thus C compacti es C The uniformization theorem above is remarkable since it says that any simply connected domain in C which is not all of C is isomorphic to H Exercise 3 Show directly that the open unit disk A z E C lt 1 is isomorphic to H hint use the map 2 gt gt I 5 Proposition 22 The holomorphic automorphisms ofC are 2 gt gt azb for some ab 6 C with a 31 0 Proof Clearly z gt gt a2 b is biholomorphic if a 31 0 Conversely7 suppose f E AutC By composing with z gt gt a2 b for appropriate a and b we can assume f0 0 and f 0 1 Thus fz 2 1222 a323 locally around 0 If fz is not a polynomial then f1z has an essential singularity at z 0 which by Picard7s theorem means that f1z it has in nitely many solutions in 2 around 0 and hence is not injective Thus fz is a polynomial But if degf gt 1 then for a general a E C the equation fz a has more than one solution meaning f is not injective again Whence fz z D Exercise 4 Use this to show that A b AutC 2 H 061 ad i be 7s 0dI d e C 2 PGL2CC c recall that we can think of C as C with a point added at in nity Proposition 23 AutH PSL2R where lt Z Z acts by z 32433 21 Riemann Surfaces of Genus One What are the complex structures one can put on the genus one surface T2 T2 is the quotient of R2 by Z992 via the free action R2 by ab Ly x a7y b If T2 has a complex structure then R2 which is simply connected and hence the univeral cover of T27 inherits a complex structure from T2 By the Uniformization Theorem we know R2 has two possible complex structures7 namely C and H Proposition 24 The universal cover of a Riemann surface of genus one is C Proof Recall that an action of G on X is discrete if for any point p E X you can nd a ball B around p such that G p B p Note that the action of Z 69 Z on R2 described above is discrete On the other hand7 the lemma below shows that there is no discrete Z 69 Z action on H Thus the universal cover of a genus one Riemann surface cannot be H and must be C D Lemma 25 Any subgroup Z 69 Z C AutH acts non discretely on H Proof Denote by a and b the generators of ZGBZ Suppose a is not diagonalizable Then we can conjugate it to the form lt 1 0 i for 0 f Since a and b commute it means b lt 1 some t E R Thus the action on H is not discrete for otherwise ma nb for some m7 n E Z So we can assume a and b are diagonalizable Since they commute they are simultaneously diagonalizable But then a lt 3 a and b g 2 so again they must generate a nondiscrete subgroup of Autllll for otherwise ma nb for some 77171 E Z Hence any genus one Riemann surface is isomorphic to C modulo the action of some xed point free subgroup Z 69 Z C AutC The xed point free automorphisms of C are the translations t1 z gt gt z b Thus any genus one Riemann surface is obtained by quotienting C by two automorphisms t1 and ty for some b7b E C Here t1 and tb correspond to the generators 10 and 01 of Z 69 Z Note that b and b must be pointing in different directions ie they must be linearly independent over R or else the quotient is not a genus one Riemann surface Rotating and scaling C by an appropriate factor allows one to assume that b 1 and b E H z E C Imz gt 0 Thus we get a map from H to the space of Riemann surfaces of genus one given by 739 gt gt Clt17 Tgt We will denote the Riemann surface corresponding to 739 E H by E E stands for elliptic curve which is another name for a genus one Riemann surface Proposition 26 The map from H to the set of Riemann surfaces is onto Two points give the same Riemann surface i they di er by the action of an element of SL2Z Proof Surjectivity of the map follows from the description above The action of 5122 Cl 2 a7b7c7dEZ7ad7bc1 on H is by 2 gt gt 3243 To understand where this action comes from recall that t1 and tb corresponded to the basis elements 10 and 01 of Z 69 Z Of course7 picking different basis elements of Z 69 Z would give the same Riemann surface Since AutZ 69 Z SL2Z changing basis for Z 69 Z corresponds to acting on H by SL2Z To show that two points in H which do not differ by an element of SL2Z correspond to different genus one Riemann surfaces let7s suppose that Clt17 Tgt and Clt17 T gt are isomorphic via some isomorphism f Then this map lifts to an isomorphism f C a C such that the lattice lt17Tgt maps to the lattice lt1T gt Thus a br and c dT where a7b7c7d E Z satisfy ad 7 be 1 since the map must be invertible Hence 739 and 7quot differ the action of an element of SL2Z D Corollary 27 The moduli space M1 ie the parameter space of genus one Riemann surfaces is llllSL2Z which has complep dimension one and happens to be isomorphic to C What are AutET7 An automorphism of ET is the same as giving an automorphism of C which commutes with the projection map p C a ET Proposition 28 E has automorphisms z gt gt zb translations andz gt gt 72 involution For general 739 these are all the automorphisms For two special values OfT we have the following eptra automorphisms ori39zgt gtiz o 1 lz gin32 Aside Just as a Riemann surface of genus one is the quotient of C by ZGBZ it turns out that a Riemann surface 0 of genus g 2 2 is the quotient of H by the xed point free action of a group As before this group is the fundamental group 7T1C of 0 However describing all the possible actions of 7T1C on H is quite dif cult The moduli space of genus g Riemann surfaces is denoted Mg and turns out to have have a complex structure of complex dimension 3973 In other words there is a 3973 dimensional space parametriZing Riemann surfaces of genus g 2 2 Recall that the moduli space of genus one Riemann surfaces is one dimensional corollary 27 whereas M0 is only a point since there is a unique complex structure on 82 22 Constructing Riemann Surfaces as Curves in P2 One natural way of describing manifolds is as the zero set of a polynomial For instance the n dimensional sphere is the zero set ofp 3 x3 7 1 in Rn One can even take the zero set of more than one polynomial For example the zero set ofp and 0 in other words all points z 0 xn satisfying p 0 x0 gives us the n 7 1 dimensional sphere Similarly an effective way to construct complex manifolds is as the zero set of polynomials in C For example 23 212 7 1 0 is isomorphic as a complex manifold to C Unfortunately since C is not compact all complex manifolds constructed this way will not be compact To remedy this we compactify C to get the projective space ll 221 Projective Spaces The projective space llV C lan is de ned as the quotient of CW 7 0 by C C 7 0 where the action is given by A 20 2 A20 Azn Just as a point in C is given in the usual coordinates as an n tuple 21 2 a point in llW is encoded in homogeneous coordinates as an n 1 tuple 20 2 where one keeps in mind that 20 2 and A20 Azn A 31 0 represent the same point Let7s examine P1 in greater detail A typical point of ll 1 is 2021 If 20 31 0 then the point has can be uniquely represented as 1t where t 2120 Thus the locus of points where 20 31 0 is an open subset U0 C lP l which is diffeomorphic to C via the map 2021 gt gt 2120 The complement of U0 is just the point 01 lntuitively P1 is just C with an extra point added at in nity so if we believe that lP l should be a complex manifold then it ought to be the Riemann sphere Let7s check this directly Consider the open subset U1 C llD1 consisting of points 2021 where 21 7E 0 Then as with U0 there is a diffeomorphism U1 C given by 2021 gt gt 2021 The intersection U0 U1 is the locus of points 2021 where 20 31 0 and 21 7E 0 Thus llD1 is obtained by glueing C to C along C by using the map t 4 gt gt 7 Since t gt gt 1t is a holomorphic map over C this gives P1 a complex structure Notice that this is the same glueing map as the one used in section 2 One can similarly study P2 A typical point of P2 is 20 21 22 Then one can consider open subsets U C P2 consisting of those points where z 7 0 As before we nd that U L C2 form holomorphic charts for P2 Notice that the complement of U is now ll Thus P2 is obtained by glueing to C2 a copy of ll Exercise 5 Check that ll has a strati cation HOSiSnC Proposition 29 Autllm E PGLn1C where the action is by left multiplication on 207quot397an39 6 Proof Since ll is Cn 7 0 modulo 3 any automorphism of ll lifts to an automorphism of Cn Conversely any automorphism of CW which commutes with the projection Cn 7 0 7 ll descends to an automorphism of ll The result now follows from the fact that up to translations the automorphism group of Cn is GLn1CC D The important thing for us is that ll is compact In the case of C the zero sets of polynomials give us non compact complex manifolds The equivalent of a polynomial in the setting of ll is a homogeneous polynomial De nition 210 Check that a polynomial f20 2 is homogeneous iffor any A E C7 0 we have f20 A2 f20 2 In other words we want to give a function on CC 1 which is invariant under the 3 action 20 2 gt gt A20 Azn and thus descends to a well de ned function on the quotient WMCi ll Exercise 6 A polynomial f20 2 is homogeneous if and only if every monomial term has the same degree n which is by de nition the degree of f Example 20 z is not homogenous whereas 28 2 is homogeneous of degree 3 If f20 2 is a general homogeneous polynomial then its zero set in ll carves out a complex submanifold of dimension n 7 1 For instance if f202122 22 then the zero set are the points in P2 of the form 20210l which as we saw is the Riemann sphere 82 Since ll is compact the manifold we get will also be compact since it is a closed subset of a compact space Exercise 7 Show that the zero set of f202122 aozo alzl a222 in P2 where the constants as are not all zero is isomorphic to the Riemann sphere A hint use the action of PGL3C on P2 Recall that our goal is to construct examples of Riemann surfaces To do this we can consider the zero set of a homogeneous polynomial f202122 of degree d For d 2 3 the Riemann surface we get will depend on f but its genus will not So our next question is if f has degree d what is the genus of f20 21 22 0 To answer this we need a new tool 222 The Euler Characteristic Let M be a topological manifold and consider an arbitrary triangulation of M The Euler characteristic xM of M is de ned as the number of vertices in the triangulation minus the number of edges plus the number of faces and so on Example Viewing the sphere 52 as a tetrahedron gives us a triangulation of 2 with four vertices six edges and four faces Thus xSZ 4 7 6 4 2 Viewing 52 as the cube gives us eight vertices twelve edges and six faces Again we get xSZ 8 7 12 6 2 the latter is not technically a triangulation so we should not use it but we still get the right answer since each face is a polygon Theorem 211 The Euler characteristic xM does not depend on the speci c triangulation used Proof Given two triangulations of M one can subdivide them further to obtain a common subtriangulation The result follows from the following result Exercise 8 Show that xM remains unchanged after taking subtriangulations D 7 Thus the Euler characteristic MM is an invariant of M It provides a way to distinguish between manifolds namely if xM1 7E XltM2gt then they are not homeomorphic In the case of real orientable manifolds of dimension two this invariant is enough to distinguish them Exercise 9 Show using your favourite triangulation that a Riemann surface of genus g has Euler characteristic 2 7 29 223 Riemann Hurwitz Theorem Suppose f C1 7 Cg is a holomorphic map between two Riemann surfaces Locally around each point p 6 C1 the map p gt gt fp can be written as 2 gt gt anz an zn1 Changing variables one can rewrite it as 2 7 z The integer n is called the rami cation indeco of f at p and denoted ep If n 1 then we say f is unrami ed at p otherwise we say p is a rami cation point A point q E Cg is a branching point if f 1q contains a rami cation point Exercise 10 What does the map 2 gt gt 2 from A to A look like identify the branching points the rami cation points and their indices For any holomorphic map f C1 7 Cg the target Cg contains only nitely many branching points If q E Cg is not a branching point then its preimage f 1q will contain d points where d is called the degree of the map f If q E Cg is a branching point then f 1q will contain less than d points The precise number of points in f 1q is related to the rami cation indices as follows Proposition 212 For any q E Cg we have Zpef1qep d Example Consider the map f 1P1 7 ll given by z gt gt 2d The preimage of a point it E C C lP l are the points z E C satisfying 2 1 it Thus for w 51 0 the preimage contains d points and hence degf d The branching points are 0 oo E El The preimage of 0 is 0 where the rami cation index is d Similarly the preimage of 00 is 00 where the rami cation index is also d check this by writing down the map f in a local neighbourhood of 00 Given f C1 7 Cg the Riemann Hurwitz theorem tells us how to relate the Euler characteristics xC1 and X02 in terms of the degree degf and the rami cation indices Theorem 213 Riemann Hurwitz xC1 dxCg 7 21601 ep 7 1 Proof Choose a triangulation of Cg such that ifp 6 C1 is a rami cation point then fp is a vertex With this added condition the preimage of this triangulation gives a triangulation of C1 Denote by ie f the number of edges vertices and faces in the triangulation of Cg so that XltOggt i 7 e f The number of edges and faces in the triangulation of C1 is de and df respectively The number of vertices is do 7 Zpeclep 7 1 Thus X02d1 2ep71 7dedfdo7ef 7 2ep71 17601 17601 D Notice that the sum on the right side is nite since all but a nite number of points have rami cation index one One can use this result to compute the genus of C1 given the genus of Cg and knowledge about the map f C1 7 Cg Example Consider again the map f 1P1 7 ll given by z gt gt 2d We saw that there are two rami cation points both of which have index d Thus 21611 ep7 1 d7 1 d7 1 8 2d 7 2 From the Riemann Hurwitz equation we nd xlP11 7 d 2 7 2d and hence MP1 2 reaf rming our earlier calculation Example In algebraic geometry Riemann surfaces are also called curves Consider the Fermat curve C in P2 given by x y z 0 for some n E N The map P2 7 P1 given by zyz 7 z is not well de ned at 001 why but since 001 is not a point on C we get a well de ned map 7139 C 7 lP l The preimage of a point 0 20 6 ll are all point x0y20 E C The number of such points is the number of solutions in y to the equation y 7 7 23 This number is in general n and thus 7139 has degree n On the other hand the map is not n 1 precisely when 7 7 23 0 or equivalently x3 723 The points in P1 which satisfy this equation are precisely the points of the form Q1 where C 71 Thus there are precisely n points in P1 over which the ber of 7139 is rami ed and each time this happens the ber contains only one point instead of n Hence by Riemann Hurwitz MC nxll 1 7 nn 7 1 2n 7 nn 7 1 7712 3n Thus the genus of C is g since MC 2 7 2g Proposition 214 The genus of a general curve of degree n in P2 is Proof The proof is along the same lines as above except that counting the rami cation points in general is a little harder A good way to visually justify this result is as follows Suppose the curve is given by a degree n homogeneous polynomial which factors completely into different monomials for example for n 4 we could have fyz y2z 7 y Each factor corresponds to a copy of P1 and thus the product describes the union of n such Phs Moreover every two Phs meet in precisely one point Thus we have n spheres every pair of which meet in exactly one point If one deforms the equation f slightly then one gets a nice smooth Riemann surface and the points in the original curve correspond to small loops in this Riemann surface which have been pinched to a point It is then easy to see that in fact the genus of this Riemann surface is For example if one takes a torus genus one Riemann surface and pinches three non intersecting loops one gets three spheres each two of which intersect at a point Even though this gives us an effective way of constructing Riemann surfaces as complex submanifolds in P2 not all Riemann surfaces can arise this way In particular since 2 has no solution we cannot construct genus two Riemann surface On the other hand we can generalize this construction by considering the zero set in P3 of two polynomials f and g If f and g are chosen to be general this construction yields a Riemann surface This way one can construct examples of Riemann surfaces of arbitrary genera 23 Constructing Riemann Surfaces as Covers Given a curve 0 one can try to con struct a new curve as a cover of C To describe such a cover it is enough to identify the degree d of the cover the branching points p1 pm of C over which there is rami cation and a nal piece of data known as monodromy This data tells one how different sheets of the cover are glued together around each rami cation point We explain further using an example 9 Example The simplest case is when G 1P1 and the degree is d 2 In this case there are two sheets and each rami cation point interchanges these two sheets In other words locally around each rami cation point the map looks like 2 gt gt 22 Notice that there is no extra monodromy data necessary The only thing we need to specify are the branching points p1 pm Lets take in 6 branching points Exercise 11 Check that if d 2 and the number of branching points is m 6 then the cover will have genus 2 Since the automorphism group of 1P1 namely PGL2C acts triply transitive we can take p1 07102 1 and p3 00 Each choice ofp4p5p6 gives us a curve of genus two Since there are no automorphisms of 1P1 xing 0 1 00 these curves turn out to be non isomorphic to each other Thus we get a three dimensional family of genus two curves Conversely any genus two curve can be written as a double cover a two to one cover of If This explains why the moduli space of curves of genus two is 3 dimensional As a consequence of this description any Riemann surface of genus two has an involution namely the map interchanging the two sheets This map is called the hyperelliptic involution Example Consider the triple cover of 1P1 branched over three points p1p2p3 This time we do need to give the monodromy data There are three sheets which we will number 1 2 3 We take the monodromy around p1 to be the permutation 123 In other words over p1 there is a rami cation point of order 3 which interchanges the sheets 1 a 2 a 3 a 1 The monodromy around p2 we take to be 12 In other words there are two points lying above p2 One of the points lies on sheet 3 and is unrami ed while the other is rami ed of order two and interchanges sheets 1 a 2 a 1 Finally the monodromy around p3 is 13 Notice that 1231213 id This is a necessary condition since the monodromy around all three points should be trivial Exercise 12 What is the genus of the resulting triple cover in the example above Project The problem of counting all the possible covers with xed branching points p1 Pm E 1P1 is known as Hurwitz7s problem Since describing such covers is the same as giving the monodromy data this is apriori a combinatorial problem involving the symmetric group S More recently this problem has been shown to have amazing and surprising connexions with the geometry of the moduli space of curves My A potential project may be to learn a little about the classical Hurwitz problem and explain in broad terms this recent connexion 24 Constructing Riemann Surfaces by Glueing This method can be used to obtain very concrete examples of interesting Riemann surfaces with lots of automorphisms Note that in general Riemann surfaces of genera g 2 2 do not have many automorphisms Theorem 215 o A general Riemann surface of genus 2 3 has no nontrivial automorphisms o A general Riemann surface of genus two has only one nontrivial automorphism the hyperelliptic involution o The order of the automorphism group of any Riemann surface of genus g 2 2 is at most 84g 71 Once again we illustrate this glueing construction using an example 10 Example Start with two regular pentagons in CC and identify opposite sides by trans lating as shown in gure 1 to get a surface S P P P P FIGURE 1 The identi cation shown yields a Riemann surface of genus 2 with automorphism group Z10Z Exercise 13 Check that topologically this gives us a surface of genus two Note that the vertices of the pentagons get identi ed to one point P Away from this point the surface inherits the complex structure from C Exercise 14 Why is the complex structure well de ned at points on the sides of the pentagons To give S a complex structure we also need to build a holomorphic chart in a neighbour hood U around P Note that there is a total angle of 2quot 10 47139 accumulating at P Thus we get a chart around P using the map p E U a C given by z gt gt Since the pentagons are regular one can rotate them by 27r5 to get an automorphism Oz of S One can also exchange the pentagons by using a rotation by 7139 to get an automorphism of order two This involution is the hyperelliptic involution from earlier It turns out these are all the automorphisms of S so AutS E Z10Z One can also obtain S as a double cover of 1P1 branched over the points 07177C27C37Q 4 where C is a fth root of unity The map 1P1 a ll given by z gt gt z xes as a set these six points and lifts to the cover to give the automorphism 04 E AutS we saw above One can apply the same glueing procedure to other shapes to obtain explicit constructions of other Riemann surfaces Exercise 15 What is the genus of the Riemann surface obtained by identifying opposite sides of an L shaped gure Curt McMullen studies such Riemann surfaces in order to understand the dynamics of ball trajectories on L shaped billiard tables 3 TOPOLOGICAL VECTOR BUNDLES Let M be a topological manifold A real vector bundle V over M of dimension n is a family of real vector spaces of dimension n parametrized by M More precisely7 V is a manifold with a map 7139 V a M such that around each point p E M there is an open neighbourhood p E U such that 7T 1U E U gtlt R A vector bundle of dimension one is called a line bundle Example The simplest vector bundle over M is the trivial bundle V R gtlt M a M which we denote In 11 By de nition7 a vector bundle is locally trivial Le over small enough open sets U C M7 it is isomorphic to R gtltU Thus7 to describe a vector bundle V over M it is enough to give local charts Ua over which V is trivial together with transition functions fa Ua U a GLAR which tell you how to glue Ua gtlt R to U gtlt R over Ua U Le the point p11 in Ua gtlt R is identi ed with p7 fa v in U gtlt R As before7 these transition functions must satisfy o fa f5 on Um m U 0 f yofa fwy OH Ua U Uw Example Suppose M Sl x2 y2 1 Along with the trivial line bundle 1 M gtlt R there7s also the Mo39bius line bundle M1 which we describe using the two charts U0 817071 and U1 81707 71 The transition function f01 UO Ul a GL1R 1 is given by my gt gt 71 ifs lt 0 and my gt gt 1 if z gt 0 The resulting line bundle looks like an in nite Mo39bius strip As this example illustrates7 vector bundles are boring locally since they are trivial but become interesting globally because of the possibility for twisting A morpthm between two vector bundles V1 L3 M and V2 71gt M is a continuous map 9 V1 a V2 which sends bers to bers Le 7T2 o g 7T1 using a anear berwise action V1p R a R V2p We impose this condition of linearity since the bers are vector spaces and not just topological spaces V1 and V2 are isomorphic if there is an invertible morphism g V1 a V2 Clearly7 if V1 2 V2 then dimV1 dimV2 but the converse is false A seethm of a vector bundle 7T V a M is a continuous map 039 M a V such that 7139 o 039 LdM There is always a canonical section called the zero section which is given by p gt gt p7 0 We will denote by l M7 V the vector space of all sections of V Exercise 16 Show that 1 has a section which does not intersect the zero section On the other hand7 show that every section of the Mo39bius bundle M1 must intersect the zero section Conclude that 1 and M1 are not isomorphic All the de nitions from vector spaces carry over naturally to vector bundles We summa rize them below 0 subbundles V C W is a subbundle if there is an injective morphism V Lgt W This means that in each ber lVp over p we can pick a vector subspace V C lVp such that the choices vary continuously with p 0 direct sums if V1 and V2 are described by transition functions fal Ua U a GLmR and f3 Ua U a GLAR then the direct sum V1 69 V2 has transition functions fig 0 0 fig Ua U aGLmMGR o tensor product similarly7 V1 8 V2 has transition functions fig 8 f3 Ua U a GLWLUR where this is short hand for the action 111 02 gt gt fi vl faz g Exercise 17 Check that the operations 69 and 8 give the set of vector bundles over M the structure of a ring where the unit is the trivial line bundle 1 Exercise 18 Consider the subbundles V17 V2 C 2 given by the linear span of cos 07 sin 9 gtlt 7 sin 927 cos 02 and cos 07 sin 9 gtlt cos 927 sin 927 respectively 1 Show that V1 V2 g M1 the Mo39bius line bundle Hint draw a picture 12 2 Show that V1 69 V2 2 2 thus concluding that the sum of two nontrivial bundles can be trivial Exercise 19 Show that a vector bundle V is invertible ie there exists a vector bundle V such that V 8 V 1 if and only if V is a line bundle ie dimV 1 A complecc vector bundle V a M is the same as a real vector bundle except that locally the trivialization is 7T 1U E U gtlt C The theory of real and complex vector bundles is very similar just like the theory of real and complex vector spaces is very similar On the other hand there are some differences For example consider the complex Mobius bundle over 5 1 de ned as above using the same transition functions f01 3 U0 U1 gt 3 Exercise 20 Show that the complex Mobius line bundle is in fact trivial hint nd a section which does not intersect the zero section and use the following result Lemma 31 A real or complem line bundle L a M is trivial if and only if there is a nonvanishing section ofL ie a section which does not intersect the zero section Proof Exercise 21 D Just as a real vector space V has a dual V H0mV R a real vector bundle V has a dual vector bundle V If V is given by transition functions fag E GLR then V has transition functions girl 6 GLR Similarly one de nes the dual of a complex vector bundle Fact We will show later that if V is a real vector bundle then V g V but if V is a complex bundle then this need not be true However it is always true that V g V 31 The Tangent and Cotangent Bundles Every real manifold M of dimension n comes equipped with an intrinsic n dimensional vector bundle TM called the tangent bundle The dual of this bundle is the cotangent bundle TA Strangely it is more natural to describe the cotangent bundle so we shall do that rst The bers of TXIp over a point p E M should correspond to the vector space rap11127 where 111 C C 0 is the ideal consisting of functions vanishing at p It is easy to check that rap1112 is a vector space of dimension n However this does not tell us how to glue the bers together To do this consider a chart U0 around p with coordinates x1 z I will be trivial over U0 and we choose as basis n vectors which we call dzl dxn Now take another chart U1 around p with coordinates y1 y lnside U0 U1 the two charts give us a map R a R so that we can think of the y as functions yi1 xn in the xi We also have a basis for TA over U1 given by dyl dy Then the transition information for the cotangent bundle over U0 U1 is given by n 3 11 7 In other words the transition function is f01 E GLR The fact that the transition functions satisfy the glueing condition fag o f f0W is a result of the chain rule 13 The tangent bundle TM is de ned as the dual of the cotangent bundle TA The local basis elements for TM are denoted air The transition functions are then 6 i n 6y 6 9 j1 Exercise 22 A somewhat tedious but important exercise everyone should do at least once in their lifetime is to check that these transition functions do in fact satisfy the glueing conditions 32 Interlude Categories Complexes and Exact Sequences A categoryC consists of a class of objects ObC and a class of maps morC between the objects For a b e E ObC there is a composition operation m0rab gtlt morbc a morae which is associative Moreover for each x E ObC there is an identity morphism 11 z a x which satis es lbof f fola for any f E m0rab Since this is abstract nonsense somesome examples one should keep in mind are Example The category of vector spaces over some eld k The objects are vector spaces over k and the morphisms are linear maps between vector spaces Example The category of abelian groups The objects are abelian groups and the mor phisms are group homomorphisms between them Example The category of topological spaces The objects are topological spaces and the morphisms are continuous maps between them A category does not necessarily have kernels and cokernels For example one cannot de ne the kernel of a map in the category of topological spaces Since we want to be able to take kernels cokernels and direct sums we will work in an enriched category known as an abelian category Clearly the categories of vector spaces and abelian groups are abelian categories Consider a sequence of maps A A0 gtA1 gtA2auaAn4figtAw A is a eomples if for each 239 we have Imf1 C Kerf In this case we can talk of the homology KerfiImfi1 lf Imfi1 Kerf then the sequence is called emaet Clearly A is exact if and only if 0 for all 239 A short emaet sequence is an exact sequence of the form oaALBLan The fact that its exact translates into 0 239 is injective o p is surjective 0 7712 Kerp A short exact sequence is said to split if B E A 69 0 Example In the category of abelian groups consider the short exact sequence 0 ZQZ Z4Z Z2Z 0 14 where the rst map is a gt gt 2a and the second map is the natural projection It is easy to check this is a short exact sequence But since Z2 Z2 Z4 the sequence does not split Exercise 23 Show that in the category of complex vector spaces any short exact sequence splits hint de ne the orthogonal complement Proposition 32 A short exact sequence 0 a A B i O a 0 splits if and only if there epists a map f B a A such that f oi idA or equivalently a map 9 O a B such that p o g idc Proof Exercise 24 D A map F between two categories C and D is called a functor It takes objects to objects and morphisms to morphism such that if f7g7 f o g E morC then Ff o g Ff o 33 Metrics on Vector Bundles Consider a manifold M with an an open cover U0 A partition of unity with respect to Ua is a collection of smooth7 nonnegative functions wa M a R such that wa is supported in the interior of Ua and Ea wa 1 Theorem 33 Any open cover of a manifold has a partition of unity ln fact7 this result holds for paracompact spaces a space is called paracompact if every open cover has a locally nite open re nement Just as you have inner products for real vector spaces or Hermitian metrics for complex vector spaces there are analogous concepts for vector bundles An inner product 7 V on a vector bundle V a M is a map V gtltM V a RJV which restricted to any ber V is an inner product 7 Vp V gtlt lg a Rf We know how to construct an inner product on a xed vector space The following theorem tells us that we can construct an inner product on any vector bundle over a manifold Proposition 34 IfV a M is a real complep vector bundle over a manifold M then V admits an inner Hermitian product Proof Choose an open cover Ua of M over which V is trivial For each 04 construct an inner product 7 0 for VUa By the theorem above we can nd a partition of unity wa subordinate to U0 Then Ea wax7 0 gives an inner product on V we use that the sum of two inner products on a vector space is also an inner product since inner products are positive de nite This result is a useful tool We use it next Proposition 35 Any short eccact sequence of vector bundles 0 a El E 3 E2 a 0 over M splits ie E g E1 69 E2 Proof Recall that by proposition 32 it is enough to nd a map f E a E1 such that f oi idEl Since E1 4 E is an injection E1 C E is a subbundle Pick an inner product 7 on E and take f to be the orthogonal projection E a E1 with respect to 7 D Corollary 36 Any short exact sequence of vector bundles 0 a El a E a E2 a 0 splits as E 2 E1 69 ElL with E2 2 ElL where L is de ned with respect to any inner product on E A good picture to have in mind is that of the Mo39bius bundle M1 over 1 and the associated short exact sequence 0 a M1 a 2 a M1 a 0 15 34 The Degree of a Line Bundle Let L be a complex line bundle on a Riemann surface 0 Consider a general section 039 O a L We can produce such a section by giving it locally and then glueing it together using a partition of unity Locally the line bundle L is trivial so it looks like C gtlt A a A where A is the open unit disk In this local picture 039 is just a map A a C By perturbing 039 we can insist that it is transverse to the zero section Then locally the inverse image of 0 E C under the map 039 A a C is a nite number of points Each point 1 E C where 039 intersects the zero section is called a zero of 039 Around each such point 1 the section 039 is a map 039 A a C where 1 0 E A and 00 0 The differential d0 TOA a TOC is a nonsingular two by two matrix We denote by 59711 the sign of the determinant of this matrix Notice that there was an ambiguity since the map 039 A a C is de ned up to post multiplication by C Fortunately multiplying by a complex number does not change the sign of det d0 Of course this would not be true if we were dealing with a real vector bundle De nition The degree of V is degV 2p 59711 E Z where the sum is over all points 1 where a transverse section a is zero The degree would not be well de ned if we just counted the number of zeroes of a general section This is because one can nd two sections which have different number of zeroes The picture to keep in mind is that from gure 2 If we count with sign both sections give the same degree but without sign we would only get a well de ned degree modulo two V 7 U FIGURE 2 The section on the left has two zeroes with signs 1 and 71 The section on the right has no zeroes Warning If we just count points then we get a well de ned element of Z2Z This is what we do with real line bundles Fortunately in the case of complex line bundles we are able to de ne 59711 and get a well de ned map to Z Example The trivial line bundle has degree 0 This is because we can nd a section which is nonzero everywhere Example The tangent line bundle on P1 has degree 2 This corresponds to the famous saying that you cannot comb the hair on a coconut 76 the tangent bundle of 2 is not trivial The more precise saying ought to be that you can comb the hairs on a coconut everywhere except at two points the north pole and the south pole At these points the matrix d0 describes a rotation and has determinant one Whence both points contribute 1 to the degree Exercise 25 Show that the degree of the tangent bundle on a Riemann surface of genus 9 is 2 7 29 Exercise 26 Show that degL1 8 L2 degL1 degL2 hint if U 1 12 are sections of L then 01 02 is a section of L1 8 L2 16 Fact degL idegL where L is the dual line bundle of L Theorem 37 Let L1 and L2 be two complecc line bundles on a Riemann surface Then L1 L2 if and only if degLl degLg We will prove this fact later For the time being lets try to improve our intuition To do this we rst need to know the following fundamental result Proposition 38 A vector bundle V a B over a contractible space E eg a dish is trivial Example Let7s construct all the complex line bundles on 82 Think of 52 as two disks D1 and D2 glued along their circumferences By the above result the restriction of any line bundle L on 52 to D1 and D2 is trivial Thus to describe L we simply need indicate how to glue the trivial line bundles over D1 to the trivial line bundle over D2 along the boundaries 3D1 3D2 81 This is equivalent to giving the clutching map f 81 a GL1C 3 If you have such a map you can consider its winding number It turns out this number is in fact the degree of the line bundle L The theorem below tells you that if you have two line bundles with the same winding number then they are isomorphic since you can interpolate between the two clutching maps to obtain a family of line bundles Theorem 39 IfV a X gtlt 01 is afarnily of vector bundles of vector bundles overX then V0 V1 where V is the vector bundle restricted to X gtlt Project The concept of a degree is a special case of a construction known as Chern classes A complex vector bundle V of dimension it over some manifold X yields n Chern classes c1V cV These classes are not integers but rather cohomology classes ciV 6 HM X Z where H2iX Z is the 2i th cohomology group of X In the case X has dimension two Riemann surfaces we nd that H2XZ Z so for a line bundle L we get degL c1L e H2X e Z 35 The Determinantal Line Bundle Suppose V a C is a vector bundle of dimension it over a Riemann surfaces 0 with transition functions fa Ua U a GLC The associated determinantal line bundle det V is de ned by the transition functions det fa Ua U a Ci Exercise 27 Check that det fa actually de nes a line bundle Proposition 310 IfO a El a E a E2 a 0 is a short eicact sequence of vector bundles then detE detE1 detE2 Proof Since we know the sequence splits we have E E E1 69 E2 In particular this means that the transition functions for E are just f1 f2 where f i 1 2 is the transition function of E Then detf1 69 f2 detf1 detf2 which by de nition are the transition functions for detE1 detE2 D We extend the de nition of degree to arbitrary vector bundles V a C by de ning degV degdetV Corollary 311 IfO a El a E a E2 a 0 is a short eicact sequence of vector bundles then degE degEll f degEzl39 17 Proof This follows immediately from the proposition above and the fact that for line bundles L1 and L2 we have degL1 8 L2 degL1 degL2 We say a real vector bundle V is orientable if detV is a trivial line bundle By de nition7 a manifold M is orientable if its tangent bundle is orientable Given a complex vector bundle V of complex dimension n one can view it as a real bundle of dimension 2n This is achieved by embedding GLnCC gt GL2nR This embedding is r cos 6 r sin 6 7r sin 6 r cos 6 to be orientable if its associated real vector bundle is orientable done componentwise via 2 rem gt gt lt A complex vector bundle V is said Proposition 312 Any complecc vector bundle is orientable Proof A not completely trivial exercise in linear algebra tells you that the following dia gram commutes GLACC gt GL2nR det det Cs GL2 ii Rs This means that it is enough to show any complex line bundle L is orientable If the transition functions of L are fag rem where r76 are functions then the transition functions for the r cos 6 r sin 6 corresponding two dimensional real vector bundle Us l are fgffl 7T Sing TCOS 6 Then the determinantal real line bundle det Us l has transition functions det fulfil r2 E R5 The key point is that r2 gt 0 so these maps can be deformed to constant maps to RJF and hence describe the trivial line bundle D Corollary 313 A compled manifold is always orientable 36 Classi cation of Topological Vector Bundles on Riemann Surfaces Lemma 314 IfV a X is a vector bundle and o a non vanishing section then V V GBII where dimV dimV 7 1 and 1 is the trivial line bundle Proof Since a is never zero the subspace it spans de nes a subbundle of V Since a is a non vanishing section of this subbundle it follows the subbundle is the trivial line bundle 1 Putting a metric on V we take V If Then V V 69 1 Proposition 315 Let V a M be a vector bundle over a manifold M of dimension dimRM m o IfV is real and dimRV gt m then there eccists a non vanishing section 0 IfV is compled and dide gt m2 then there CCElStS a non vanishing section Proof In both cases7 since the real dimension of the ber is bigger than the real dimension of the base7 one can take a general section and perturb it locally around vanishing points so that it no longer vanishes Corollary 316 IfV is a compled vector bundle on a Riemann surface then V E detV 691m where m dimV 7 1 18 Corollary 317 The ring R of vector bundles on a Riemann surface is isomorphic to lel czt Proof From the corollary above a vector bundle V is uniquely determined by the pair dimm dew Exercise 28 Show that dimV1 8 V2 dimV1 dimV2 and degV1 8 V2 dimV1 degVg dimV2 degV1 hint use that V1 and V2 split into the direct sum of line bundles Consequently the operations 69 and 8 on such pairs are given by a be d ae bd and ab ed ac ad be From this it follows that the map R a given by ab gt gt a bx is a ring isomorphism D Finally its worth mentioning the following cancellation law Proposition 318 IfV 69 In 2 V 69 In then V E V 37 Holomorphic Vector Bundles A holomorphic vector bundle is a complex vector bundle where the transition functions U04 a GLnCC are not just smooth but also holomor phic It is important to realize that the concept of a holomorphic vector bundle only makes sense over a complex manifold for example Riemann surface The total space V a M of a holomorphic vector bundle is a complex manifold though it is not compact Two holo morphic vector bundles V L M and V L M are isomorphic if there exists a holomorphic map f V a V such that f on 7T zle maps the ber Vn over p E M to the ber V also over p E M and which acts linearly on the bers zle the restricted map fp V a V is linear Example The trivial vector bundle In is holomorphic Example If M is a complex manifold of complex dimension n then the tangent bundle TM is a holomorphic vector bundle of dimension n This is because the transition functions describing M are holomorphic by de nition and hence their derivatives which are used to used to describe the transition functions for TM are also holomorphic Exercise 29 Show that V is holomorphic if and only if V is holomorphic 371 Holomorphz39e Line Bundles over P1 Recall that the complex line bundles over P1 are in bijection with Z via the degree map Let us give the complex line bundle Ln of degree n E Z a holomorphic structure As a complex manifold we view P1 as two copies U0 and U1 of C glued together along Ci via the map f01 z gt gt 12 To build Ln as a holomorphic line bundle we take it to be trivial over both copies of C and hence we need only give the glueing map gn 3 a GL1C g 3 Since we want Ln to be holomorphic gn must also be holomorphic We take gn 3 a 3 to be 2 gt gt 2 We will check later that in fact degLn n For now notice that L0 is indeed the trivial line bundle since the transition function go is the map 2 gt gt 1 Moreover the transition function for Ln 8 Ln is gm gn gmn and thus Ln 8 Ln g Lmn This is expected since degLm 8 Ln degLm degLn m n degLnn What is the holomorphic tangent bundle T Tum of P17 Pick coordinates z and w over U0 and U1 such that over U0 U1 C the glueing is made via w 12 By de nition 19 TlUO 2 U0 gtlt C and Tle1 2 U1 gtlt C The glueing over U0 U1 is then 6 d 1 6 1 6 62 N dz 6w 7 226w This means the transition function for T is z gt gt 7122 which means Tm g L2 Thus we see again that the tangent bundle of lP l has degree 2 Exercise 30 Show that the holomorphic line bundle with transition function 7122 is in fact isomorphic to L2 where the transition function is 122 Since L g L it follows that the holomorphic cotangent line bundle TB of lP l is isomor phic to L4 Warning Though every complex line bundle over a Riemann surface 0 can be given a holomorphic structure it is not true that this structure is unique The picture one should keep in mind is the sequence of maps d holomorphic vector bundles over 0 a complex vector bundles over 0 i Z By what we said earlier7 the second map is an isomorphism The rst map is surjective but not necessarily injective If C P1 the rst map is in fact an isomorphism but if C has genus g 2 1 then it is not injective Finally7 be aware that over arbitrary complex manifolds7 the rst map may be neither injective nor surjective For example7 if C is a genus one Riemann surface then there exist complex line bundles on C X C which does not have a holomorphic structure ie there is no holomorphic line bundle which is topologically isomorphic to it Warning It is not true that every holomorphic vector bundle over a Riemann surface of genus g 2 1 splits as a direct sum of line bundles The earlier proof proposition 35 which worked for complex vector bundles breaks down in the case of holomorphic vector bundles because the partition of unity and subsequently the inner product are not holomorphic they are only smooth 38 Sections of Holomorphic Vector Bundles Let 7T V a M be a holomorphic vector bundle over a complex manifold M A holomorphic section or just section for short is a holomorphic map a M a V such that 0 on idM We denote by l M7 Vh or PV for short the vector space of all holomorphic sections Theorem 319 IfM is a compact complecc manifold and V a holomorphic vector bundle then PMVh01 is a nite dimensional vector space Example Let 1 be the trivial line bundle over a Riemann surface C If we view In as a topological complex line bundle then FCIf0p is the space of all 0 maps 0 a C and hence has in nite dimension On the other hand7 if we view 1 as a holomorphic line bundle then PCIwl is the space of holomorphic maps 0 a C Since G is compact we know any such holomorphic map must be constant Thus PCIwl E C Example Let7s describe PllD1L2 where L2 is the degree two line bundle de ned in section 371 Recall that lP l is covered by U0 g C and U1 g C which have local coordinates z and it Over U0 the bundle L2 is trivial so a holomorphic section is the same as a holomorphic map C a C given by z gt gt Similarly7 over U1 a section is the same as a holomorphic map it gt gt gw We need these two local sections to agree on the overlap U0 U1 C We 20 know 2 110 and the glueing of L2 over U0 U1 is done by 122 Thus7 in the trivialization over U17 the section fz is w2f1w This must equal the holomorphic function gw over U0 U1 Whence PllD17L2 is isomorphic to the space of holomorphic functions fz such that w2f1w is also a holomorphic function These functions are precisely the polynomials of degree at most two ie ofthe form abzczz Whence dimPllD17L2 3 Exercise 31 Check that PllD1Ln corresponds to vector space of polynomials a0 H112 anz ln particular7 conclude that dimFlP17Ln 0 if n lt 0 Proposition 320 IfL is a holornorphic line bundle on a Riemann surface with degL lt 0 then dimPL 0 ie the only holornorphic section is the zero section Proof Suppose L has a nonzero holomorphic section a and let p be a point where L vanishes We will show that such a p always contribues 1 to degL and hence degL 2 0 contradiction Locally around p7 the section looks like a map C a C given by z gt gt fz where f0 0 If we write fz a7y ibz7y then p contributes 1 to degL if and only if det gm Zy gt 0 But since f is holomorphic we know am by and ay ibm so that z y the determinant is a b gt 0 D Exercise 32 In the proof of this proposition we saw that if a holomorphic section of L vanishes at n points then degL n Use this to show that degLn n hint use the earlier description of UL for n gt 0 4 SHEAVES For a basic introduction to sheaves see Ha chapter ll section 1 Fix a topological space X A presheaf of abelian groups on X consists of the following data 0 to each open set U C X an associated abelian group fU o for each V C U restriction maps pUV fU a fV which satisfy pUU id and pUW pVW 0 pm whenever W C V C U Usually we will write the restriction pUVs of s E fU to fV as sly A sheafis a presheaf with the following added condition Let Uihd be an open cover of U and suppose that s E fUl satisfy silUmUj slemUJ for every i7j E I Then there exists a unique 5 E fU such that slUi s for alli E I We now give several examples of sheaves in order to develop our intuition It is a good exercise to check these are indeed sheaves Example Locally constant sheaf Z is de ned by assigning Z to each connected component of U In other words7 ZU is isomorphic to Z990 where c is the number of connected components of U Notice that for U C X the map ZX a ZU is not necessarily onto Similarly one de nes the locally constant sheaves R and C Example The sheaf CWGR assigns to each open set U the group of 0quot0 functions f U a R Similarly one de nes C C Example Let X be a complex manifold The sheaf 0 OX associates to an open U the group of holomorphic functions f U a C Notice that if X is compact this means OXX C Similarly one can de ne O by associating to U the group of holomorphic 21 maps f U a C5 Example If E is a holomorphic vector bundle then one can de ne the associated sheaf shE by letting shEU be the group of holomorphic sections of EU For example7 OX shIl We will usually write shE as E since there is rarely any ambiguity between viewing E as a vector bundle and viewing it as a sheaf Example If f and g are sheaves on X then 69 9 is the sheaf given by f gU fU 69 9U and 1789 is the sheaf given by f gU fU z 9U Notice that by de nition if E1 and E2 are vector bundles on X then shE1 69 E2 shE1 69 shE2 The same is true of 8 but only if we tensor over OXU rather than Z see section 44 Example The dual sheaf f of f is de ned by fU H0771TU7 OXU C if p E U 0 if p Z U Example Let C be a Riemann surface Then 007p associates to U the group of holomorphic maps U a C which vanish at p If C is compact then 007p0 0 Similarly one de nes 001 by associating to U the group of holomorphic maps U a C which have at most a pole of order one at p This de nition generalizes to give 00kp for any h E Z For instance7 suppose C P1 and p 0 6 P1 Then FP1O2p a7b7c E C and in general dimFlP1Okp k 1 One can also de ne sheaves such as 901 q for points p7q E C in much the same way Example Suppose L is a line bundle on a Riemann surface X Then L7p associates to U the group of holomorphic sections of LlU which vanish at p Notice that if p Z U then LltepgtltUgt LltUgt Example Consider f given by fU Example Let p E X be a point The skyscraper sheaf C17 is de ned by C lfp E Q7 U 7g X Then f is a presheaf but 0 otherw1se not a sheaf However7 if you remove the condition that fU C only if U 31 X then you get the skyscraper sheaf Cp which is a sheaf We mention the following fact without proof Fact There exists a unique way to turn a presheaf into a sheaf The process is called shea catioh Recall that a presheaf is not a sheaf if there is no unique lift 5 satisfying siUi 5 whenever siiymyj slemUj Shea cation essentially throws in such 5 whenever necessary and removes such 5 if the lift is not unique see Ha Notation If f is a sheaf on X the group fX is called the space of global sections of f We often denote fX by Hf Elements of Hf are called global sections of 7 401 Morphisms of Sheaves Let f1 and 72 be two sheaves on X A morphism o 71 a 72 consists of maps by f1U a f2U for every open U C X which commute with restrictions 7 ie for any V C U the following diagram commutes MU k MU pUV PUV am A am Example If f X a R is a 0 map then we have a morphism C R a C R given by 9 H fa 22 Example Let C be a Riemann surface and f C a C a rnerornorphic map with a pole at p E C of order k E Z3 Then we get a map bf 007kp a 90 given by g gt gt fg More generally7 we get a map bf 0071 7 kp a 00np for any n E Z Example Given a point p E X there is a morphism of sheaves OX a CF given by f gt gt fp The kernel of a map b 71 a 72 is the sheaf ker which associates to U the group ker gtU f1U a f2U On the other hand7 U gt gt irn U 71U a f2U only de nes a presheaf The sheaf irn is de ned to be the shea cation of this presheaf As before7 we say that a sequence of sheaves fF1 fl Hl a is exact if ker i irn 1 for all z Example Fix a point p 6 P1 The sequence 0 a OP17p a 011m a 17 a 0 is exact Sirnilarly7 0 a 0314721 0 OP17p CZ 0 is exact On the other hand7 even though the map OP17p a CF is surjective the map on global sections PP17Op17p a PP17Cp E C is not surjective since PP1OP17p 0 What is happening is that U gt gt coker OP172pU a OP17pU is only a presheaf and we need to shea fy to get CF After doing this7 the map OP17pC a CPU is no longer surjective This might seem like an inconvenience7 but its this precise phenomenon that allows us to develop the theory of sheaf cohornology 41 Cech Cohomology Cech cohornology is a cohornology theory for sheaves Chapter III section 4 of Ha contains a short note on the subject The rest of chapter III of Ha as well as I provide an introduction to the derived functor approach to sheaf cohornology Although we will not talk about derived functors this is a very important concept and a viable subject for a project Fix a complex manifold X and a sheaf f on X The Cech cornplex 0mm 00 L0 01 102 2 with respect to a cover Uahd of X is de ned as follows We let 00 Had fUa so that an element of CO has the form fa with fa E fUa Next 01 Ha d fUa U where an element looks like fa Similarly CZ H fUa U U7 and so on The differentials are 0 do CO a 01 given by fa gt gt faijU 7 f ijU E fUa U d1 01 a 02 given by fa H f 7 fm fa e HUQ m U 7 U7 0 dn C a 0 is de ned similarly as an alternating surn 049761 Thanks to the alternating signs dHl 0 dl 0 so that we get a complex The cohornology of this complex gives us the Cech cohornology groups H X77 kerdnirndn1 The last thing we need to do is understand how this de nition depends on the open cover U0 It turns out the cohornology groups do not depend on the cover if the cover is chosen ne enough This is expressed more precisely by the following two results 23 Theorem 41 Leray IfUa and V are two open couers ofX for which H Um mumUmf 0 H V 1mmv f for n gt 0 then the Cech cohomology groups calculated with respect to the covers U0 and V are the same Theorem 42 If each component ofU is a conuecc domain then for any open couer ofU the higher Cech cohomology is zero Thus in order to calculate Cech cohornology it is enough to choose an open cover of X where all the open sets U01 U0 are convex domains Exercise 33 Show that H0X PX The higher Cech cohornology groups do not generally have such nice interpretations How ever we shall see that the group H1X 0 does have a nice interpretation 7 it is the group parametrizing holornorphic line bundles on X But rst let7s calculate the Cech cohomology of some simple spaces and sheaves Example Let7s compute the Cech cohomology of the constant sheaf Z on 81 We cover SI using arcs U1 U2 U3 which overlap pairwise but where U1 Ug U3 0 Since the arcs as well as their intersections are convex domains this cover is good enough to use for computing cohornology Now CO Z3 abc and Cl Z9 where the differential do 00 a Cl is given by dabc 0a7 ba 7cb7 a0b7 cc7 ac 7b0 This means that H0X Z ltaaagt E Z as expected since FSlZ Z The kernel of d1 contains elements of the form 0xy7z0z7y7z0 6 Cl and thus H1SIZ g Z Similarly one can check that HiSl Z 0 for i gt 1 Exercise 34 Calculate HiSlZ using a cover of 1 by two open sets instead of three this should be simpler than the computation above so you can ll in all the details Exercise 35 Show H lC Z E HiSlZ by a direct calculation of the left hand side De nition 43 D C C is a domain of holomorphy iffor each D containing D there epists some holomorphic function f on D which does not emtend holomorphically to D Example Any domain in C eg C is a domain of holomorphy However C2 7 0 C C2 is not a domain of holomorphy since by Hart0g7s theorem any holornorphic function on C2 70 extends to a holornorphic function on C2 Theorem 44 IfU is a domain of holomorphy then for any holomorphic vector bundle V on U the higher Cech cohomology is zero Thus in order to calculate Cech cohomology of holomorphic vector bundles it is enough to choose an open cover of X where all the open sets are domains of holomorphy Aside This whole issue of choosing an open cover might seem a little strange The right way to think about it is that you want to use open sets which have zero higher cohornology This approach suffers from the chicken and egg problem in that we cannot de ne Cech cohornology until we choose a cover and yet we cannot choose a cover until we know its cohornology Despite this conundrum the derived functor de nition of sheaf cohornology tells us this is the correct philosophical way to think Likewise in practice7 this is the most useful method to use in order to decide on a cover Example Consider C llD1 with the trivial holomorphic line bundle Recall that the corresponding sheaf is denoted 90 As an open cover we use the usual open sets U0 lP l foo and U1 P1 70 which are domains in C We know H0C7 Oc C To compute H1C7OC we need to understand holomorphic maps U0 U1 C a C Any such map looks like zgt gt anz a1z 1a0a121anz and hence can be written as f 7 y where f a0 a12 l is holomorphic on U0 and g a12 1 12272 is holomorphic on U1 This means that H1llD17Op1 0 The higher Cech cohomology groups also vanish since the cover contains only two open sets We have the following cohomology groups when G is a Riemann surface of genus 9 Z 20 c 23970 c 29 39 c H O Z Z Z 1 HlO 00 9 i1 7 Z F2 7 0 gt2 0 23923 7 Proposition 45 We have the following vanishing results for cohomology o ifX is a topological manifold of real dimension n then HiX7Z 0 fori gt n o Grothendieck ifX is a complecc manifold of compled dimension n and V any holo morphic vector bundle on X then H lX7 V 0 fori gt n 411 Fine Sheaves Fine sheaves are a class of sheaves which have the useful property that they have zero higher cohomology To de ne ne sheaves we rst need the concept of support The support suppf of a sheaf f on X consists of those points z E X where for any open set U containing z one can nd an open z E V C U such that 7V 31 0 Similarly7 the support supp of a morphism of sheaves o 71 a 1 consists of points z E X such that for any open set U containing z one can nd an open subset z E V C U such that by f1V a f2V is nonzero Example The skyscraper sheaf Cp for p E X has support suppCp On the other hand7 suppOX X De nition 46 A sheaf over X is ne iffor every locally nite cover Ua ofX by open sets there eccist morphisms ta f a f such that 0 supplt a C U0 0 Eu bu id Example The main examples of ne sheaves for us are C R and 0 0C To de ne ta one takes a partition of unity that with respect to the open cover Ua Then ta is multiplication by ya ie takes a function f to waf Example Z OX and 0 0C are not generally ne sheaves For instance7 if X is a compact complex manifold then the only morphisms OX a OX is multiplication by a constant All of these have support all of X so if we pick any nontrivial cover Ua of X we cannot nd the required morphisms ta with support in U0 Theorem 47 If is a ne sheaf on X then HiX7f 0 fori gt 0 25 Proof Pick a cover Ua of X that we can use it to compute the Cech cohomology of f Denote by Cquot the Cech complex of f with respect to Ua Since f is ne we can nd ta f a f such that supp gta C Ua and Ea ta 2d Since each ta is a sheaf morphism it commutes with the differential maps of 039 giving us a complex aC39 for each a Since Ea ta 2d we have that aHi aC39 In other words7 the morphisms ta break up the cohomology PPX7 into the direct sum of smaller pieces On the other hand7 each 50 computes the cohomology of tax C f Since ta is supported on Ua the sheaf 1507 is also supported on Ua and hence PP39X7 1501 PliUm a7an Since we chose the UDs small enough these groups are zero for 2 gt 0 which means that PPX aHi af 0 for 2 gt 0 D 412 The Exponential Sequence A map of complexes E a F39 consists of maps fl El a Fl such that dofl fi1od239e the maps fi commute with the differential A sequence of maps 0 a E a F a G a 0 is short exact if for each 2 the sequence 0 a El a Fl a Gl a 0 is exact Lemma 48 snake lemma A short exact sequence of complexes 0 a E a F a G a 0 leads to a long exact sequence of cohomology u mwy mwy m Hmwyim Proof sketch We describe how to obtain the connecting morphism a Hi1F39 the rest of the maps are obvious Suppose u E kerGl a 6 Since Fl a Gi is onto we can nd an element 22 6 Fl which maps onto u Now consider 2 du Since du 0 the image of 2 under the map Fltl a G 1 is zero all squares are commuting Thus there must be some 2 6 EH 1 which maps to 2 6 Fl We map 22 to 2 It now remains to check that this map is well de ned 239e that 0 E Gi maps to zero and that the rest of the maps are well de ned Finally one needs to check that the long sequence is exact D Corollary 49 IfO a S a f a g a 0 2s a short exact sequence of sheaves on a topological space X then the following sequence is exact maa m mm WWXa m Proof Choose an open cover of X which we can use to compute the Cech cohomology groups of S f and 9 From this we get a short exact sequence of complexes 0 a 08 a Cf a 09 a 0 because 0 a S a f a g a 0 is exact The snake lemma gives us the long exact sequence of cohomology groups B Corollary 410 IfO a 8 a f a g a 0 2s a short exact sequence of sheaves then x8 7 xf Xg 0 where x4 Zi71iranhHiX4 denotes the Euler character239st239c of the sheaf A Proof A general fact from algebra says that if 0 a A1 a a An a 0 is an exact sequence of abelian groups then Zi71 rankAZ 0 The result follows since by the previous corollary mmaa m ammm m xa m is an exact sequence D 26 Example Fix a complex manifold X and consider the esponential sequence 0 a Z a OX 2 0 a 0 The rst map takes a E Z to the constant map in OX The second map is the exponential f gt gt exp27rif Since expt E 3 for any t E 3 this map is well de ned Proposition 411 The esponential sequence is exact Proof The question of exactness of sheaves is a local one Given any holomorphic map f U a 3 we can take the logarithm to obtain a holomorphic map logf U a 3 This shows exp OX a O is onto Finally7 we have exactness in the middle since the elements of OX which are mapped by exp to one the identity in 0 are precisely the integers because exp27rit 1 exactly when t E Z D Since H0X7Z Z HoX7 OX 3 and H0XO 3 we get a long exact sequence of abelian groups 0 gt Z gt c E o a H1XZ a H1XOX H1O a H2XZ a Since the map 3 3 is onto the sequence 0 a Z a 3 3 a 0 is in fact exact So the sequence above splits to give us the long exact sequence 0 a H1XZ a H1XOX H1O a H2XZ a H2XOX a If X is a Riemann surface of genus 9 we saw that H1X7Z Z29 H1XOX 397 H2X7Z Z and H2XOX 0 Thus we get the long exact sequence 0gtZ29gtC9gt H1XO Z O Example Similarly we can consider the exponential sequence 0 a Z a owe o ltc a 0 Since 0 03 is a ne sheaf see section 411 we know HiXC 3 0 for i gt 0 Then the corresponding long exact sequence implies H1XC 3 H2XZ So if X is a Riemann surface we get H1XC 3 E Z 42 Line Bundles and Cech Cohomology Theorem 412 IfX is a comples manifold then H1XO parametrizes holornorphic line bundles on X Proof Pick a cover Ua of X and denote by 00 0 01 Lb O2 a the Cech complex of 0 Then kerd1 Hm fag fag f fm 1 where fag U0 m U a 3 Also imd0 Hum fag fag gaggl for some ga Ua a 3 Choose Ua ne enough so that any line bundle is trivial over any U041 UM lf L is a line bundle choose a trivialization ta L Ua a 3 gtlt Ua for each a Denote the glueing functions of L with respect to this trivialization by fag Then the glueing condition implies fag f v f0W which means fag E kerd1 However7 this element depends on the trivialization ta we picked for L Changing the trivialization by some ga Ua a 3 gives us 27 t ta g0 Then the new glueing functions are fu t gujlfa g which differ from the old glueing functions by an element of imd0 Thus7 given a line bundle L we get a well de ned element of kerd1imd0 H lX7 Conversely7 given an element of H1XO we can reverse the description above to construct back L Corollary 413 For each n E Z the space of degree n line bundles on a Riemann surface 0 of genus g is isomorphic to an n dimensional torus CgA where A C C9 is an integral lattice of rank 2g Proof Lets go back to the short exact sequence 0gtZ29gtC9gt H1COE Z 0 Since the sequence is exact7 the kernel of the degree map is 9 modulo the image of Z29 inside 9 which is a lattice A of rank 2g D Aside The map P11 C7 05 a Z is none other than the degree map de ned earlier we wont prove this here In fact7 the de nition of degree is usually provided by this map and not in terms of counting zeros of a general section as we did in section 34 The total space PPXO of holomorphic line bundles on X is called the Picard group and denoted PicX The group operation is the tensor 8 and the inverse of a line bundle L is the dual line bundle Li recall that L L E 1 PicdX C PicX denotes the subspace of line bundles of degree d Pic0X is called the Jacobian and denoted JX Notice that PicdX Pic5X a Picd5X so that JX is a subgroup of PicX If X C is a Riemann surface of genus g then the corollary above implies PicC JC gtlt Z and also that JC is a compact complex manifold jgA of dimension g As the complex structure of C varies7 the lattice A E 9 moves and we get different tori Even though these tori are topologically the same they are homeomorphic to Sl29 as complex manifolds they are different A classical theorem of Torelli essentially says that one can recover the Riemann surface 0 from its Jacobian JC The complex and real topological analogues of the sheaf O are the sheaves 0 0C and C R Consequently7 we have the following analogue of theorem 412 Theorem 414 IfX is a topological manifold then Pl lX7 0 0CC parametrizes compled topological line bundles and PPXC R parametrizes real line bundles on X Proof The proof is precisely the same as that of theorem 412 D Theorem 415 H1X 0000 H2XZ and H1XO R 2 H1XZ2 Proof The exponential sequence for 0 0C is 0 a Z a 0 0CC C C a 0 leading to the long exact sequence 4 H1X owe a H1X omen E H2XZ a H2XO C a Since 0 0CC is a ne sheaf we have HiXC CC 0 for i gt 0 implying that the map c d c H1XC CC i H2X7 Z is an isomorphism 28 exp The exponential map for C R is 0 7 0 7 0 0R 7 C RgtO 7 0 This shows that C R g C RgtO On the other hand7 we also have the short exact sequence 0 7gt Z2 7gt CWR g CWRgtO 7gt 0 where mf f2 Since C X RgtO C X R is ne the higher cohomology vanishes and the long exact sequence becomes 0 7 22 7 momma 2 r0 ngt0 7 H1XZ2 7 H1X O R 7 0 But for any map 9 X 7 RgtO we have a well de ned square root This means PC R g PC RgtO is surjective and hence H1XZ2 g H1XC R D Corollary 416 IfC is a Riemann surface of genus 9 then complecc and real line bundles are parametrized via the degree map by Z and Z229 respectively Proof This follows immediately from the previous proposition and the fact that HZC7 Z Z and H1O7Z2 Z229 D Exercise 36 Show that H1Sl7C OC 0 and conclude that all complex line bundles on Sl are trivial Exercise 37 Show that up to isomorphism there are exactly two real line bundles on 817 namely the trivial and the Mo39bius line bundle 43 RiemannRoch and Serre Duality Let L be a holomorphic line bundle on a Rie mann surface 0 of genus 9 What is the dimension ofthe space of global sections H0 07 L UL To answer this question we rst calculate the Euler characteristic MC L Zi71ihiL where hlL ranlltHlC7 L Since dimCC 1 Grothendieck7s theorem 45 tells us hiL 0 for i gt 1 Thus MC L h0L 7 h1L Let7s rst consider an example Fix a point p E C and consider the sheaf OC np for some n E Z later we will show this sheaf corresponds to a line bundle of degree To calculate MC Ocnp we have the short exact sequence of sheaves 0 7gt 0071 7 lp 7gt 000110 7gt 17 7gt 0 where CF is the sky scraper sheaf at p We can think of CF as a complex line bundle on a point Since a point is a zero dimensional complex manifold Grothendieck7s theorem tells us that hiC7Cp 0 for i gt 0 Also7 h0C7Cp 1 so that xCCp 1 Thus7 by corollary 4107 xCOCnp MC 0071 7 1p 1 Repeating in this way we get Exercise 38 Show by direct computation that hlC7 CF 0 for i gt 0 Theorem 417 Riemann Roch IfV is a holomorphic vector bundle on a Riemann surface 0 of genus 9 then MC V degV17 granhV In particular ifL is a line bundle of degree d then MC L d 7 g 1 We will give a proof in the next section What is remarkable about this result is that MC L only depends on the topological isomorphism class of L For example7 if C is a curve 1 ifpq Thus ho of genus g gt 0 and p7q E C are two points then h0OCp 7 q 0 ifp 7g q 29 and consequently also hl depends not only on the topological type degOcp 7 q 0 for any pq E C but also on the holomorphic type On the other hand the difference XC L h0L 7 h1L is a topological invariant Now that we understand xC L we turn our attention to computing h1 This problem has a partial answer in the form of Serre duality Theorem 418 Serre duality Let V be a vector bundle on C and denote by K0 the cotangent bundle ofC this is also known as the canonical line bundle Then H1C V H0C V 8 KCV In particular h1V h0V 8 KC Example Suppose C llD1 so that KC L2 OP172 see section 371 If n 2 0 then h1Ln h0L L2 h0Ln2 which is zero since degLn2 7n 7 2 lt 0 Thus h0Ln xLn n701 compare with section 38 Thus if n lt 0 then h0Ln 0 and h1Ln 7n 71 Example If C is a Riemann surface of genus 9 then h1KC h0KE KC h0OC 1 But xKC 2g 7 2 7 9 1 g 71 which means h0KC 9 Note that showing directly from de nition that the canonical line bundle has a 9 dimensional space of sections is somewhat dif cult Exercise 39 Let C be a Riemann surface of genus g and L a line bundle of degree 29 7 2 Show that h0L 2 g 7 1 if and only if L KC hint use that if L is a line bundle of degree zero then h0L 0 unless L g 00 Example Suppose L is a line bundle on a Riemann surface 0 of genus 9 such that degL gt 29 7 2 Then degL 8 KC lt 2 7 2g 29 7 2 0 so that h1L 0 and h0L xL degL 7 g 1 More generally we have the following useful result Proposition 419 Let V be a holomorphic vector bundle on G Then h0V 8 L gt 0 if degL gtgt 0 and h0V X L 0 if degL ltlt 0 Proof The rst assertion follows from Riemann Roch since degV 3 L degdetV o L degdetV L dimltVgt degV degL dimV which tends to in nity as degL tends to in nity To prove the second assertion we use that H0V is nite dimensional Then we can consider among all sections of V the one which vanishes to highest possible order d But then H0V OC7np 0 for any n gt d and point p E C In the next section we will show that 007np corresponds to a line bundle of degree 7n and that conversely any line bundle is essentially of this form theorem 427 D This proposition says that any line bundle on a Riemann surface 0 has a meromorphic section Although rather believable this is not a trivial statement to prove and requires some analysis It would seem like we gave a proof above but in the process we used Riemann Roch whose proof will turn out to rely on this proposition In fact if one were to trace through the arguments carefully one would nd that proving the existence of meromorphic sections of line bundles over Riemann surfaces is the main and almost only serious technical obstacle 44 Vector bundles locally free sheaves and divisors To a holomorphic vector bundle V over X we can associate its sheaf of sections which we also denote V Given a local section 5 E VU we can multiply it by a holomorphic function f E OXU to obtain a new 30 section f s E VU Thus VU has the extra structure of a OXU module Since this holds for any open U we nd that V is in fact a sheaf of OX modules De nition 420 A sheaf is an OX module iffor any open U C X the group fU is an OXU module such that for any V C U the action commutes with the restriction map ie forf E OXU ands E fU we have fslV flVslV De nition 421 A morphism of sheaves b 1 a 1 is a map of OX modules if it com mutes with the OX action ie for f E OXU and s E f1U we have bf s f s We denote the space of OX module morphisms 71 a 72 by Homoxf1f2 The problem with the category of sheaves is that its too big and in particular contains too many morphisms So from now on we will work only in the category of OX modules where the objects are OX modules and the morphisms are OX module maps between them Consequently a few of our old de nitions must be adjusted to work in the category of OX modules rather than the category of sheaves For example the dual sheaf f is now de ned by fU HomoxfU OXU where we consider just OX module morphisms rather than all sheaf morphisms Also the tensor product of f1 and 72 is 1 OX 72 rather than f1 lt81 72 Lemma 422 ff is a sheaf of OX modules then HomOXOXf H0Xf PCT More generally ifL is a line bundle then HomOXL f H0Xf ox L Pf 0x L Proof Suppose the constant section 1 E NOX maps to s E Hf If a E OXU then since the map OX a f commutes with restrictions and the OX action we have that a a1lU gt gt aslU So to give f E HomoXOXf it is enough to give the image 5 E Hf of 1 E FOX Conversely any such 5 E PCT gives us a morphism Given a map L a f we can tensor both sides by L to get a map L OX L 2 OX a f 0x L and vice versa Then by the previous argument such maps correspond to elements of Hf OX U D Recall that an R module M is free of rank n if M g R99 Since any rank n vector bundle V is locally trivial we can nd open sets U where VlU 0 So over such sets V is isomorphic to a free OX module of rank n Motivated by this observation we de ne De nition 423 An OX module f is locally free of rank n if every pointp E X is contained in an open neighbourhoodp E U where fly is a free OXU module of rank n Theorem 424 Any holomorphic vector bundle V of rank n is isomorphic to a locally free sheaf of rank n and conversely Moreover any morphism of vector bundles is a morphism of OX modules althought the converse is not true Proof We saw already that if V is a vector bundle of rank n then the corresponding sheaf is locally free of rank n Conversely suppose V is a locally free sheaf of rank n For simplicity of notation let7s assume n 1 By de nition we have open sets U0 and isomorphisms f0 Van a OXUD So over Ua U we can consider faof l to get a map OXUD U a OXUD U Since this map is a morphism of OX modules following the lemma above we just need to give the map 1 gt gt a E OXUD U Since this morphism is invertible we must 31 have a E OUa U This a represents the glueing information allowing us to recover the line bundle The only thing to check in the second assertion is that a map of vector bundles commutes with the action of multiplication by a function This is true since the berwise map is linear and a linear map 3 a C commutes with the C action The reason why the converse is not true is that the berwise map may drop in rank For example7 consider the map between the trivial line bundles over CC given by multiplication by z This map is an isomorphism of bers over 2 7E 0 but drops rank at z 0 The quotient is then the skyscraper sheaf CO at z 0 which is not a vector bundle Hence we do not allow this in the case of vector bundles but do allow this map when working with sheaves because the skyscraper sheaf is a perfectly good sheaf even though its not locally free Exercise 40 Show that the tensor product 71 OX 72 of two locally free sheaves f1 and 72 of ranks n1 and n2 is a locally free sheaf of rank nlng hint this is a local question so you can assume that both E and 72 are trivial There is a really good description we can give of locally free sheaves of rank one 1e line bundles in terms of divisors De nition 425 A divisor D on a Riemann surface 0 is a formal nite Z linear sum of points D 21 aipl where al 6 Z and pl 6 C The degree ofD is degD ai Recall that in section 4 we de ned sheaves 00hp for any h E Z and p E P by taking 00kpU to be sections of U which vanish to order k at p if h S 0 or have poles of order at most h at p if h gt 0 We extend this de nition to give us sheaves OCD for any divisor D aipi Exercise 41 Show that OCD1 D2 g OCD1 OC 00D2 Exercise 42 Show that Ocp g 0010 hint it is enough to show that the map 001 8 00710 a 90 given by f1 f2 gt gt f1f2 is an isomorphism and conclude that OCD E 007D Lemma 426 Let D be a divisor on a Riemann surface 0 The sheaf OCD is a locally free sheaf of rank one and degree degD Proof Since 902 aipi E iOCaZpi it is enough to prove the assertion for D np where n E Z and p E 0 Away from p this sheaf is isomorphic to 90 so we just need to prove that in a small enough neighbourhood U around p the sheaf is isomorphic to OCU lf 2 is a local coordinate at p then the isomorphism Ocnp U a OCU is given by g gt gt 27p g It remains to show that degOc n Since 00np E 9077110 we can assume that n 2 0 Thus we have the global section 1 E FOcnp Let U be a small neighbourhood of p Then 00np is trivialized over U by g gt gt z 7 p 9 So the section 1 around p locally looks like 2 7 p Away from p the line bundle Ocnp is trivial so the section 1 does not intersect the zero section Hence 1 vanishes only at p where it vanishes to order n Thus7 by our original de nition of degree from section 347 degOCnp n Theorem 427 Any locally free sheaf L of rank one is isomorphic to OCD for some divisor D 32 Proof Suppose L has a global section 5 which vanishes along some divisor D aipl where al gt 0 ie it vanishes to order al at pi Then this gives a section of L OC 007D which does not vanish Thus L 800 007D 2 90 which means L OCD More generally7 if L has no section then by proposition 419 we know that L OC Ocnp has a section and so L 800 Ocnp E OCD for some divisor D Whence L E OCD 7 np D We still need to understand when 90 D OCD To do this we introduce the concept of principal divisor Consider a map f O 7 ll To f we associate the divisor f 107f 1oo which we denote A divisor D is a principal divisor if D f for some f C 7 ll Two divisors D and D are called linearly equivalent which we denote D E D if D 7 D is a principal divisor Theorem 428 OCD OCD if and only ifD E D Example We know that on lP l there is only one holomorphic line bundle of degree d for each d E Z up to isomorphism So if p7q 6 ll then we must have OP1dp g OP1dq for every d E Z This means that dp E do so by the theorem above there must be some map f lP l 7 ll such that f 10 7 f 1oo dp 7 dq If we take p 0 and q 00 then we can write down this map explicitly as 2 gt7 2 Example If C is a Riemann surface of genus g gt 0 then 001 001 for any p 7E q E C To see this we must show that p 7 q is not a principal divisor But ifp 7 q f for some f O 7 ll then this map f would be of degree one and hence an isomorphism contradiction since 0 ll To summarize7 we have the following three equivalent descriptions line bundles gt locally free sheaves of rank one gt divisors up to isomorphism up to isomorphism up to linear equivalence 45 A proof of RiemannRoch for curves We need a technical lemma To appreciate the lemma recall that if E 7 F is an inclusion of locally free sheaves then the quotient need not be locally free The standard example is the exact sequence 0 7 OC 7 Ocp 7 CI 7 0 The reason this happens is that locally in the trivialization around p the map on the ber over 2 is multiplication by z 7p because the isomorphism 001 U 7 OCU is given by 9 gt7 2 7 pg Thus at p the rank of the map drops from one to zero and that is why at p we get a cokernel which shows up as the skyscraper sheaf 17 In our case we have a vector bundle V and we want to nd a line subbundle L 7 V so that its quotient is locally free this way we can study V by studying L and its quotient ln lemma 422 we learned that QC module morphisms L 7 V correspond to global sections 5 E PV OC L As in the example above7 the quotient sheaf will be locally free if and only if this section is nowhere zero Lemma 429 Let V be a vector bundle on a Riemann surface 0 If L is a line bundle of highest possible degree such that there eccists a non zero rnap L 7 V then quotient sheaf is locally free Proof A map L 7 V corresponds to a section in s E H0CV 800 L If degL is large enough then there are no global sections and if degL is really negative then there must be 33 some global sections 419 Thus such an L of highest degree exists Suppose that s vanishes along some nonzero divisor D Lilpl with al E Z Then H0C7 V OC L 800 007D contains a section which means there is a map L 800 OCD 7 V This is a contradiction since degL OC OCD gt degL Theorem 430 Riemann Roch IfV is a holomorphic vector bundle on a Riemann surface 0 of genus 9 then C7 V degV 1 7 granhV Proof We rst reduce to the case when rankV 1 By the lemma 429 we can nd a short exact sequence 0 7 L 7 V 7 W 7 0 where L is a line bundle and W is a vector bundle Since L and W have ranks less than the rank of V we can assume by induction that Riemann Roch holds for them Hence xL degL17 g and xW degW17 grankW On the other hand7 xV xL xW and degV degL degW and rankV rankL rankW so the result follows It remains to prove Riemann Roch for line bundles L We know L 00D for some divisor D llpl so we proceed by induction on loll Suppose without loss of generality that p1 gt 0 so that have the short exact sequence 0 7 OCD 7 pl 7 OCD 7 CF By induction xOCD 7 101 degD 7 1 7 g 1 and MCI 1 so that xOCD xOCD 7101 xCCp degD 7 g 1 The base case for this induction is L 90 which follows by direct computation since h0OC 1 and h1Oc g D 5 CLASSIFYING VECTOR BUNDLES ON RIEMANN SURFACES 51 Grothendieckls classi cation of vector bundles on ll We know for each d E Z there exists a unique holomorphic line bundle on llD1 of degree d which we denote OP1d Theorem 51 Grothendieck Any holomorphic rank n vector bundle on llD1 is uniquely isomorphic to a direct sum of line bundles 691OP1dZ where d1 3 S dn and dl E Z Proof First we show that any rank n vector bundle V on P1 splits as a direct sum of line bundles Using lemma 429 we take a line bundle L OP1d of highest degree so that we have a short exact sequence 0 7 L 7 V 7 V 7 0 where V is also a vector bundle By induction we can assume V splits as 110P1di Claim dl S d Suppose dl gt d for some i Tensoring the short exact sequence with OP17d 7 1 we get another short exact sequence see lemma 53 below 0 7 OP171 7 V 011 OP17d i 1 7 eaygfowdi i d71 7 0 Since H0O 171 H1O 171 0 if we look at the long exact sequence in cohomology we get H0V Onp1 OP17d 7 1 7 HO1IOP1di 7 d 7 1 which is nonzero since dl 7 d 7 1 2 0 But this means V has a section of degree d 1 which is a contradiction Consider now proposition 52 below where we take 8 L7 V and g V Then H1g 0P18 H1e iOP1d7di H0iOP1d7di OP1KP1 H0i01p1di7d72 0 since K m OP172 and dl 7d 7 2 lt 0 Thus the sequence splits and hence V splits as the direct sum of line bundles Finally7 suppose iOP1al E iOP1bZ for some ahbi E Z with 11 lt lt an and b1 lt 3 bn If on lt bn then tensoring both sides by OP17bn we get iOP1ai 7 bn E iOP1bi 7 b The left side has no sections since 11 lt bn whereas the right side has at least one section coming from the Owl bn 7 bn E OHM summand Thus an bn Next7 if 1W1 lt bikl then tensor by OP17bn1 and look at global sections again to conclude7 as before7 that alkl bikl Continuing this way we show 11 bl for all i thus completing the proof Proposition 52 IfO a S f i g a 0 is a short eacact sequence of vector bundles on a complecc manifold X and H1g OX S 0 then the sequence splits ie 7 E S 69 9 Proof By proposition 32 it is enough to nd a map 9 g a f such that pog idg This can be done if the following map induced by p is surjective Homox 977 i HomoX g 9 since we can take as 9 any preimage of idg E Homox 99 But H0m0xg77 HOWOX 9593k OX OX Similarly7 HomOX 99 rg OX Q So we need to show that Pg OX f a Pg OX g is surjective To do this we rst tensor 0 a S a f a g a 0 by 9 to obtain a short exact sequence see lemma 53 below whose associated long exact sequence of cohomology is 0 a PQ ox 8 a PQ ox f a N 0X Q a H1Q ox 8 a Since H1g OX S 0 this implies Pg OX f a Pg OX g is surjective and the result follows E Lemma 53 IfO a S a f a g a 0 is a short eacact sequence on a complem manifoldX and V a vector bundle then 0 a S OX V a f 0x V a g OX V a 0 is also short eccact Proof A sequence is exact if it is locally exact for a ne enough open cover Ua of X Suppose V has rank one We can pick such an open cover where Van is triVial and hence isomorphic to OXUa Then over Ua we have 8Ua OXUQ VUa g 8Ua and similarly with f and 9 so the sequence remains exact The same thing happens if V has rank greater than one Of course7 in general a short exact sequence does not necessarily remain exact after ten soring For example7 consider the sequence of abelian groups 0 a Z L Z a Z2 a 0 If we tensor over Z with Z2 then we get 0 a Z2 g Z2 a Z2 a 0 which is no longer exact 52 Atiyah s classi cation of vector bundles on elliptic curves Let C be a Riemann surface of genus one First we take a closer look at line bundles on C We know that the space JC of degree zero holomorphic line bundles on C is a torus of dimension one This means that JC is also a genus one Riemann surface Proposition 54 IfC is a genus one Riemann surface then JC E C E PicdC for any d e Z Proof Recall that Pied denotes the space of line bundles on C of degree d If E is a line bundle of degree d then we have an isomorphism JC a PicdC given by L gt gt L 800 E L is a degree zero line bundle The inverse of this map is L gt gt L 800 E because E 800 E g 00 Thus all PicdC are isomorphic as d varies 35 A proof that JC E C can be obtained directly from the description of JC as the quotient of H1COC E C by a lattice A However7 we give a more direct proof by showing that the map 0 a Pic1C given by p gt gt 001 is an isomorphism To see that the map is injective we need to show 001 001 unless p q This we saw to hold true for any Riemann surface 0 of genus g gt 0 see the second example at the end of section 44 Next we need to show that any line bundle L of degree one is isomorphic to 001 for some point p E C By Riemann Roch7 if degL 1 then h0L 1 since by Serre duality h1L h0L 0 KC 90 because 0 has genus one So L has a global section which vanishes at exactly one point p since degL 1 Consequently7 as we saw in the proof of theorem 427 this means L 2 001 Aside One can check that the group structure on JC coming from 8 is the same as the natural group structure on 0 coming from the fact that G g CA For C a Riemann surface of arbitrary genus7 the map 0 a Pic1C used in the proof above is called the Abel Jacobi map The following is a direct corollary of the proof Corollary 55 Let C be a Riemann surface of genus at least one Then the Abel Jacobi map 0 a Pic1C given byp gt gt 001 is injective Moreover ifC has genus one then it is an isomorphism De nition 56 A vector bundle V is called indecomposable if it cannot be written as a direct sum V GBV ofnonzero subbundles V is irreducible ifit does not contain any nonzero subbundles V g V Warning A holomorphic vector bundle V can be indecomposable yet not irreducible This is because there might be some V g V but no complement V so that V V GBV Of course7 in the case of complex or real vector bundles the two notions coincide proposition 35 Denote by Bunr7 d the space of indecomposable vector bundles of rank r and degree d on C The proposition above asserts that Bun17 d g C The determinantal map V gt gt detV de nes a map Bunrd a Bun1d Theorem 57 Atiyah Bunr7 d g C and the map det Bunr7 d a Bun17 d has degree h2 where h gcdrd Moreover V E Bunrd is irreducible if and only ifh 1 Proof The proof is similar to that of Grothendieck7s theorem 51 An interesting reference is Atiyah7s original paper D REFERENCES A Mi F ATIYAH7 Vector bundles over an elliptic curve7 Proc London Math Soc 3 7 1957 41474521 Ha R HARTSHORNE7 Algebraic geometry 1 B lVERSEN7 Cohomology of sheaves l R MIRANDA7 Algebraic curves and Riemann surfaces 2


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