VECTOR BUNDLES ON RIEMANN SURFACES

SABIN CAUTIS

Contents

1. Differentiable Manifolds 22. Complex Manifolds 32.1. Riemann Surfaces of Genus One 42.2. Constructing Riemann Surfaces as Curves in P2 62.3. Constructing Riemann Surfaces as Covers 92.4. Constructing Riemann Surfaces by Glueing 103. Topological Vector Bundles 113.1. The Tangent and Cotangent Bundles 133.2. Interlude: Categories, Complexes and Exact Sequences 143.3. Metrics on Vector Bundles 153.4. The Degree of a Line Bundle 163.5. The Determinantal Line Bundle 173.6. Classification of Topological Vector Bundles on Riemann Surfaces 183.7. Holomorphic Vector Bundles 193.8. Sections of Holomorphic Vector Bundles 204. Sheaves 214.1. Cech Cohomology 234.2. Line Bundles and Cech Cohomology 274.3. Riemann-Roch and Serre Duality 294.4. Vector bundles, locally free sheaves and divisors 304.5. A proof of Riemann-Roch for curves 335. Classifying vector bundles on Riemann surfaces 345.1. Grothendieck’s classification of vector bundles on P1 345.2. Atiyah’s classification of vector bundles on elliptic curves 35References 36

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1. Differentiable Manifolds

Two topological spaces X and Y are homeomorphic if there exist continuous maps f :X → Y and g : Y → X such that g ◦ f = idX and f ◦ g = idY . This is denoted X ∼= Y .

Exercise 1. Show that S1 and the open unit interval (0, 1) are not homeomorphic.Exercise 2. Show that (0, 1) and the real line R are homeomorphic.Warning: to show X ∼= Y it is not enough to find a continuous map f : X → Y which is

1-1 and onto because the inverse map f−1 may not be continuous. For example, the naturalinclusion f : (0, 1] → S1 is continuous and bijective but the inverse f−1 is not continuous atf(1).

A manifold of dimension n is a Hausdorff topological space M such that every point p ∈Mhas a neighbourhood p ∈ U ⊂ M which is homeomorphic to the open unit ball in Rn. Inother words, M locally looks like Rn.

Example. The real line R, the unit circle S1 and the open interval (0, 1) are one-dimensional manifolds. The semiopen interval (0, 1] is not a manifold since there is no openneighbourhood of 1 homeomorphic to an open interval in R.

Example. The sphere S2, the torus S1 × S1, the open cylinder (0, 1)× S1 as well as anyopen subset of R2 are two-dimensional manifolds.

An open cover of a topological space X is a collection of open subspaces Uα such that∪αUα = X. A manifold M is compact if every open cover of M has a finite subcover orequivalently if every sequence of points {pi} ⊂ M contains a subsequence which convergesto a point p∞ ∈M .

By definition, an n-dimensional manifold M is covered by open sets Uα which are home-omorphic via maps fα : Uα → Rn to an open ball in Rn. Denote Uαβ = Uα ∩ Uβ andfαβ = fβ ◦ f−1

α : fα(Uαβ) → fβ(Uαβ). One should think of M as being built from these ballsby glueing Uα to Uβ along Uαβ as dictated by fαβ (i.e. p is identified with fαβ(p)).

The open sets fα(Uα) ⊂ Rn are called charts and the functions fαβ between them arecalled transitionfunctions. By construction, the transition functions satisfy:

• fαβ = f−1βα on Uαβ = Uα ∩ Uβ

• fβγ ◦ fαβ = fαγ on Uαβγ = Uα ∩ Uβ ∩ Uγ

Conversely, one can reconstruct M given the open cover {Uα} together with transition func-tions fαβ satisfying these two conditions.

Example. Consider S1 = {(x, y) : x2 + y2 = 1} ⊂ R2. We can use as an open cover thesets U1 = S1 − (0, 1) and U2 = S1 − (0,−1). Then we have homeomorphisms fi : Ui → Rgiven by f1(x, y) = −x

y−1and f2(x, y) = x

y+1. One can check that f−1

1 : R → U1 is given by

t 7→(

2tt2+1

, t2−1t2+1

)

. Then f21 = f2 ◦ f−11 : R∗ → R∗ is given by t 7→ 1

t. Thus S1 is obtained by

glueing R to itself along R∗ = R− 0 using the transition function f21(t) = 1t.

A manifold has the extra structure of a differentiable manifold if all the maps (all thetransition functions) are not only continuous but also C∞ (infinitely differentiable). Indimensions at most three, these two notions coincide – in other words, a topological manifoldof dimension one, two or three has a unique differentiable structure. This is not true in higherdimensions. For example, by a result of Milnor, the seven sphere S7 has exactly 28 differentdifferentiable structures.

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The following is a short summary of classification results for compact manifolds:

• The only compact 0-dimensional manifold is the point.• The only compact 1-dimensional manifold is the unit circle S1.• There are two types of surfaces: orientable and non-orientable. The orientable sur-

faces are classified by their genus g. In other words, for each g ∈ Z≥0 there is exactlyone surface of that genus. The sphere has genus zero, the torus T 2 = S1 × S1 hasgenus one, the double torus (containing two holes) has genus two and so on.

• 3-dimensional manifolds are classified by the geometrization conjecture of Thurston.Recently, a proof was proposed by Perelman and it is being verified at the moment.

• The classification of 4-manifolds was essentially established by Freedman (althoughthe classification of differentiable 4-manifolds remains open).

• Manifolds of dimension at least 5 (both topological as well as differentiable) have inprinciple been classified.

2. Complex Manifolds

A differentiable manifold of dimension n is modelled on Rn and C∞ functions. Analogously,we can use Cn together with holomorphic transition functions to construct complex manifoldsof (complex) dimension n (real dimension 2n). Locally, a complex manifold will have chartsUα with maps fα : Uα → Cn such that the transition functions fαβ from fα(Uα ∩ Uβ) ⊂ Cn

to fβ(Uα ∩ Uβ) ⊂ Cn are biholomorphic (instead of just homeomorphisms).Example. The 2-sphere S2 has the structure of a complex manifold since S2 can be

gotten by glueing two copies of C along C∗ = C − 0 using the transition function z 7→ 1z.

This construction is analogous to the way we constructed the circle S1 by glueing togethertwo copies of R along R∗ = R− 0.

Given two complex manifolds M1 and M2 a morphism f : M1 →M2 is a continuous mapsuch that its restriction to charts is holomorphic. We say f is an isomorphism if it has aholomorphic inverse.

Before continuing we mention a few useful results about holomorphic maps:

• (Open Mapping Theorem) A nonconstant holomorphic map C→ C maps open setsto open sets.

• A holomorphic map which is injective and surjective is biholomorphic (i.e. it has aninverse which is also holomorphic).

A compact complex manifold of (complex) dimension one is known as a Riemann surface.In a way, these are the simplest compact, complex manifolds and the source of some incrediblybeautiful mathematics.

It turns out a complex manifold is automatically orientable (see section 3.5). Thus Rie-mann surfaces are topologically classified by their genus.

The sphere (genus g = 0) has only one complex structure. In other words, two Riemannsurfaces which are homeomorphic to the 2-sphere then are actually isomorphic as Riemannsurfaces (i.e. there is a biholomorphic map between them). This result follows from thefollowing more general theorem.

Theorem 2.1 (Uniformization Theorem). Any simply connected complex manifold of di-mension one is isomorphic to precisely one of the following spaces:

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• the Riemann sphere C• the complex plane C• the hyperbolic plane H = {z ∈ C : Im(z) > 0}

Intuitively, we think of C as C with a point added at infinity (thus C compactifies C). Theuniformization theorem above is remarkable since it says that any simply connected domainin C (which is not all of C) is isomorphic to H.

Exercise 3. Show directly that the open unit disk ∆ = {z ∈ C : |z| < 1} is isomorphicto H (hint: use the map z 7→ z−i

iz−1).

Proposition 2.2. The (holomorphic) automorphisms of C are z 7→ az+ b for some a, b ∈ C(with a 6= 0).

Proof. Clearly z 7→ az + b is biholomorphic if a 6= 0. Conversely, suppose f ∈ Aut(C). Bycomposing with z 7→ az + b for appropriate a and b we can assume f(0) = 0 and f ′(0) = 1.Thus f(z) = z + a2z

2 + a3z3 + . . . locally around 0. If f(z) is not a polynomial then f(1/z)

has an essential singularity at z = 0 which by Picard’s theorem means that f(1/z) = w hasinfinitely many solutions in z around 0 and hence is not injective. Thus f(z) is a polynomial.But if deg(f) > 1 then for a general a ∈ C the equation f(z) = a has more than one solutionmeaning f is not injective again. Whence f(z) = z. �

Exercise 4. Use this to show that

Aut(C) = {z 7→ az + b

cz + d: ad− bc 6= 0}/{dI : d ∈ C} ∼= PGL2(C)

(recall that we can think of C as C with a point added at infinity).

Proposition 2.3. Aut(H) = PSL2(R) where

(

a bc d

)

acts by z = az+bcz+d

.

2.1. Riemann Surfaces of Genus One. What are the complex structures one can puton the genus one surface T 2? T 2 is the quotient of R2 by Z⊕2 via the free action R2 by(a, b) · (x, y) = (x + a, y + b). If T 2 has a complex structure then R2, which is simplyconnected and hence the univeral cover of T 2, inherits a complex structure from T 2. By theUniformization Theorem we know R2 has two possible complex structures, namely C and H.

Proposition 2.4. 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 ∈ X you can find a ballB around p such that G · p∩B = p. Note that the action of Z⊕Z on R2 described above isdiscrete. On the other hand, the lemma below shows that there is no discrete Z⊕ Z actionon H. Thus the universal cover of a genus one Riemann surface cannot be H and must beC. �

Lemma 2.5. Any subgroup Z⊕ Z ⊂ Aut(H) acts non-discretely on H.

Proof. Denote by a and b the generators of Z⊕Z. Suppose a is not diagonalizable. Then we

can conjugate it to the form

(

1 s0 1

)

. Since a and b commute it means b =

(

1 t0 1

)

for

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some t ∈ R. Thus the action on H is not discrete for otherwise ma = nb for some m,n ∈ Z.So we can assume a and b are diagonalizable. Since they commute they are simultaneously

diagonalizable. But then a =

(

λ 00 λ

)

and b =

(

µ 00 µ

)

so again they must generate a

nondiscrete subgroup of Aut(H) for otherwise ma = nb for some m,n ∈ Z. �

Hence any genus one Riemann surface is isomorphic to C modulo the action of some fixed-point free subgroup Z ⊕ Z ⊂ Aut(C). The fixed-point free automorphisms of C are thetranslations tb : z 7→ z + b. Thus any genus one Riemann surface is obtained by quotientingC by two automorphisms tb and tb′ for some b, b′ ∈ C. Here tb and tb′ correspond to thegenerators (1, 0) and (0, 1) of Z ⊕ Z. Note that b and b′ must be pointing in differentdirections (i.e. they must be linearly independent over R) or else the quotient is not a genusone Riemann surface. Rotating and scaling C by an appropriate factor allows one to assumethat b′ = 1 and b ∈ H = {z ∈ C : Im(z) > 0}. Thus we get a map from H to the space ofRiemann surfaces of genus one given by τ 7→ C/〈1, τ〉. We will denote the Riemann surfacecorresponding to τ ∈ H by Eτ (E stands for elliptic curve which is another name for a genusone Riemann surface).

Proposition 2.6. The map from H to the set of Riemann surfaces is onto. Two points givethe same Riemann surface iff. they differ by the action of an element of SL2(Z).

Proof. Surjectivity of the map follows from the description above. The action of

SL2(Z) = {(

a bc d

)

: a, b, c, d ∈ Z, ad− bc = 1}

on H is by z 7→ az+bcz+d

. To understand where this action comes from recall that tb and tb′

corresponded to the basis elements (1, 0) and (0, 1) of Z ⊕ Z. Of course, picking differentbasis elements of Z⊕ Z would give the same Riemann surface. Since Aut(Z⊕ Z) = SL2(Z)changing basis for Z⊕ Z corresponds to acting on H by SL2(Z).

To show that two points in H which do not differ by an element of SL2(Z) correspond todifferent genus one Riemann surfaces let’s suppose that C/〈1, τ〉 and C/〈1, τ ′〉 are isomorphic

via some isomorphism f . Then this map lifts to an isomorphism f : C → C such that thelattice 〈1, τ〉 maps to the lattice 〈1, τ ′〉. Thus f(1) = a + bτ ′ and f(τ) = c + dτ ′ wherea, b, c, d ∈ Z satisfy ad − bc = 1 (since the map must be invertible). Hence τ and τ ′ differthe action of an element of SL2(Z). �

Corollary 2.7. The moduli space M1 (i.e. the parameter space) of genus one Riemannsurfaces is H/SL2(Z) (which has complex dimension one and happens to be isomorphic toC).

What are Aut(Eτ )? An automorphism of Eτ is the same as giving an automorphism of Cwhich commutes with the projection map p : C→ Eτ .

Proposition 2.8. Eτ has automorphisms z 7→ z+b (translations) and z 7→ −z (involution).For general τ these are all the automorphisms. For two special values of τ we have thefollowing extra automorphisms:

• τ = i: z 7→ iz5

• τ = 1+i√

32

: z 7→ eiπ/3z

Aside: Just as a Riemann surface of genus one is the quotient of C by Z⊕Z, it turns outthat a Riemann surface C of genus g ≥ 2 is the quotient of H by the fixed point free action ofa group. As before, this group is the fundamental group π1(C) of C. However, describing allthe possible actions of π1(C) on H is quite difficult. The moduli space of genus g Riemannsurfaces is denoted Mg and turns out to have have a complex structure of complex dimension3g−3. In other words, there is a 3g−3 dimensional space parametrizing Riemann surfaces ofgenus g ≥ 2. Recall that the moduli space of genus one Riemann surfaces is one dimensional(corollary 2.7) whereas M0 is only a point since there is a unique complex structure on S2.

2.2. Constructing Riemann Surfaces as Curves in P2. One natural way of describingmanifolds is as the zero set of a polynomial. For instance, the n-dimensional sphere is thezero set of p = x2

0 + · · · + x2n − 1 in Rn+1. One can even take the zero set of more than one

polynomial. For example, the zero set of p and x0 (in other words, all points x = (x0, . . . , xn)satisfying p(x) = 0 = x0) gives us the (n − 1)-dimensional sphere. Similarly, an effectiveway to construct complex manifolds is as the zero set of polynomials in Cn. For example,z20 + z2

1 − 1 = 0 is isomorphic (as a complex manifold) to C. Unfortunately, since Cn is notcompact all complex manifolds constructed this way will not be compact. To remedy thiswe compactify Cn to get the projective space Pn.

2.2.1. Projective Spaces. The projective space Pn(C) = Pn is defined as the quotient ofCn+1 − 0 by C∗ = C− 0 where the action is given by λ · (z0, . . . , zn) = (λz0, . . . , λzn). Justas a point in Cn is given in the usual coordinates as an n-tuple (z1, . . . , zn), a point in Pn isencoded in homogeneous coordinates as an n + 1-tuple [z0, . . . , zn] where one keeps in mindthat [z0, . . . , zn] and [λz0, . . . , λzn] (λ 6= 0) represent the same point.

Let’s examine P1 in greater detail. A typical point of P1 is [z0, z1]. If z0 6= 0 then the pointhas can be uniquely represented as [1, t] where t = z1/z0. Thus the locus of points wherez0 6= 0 is an open subset U0 ⊂ P1 which is diffeomorphic to C via the map [z0, z1] 7→ z1/z0.The complement of U0 is just the point [0, 1]. Intuitively, P1 is just C with an extra pointadded at infinity, so if we believe that P1 should be a complex manifold then it ought to bethe Riemann sphere C.

Let’s check this directly. Consider the open subset U1 ⊂ P1 consisting of points [z0, z1]

where z1 6= 0. Then, as with U0, there is a diffeomorphism U1∼−→ C given by [z0, z1] 7→ z0/z1.

The intersection U0 ∩ U1 is the locus of points [z0, z1] where z0 6= 0 and z1 6= 0. Thus, P1 isobtained by glueing C to C along C∗ by using the map t = z0

z1

7→ z1

z0

= 1t. Since t 7→ 1/t is

a holomorphic map over C∗ this gives P1 a complex structure. Notice that this is the sameglueing map as the one used in section 2.

One can similarly study P2. A typical point of P2 is [z0, z1, z2]. Then one can consider open

subsets Ui ⊂ P2 consisting of those points where zi 6= 0. As before, we find that Ui∼−→ C2

form holomorphic charts for P2. Notice that the complement of Ui is now P1. Thus P2 isobtained by glueing to C2 a copy of P1.

Exercise 5. Check that Pn has a stratification ∐0≤i≤nCn.

Proposition 2.9. Aut(Pn) ∼= PGLn+1(C) where the action is by left multiplication on[z0, . . . , zn].

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Proof. Since Pn is Cn+1−0 modulo C∗, any automorphism of Pn lifts to an automorphism ofCn+1. Conversely, any automorphism of Cn+1 which commutes with the projection Cn+1 −0 → Pn descends to an automorphism of Pn. The result now follows from the fact that upto translations the automorphism group of Cn+1 is GLn+1(C). �

The important thing for us is that Pn is compact. In the case of Cn the zero sets ofpolynomials give us (non-compact) complex manifolds. The equivalent of a polynomial inthe setting of Pn is a homogeneous polynomial.

Definition 2.10. Check that a polynomial f(z0, . . . , zn) is homogeneous if for any λ ∈ C−0we have f(λz0, . . . , λzn) = f(z0, . . . , zn).

In other words, we want to give a function on Cn+1 which is invariant under the C∗ action(z0, . . . , zn) 7→ (λz0, . . . , λzn) and thus descends to a well defined function on the quotientCn+1/C∗ = Pn.

Exercise 6. A polynomial f(z0, . . . , zn) is homogeneous if and only if every monomialterm has the same degree n (which is by definition the degree of f).

Example. z0 + z21 is not homogenous whereas z3

0 + z31 is homogeneous of degree 3.

If f(z0, . . . , zn) is a general homogeneous polynomial then its zero set in Pn carves out acomplex submanifold of dimension n − 1. For instance, if f(z0, z1, z2) = z2 then the zeroset are the points in P2 of the form [z0, z1, 0] which as we saw is the Riemann sphere S2.Since Pn is compact, the manifold we get will also be compact (since it is a closed subset ofa compact space).

Exercise 7. Show that the zero set of f(z0, z1, z2) = a0z0 + a1z1 + a2z2 in P2 (where the

constants ai’s are not all zero) is isomorphic to the Riemann sphere C (hint: use the actionof PGL3(C) on P2).

Recall that our goal is to construct examples of Riemann surfaces. To do this we canconsider the zero set of a homogeneous polynomial f(z0, z1, z2) of degree d. For d ≥ 3 theRiemann surface we get will depend on f but it’s genus will not. So our next question is: iff has degree d what is the genus of f(z0, z1, z2) = 0. To answer this we need a new tool.

2.2.2. The Euler Characteristic. LetM be a (topological) manifold and consider an arbitrarytriangulation of M . The Euler characteristic χ(M) of M is defined as the number of verticesin the triangulation minus the number of edges plus the number of faces and so on.

Example. Viewing the sphere S2 as a tetrahedron gives us a triangulation of S2 withfour vertices, six edges and four faces. Thus χ(S2) = 4− 6 + 4 = 2. Viewing S2 as the cubegives us eight vertices, twelve edges and six faces. Again, we get χ(S2) = 8 − 12 + 6 = 2(the latter is not technically a triangulation so we should not use it but we still get the rightanswer since each face is a polygon).

Theorem 2.11. The Euler characteristic χ(M) does not depend on the specific triangulationused.

Proof. Given two triangulations of M one can subdivide them further to obtain a commonsubtriangulation. The result follows from the following result:

Exercise 8. Show that χ(M) remains unchanged after taking subtriangulations. �

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Thus the Euler characteristic χ(M) is an invariant of M . It provides a way to distinguishbetween manifolds: namely, if χ(M1) 6= χ(M2) then they are not homeomorphic. In the caseof 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 ghas Euler characteristic 2 − 2g.

2.2.3. Riemann-Hurwitz Theorem. Suppose f : C1 → C2 is a holomorphic map between twoRiemann surfaces. Locally around each point p ∈ C1 the map p 7→ f(p) can be written asz 7→ anz

n + an+1zn+1 + . . . . Changing variables one can rewrite it as z → zn. The integer n

is called the ramification index of f at p and denoted ep. If n = 1 then we say f is unramifiedat p, otherwise we say p is a ramification point. A point q ∈ C2 is a branching point if f−1(q)contains a ramification point.

Exercise 10. What does the map z 7→ zn from ∆ to ∆ look like? (identify the branchingpoints, the ramification points and their indices)

For any holomorphic map f : C1 → C2 the target C2 contains only finitely many branchingpoints. If q ∈ C2 is not a branching point then its preimage f−1(q) will contain d points,where d is called the degree of the map f . If q ∈ C2 is a branching point then f−1(q)will contain less than d points. The precise number of points in f−1(q) is related to theramification indices as follows:

Proposition 2.12. For any q ∈ C2 we have∑

p∈f−1(q) ep = d.

Example. Consider the map f : P1 → P1 given by z 7→ zd. The preimage of a pointw ∈ C ⊂ P1 are the points z ∈ C satisfying zd = w. Thus for w 6= 0 the preimage containsd points and hence deg(f) = d. The branching points are 0,∞ ∈ P1. The preimage of 0 is 0where the ramification index is d. Similarly, the preimage of ∞ is ∞ where the ramificationindex is also d (check this by writing down the map f in a local neighbourhood of ∞).

Given f : C1 → C2, the Riemann-Hurwitz theorem tells us how to relate the Eulercharacteristics χ(C1) and χ(C2) in terms of the degree deg(f) and the ramification indices.

Theorem 2.13 (Riemann-Hurwitz). χ(C1) = dχ(C2) −∑

p∈C1(ep − 1).

Proof. Choose a triangulation of C2 such that if p ∈ C1 is a ramification point then f(p) is avertex. With this added condition, the preimage of this triangulation gives a triangulationof C1. Denote by v, e, f the number of edges, vertices and faces in the triangulation of C2

(so that χ(C2) = v − e+ f). The number of edges and faces in the triangulation of C1 is deand df respectively. The number of vertices is dv − ∑

p∈C1(ep − 1). Thus

χ(C2) = dv −∑

p∈C1

(ep − 1) − de+ df = d(v − e+ f) −∑

p∈C1

(ep − 1).

�

Notice that the sum on the right side is finite since all but a finite number of points haveramification index one. One can use this result to compute the genus of C1 given the genusof C2 and knowledge about the map f : C1 → C2.

Example. Consider again the map f : P1 → P1 given by z 7→ zd. We saw that there aretwo ramification points, both of which have index d. Thus

∑

p∈P1(ep−1) = (d−1)+(d−1) =8

2d − 2. From the Riemann-Hurwitz equation we find χ(P1)(1 − d) = 2 − 2d and henceχ(P1) = 2 (reaffirming our earlier calculation).

Example. In algebraic geometry, Riemann surfaces are also called curves. Consider theFermat curve C in P2 given by xn + yn + zn = 0 for some n ∈ N. The map P2 → P1 givenby [x, y, z] → [x, z] is not well defined at [0, 0, 1] (why?) but since [0, 0, 1] is not a point onC we get a well defined map π : C → P1. The preimage of a point [x0, z0] ∈ P1 are all point[x0, y, z0] ∈ C. The number of such points is the number of solutions in y to the equationyn = −xn

0 − zn0 . This number is in general n and thus π has degree n. On the other hand,

the map is not n : 1 precisely when −xn0 − zn

0 = 0 or equivalently xn0 = −zn

0 . The pointsin P1 which satisfy this equation are precisely the points of the form [ζ, 1] where ζn = −1.Thus there are precisely n points in P1 over which the fiber of π is ramified and each timethis happens the fiber contains only one point instead of n. Hence, by Riemann-Hurwitz:

χ(C) = nχ(P1) − n(n− 1) = 2n− n(n− 1) = −n2 + 3n.

Thus, the genus of C is g =(

n−12

)

since χ(C) = 2 − 2g.

Proposition 2.14. The genus of a general curve of degree n in P2 is(

n−12

)

.

Proof. The proof is along the same lines as above except that counting the ramificationpoints in general is a little harder. �

A good way to visually justify this result is as follows. Suppose the curve is given by adegree n homogeneous polynomial which factors completely into different monomials (forexample, for n = 4 we could have f(x, y, z) = xyz(x− y+ z)). Each factor corresponds to acopy of P1 and thus the product describes the union of n such P1’s. Moreover, every two P1’smeet in precisely one point. Thus we have n spheres every pair of which meet in exactly onepoint. If one deforms the equation f slightly then one gets a nice (smooth) Riemann surfaceand the points in the original curve correspond to

(

n2

)

small loops in this Riemann surfacewhich have been pinched to a point. It is then easy to see that in fact the genus of thisRiemann surface is

(

n−12

)

. 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 intersectat a point.

Even though this gives us an effective way of constructing Riemann surfaces as complexsubmanifolds in P2 not all Riemann surfaces can arise this way. In particular, since

(

n−12

)

= 2has no solution, we cannot construct genus two Riemann surface. On the other hand we cangeneralize this construction by considering the zero set in P3 of two polynomials f and g. Iff and g are chosen to be general this construction yields a Riemann surface. This way onecan construct examples of Riemann surfaces of arbitrary genera.

2.3. Constructing Riemann Surfaces as Covers. Given a curve C one can try to con-struct a new curve as a cover of C. To describe such a cover it is enough to identify thedegree d of the cover, the branching points p1, . . . , pm of C over which there is ramificationand a final piece of data known as monodromy. This data tells one how different sheets ofthe cover are glued together around each ramification point. We explain further using anexample.

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Example. The simplest case is when C = P1 and the degree is d = 2. In this case thereare two sheets and each ramification point interchanges these two sheets. In other words,locally around each ramification point the map looks like z 7→ z2. Notice that there is noextra monodromy data necessary. The only thing we need to specify are the branching pointsp1, . . . , pm. Let’s take m = 6 branching points.

Exercise 11. Check that if d = 2 and the number of branching points is m = 6 then thecover will have genus 2.

Since the automorphism group of P1 (namely PGL2(C)) acts triply transitive we can takep1 = 0, p2 = 1 and p3 = ∞. Each choice of p4, p5, p6 gives us a curve of genus two. Since thereare no automorphisms of P1 fixing 0, 1,∞ these curves turn out to be non-isomorphic to eachother. Thus we get a three dimensional family of genus two curves. Conversely, any genustwo curve can be written as a double cover (a two-to-one cover) of P1. This explains why themoduli 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 twosheets). This map is called the hyperelliptic involution.

Example. Consider the triple cover of P1 branched over three points p1, p2, p3. This timewe 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 ramification point of order 3 which interchanges the sheets 1 → 2 → 3 → 1. Themonodromy around p2 we take to be (12). In other words, there are two points lying abovep2. One of the points lies on sheet 3 and is unramified while the other is ramified of ordertwo and interchanges sheets 1 → 2 → 1. Finally, the monodromy around p3 is (13). Noticethat (123)(12)(13) = id. This is a necessary condition since the monodromy around all threepoints 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 fixed branching points

p1, . . . , pm ∈ P1 is known as Hurwitz’s problem. Since describing such covers is the same asgiving the monodromy data this is apriori a combinatorial problem involving the symmetricgroup Sn. More recently, this problem has been shown to have amazing and surprisingconnexions with the geometry of the moduli space of curves Mg. A potential project maybe to learn a little about the classical Hurwitz problem and explain (in broad terms) thisrecent connexion.

2.4. Constructing Riemann Surfaces by Glueing. This method can be used to obtainvery concrete examples of interesting Riemann surfaces with lots of automorphisms. Notethat in general, Riemann surfaces of genera g ≥ 2 do not have many automorphisms:

Theorem 2.15.

• A general Riemann surface of genus ≥ 3 has no nontrivial automorphisms.• A general Riemann surface of genus two has only one nontrivial automorphism (the

hyperelliptic involution).• The order of the automorphism group of any Riemann surface of genus g ≥ 2 is at

most 84(g − 1).

Once again, we illustrate this glueing construction using an example.10

Example. Start with two regular pentagons in C and identify opposite sides (by trans-lating) as shown in figure 1 to get a surface S.

����

����

����

����

����

����

����

����

PP

P

P

PP

P

P

Figure 1. The identification shown yields a Riemann surface of genus 2 withautomorphism group Z/10Z.

Exercise 13. Check that topologically this gives us a surface of genus two.Note that the vertices of the pentagons get identified to one point P . Away from this

point the surface inherits the complex structure from C.Exercise 14. Why is the complex structure well defined 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 2π5· 10 = 4π accumulating at P . Thus

we get a chart around P using the map p ∈ U → C given by z 7→ √z.

Since the pentagons are regular one can rotate them by 2π/5 to get an automorphism αof S. One can also exchange the pentagons by using a rotation by π to get an automorphismof order two. This involution is the hyperelliptic involution from earlier. It turns out theseare all the automorphisms of S so Aut(S) ∼= Z/10Z.

One can also obtain S as a double cover of P1 branched over the points 0, 1, ζ, ζ2, ζ3, ζ4

where ζ is a fifth root of unity. The map P1 → P1 given by z 7→ ζz fixes (as a set) these sixpoints and lifts to the cover to give the automorphism α ∈ Aut(S) we saw above. One canapply the same glueing procedure to other shapes to obtain explicit constructions of otherRiemann surfaces.

Exercise 15. What is the genus of the Riemann surface obtained by identifying oppositesides of an L-shaped figure? Curt McMullen studies such Riemann surfaces in order tounderstand 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 familyof real vector spaces of dimension n parametrized by M . More precisely, V is a manifoldwith a map π : V →M such that around each point p ∈M there is an open neighbourhoodp ∈ U such that π−1(U) ∼= U ×Rn. A vector bundle of dimension one is called a line bundle.

Example. The simplest vector bundle over M is the trivial bundle V = Rn ×M → Mwhich we denote In.

11

By definition, a vector bundle is locally trivial (i.e. over small enough open sets U ⊂M , itis isomorphic to Rn×U). Thus, to describe a vector bundle V overM it is enough to give localcharts Uα over which V is trivial together with transition functions fαβ : Uα ∩Uβ → GLn(R)which tell you how to glue Uα ×Rn to Uβ ×Rn over Uα ∩Uβ (i.e. the point (p, v) in Uα ×Rn

is identified with (p, fαβv) in Uβ × Rn). As before, these transition functions must satisfy:

• fαβ = f−1βα on Uα ∩ Uβ

• fβγ ◦ fαβ = fαγ on Uα ∩ Uβ ∩ Uγ

Example. Suppose M = S1 = {(x, y) : x2 + y2 = 1}. Along with the trivial line bundleI1 = M × R there’s also the Mobius line bundle M1 which we describe using the two chartsU0 = S1−(0, 1) and U1 = S1−(0,−1). The transition function f01 : U0∩U1 → GL1(R) = R∗

is given by (x, y) 7→ −1 if x < 0 and (x, y) 7→ 1 if x > 0. The resulting line bundle looks likean (infinite) Mobius strip. As this example illustrates, vector bundles are boring locally (sincethey are trivial) but become interesting globally because of the possibility for “twisting”.

A morphism between two vector bundles V1π1−→ M and V2

π2−→ M is a continuous mapg : V1 → V2 which sends fibers to fibers (i.e. π2 ◦ g = π1) using a linear fiberwise action(V1)p

∼= Rn → Rn ∼= (V2)p. We impose this condition of linearity since the fibers are vectorspaces and not just topological spaces. V1 and V2 are isomorphic if there is an invertiblemorphism g : V1 → V2. Clearly, if V1

∼= V2 then dim(V1) = dim(V2) but the converse is false.A section of a vector bundle π : V → M is a (continuous) map σ : M → V such that

π ◦ σ = idM . There is always a canonical section called the zero section which is given byp 7→ (p, 0). We will denote by Γ(M,V ) the vector space of all sections of V .

Exercise 16. Show that I1 has a section which does not intersect the zero section. Onthe other hand, show that every section of the Mobius bundle M1 must intersect the zerosection. Conclude that I1 and M1 are not isomorphic.

All the definitions from vector spaces carry over naturally to vector bundles. We summa-rize them below:

• subbundles: V ⊂ W is a subbundle if there is an injective morphism V → W . Thismeans that in each fiber Wp over p we can pick a vector subspace Vp ⊂ Wp such thatthe choices vary continuously with p.

• direct sums: if V1 and V2 are described by transition functions f1αβ : Uα ∩ Uβ →

GLm(R) and f2αβ : Uα ∩ Uβ → GLn(R) then the direct sum V1 ⊕ V2 has transition

functions(

f1αβ 00 f2

αβ

)

: Uα ∩ Uβ → GLm+n(R).

• tensor product: similarly, V1 ⊗ V2 has transition functions f1αβ ⊗ f2

αβ : (Uα ∩ Uβ) →GLmn(R) where this is short hand for the action v1 ⊗ v2 7→ f1

αβv1 ⊗ f2αβv2.

Exercise 17. Check that the operations ⊕ and ⊗ give the set of vector bundles over Mthe structure of a ring where the unit is the trivial line bundle I1.

Exercise 18. Consider the subbundles V1, V2 ⊂ I2 given by the linear span of (cos θ, sin θ)×(− sin θ/2, cos θ/2) and (cos θ, sin θ) × (cos θ/2, sin θ/2), respectively.

(1) Show that V1∼= V2

∼= M1 (the Mobius line bundle). (Hint: draw a picture).12

(2) Show that V1 ⊕ V2∼= I2, thus concluding that the sum of two nontrivial bundles can

be trivial.

Exercise 19. Show that a vector bundle V is invertible (i.e. there exists a vector bundleV ′ such that V ⊗ V ′ ∼= I1) if and only if V is a line bundle (i.e. dim(V ) = 1).

A complex vector bundle V → M is the same as a real vector bundle except that locallythe trivialization is π−1(U) ∼= U×Cn. The theory of real and complex vector bundles is verysimilar (just like the theory of real and complex vector spaces is very similar). On the otherhand, there are some differences. For example, consider the complex Mobius bundle over S1

defined as above using the same transition functions

f01 : U0 ∩ U1 → GL1(R) → GL1(C) = C∗

Exercise 20. Show that the complex Mobius line bundle is in fact trivial (hint: find asection which does not intersect the zero section and use the following result).

Lemma 3.1. A real (or complex) line bundle L → M is trivial if and only if there is anonvanishing section of L (i.e. a section which does not intersect the zero section).

Proof. Exercise 21. . �

Just as a real vector space V has a dual V ∗ = Hom(V,R), a real vector bundle V hasa dual vector bundle V ∗. If V is given by transition functions fαβ ∈ GLn(R) then V ∗ hastransition functions (f t

αβ)−1 ∈ GLn(R). Similarly, one defines the dual of a complex vectorbundle.

Fact: We will show later that if V is a real vector bundle then V ∼= V ∗ but if V is acomplex bundle then this need not be true. However, it is always true that V ∗∗ ∼= V .

3.1. The Tangent and Cotangent Bundles. Every real manifold M of dimension ncomes equipped with an intrinsic n dimensional vector bundle TM called the tangent bundle.The dual of this bundle is the cotangent bundle T ∗

M . Strangely, it is more natural to describethe cotangent bundle so we shall do that first.

The fibers of (T ∗M)p over a point p ∈ M should correspond to the vector space mp/m

2p

where mp ⊂ C∞ is the ideal consisting of functions vanishing at p. It is easy to check thatmp/m

2p is a vector space of dimension n. However, this does not tell us how to glue the fibers

together. To do this, consider a chart U0 around p with coordinates x1, . . . , xn. T ∗M will

be trivial over U0 and we choose as basis n vectors which we call dx1, . . . , dxn. Now takeanother chart U1 around p with coordinates y1, . . . , yn. Inside U0 ∩ U1 the two charts giveus a map Rn = (xi) → (yi) = Rn so that we can think of the yi as functions yi(x1, . . . , xn)in the xi. We also have a basis for T ∗

M over U1 given by dy1, . . . , dyn. Then the transitioninformation for the cotangent bundle over U0 ∩ U1 is given by:

dyi =n

∑

j=1

∂yi

∂xjdxj.

In other words, the transition function is f01 =(

∂yi(x1,...,xn)∂xj

)

∈ GLn(R). The fact that the

transition functions satisfy the glueing condition fαβ ◦ fβγ = fαγ is a result of the chain rule.13

The tangent bundle TM is defined as the dual of the cotangent bundle T ∗M . The local basis

elements for TM are denoted ∂∂xi

. The transition functions are then

∂

∂xi=

n∑

j=1

∂yj

∂xi

∂

∂yj.

Exercise 22. A somewhat tedious but important exercise everyone should do at leastonce in their lifetime is to check that these transition functions do in fact satisfy the glueingconditions.

3.2. Interlude: Categories, Complexes and Exact Sequences. A category C consistsof a class of objectsOb(C) and a class of mapsmor(C) between the objects. For a, b, c ∈ Ob(C)there is a composition operation mor(a, b) × mor(b, c) → mor(a, c) which is associative.Moreover, for each x ∈ Ob(C) there is an identity morphism 1x : x → x which satisfies1b ◦ f = f = f ◦ 1a for any f ∈ mor(a, b).

Since this is abstract nonsense, somesome examples one should keep in mind are:Example. The category of vector spaces over some field k. The objects are vector spacesover 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 themorphisms are continuous maps between them.

A category does not necessarily have kernels and cokernels. For example, one cannotdefine the kernel of a map in the category of topological spaces. Since we want to be ableto take kernels, cokernels and direct sums we will work in an enriched category known asan abelian category. Clearly, the categories of vector spaces and abelian groups are abeliancategories.

Consider a sequence of maps A·:

A0f1−→ A1

f2−→ A2 → · · · → An−1fn−→ An.

A· is a complex if for each i we have Im(fi−1) ⊂ Ker(fi). In this case we can talk of thehomology Hi(A·) = Ker(fi)/Im(fi−1). If Im(fi−1) = Ker(fi) then the sequence is calledexact. Clearly, A· is exact if and only if Hi(A·) = 0 for all i.

A short exact sequence is an exact sequence of the form:

0 → Ai−→ B

p−→ C → 0.

The fact that it’s exact translates into:

• i is injective• p is surjective• Im(i) = Ker(p)

A short exact sequence is said to split if B ∼= A⊕ C.Example. In the category of abelian groups consider the short exact sequence

0 → Z/2Z→ Z/4Z→ Z/2Z→ 014

where the first map is a 7→ 2a and the second map is the natural projection. It is easy tocheck this is a short exact sequence. But since Z/2⊕Z/2 6∼= Z/4 the sequence does not split.

Exercise 23. Show that in the category of (complex) vector spaces any short exactsequence splits (hint: define the orthogonal complement).

Proposition 3.2. A short exact sequence 0 → Ai−→ B

p−→ C → 0 splits if and only if thereexists a map f : B → A such that f ◦ i = idA (or equivalently, a map g : C → B such thatp ◦ g = idC).

Proof. Exercise 24. . �

A map F between two categories C and D is called a functor. It takes objects to objectsand morphisms to morphism such that if f, g, f ◦ g ∈ mor(C) then F (f ◦ g) = F (f) ◦ F (g).

3.3. Metrics on Vector Bundles. Consider a manifold M with an an open cover Uα.A partition of unity with respect to Uα is a collection of smooth, nonnegative functionsψα : M → R such that ψα is supported in the interior of Uα and

∑

α ψα = 1.

Theorem 3.3. Any open cover of a manifold has a partition of unity.

In fact, this result holds for paracompact spaces (a space is called paracompact if everyopen cover has a locally finite open refinement).

Just as you have inner products 〈·, ·〉 for real vector spaces (or Hermitian metrics forcomplex vector spaces) there are analogous concepts for vector bundles. An inner product〈·, ·〉V on a vector bundle V → M is a map V ×M V → R+ which restricted to any fiber Vp

is an inner product 〈·, ·〉Vp: Vp × Vp → R+.

We know how to construct an inner product on a fixed vector space. The following theoremtells us that we can construct an inner product on any vector bundle over a manifold.

Proposition 3.4. If V → M is a real (complex) vector bundle over a manifold M then Vadmits an inner (Hermitian) product.

Proof. Choose an open cover Uα of M over which V is trivial. For each α construct aninner product 〈·, ·〉α for VUα

. By the theorem above we can find a partition of unity ψα

subordinate to Uα. Then∑

α ψα〈·, ·〉α gives an inner product on V (we use that the sumof two inner products on a vector space is also an inner product since inner products arepositive definite). �

This result is a useful tool. We use it next:

Proposition 3.5. Any short exact sequence of vector bundles 0 → E1i−→ E

p−→ E2 → 0 overM splits (i.e. E ∼= E1 ⊕ E2).

Proof. Recall that by proposition (3.2) it is enough to find a map f : E → E1 such that

f ◦ i = idE1. Since E1

i−→ E is an injection E1 ⊂ E is a subbundle. Pick an inner product〈·, ·〉E on E and take f to be the orthogonal projection E → E1 with respect to 〈·, ·〉E . �

Corollary 3.6. Any short exact sequence of vector bundles 0 → E1 → E → E2 → 0 splitsas E ∼= E1 ⊕ E⊥

1 with E2∼= E⊥

1 where ⊥ is defined with respect to any inner product on E.

A good picture to have in mind is that of the Mobius bundle M1 over S1 and the associatedshort exact sequence 0 →M1 → I2 →M1 → 0.

15

3.4. The Degree of a Line Bundle. Let L be a complex line bundle on a Riemann surfaceC. Consider a general section σ : C → L. We can produce such a section by giving it locallyand then glueing it together using a partition of unity. Locally, the line bundle L is trivialso it looks like C× ∆ → ∆ where ∆ is the open unit disk. In this local picture, σ is just amap ∆ → C.

By perturbing σ we can insist that it is transverse to the zero section. Then locally, theinverse image of 0 ∈ C under the map σ : ∆ → C is a finite number of points. Each pointp ∈ C where σ intersects the zero section is called a zero of σ. Around each such pointp the section σ is a map σ : ∆ → C where p = 0 ∈ ∆ and σ(0) = 0. The differentialdσ : T0∆ → T0C is a nonsingular two-by-two matrix. We denote by sgn(p) the sign of thedeterminant of this matrix. Notice that there was an ambiguity since the map σ : ∆ → Cis defined up to post-multiplication by C∗. Fortunately, multiplying by a complex numberdoes not change the sign of det dσ. Of course, this would not be true if we were dealing witha real vector bundle.

Definition: The degree of V is deg(V ) =∑

p sgn(p) ∈ Z where the sum is over all points

p where a (transverse) section σ is zero.The degree would not be well defined if we just counted the number of zeroes of a general

section. This is because one can find two sections which have different number of zeroes.The picture to keep in mind is that from figure 2. If we count with sign, both sections givethe same degree but without sign we would only get a well defined degree modulo two.

+−

Figure 2. The section on the left has two zeroes with signs +1 and −1. Thesection on the right has no zeroes.

Warning: If we just count points then we get a well defined element of Z/2Z. This iswhat we do with real line bundles. Fortunately, in the case of complex line bundles we areable to define sgn(p) and get a well defined map to Z.

Example. The trivial line bundle has degree 0. This is because we can find a sectionwhich is nonzero everywhere.

Example. The tangent line bundle on P1 has degree 2. This corresponds to the famoussaying that you cannot comb the hair on a coconut (i.e. the tangent bundle of S2 is nottrivial). The more precise saying ought to be that you can comb the hairs on a coconuteverywhere except at two points (the north pole and the south pole). At these points thematrix dσ describes a rotation and has determinant one. Whence both points contribute +1to the degree.

Exercise 25. Show that the degree of the tangent bundle on a Riemann surface of genusg is 2 − 2g.

Exercise 26. Show that deg(L1 ⊗ L2) = deg(L1) + deg(L2) (hint: if σi (i = 1, 2) aresections of Li then σ1 · σ2 is a section of L1 ⊗ L2).

16

Fact: deg(L∗) = −deg(L) where L∗ is the dual line bundle of L.

Theorem 3.7. Let L1 and L2 be two complex line bundles on a Riemann surface. ThenL1

∼= L2 if and only if deg(L1) = deg(L2).

We will prove this fact later. For the time being let’s try to improve our intuition. To dothis we first need to know the following fundamental result.

Proposition 3.8. A vector bundle V → B over a contractible space B (e.g. a disk) istrivial.

Example. Let’s construct all the complex line bundles on S2. Think of S2 as two disksD1 and D2 glued along their circumferences. By the above result, the restriction of any linebundle L on S2 to D1 and D2 is trivial. Thus to describe L we simply need indicate how toglue the trivial line bundles over D1 to the trivial line bundle over D2 along the boundaries∂D1 = ∂D2 = S1. This is equivalent to giving the clutching map f : S1 → GL1(C) = C∗. Ifyou have such a map you can consider its winding number. It turns out this number is infact the degree of the line bundle L. The theorem below tells you that if you have two linebundles with the same winding number then they are isomorphic (since you can interpolatebetween the two clutching maps to obtain a family of line bundles).

Theorem 3.9. If V → X× [0, 1] is a family of vector bundles of vector bundles over X thenV0

∼= V1 (where Vi is the vector bundle restricted to X × {i}).Project: The concept of a degree is a special case of a construction known as Chern

classes. A complex vector bundle V of dimension n over some manifold X yields n Chernclasses c1(V ), . . . , cn(V ). These classes are not integers but rather cohomology classes ci(V ) ∈H2i(X,Z) where H2i(X,Z) is the 2i-th cohomology group of X. In the case X has dimensiontwo (Riemann surfaces) we find that H2(X,Z) = Z so for a line bundle L we get deg(L) =c1(L) ∈ H2(X) ∈ Z.

3.5. The Determinantal Line Bundle. Suppose V → C is a vector bundle of dimension nover a Riemann surfaces C with transition functions fαβ : Uα∩Uβ → GLn(C). The associateddeterminantal line bundle detV is defined by the transition functions det fαβ : Uα∩Uβ → C∗.

Exercise 27. Check that det fαβ actually defines a line bundle.

Proposition 3.10. If 0 → E1 → E → E2 → 0 is a short exact sequence of vector bundlesthen det(E) ∼= det(E1) ⊗ det(E2).

Proof. Since we know the sequence splits we have E ∼= E1 ⊕ E2. In particular, this meansthat the transition functions for E are just f1⊕f2 where fi (i = 1, 2) is the transition functionof Ei. Then det(f1 ⊕ f2) = det(f1) det(f2) which (by definition) are the transition functionsfor det(E1) ⊗ det(E2). �

We extend the definition of degree to arbitrary vector bundles V → C by defining deg(V ) =deg(det(V )).

Corollary 3.11. If 0 → E1 → E → E2 → 0 is a short exact sequence of vector bundles thendeg(E) = deg(E1) + deg(E2).

17

Proof. This follows immediately from the proposition above and the fact that for line bundlesL1 and L2 we have deg(L1 ⊗ L2) = deg(L1) + deg(L2). �

We say a real vector bundle V is orientable if det(V ) is a trivial line bundle. By definition,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 bundleof dimension 2n. This is achieved by embedding GLn(C) → GL2n(R). This embedding is

done componentwise via z = reiθ 7→(

r cos θ r sin θ−r sin θ r cos θ

)

. A complex vector bundle V is said

to be orientable if its associated real vector bundle is orientable.

Proposition 3.12. Any complex vector bundle is orientable.

Proof. A (not completely trivial) exercise in linear algebra tells you that the following dia-gram commutes:

GLn(C) //

det

��

GL2n(R)det

$$H

H

H

H

H

H

H

H

H

H

C∗ // GL2(R)det

// R∗

This means that it is enough to show any complex line bundle L is orientable. If the transitionfunctions of L are fαβ = reiθ (where r, θ are functions) then the transition functions for the

corresponding two dimensional real vector bundle Lreal are f realαβ =

(

r cos θ r sin θ−r sin θ r cos θ

)

. Then

the determinantal (real) line bundle detLreal has transition functions det f realαβ = r2 ∈ R∗.

The key point is that r2 > 0 so these maps can be deformed to constant maps to R+ andhence describe the trivial line bundle. �

Corollary 3.13. A complex manifold is always orientable.

3.6. Classification of Topological Vector Bundles on Riemann Surfaces.

Lemma 3.14. If V → X is a vector bundle and σ a non-vanishing section then V ∼= V ′⊕ I1where dim(V ′) = dim(V ) − 1 and I1 is the trivial line bundle.

Proof. Since σ is never zero the subspace it spans defines a subbundle of V . Since σ is anon-vanishing section of this subbundle it follows the subbundle is the trivial line bundle I1.Putting a metric on V we take V ′ = I⊥1 . Then V = V ′ ⊕ I1. �

Proposition 3.15. Let V → M be a vector bundle over a manifold M of dimensiondimR(M) = m.

• If V is real and dimR(V ) > m then there exists a non-vanishing section.• If V is complex and dimC(V ) > m/2 then there exists a non-vanishing section.

Proof. In both cases, since the real dimension of the fiber is bigger than the real dimensionof the base, one can take a general section and perturb it locally around vanishing points sothat it no longer vanishes. �

Corollary 3.16. If V is a complex vector bundle on a Riemann surface then V ∼= det(V )⊕Imwhere m = dim(V ) − 1.

18

Corollary 3.17. The ring R of vector bundles on a Riemann surface is isomorphic toZ[x]/(x2).

Proof. From the corollary above, a vector bundle V is uniquely determined by the pair(dim(V ), deg(V )).

Exercise 28. Show that dim(V1 ⊗ V2) = dim(V1) · dim(V2) and deg(V1 ⊗ V2) = dim(V1) ·deg(V2)+dim(V2)·deg(V1) (hint: use that V1 and V2 split into the direct sum of line bundles).

Consequently, the operations ⊕ and ⊗ on such pairs are given by (a, b)+(c, d) = (a+c, b+d)and (a, b) · (c, d) = (ac, ad+ bc). From this, it follows that the map R → Z[x]/(x2) given by(a, b) 7→ a+ bx is a ring isomorphism. �

Finally, it’s worth mentioning the following cancellation law.

Proposition 3.18. If V ⊕ In ∼= V ′ ⊕ In then V ∼= V ′.

3.7. Holomorphic Vector Bundles. A holomorphic vector bundle is a complex vectorbundle where the transition functions Uαβ → GLn(C) are not just smooth but also holomor-phic. It is important to realize that the concept of a holomorphic vector bundle only makessense over a complex manifold (for example, Riemann surface). The total space V → M ofa holomorphic vector bundle is a complex manifold (though it is not compact). Two holo-

morphic vector bundles Vπ−→ M and V ′ π′

−→ M are isomorphic if there exists a holomorphicmap f : V → V ′ such that f ◦ π′ = π (i.e. maps the fiber Vp over p ∈M to the fiber V ′

p alsoover p ∈ M) and which acts linearly on the fibers (i.e. the restricted map fp : Vp → V ′

p islinear).

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 functionsdescribing M are holomorphic (by definition) and hence their derivatives, which are used toused to describe the transition functions for TM , are also holomorphic.

Exercise 29. Show that V is holomorphic if and only if V ∗ is holomorphic.

3.7.1. Holomorphic Line Bundles over P1. Recall that the complex line bundles over P1 arein bijection with Z via the degree map. Let us give the complex line bundle Ln of degreen ∈ Z a holomorphic structure. As a complex manifold we view P1 as two copies U0 andU1 of C glued together along C∗ via the map f01 : z 7→ 1/z. To build Ln as a holomorphicline bundle we take it to be trivial over both copies of C and hence we need only give theglueing map gn : C∗ → GL1(C) ∼= C∗. Since we want Ln to be holomorphic gn must also beholomorphic. We take gn : C∗ → C∗ to be z 7→ z−n.

We will check later that in fact deg(Ln) = n. For now, notice that L0 is indeed the trivialline bundle since the transition function g0 is the map z 7→ 1. Moreover, the transitionfunction for Lm ⊗ Ln is gm · gn = gm+n and thus Lm ⊗ Ln

∼= Lm+n. This is expected sincedeg(Lm ⊗ Ln) = deg(Lm) + deg(Ln) = m+ n = deg(Lm+n).

What is the holomorphic tangent bundle T = TP1 of P1? Pick coordinates z and w overU0 and U1 such that over U0 ∩ U1 = C∗ the glueing is made via w = 1/z. By definition,

19

T |U0

∼= U0 × C〈 ∂∂z〉 and T |U1

∼= U1 × C〈 ∂∂w

〉. The glueing over U0 ∩ U1 is then

∂

∂z∼ d

dz

(

1

z

)

∂

∂w= − 1

z2

∂

∂w.

This means the transition function for T is z 7→ −1/z2 which means TP1∼= L2. Thus we see

again that the tangent bundle of P1 has degree 2.Exercise 30. Show that the holomorphic line bundle with transition function −1/z2 is

in fact isomorphic to L2 where the transition function is 1/z2.Since L∗

n∼= L−n it follows that the holomorphic cotangent line bundle T ∗

P1 of P1 is isomor-phic to L−2.

Warning: Though every complex line bundle over a Riemann surface C can be given aholomorphic structure it is not true that this structure is unique. The picture one shouldkeep in mind is the sequence of maps

holomorphic vector bundles over C → complex vector bundles over Cdeg−−→ Z

By what we said earlier, the second map is an isomorphism. The first map is surjective butnot necessarily injective. If C = P1 the first map is in fact an isomorphism but if C has genusg ≥ 1 then it is not injective. Finally, be aware that over arbitrary complex manifolds, thefirst map may be neither injective nor surjective. For example, if C is a genus one Riemannsurface then there exist complex line bundles on C × C which does not have a holomorphicstructure (i.e. 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 ofgenus g ≥ 1 splits as a direct sum of line bundles. The earlier proof (proposition 3.5) whichworked for complex vector bundles breaks down in the case of holomorphic vector bundlesbecause the partition of unity and subsequently the inner product are not holomorphic (theyare only smooth).

3.8. Sections of Holomorphic Vector Bundles. Let π : V → M be a holomorphicvector bundle over a complex manifold M . A holomorphic section (or just section for short)is a holomorphic map σ : M → V such that σ ◦π = idM . We denote by Γ(M,V hol) (or Γ(V )for short) the vector space of all holomorphic sections.

Theorem 3.19. If M is a compact complex manifold and V a holomorphic vector bundlethen Γ(M,V hol) is a finite dimensional vector space.

Example. Let I1 be the trivial line bundle over a Riemann surface C. If we view In asa (topological) complex line bundle then Γ(C, I top

1 ) is the space of all C∞ maps C → C andhence has infinite dimension. On the other hand, if we view I1 as a holomorphic line bundlethen Γ(C, Ihol

1 ) is the space of holomorphic maps C → C. Since C is compact we know anysuch holomorphic map must be constant. Thus Γ(C, Ihol

1 ) ∼= C.Example. Let’s describe Γ(P1, L2) where L2 is the degree two line bundle defined in

section 3.7.1. Recall that P1 is covered by U0∼= C and U1

∼= C which have local coordinates zand w. Over U0 the bundle L2 is trivial so a holomorphic section is the same as a holomorphicmap C → C given by z 7→ f(z). Similarly, over U1 a section is the same as a holomorphicmap w 7→ g(w). We need these two local sections to agree on the overlap U0 ∩U1

∼= C∗. We20

know z = 1/w and the glueing of L2 over U0 ∩U1 is done by 1/z2. Thus, in the trivializationover U1, the section f(z) is 1

z2f(z) = w2f(1/w). This must equal the holomorphic functiong(w) over U0 ∩ U1. Whence Γ(P1, L2) is isomorphic to the space of holomorphic functionsf(z) such that w2f(1/w) is also a holomorphic function. These functions are precisely thepolynomials of degree at most two (i.e. of the form a+bz+cz2). Whence dim(Γ(P1, L2)) = 3.

Exercise 31. Check that Γ(P1, Ln) corresponds to vector space of polynomials a0 +a1z+· · · + anz

n. In particular, conclude that dim(Γ(P1, Ln)) = 0 if n < 0.

Proposition 3.20. If L is a holomorphic line bundle on a Riemann surface with deg(L) < 0then dim(Γ(L)) = 0 (i.e. the only holomorphic section is the zero section).

Proof. Suppose L has a nonzero (holomorphic) section σ and let p be a point where Lvanishes. We will show that such a p always contribues +1 to deg(L) and hence deg(L) ≥ 0(contradiction). Locally around p, the section looks like a map C → C given by z 7→ f(z)where f(0) = 0. If we write f(z) = a(x, y) + ib(x, y) then p contributes +1 to deg(L) if and

only if det

(

ax ay

bx by

)

> 0. But since f is holomorphic we know ax = by and ay = −bx so that

the determinant is a2x + b2x > 0. �

Exercise 32. In the proof of this proposition we saw that if a holomorphic section of Lvanishes at n points then deg(L) = n. Use this to show that deg(Ln) = n (hint: use theearlier description of Γ(Ln) for n > 0).

4. Sheaves

For a basic introduction to sheaves see [Ha] chapter II section 1.Fix a topological space X. A presheaf F of abelian groups on X consists of the following

data:

• to each open set U ⊂ X an associated abelian group F(U)• for each V ⊂ U restriction maps ρUV : F(U) → F(V ) which satisfy ρUU = id andρUW = ρV W ◦ ρUV whenever W ⊂ V ⊂ U .

Usually we will write the restriction ρUV (s) of s ∈ F(U) to F(V ) as s|V . A sheaf is a presheafwith the following added condition. Let {Ui}i∈I be an open cover of U and suppose thatsi ∈ F(Ui) satisfy si|Ui∩Uj

= sj|Ui∩Ujfor every i, j ∈ I. Then there exists a unique s ∈ F(U)

such that s|Ui= si for all i ∈ I.

We now give several examples of sheaves in order to develop our intuition. It is a goodexercise to check these are indeed sheaves:Example. Locally constant sheaf Z is defined by assigning Z to each connected componentof U . In other words, Z(U) is isomorphic to Z⊕c where c is the number of connectedcomponents of U . Notice that for U ⊂ X the map Z(X) → Z(U) is not necessarily onto.Similarly one defines the locally constant sheaves R and C.Example. The sheaf C∞(R) assigns to each open set U the group of C∞ functions f : U →R. Similarly one defines C∞(C).Example. Let X be a complex manifold. The sheaf Ohol

X = OX associates to an open Uthe group of holomorphic functions f : U → C. Notice that if X is compact this meansOX(X) ∼= C. Similarly one can define O∗

X by associating to U the group of holomorphic21

maps f : U → C∗.Example. If E is a holomorphic vector bundle then one can define the associated sheafsh(E) by letting sh(E)(U) be the group of holomorphic sections of E|U . For example,OX = sh(I1). We will usually write sh(E) as E since there is rarely any ambiguity betweenviewing E as a vector bundle and viewing it as a sheaf.Example. If F and G are sheaves on X then F ⊕ G is the sheaf given by (F ⊕ G)(U) =F(U)⊕G(U) and F ⊗G is the sheaf given by (F ⊗G)(U) = F(U)⊗Z G(U). Notice that (bydefinition) if E1 and E2 are vector bundles on X then sh(E1 ⊕E2) = sh(E1)⊕ sh(E2). Thesame is true of ⊗ but only if we tensor over OX(U) rather than Z (see section 4.4).Example. The dual sheaf F∗ of F is defined by F∗(U) = Hom(F(U),OX(U)).

Example. Let p ∈ X be a point. The skyscraper sheaf Cp is defined by

{

C if p ∈ U0 if p 6∈ U

Example. Let C be a Riemann surface. Then OC(−p) associates to U the group ofholomorphic maps U → C which vanish at p. If C is compact then OC(−p)(C) = 0. Similarlyone defines OC(p) by associating to U the group of holomorphic maps U → C which have atmost a pole of order one at p. This definition generalizes to give OC(kp) for any k ∈ Z. For

instance, suppose C = P1 and p = 0 ∈ P1. Then Γ(P1,O(2p)) = {ax2+bxy+cy2

x2 : a, b, c ∈ C}and in general dim(Γ(P1,O(kp))) = k + 1. One can also define sheaves such as OC(p + q)for points p, q ∈ C in much the same way.Example. Suppose L is a line bundle on a Riemann surface X. Then L(−p) associates toU the group of holomorphic sections of L|U which vanish at p. Notice that if p 6∈ U thenL(−p)(U) = L(U).

Example. Consider F given by F(U) =

{

C if p ∈ U,U 6= X0 otherwise

. Then F is a presheaf but

not a sheaf. However, if you remove the condition that F(U) = C only if U 6= X then youget 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 calledsheafification. Recall that a presheaf is not a sheaf if there is no unique lift s satisfyings|Ui

= si whenever si|Ui∩Uj= sj|Ui∩Uj

. Sheafification essentially throws in such s whenevernecessary and removes such s if the lift is not unique (see [Ha]).

Notation: If F is a sheaf on X the group F(X) is called the space of global sections ofF . We often denote F(X) by Γ(F). Elements of Γ(F) are called global sections of F .

4.0.1. Morphisms of Sheaves. Let F1 and F2 be two sheaves on X. A morphism φ : F1 → F2

consists of maps φU : F1(U) → F2(U) for every open U ⊂ X which commute with restrictions– i.e. for any V ⊂ U the following diagram commutes:

F1(U)φU

//

ρUV

��

F2(U)

ρUV

��

F1(V )φV

// F2(V )

Example. If f : X → R is a C∞ map then we have a morphism C∞(R) → C∞(R) given byg 7→ fg.

22

Example. Let C be a Riemann surface and f : C → C a meromorphic map with a pole atp ∈ C of order k ∈ Z+. Then we get a map φf : OC(−kp) → OC given by g 7→ fg. Moregenerally, we get a map φf : OC((n− k)p) → OC(np) for any n ∈ Z.Example. Given a point p ∈ X there is a morphism of sheaves OX → Cp given by f 7→ f(p).

The kernel of a map φ : F1 → F2 is the sheaf ker(φ) which associates to U the groupkerφU : F1(U) → F2(U). On the other hand, U 7→ imφU : F1(U) → F2(U) only defines apresheaf. The sheaf im(φ) is defined to be the sheafification of this presheaf. As before, we

say that a sequence of sheaves · · · Fi−1φi−1−−→ Fi

φi−→ Fi+1 → · · · is exact if ker(φi) = im(φi−1)for all i.

Example. Fix a point p ∈ P1. The sequence

0 → OP1(−p) → OP1 → Cp → 0

is exact. Similarly,

0 → OP1(−2p) → OP1(−p) → Cp → 0

is exact. On the other hand, even though the map OP1(−p) → Cp is surjective the map onglobal sections Γ(P1,OP1(−p)) → Γ(P1,Cp) ∼= C is not surjective since Γ(P1,OP1(−p)) = 0.What is happening is that

U 7→ coker (OP1(−2p)(U) → OP1(−p)(U))

is only a presheaf and we need to sheafify to get Cp. After doing this, the map OP1(−p)(C) →Cp(U) is no longer surjective. This might seem like an inconvenience, but it’s this precisephenomenon that allows us to develop the theory of sheaf cohomology.

4.1. Cech Cohomology. Cech cohomology is a cohomology theory for sheaves. ChapterIII 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 cohomology.Although we will not talk about derived functors this is a very important concept and aviable subject for a project.

Fix a complex manifold X and a sheaf F on X. The Cech complex

C ·(F , Uα) = C0 d0−→ C1 d1−→ C2 d2−→ · · ·with respect to a cover {Uα}α∈I of X is defined as follows. We let C0 =

∏

α∈I F(Uα) sothat an element of C0 has the form {fα} with fα ∈ F(Uα). Next C1 =

∏

α,β∈I F(Uα ∩ Uβ)

where an element looks like {fα,β}. Similarly C2 =∏

α,β,γ∈I F(Uα ∩Uβ ∩Uγ) and so on. Thedifferentials are:

• d0 : C0 → C1 given by {fα} 7→ {fα|Uα∩Uβ− fβ|Uα∩Uβ

∈ F(Uα ∩ Uβ)}• d1 : C1 → C2 given by {fα,β} 7→ {fβ,γ − fα,γ + fα,β ∈ F(Uα ∩ Uβ ∩ Uγ)}• dn : Cn → Cn+1 is defined similarly as an alternating sum.

Thanks to the alternating signs di+1 ◦ di = 0 so that we get a complex. The cohomologyof this complex gives us the Cech cohomology groups Hn(X,F) = ker(dn)/im(dn−1). Thelast thing we need to do is understand how this definition depends on the open cover Uα.It turns out the cohomology groups do not depend on the cover if the cover is chosen fineenough. This is expressed more precisely by the following two results.

23

Theorem 4.1 (Leray). If {Uα} and {Vβ} are two open covers of X for which

Hn(Uα1∩ · · · ∩ Uαi

,F) = 0 = Hn(Vβ1∩ · · · ∩ Vβj

,F)

for n > 0 then the Cech cohomology groups calculated with respect to the covers {Uα} and{Vβ} are the same.

Theorem 4.2. If each component of U is a convex domain then for any open cover of Uthe higher Cech cohomology is zero.

Thus, in order to calculate Cech cohomology, it is enough to choose an open cover of Xwhere all the open sets Uα1

∩ · · · ∩ Uαiare convex domains.

Exercise 33. Show that H0(X,F) = Γ(X,F).The higher Cech cohomology groups do not generally have such nice interpretations. How-

ever, we shall see that the group H1(X,O∗X) does have a nice interpretation – it is the group

parametrizing holomorphic line bundles on X. But first, let’s calculate the Cech cohomologyof some simple spaces and sheaves.

Example. Let’s compute the Cech cohomology of the constant sheaf Z on S1. We coverS1 using arcs U1, U2, U3 which overlap pairwise but where U1∩U2∩U3 = ∅. Since the arcs aswell as their intersections are convex domains this cover is good enough to use for computingcohomology. Now C0 = Z3 = (a, b, c) and C1 = Z9 where the differential d0 : C0 → C1 isgiven by

d(a, b, c) = (0, a− b, a− c, b− a, 0, b− c, c− a, c− b, 0).

This means that H0(X,Z) = 〈(a, a, a)〉 ∼= Z (as expected since Γ(S1,Z) = Z). The kernel ofd1 contains elements of the form (0, x, y,−x, 0, z,−y,−z, 0) ∈ C1 and thus H1(S1,Z) ∼= Z.Similarly one can check that H i(S1,Z) = 0 for i > 1.

Exercise 34. Calculate H i(S1,Z) using a cover of S1 by two open sets instead of three(this should be simpler than the computation above so you can fill in all the details).

Exercise 35. Show H i(C∗,Z) ∼= H i(S1,Z) by a direct calculation of the left hand side.

Definition 4.3. D ⊂ Cm is a domain of holomorphy if for each D′ containing D there existssome holomorphic function f on D which does not extend holomorphically to D′.

Example. Any domain in C (e.g. C∗) is a domain of holomorphy. However, C2 − 0 ⊂ C2

is not a domain of holomorphy since by Hartog’s theorem any holomorphic function on C2−0extends to a holomorphic function on C2.

Theorem 4.4. If U is a domain of holomorphy then for any holomorphic vector bundle Von U the higher Cech cohomology is zero.

Thus, in order to calculate Cech cohomology of holomorphic vector bundles, it is enoughto 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 rightway to think about it is that you want to use open sets which have zero higher cohomology.This approach suffers from the chicken and egg problem in that we cannot define Cechcohomology until we choose a cover and yet we cannot choose a cover until we know itscohomology. Despite this conundrum, the derived functor definition of sheaf cohomology

24

tells us this is the correct philosophical way to think. Likewise in practice, this is the mostuseful method to use in order to decide on a cover.

Example. Consider C = P1 with the trivial holomorphic line bundle. Recall that thecorresponding sheaf is denoted OC . As an open cover we use the usual open sets U0 = P1−∞and U1 = P1−0 which are domains in C. We know H0(C,OC) = C. To compute H1(C,OC)we need to understand holomorphic maps U0 ∩ U1 = C∗ → C. Any such map looks like

z 7→ · · · + a−nz−n + · · · + a−1z

−1 + a0 + a1z1 + · · · + anz

n + . . .

and hence can be written as f − g where f = a0 + a1z1 + . . . is holomorphic on U0 and

−g = a−1z−1 + a−2z

−2 + . . . is holomorphic on U1. This means that H1(P1,OP1) = 0. Thehigher Cech cohomology groups also vanish since the cover contains only two open sets.

We have the following cohomology groups when C is a Riemann surface of genus g:

H i(C,Z) =

Z i = 0Z2g i = 1Z i = 20 i ≥ 3

H i(C,OC) =

C i = 0Cg i = 10 i ≥ 2

Proposition 4.5. We have the following vanishing results for cohomology:

• if X is a topological manifold of real dimension n then H i(X,Z) = 0 for i > n• (Grothendieck) if X is a complex manifold of complex dimension n and V any holo-

morphic vector bundle on X then H i(X,V ) = 0 for i > n.

4.1.1. Fine Sheaves. Fine sheaves are a class of sheaves which have the useful property thatthey have zero higher cohomology. To define fine sheaves we first need the concept of support.The support supp(F) of a sheaf F on X consists of those points x ∈ X where for any openset U containing x one can find an open x ∈ V ⊂ U such that F(V ) 6= 0. Similarly, thesupport supp(φ) of a morphism of sheaves φ : F1 → F2 consists of points x ∈ X suchthat for any open set U containing x one can find an open subset x ∈ V ⊂ U such thatφV : F1(V ) → F2(V ) is nonzero.

Example. The skyscraper sheaf Cp for p ∈ X has support supp(Cp) = {p}. On the otherhand, supp(OX) = X.

Definition 4.6. A sheaf F over X is fine if for every locally finite cover {Uα} of X by opensets there exist morphisms φα : F → F such that

• supp(φα) ⊂ Uα

• ∑

α φα = id

Example. The main examples of fine sheaves for us are C∞(R) and C∞(C). To defineφα one takes a partition of unity {ψα} with respect to the open cover {Uα}. Then φα ismultiplication by ψα (i.e. takes a function f to ψαf).

Example. Z, OX and C∞(C∗) are not (generally) fine sheaves. For instance, if X isa compact complex manifold then the only morphisms OX → OX is multiplication by aconstant. All of these have support all of X so if we pick any nontrivial cover {Uα} of X wecannot find the required morphisms φα with support in Uα.

Theorem 4.7. If F is a fine sheaf on X then H i(X,F) = 0 for i > 0.

25

Proof. Pick a cover {Uα} of X that we can use it to compute the Cech cohomology of F .Denote by C · the Cech complex of F with respect to {Uα}. Since F is fine we can findφα : F → F such that supp(φα) ⊂ Uα and

∑

α φα = id. Since each φα is a sheaf morphismit commutes with the differential maps of C · giving us a complex φα(C ·) for each α. Since∑

α φα = id we have that H i(C ·) = ⊕αHi(φα(C ·)). In other words, the morphisms φα break

up the cohomology H i(C ·) = H i(X,F) into the direct sum of smaller pieces.On the other hand, each φα(C ·) computes the cohomology of φα(F) ⊂ F . Since φα

is supported on Uα the sheaf φα(F) is also supported on Uα and hence H i(X,φα(F)) =H i(Uα, φα(F)|Uα

). Since we chose the Uα’s small enough these groups are zero for i > 0which means that H i(X,F) = ⊕αH

i(φα(F)) = 0 for i > 0. �

4.1.2. The Exponential Sequence. A map of complexes E· → F · consists of maps fi : Ei → F i

such that d◦fi = fi+1◦d (i.e. the maps fi commute with the differential). A sequence of maps0 → E· → F · → G· → 0 is short exact if for each i the sequence 0 → Ei → F i → Gi → 0 isexact.

Lemma 4.8 (snake lemma). A short exact sequence of complexes 0 → E· → F · → G· → 0leads to a long exact sequence of cohomology:

· · · → H i(E·) → H i(F ·) → H i(G·) → H i+1(E·) → . . .

Proof. (sketch) We describe how to obtain the connecting morphism H i(G·) → H i+1(F ·)(the rest of the maps are obvious). Suppose u ∈ ker(Gi → Gi+1). Since F i → Gi is ontowe can find an element u′ ∈ F i which maps onto u. Now consider v = d(u′). Since du = 0the image of v under the map F i+1 → Gi+1 is zero (all squares are commuting). Thus theremust be some v′ ∈ Ei+1 which maps to v ∈ F i+1. We map u to v′. It now remains to checkthat this map is well defined (i.e. that 0 ∈ Gi maps to zero) and that the rest of the mapsare well defined. Finally one needs to check that the long sequence is exact. �

Corollary 4.9. If 0 → E → F → G → 0 is a short exact sequence of sheaves on a topologicalspace X then the following sequence is exact

· · · → H i(X, E) → H i(X,F) → H i(X,G) → H i+1(X, E) → . . .

Proof. Choose an open cover of X which we can use to compute the Cech cohomology groupsof E ,F and G. From this we get a short exact sequence of complexes 0 → C ·(E) → C ·(F) →C ·(G) → 0 because 0 → E → F → G → 0 is exact. The snake lemma gives us the long exactsequence of cohomology groups. �

Corollary 4.10. If 0 → E → F → G → 0 is a short exact sequence of sheaves then χ(E) −χ(F) + χ(G) = 0 where χ(A) =

∑

i(−1)irank(H i(X,A)) denotes the Euler characteristic ofthe sheaf A.

Proof. A general fact from algebra says that if 0 → A1 → · · · → An → 0 is an exactsequence of abelian groups then

∑

i(−1)irank(Ai) = 0. The result follows since by theprevious corollary

· · · → H i(X, E) → H i(X,F) → H i(X,G) → H i+1(X, E) → . . .

is an exact sequence. �

26

Example. Fix a complex manifold X and consider the exponential sequence

0 → Z→ OXexp−−→ O∗

X → 0

The first map takes a ∈ Z to the constant map in OX . The second map is the exponentialf 7→ exp(2πif). Since exp(t) ∈ C∗ for any t ∈ C this map is well-defined.

Proposition 4.11. The exponential sequence is exact.

Proof. The question of exactness of sheaves is a local one. Given any holomorphic mapf : U → C∗ we can take the logarithm to obtain a holomorphic map log f : U → C. Thisshows exp : OX → O∗

X is onto. Finally, we have exactness in the middle since the elements ofOX which are mapped by exp to one (the identity in O∗

X) are precisely the integers (becauseexp(2πit) = 1 exactly when t ∈ Z). �

Since H0(X,Z) = Z, H0(X,OX) = C and H0(X,O∗X) = C∗ we get a long exact sequence

of abelian groups:

0 → Z→ Cexp−−→ C∗ → H1(X,Z) → H1(X,OX)

exp−−→ H1(O∗X) → H2(X,Z) → . . .

Since the map Cexp−−→ C∗ is onto the sequence 0 → Z → C

exp−−→ C∗ → 0 is in fact exact. Sothe sequence above splits to give us the long exact sequence:

0 → H1(X,Z) → H1(X,OX)exp−−→ H1(O∗

X) → H2(X,Z) → H2(X,OX) → . . .

If X is a Riemann surface of genus g we saw that H1(X,Z) = Z2g, H1(X,OX) = Cg,H2(X,Z) = Z and H2(X,OX) = 0. Thus we get the long exact sequence:

0 → Z2g → Cg → H1(X,O∗X) → Z→ 0

Example. Similarly we can consider the exponential sequence

0 → Z→ C∞(C)exp−−→ C∞(C∗) → 0

Since C∞(C) is a fine sheaf (see section 4.1.1) we know H i(X,C∞(C)) = 0 for i > 0. Thenthe corresponding long exact sequence implies H1(X,C∞(C∗)) ∼= H2(X,Z). So if X is aRiemann surface we get H1(X,C∞(C∗)) ∼= Z.

4.2. Line Bundles and Cech Cohomology.

Theorem 4.12. If X is a complex manifold then H1(X,O∗X) parametrizes holomorphic line

bundles on X.

Proof. Pick a cover {Uα} of X and denote by C0 d0−→ C1 d1−→ C2 → . . . the Cech complex

of O∗X . Then ker(d1) =

{

∏

α,β fαβ : fαβ · f−1αγ · fβγ = 1

}

where fαβ : Uα ∩ Uβ → C∗. Also

im(d0) ={

∏

α,β fαβ : fαβ = gαg−1β

}

for some gα : Uα → C∗.

Choose {Uα} fine enough so that any line bundle is trivial over any Uα1∩ · · · ∩ Uαn

. IfL is a line bundle choose a trivialization tα : L|Uα

→ C × Uα for each α. Denote theglueing functions of L with respect to this trivialization by fαβ. Then the glueing conditionimplies fαβ · fβγ = fαγ which means {fαβ} ∈ ker(d1). However, this element depends on thetrivialization tα we picked for L. Changing the trivialization by some gα : Uα → C∗ gives us

27

t′α = tα · gα. Then the new glueing functions are f ′αβ = g−1

α fαβgβ which differ from the oldglueing functions by an element of im(d0).

Thus, given a line bundle L we get a well defined element of ker(d1)/im(d0) = H1(X,O∗X).

Conversely, given an element of H1(X,O∗X) we can reverse the description above to construct

back L. �

Corollary 4.13. For each n ∈ Z, the space of degree n line bundles on a Riemann surfaceC of genus g is isomorphic to an n-dimensional torus Cg/Λ where Λ ⊂ Cg is an integrallattice of rank 2g.

Proof. Let’s go back to the short exact sequence:

0 → Z2g → Cg → H1(C,O∗C) → Z→ 0

Since the sequence is exact, the kernel of the degree map is Cg modulo the image of Z2g

inside Cg which is a lattice Λ of rank 2g. �

Aside: The map H1(C,O∗C) → Z is none other than the degree map defined earlier (we

won’t prove this here). In fact, the definition of degree is usually provided by this map andnot in terms of counting zeros of a general section (as we did in section 3.4).

The total space H1(X,O∗X) of holomorphic line bundles on X is called the Picard group

and denoted Pic(X). The group operation is the tensor ⊗ and the inverse of a line bundle Lis the dual line bundle L∗ (recall that L⊗L∗ ∼= I1). Pic

d(X) ⊂ Pic(X) denotes the subspaceof line bundles of degree d. Pic0(X) is called the Jacobian and denoted J(X). Notice thatPicd(X) ⊗ Pice(X) → Picd+e(X) so that J(X) is a subgroup of Pic(X).

If X = C is a Riemann surface of genus g then the corollary above implies Pic(C) ∼=J(C) × Z and also that J(C) is a compact complex manifold Cg/Λ of dimension g. As thecomplex structure of C varies, the lattice Λ ∈ Cg moves and we get different tori. Eventhough these tori are topologically the same (they are homeomorphic to (S1)2g) as complexmanifolds they are different. A classical theorem of Torelli (essentially) says that one canrecover the Riemann surface C from its Jacobian J(C).

The complex and real topological analogues of the sheaf O∗X are the sheaves C∞(C∗) and

C∞(R∗). Consequently, we have the following analogue of theorem 4.12.

Theorem 4.14. If X is a topological manifold then H1(X,C∞(C∗)) parametrizes complex(topological) line bundles and H1(X,C∞(R∗)) parametrizes real line bundles on X.

Proof. The proof is precisely the same as that of theorem 4.12. �

Theorem 4.15. H1(X,C∞(C∗)) ∼= H2(X,Z) and H1(X,C∞(R∗)) ∼= H1(X,Z/2).

Proof. The exponential sequence for C∞(C∗) is 0 → Z → C∞(C)exp−−→ C∞(C∗) → 0 leading

to the long exact sequence

· · · → H1(X,C∞(C)) → H1(X,C∞(C∗))deg−−→ H2(X,Z) → H2(X,C∞(C)) → . . .

Since C∞(C) is a fine sheaf we have H i(X,C∞(C)) = 0 for i > 0 implying that the map

H1(X,C∞(C∗))deg−−→ H2(X,Z) is an isomorphism.

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The exponential map for C∞(R) is 0 → 0 → C∞(R)exp−−→ C∞(R>0) → 0. This shows that

C∞(R) ∼= C∞(R>0). On the other hand, we also have the short exact sequence

0 → Z/2 → C∞(R∗)m−→ C∞(R>0) → 0

where m(f) = f2. Since C∞(R>0) ∼= C∞(R) is fine the higher cohomology vanishes and thelong exact sequence becomes

0 → Z/2 → Γ(C∞(R∗))m−→ Γ(C∞(R>0)) → H1(X,Z/2) → H1(X,C∞(R∗)) → 0

But for any map g : X → R>0 we have a well defined square root√g. This means

Γ(C∞(R∗))m−→ Γ(C∞(R>0)) is surjective and hence H1(X,Z/2) ∼= H1(X,C∞(R∗)). �

Corollary 4.16. If C is a Riemann surface of genus g then complex and real line bundlesare parametrized via the degree map by Z and (Z/2)2g respectively.

Proof. This follows immediately from the previous proposition and the fact that H2(C,Z) =Z and H1(C,Z/2) = (Z/2)2g. �

Exercise 36. Show that H1(S1, C∞(C∗)) = 0 and conclude that all complex line bundleson S1 are trivial.

Exercise 37. Show that up to isomorphism there are exactly two real line bundles onS1, namely the trivial and the Mobius line bundle.

4.3. Riemann-Roch and Serre Duality. Let L be a holomorphic line bundle on a Rie-mann surface C of genus g. What is the dimension of the space of global sections H0(C,L) =Γ(L)? To answer this question we first calculate the Euler characteristic χ(C,L) =

∑

i(−1)ihi(L)where hi(L) = rankH i(C,L). Since dimC(C) = 1 Grothendieck’s theorem (4.5) tells ushi(L) = 0 for i > 1. Thus χ(C,L) = h0(L) − h1(L).

Let’s first consider an example. Fix a point p ∈ C and consider the sheaf OC(np) for somen ∈ Z (later we will show this sheaf corresponds to a line bundle of degree n). To calculateχ(C,OC(np)) we have the short exact sequence of sheaves

0 → OC((n− 1)p) → OC(np) → Cp → 0

where Cp is the sky-scraper sheaf at p. We can think of Cp as a complex line bundle ona point. Since a point is a zero dimensional complex manifold Grothendieck’s theoremtells us that hi(C,Cp) = 0 for i > 0. Also, h0(C,Cp) = 1 so that χ(C,Cp) = 1. Thus,by corollary 4.10, χ(C,OC(np)) = χ(C,OC((n − 1)p)) + 1. Repeating in this way we getχ(C,OC(np)) = χ(C,OC) + n = n+ 1 − g.

Exercise 38. Show by direct computation that hi(C,Cp) = 0 for i > 0.

Theorem 4.17 (Riemann-Roch). If V is a holomorphic vector bundle on a Riemann surfaceC of genus g then χ(C, V ) = deg(V ) + (1− g)rank(V ). In particular, if L is a line bundle ofdegree d then χ(C,L) = d− g + 1.

We will give a proof in the next section. What is remarkable about this result is thatχ(C,L) only depends on the topological isomorphism class of L. For example, if C is a curve

of genus g > 0 and p, q ∈ C are two points then h0(OC(p − q)) =

{

1 if p = q0 if p 6= q

Thus h0

29

(and consequently also h1) depends not only on the topological type (deg(OC(p − q)) = 0for any p, q ∈ C) but also on the holomorphic type. On the other hand, the differenceχ(C,L) = h0(L) − h1(L) is a topological invariant.

Now that we understand χ(C,L) we turn our attention to computing h1(L). This problemhas a partial answer in the form of Serre duality:

Theorem 4.18 (Serre duality). Let V be a vector bundle on C and denote by KC thecotangent bundle of C (this is also known as the canonical line bundle). Then H1(C, V ) =H0(C, V ∗ ⊗KC)∨. In particular, h1(V ) = h0(V ∗ ⊗KC).

Example. Suppose C = P1 so that KC = L−2 = OP1(−2) (see section 3.7.1). If n ≥ 0then h1(Ln) = h0(L∗

n ⊗ L−2) = h0(L−n−2) which is zero since deg(L−n−2) = −n − 2 < 0.Thus h0(Ln) = χ(Ln) = n−0+1 (compare with section 3.8). Thus, if n < 0 then h0(Ln) = 0and h1(Ln) = −n− 1.

Example. If C is a Riemann surface of genus g then h1(KC) = h0(K∗C ⊗KC) = h0(OC) =

1. But χ(KC) = (2g − 2) − g + 1 = g − 1 which means h0(KC) = g. Note that showingdirectly from definition that the canonical line bundle has a g dimensional space of sectionsis somewhat difficult.

Exercise 39. Let C be a Riemann surface of genus g and L a line bundle of degree 2g−2.Show that h0(L) ≥ g− 1 if and only if L 6≃ KC (hint: use that if L is a line bundle of degreezero then h0(L) = 0 unless L ∼= OC).

Example. Suppose L is a line bundle on a Riemann surface C of genus g such thatdeg(L) > 2g − 2. Then deg(L∗ ⊗ KC) < 2 − 2g + 2g − 2 = 0 so that h1(L) = 0 andh0(L) = χ(L) = deg(L) − g + 1. More generally, we have the following useful result.

Proposition 4.19. Let V be a holomorphic vector bundle on C. Then h0(V ⊗ L) > 0 ifdeg(L) ≫ 0 and h0(V ⊗ L) = 0 if deg(L) ≪ 0.

Proof. The first assertion follows from Riemann-Roch since

deg(V ⊗ L) = deg(det(V ⊗ L)) = deg(det(V ) ⊗ L⊗dim(V )) = deg(V ) + deg(L) · dim(V )

which tends to infinity as deg(L) tends to infinity. To prove the second assertion we use thatH0(V ) is finite dimensional. Then we can consider among all sections of V the one whichvanishes to highest possible order d. But then H0(V ⊗OC(−np)) = 0 for any n > d and pointp ∈ C. In the next section we will show that OC(−np) corresponds to a line bundle of degree−n and that conversely any line bundle is (essentially) of this form (theorem (4.27)). �

This proposition says that any line bundle on a Riemann surface C has a meromorphicsection. Although rather believable this is not a trivial statement to prove and requires someanalysis. It would seem like we gave a proof above but in the process we used Riemann-Rochwhose proof will turn out to rely on this proposition. In fact, if one were to trace through thearguments carefully, one would find that proving the existence of meromorphic sections ofline bundles over Riemann surfaces is the main (and almost only) serious technical obstacle.

4.4. Vector bundles, locally free sheaves and divisors. To a holomorphic vector bundleV over X we can associate its sheaf of sections which we also denote V . Given a (local)section s ∈ V (U) we can multiply it by a holomorphic function f ∈ OX(U) to obtain a new

30

section f · s ∈ V (U). Thus V (U) has the extra structure of a OX(U)-module. Since thisholds for any open U we find that V is in fact a sheaf of OX-modules.

Definition 4.20. A sheaf F is an OX-module if for any open U ⊂ X the group F(U) is anOX(U)-module such that for any V ⊂ U the action commutes with the restriction map (i.e.for f ∈ OX(U) and s ∈ F(U) we have (f · s)|V = f |V · s|V ).

Definition 4.21. A morphism of sheaves φ : F1 → F2 is a map of OX-modules if it com-mutes with the OX action (i.e. for f ∈ OX(U) and s ∈ F1(U) we have φ(f · s) = f · φ(s)).We denote the space of OX-module morphisms F1 → F2 by HomOX

(F1,F2).

The problem with the category of sheaves is that it’s too big and in particular containstoo many morphisms. So from now on we will work only in the category of OX-moduleswhere the objects are OX-modules and the morphisms are OX-module maps between them.Consequently, a few of our old definitions 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 definedby F∗(U) = HomOX

(F(U),OX(U)) where we consider just OX-module morphisms ratherthan all sheaf morphisms. Also, the tensor product of F1 and F2 is F1 ⊗OX

F2 rather thanF1 ⊗Z F2.

Lemma 4.22. If F is a sheaf of OX-modules then HomOX(OX ,F) = H0(X,F) = Γ(F).

More generally, if L is a line bundle then HomOX(L,F) = H0(X,F⊗OX

L∗) = Γ(F⊗OXL∗).

Proof. Suppose the constant section 1 ∈ Γ(OX) maps to s ∈ Γ(F). If a ∈ OX(U) then sincethe map OX → F commutes with restrictions and the OX action we have that a = a ·1|U 7→a ·s|U . So to give f ∈ HomOX

(OX ,F) it is enough to give the image s ∈ Γ(F) of 1 ∈ Γ(OX).Conversely, any such s ∈ Γ(F) gives us a morphism.

Given a map L → F we can tensor both sides by L∗ to get a map L ⊗OXL∗ ∼= OX →

F⊗OXL∗ (and vice versa). Then by the previous argument such maps correspond to elements

of Γ(F ⊗OXL∗). �

Recall that an R-module M is free of rank n if M ∼= R⊕n. Since any rank n vector bundleV is locally trivial we can find open sets U where V |U ∼= O⊕n

X . So over such sets V isisomorphic to a free OX-module of rank n. Motivated by this observation we define:

Definition 4.23. An OX-module F is locally free of rank n if every point p ∈ X is containedin an open neighbourhood p ∈ U where F|U is a free OX(U) module of rank n.

Theorem 4.24. Any holomorphic vector bundle V of rank n is isomorphic to a locally freesheaf of rank n and conversely. Moreover, any morphism of vector bundles is a morphismof 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 islocally free of rank n. Conversely, suppose V is a locally free sheaf of rank n. For simplicityof notation let’s assume n = 1. By definition we have open sets Uα and isomorphismsfα : V |Uα

→ OX(Uα). So over Uα∩Uβ we can consider fα ◦f−1β to get a map OX(Uα∩Uβ) →

OX(Uα ∩ Uβ). Since this map is a morphism of OX-modules, following the lemma above wejust need to give the map 1 7→ a ∈ OX(Uα ∩ Uβ). Since this morphism is invertible we must

31

have a ∈ O∗X(Uα ∩Uβ). 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 fiberwise map is linearand a linear map Cm → Cn commutes with the C action. The reason why the converse isnot true is that the fiberwise map may drop in rank. For example, consider the map betweenthe trivial line bundles over C given by multiplication by z. This map is an isomorphism offibers over z 6= 0 but drops rank at z = 0. The quotient is then the skyscraper sheaf C0 atz = 0 which is not a vector bundle. Hence we do not allow this in the case of vector bundlesbut do allow this map when working with sheaves because the skyscraper sheaf is a perfectlygood sheaf (even though it’s not locally free). �

Exercise 40. Show that the tensor product F1 ⊗OXF2 of two locally free sheaves F1 and

F2 of ranks n1 and n2 is a locally free sheaf of rank n1n2 (hint: this is a local question soyou can assume that both F1 and F2 are trivial).

There is a really good description we can give of locally free sheaves of rank one (ı.e. linebundles) in terms of divisors.

Definition 4.25. A divisor D on a Riemann surface C is a (formal) finite Z-linear sum ofpoints D =

∑ni=1 aipi where ai ∈ Z and pi ∈ C. The degree of D is deg(D) =

∑

i ai.

Recall that in section 4 we defined sheaves OC(kp) for any k ∈ Z and p ∈ P by takingOC(kp)(U) to be sections of U which vanish to order k at p (if k ≤ 0) or have poles of orderat most k at p (if k > 0). We extend this definition to give us sheaves OC(D) for any divisorD =

∑

i aipi.Exercise 41. Show that OC(D1 +D2) ∼= OC(D1) ⊗OC

OC(D2).Exercise 42. Show that OC(p)∗ ∼= OC(−p) (hint: it is enough to show that the map

OC(p) ⊗ OC(−p) → OC given by f1 ⊗ f2 7→ f1f2 is an isomorphism) and conclude thatOC(D)∗ ∼= OC(−D).

Lemma 4.26. Let D be a divisor on a Riemann surface C. The sheaf OC(D) is a locallyfree sheaf of rank one and degree deg(D).

Proof. Since OC(∑

i aipi) ∼= ⊗iOC(aipi) it is enough to prove the assertion for D = np wheren ∈ Z and p ∈ C. Away from p this sheaf is isomorphic to OC so we just need to prove thatin a small enough neighbourhood U around p the sheaf is isomorphic to OC(U). If z is alocal coordinate at p then the isomorphism OC(np)(U) → OC(U) is given by g 7→ (z−p)n ·g.

It remains to show that deg(OC(np)) = n. Since OC(np)∗ ∼= OC(−np) we can assume thatn ≥ 0. Thus we have the global section 1 ∈ Γ(OC(np)). Let U be a small neighbourhood ofp. Then OC(np) is trivialized over U by g 7→ (z − p)n · g. So the section 1 around p locallylooks like (z − p)n. Away from p the line bundle OC(np) is trivial so the section 1 does notintersect the zero section. Hence 1 vanishes only at p where it vanishes to order n. Thus, byour original definition of degree from section 3.4, deg(OC(np)) = n. �

Theorem 4.27. Any locally free sheaf L of rank one is isomorphic to OC(D) for somedivisor D.

32

Proof. Suppose L has a global section s which vanishes along some divisorD =∑

i aipi whereai > 0 (i.e. it vanishes to order ai at pi). Then this gives a section of L⊗OC

OC(−D) whichdoes not vanish. Thus L ⊗OC

OC(−D) ∼= OC which means L ∼= OC(D). More generally, ifL has no section then by proposition 4.19 we know that L⊗OC

OC(np) has a section and soL⊗OC

OC(np) ∼= OC(D) for some divisor D. Whence L ∼= OC(D − np). �

We still need to understand when OC(D) ∼= OC(D′). To do this we introduce the concept ofprincipal divisor. Consider a map f : C → P1. To f we associate the divisor f−1(0)−f−1(∞)which we denote (f). A divisor D is a principal divisor if D = (f) for some f : C → P1.Two divisors D and D′ are called linearly equivalent which we denote D ≡ D′ if D−D′ is aprincipal divisor.

Theorem 4.28. OC(D) ∼= OC(D′) if and only if D ≡ D′.

Example. We know that on P1 there is only one holomorphic line bundle of degree d foreach d ∈ Z (up to isomorphism). So if p, q ∈ P1 then we must have OP1(dp) ∼= OP1(dq) forevery d ∈ Z. This means that dp ≡ dq so by the theorem above there must be some mapf : P1 → P1 such that f−1(0)− f−1(∞) = dp− dq. If we take p = 0 and q = ∞ then we canwrite down this map explicitly as z 7→ zd.

Example. If C is a Riemann surface of genus g > 0 then OC(p) 6∼= OC(q) for anyp 6= q ∈ C. To see this we must show that p− q is not a principal divisor. But if p− q = (f)for some f : C → P1 then this map f would be of degree one and hence an isomorphism(contradiction since C 6∼= P1).

To summarize, we have the following three equivalent descriptions:

line bundles ⇔ locally free sheaves of rank one ⇔ divisors(up to isomorphism) (up to isomorphism) (up to linear equivalence)

4.5. A proof of Riemann-Roch for curves. We need a technical lemma. To appreciatethe lemma recall that if E → F is an inclusion of locally free sheaves then the quotient neednot be locally free. The standard example is the exact sequence 0 → OC → OC(p) → Cp →0. The reason this happens is that locally in the trivialization around p the map on the fiberover z is multiplication by z − p (because the isomorphism OC(p)(U)

∼−→ OC(U) is given byg 7→ (z − p)g). Thus at p the rank of the map drops from one to zero and that is why at pwe get a cokernel which shows up as the skyscraper sheaf Cp.

In our case we have a vector bundle V and we want to find a line subbundle L → V sothat its quotient is locally free (this way we can study V by studying L and its quotient).In lemma 4.22 we learned that OC-module morphisms L→ V correspond to global sectionss ∈ Γ(V ⊗OC

L∗). As in the example above, the quotient sheaf will be locally free if and onlyif this section is nowhere zero.

Lemma 4.29. Let V be a vector bundle on a Riemann surface C. If L is a line bundle ofhighest possible degree such that there exists a non-zero map L → V then quotient sheaf islocally free.

Proof. A map L → V corresponds to a section in s ∈ H0(C, V ⊗OCL∗). If deg(L) is large

enough then there are no global sections and if deg(L) is really negative then there must be33

some global sections (4.19). Thus such an L of highest degree exists. Suppose that s vanishesalong some nonzero divisor D =

∑

i aipi with ai ∈ Z>0. Then H0(C, V ⊗OCL∗⊗OC

OC(−D))contains a section which means there is a map L⊗OC

OC(D) → V . This is a contradictionsince deg(L⊗OC

OC(D)) > deg(L). �

Theorem 4.30 (Riemann-Roch). If V is a holomorphic vector bundle on a Riemann surfaceC of genus g then χ(C, V ) = deg(V ) + (1 − g)rank(V ).

Proof. We first reduce to the case when rank(V ) = 1. By the lemma (4.29) we can find a shortexact sequence 0 → L → V → W → 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 χ(L) = deg(L)+(1−g) and χ(W ) = deg(W )+(1−g)rank(W ).On the other hand, χ(V ) = χ(L) + χ(W ) and deg(V ) = deg(L) + deg(W ) and rank(V ) =rank(L) + rank(W ) so the result follows.

It remains to prove Riemann-Roch for line bundles L. We know L ∼= OC(D) for somedivisorD =

∑

i aipi so we proceed by induction on∑

i |ai|. Suppose without loss of generalitythat p1 > 0 so that have the short exact sequence 0 → OC(D − p1) → OC(D) → Cp. Byinduction χ(OC(D − p1)) = deg(D) − 1 − g + 1 and χ(Cp) = 1 so that χ(OC(D)) =χ(OC(D− p1))+χ(Cp) = deg(D)− g+1. The base case for this induction is L = OC whichfollows by direct computation since h0(OC) = 1 and h1(OC) = g. �

5. Classifying vector bundles on Riemann surfaces

5.1. Grothendieck’s classification of vector bundles on P1. We know for each d ∈ Zthere exists a unique holomorphic line bundle on P1 of degree d which we denote OP1(d).

Theorem 5.1 (Grothendieck). Any holomorphic rank n vector bundle on P1 is uniquelyisomorphic to a direct sum of line bundles ⊕n

i=1OP1(di) where d1 ≤ · · · ≤ dn and di ∈ Z.

Proof. First we show that any rank n vector bundle V on P1 splits as a direct sum of linebundles. Using lemma (4.29) we take a line bundle L = OP1(d) of highest degree so that wehave a short exact sequence 0 → L → V → V ′ → 0 where V ′ is also a vector bundle. Byinduction we can assume V ′ splits as ⊕n−1

i=1 OP1(di).Claim: di ≤ d. Suppose di > d for some i. Tensoring the short exact sequence with

OP1(−d− 1) we get another short exact sequence (see lemma (5.3) below) 0 → OP1(−1) →V ⊗O

P1OP1(−d − 1) → ⊕n−1

i=1 OP1(di − d − 1) → 0. Since H0(OP1(−1)) = H1(OP1(−1)) = 0

if we look at the long exact sequence in cohomology we get H0(V ⊗OP1

OP1(−d − 1))∼−→

H0(⊕n−1i=1 OP1(di − d − 1)) which is nonzero since di − d − 1 ≥ 0. But this means V has a

section of degree d+ 1 which is a contradiction.Consider now proposition (5.2) below where we take E = L,F = V and G = V ′. Then

H1(G∗⊗OP1E) = H1(⊕iOP1(d−di)) = H0(⊕iOP1(d−di)

∗⊗OP1KP1) = H0(⊕iOP1(di−d−2)) = 0

since KP1 = OP1(−2) and di − d− 2 < 0. Thus the sequence splits and hence V splits as thedirect sum of line bundles.

Finally, suppose ⊕iOP1(ai) ∼= ⊕iOP1(bi) for some ai, bi ∈ Z with a1 ≤ · · · ≤ an andb1 ≤ · · · ≤ bn. If an < bn then tensoring both sides by OP1(−bn) we get ⊕iOP1(ai − bn) ∼=⊕iOP1(bi − bn). The left side has no sections since ai < bn whereas the right side has at

34

least one section coming from the OP1(bn − bn) ∼= OP1 summand. Thus an = bn. Next, if

an−1 < bn−1 then tensor by OP1(−bn−1) and look at global sections again to conclude, asbefore, that an−1 = bn−1. Continuing this way we show ai = bi for all i thus completing theproof. �

Proposition 5.2. If 0 → E i−→ F p−→ G → 0 is a short exact sequence of vector bundles on acomplex manifold X and H1(G∗ ⊗OX

E) = 0 then the sequence splits (i.e. F ∼= E ⊕ G).

Proof. By proposition (3.2) it is enough to find a map g : G → F such that p ◦ g = idG. This

can be done if the following map induced by p is surjective: HomOX(G,F)

p−→ HomOX(G,G)

since we can take as g any preimage of idG ∈ HomOX(G,G). But

HomOX(G,F) = HomOX

(OX ,G∗ ⊗OXF) = Γ(G∗ ⊗OX

F).

Similarly, HomOX(G,G) = Γ(G∗⊗OX

G). So we need to show that Γ(G∗⊗OXF) → Γ(G∗⊗OX

G) is surjective.To do this we first tensor 0 → E → F → G → 0 by G∗ to obtain a short exact sequence

(see lemma (5.3) below) whose associated long exact sequence of cohomology is

0 → Γ(G∗ ⊗OXE) → Γ(G∗ ⊗OX

F) → Γ(G∗ ⊗OXG) → H1(G∗ ⊗OX

E) → · · ·Since H1(G∗ ⊗OX

E) = 0 this implies Γ(G∗ ⊗OXF) → Γ(G∗ ⊗OX

G) is surjective and theresult follows. �

Lemma 5.3. If 0 → E → F → G → 0 is a short exact sequence on a complex manifold Xand V a vector bundle then 0 → E ⊗OX

V → F ⊗OXV → G ⊗OX

V → 0 is also short exact.

Proof. A sequence is exact if it is locally exact for a fine enough open cover Uα of X. SupposeV has rank one. We can pick such an open cover where V |Uα

is trivial and hence isomorphicto OX(Uα). Then over Uα we have E(Uα) ⊗OX(Uα) V (Uα) ∼= E(Uα) and similarly with Fand G so the sequence remains exact. The same thing happens if V has rank greater thanone. �

Of course, in general a short exact sequence does not necessarily remain exact after ten-

soring. For example, consider the sequence of abelian groups 0 → Z×2−→ Z → Z/2 → 0. If

we tensor (over Z) with Z/2 then we get 0 → Z/20−→ Z/2 → Z/2 → 0 which is no longer

exact.

5.2. Atiyah’s classification of vector bundles on elliptic curves. Let C be a Riemannsurface of genus one. First we take a closer look at line bundles on C. We know that thespace J(C) of degree zero holomorphic line bundles on C is a torus of dimension one. Thismeans that J(C) is also a genus one Riemann surface.

Proposition 5.4. If C is a genus one Riemann surface then J(C) ∼= C ∼= Picd(C) for anyd ∈ Z.

Proof. Recall that Picd denotes the space of line bundles on C of degree d. If E is a linebundle of degree d then we have an isomorphism J(C) → Picd(C) given by L 7→ L ⊗OC

E(L is a degree zero line bundle). The inverse of this map is L′ 7→ L′ ⊗OC

E∗ becauseE ⊗OC

E∗ ∼= OC . Thus all Picd(C) are isomorphic (as d varies).35

A proof that J(C) ∼= C can be obtained directly from the description of J(C) as thequotient of H1(C,OC) ∼= C by a lattice Λ. However, we give a more direct proof by showingthat the map C → Pic1(C) given by p 7→ OC(p) is an isomorphism. To see that the mapis injective we need to show OC(p) 6∼= OC(q) unless p = q. This we saw to hold true forany Riemann surface C of genus g > 0 (see the second example at the end of section (4.4).Next we need to show that any line bundle L of degree one is isomorphic to OC(p) forsome point p ∈ C. By Riemann-Roch, if deg(L) = 1 then h0(L) = 1 since by Serre dualityh1(L) = h0(L∗) = 0 (KC

∼= OC because C has genus one). So L has a global section whichvanishes at exactly one point p (since deg(L) = 1). Consequently, as we saw in the proof oftheorem (4.27) this means L ∼= OC(p).

Aside: One can check that the group structure on J(C) coming from ⊗ is the same asthe natural group structure on C coming from the fact that C ∼= C/Λ. �

For C a Riemann surface of arbitrary genus, the map C → Pic1(C) used in the proofabove is called the Abel-Jacobi map. The following is a direct corollary of the proof:

Corollary 5.5. Let C be a Riemann surface of genus at least one. Then the Abel-Jacobimap C → Pic1(C) given by p 7→ OC(p) is injective. Moreover, if C has genus one then it isan isomorphism.

Definition 5.6. A vector bundle V is called indecomposable if it cannot be written as adirect sum V ′⊕V ′′ of nonzero subbundles. V is irreducible if it does not contain any nonzerosubbundles V ′ V .

Warning: A holomorphic vector bundle V can be indecomposable yet not irreducible.This is because there might be some V ′ V but no complement V ′′ so that V = V ′⊕V ′′. Ofcourse, in the case of complex (or real) vector bundles the two notions coincide (proposition3.5).

Denote by Bun(r, d) the space of indecomposable vector bundles of rank r and degree d onC. The proposition above asserts that Bun(1, d) ∼= C. The determinantal map V 7→ det(V )defines a map Bun(r, d) → Bun(1, d).

Theorem 5.7 (Atiyah). Bun(r, d) ∼= C and the map det : Bun(r, d) → Bun(1, d) has degreeh2 where h = gcd(r, d). Moreover, V ∈ Bun(r, d) is irreducible if and only if h = 1.

Proof. The proof is similar to that of Grothendieck’s theorem (5.1). An interesting referenceis Atiyah’s original paper [A]. �

References

[A] M. F. Atiyah, Vector bundles over an elliptic curve, Proc. London Math. Soc. (3) 7 (1957) 414–452.[Ha] R. Hartshorne, Algebraic geometry.[I] B. Iversen, Cohomology of sheaves.[M] R. Miranda, Algebraic curves and Riemann surfaces.

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