VECTOR BUNDLES ON RIEMANN SURFACES
SABIN CAUTIS
Contents
1. Differentiable Manifolds
2. Complex Manifolds
2.1. Riemann Surfaces of Genus One
2.2. Constructing Riemann Surfaces as Curves in P2
2.3. Constructing Riemann Surfaces as Covers
2.4. Constructing Riemann Surfaces by Glueing
3. Topological Vector Bundles
3.1. The Tangent and Cotangent Bundles
3.2. Interlude: Categories, Complexes and Exact Sequences
3.3. Metrics on Vector Bundles
3.4. The Degree of a Line Bundle
3.5. The Determinantal Line Bundle
3.6. Classification of Topological Vector Bundles on Riemann Surfaces
3.7. Holomorphic Vector Bundles
3.8. Sections of Holomorphic Vector Bundles
4. Sheaves
4.1. Cech Cohomology
4.2. Line Bundles and Cech Cohomology
4.3. Riemann-Roch and Serre Duality
4.4. Vector bundles, locally free sheaves and divisors
4.5. A proof of Riemann-Roch for curves
5. Classifying vector bundles on Riemann surfaces
5.1. Grothendieck’s classification of vector bundles on P1
5.2. Atiyah’s classification of vector bundles on elliptic curves
References
1
2
3
4
6
9
10
11
13
14
15
16
17
18
19
20
21
23
27
29
30
33
34
34
35
36
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.
1
Exercise 1. Show that S 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 natural
inclusion f : (0, 1] → S 1 is continuous and bijective but the inverse f −1 is not continuous at
f (1).
A manifold of dimension n is a Hausdorff topological space M such that every point p ∈ M
has a neighbourhood p ∈ U ⊂ M which is homeomorphic to the open unit ball in Rn . In
other words, M locally looks like Rn .
Example. The real line R, the unit circle S 1 and the open interval (0, 1) are onedimensional manifolds. The semiopen interval (0, 1] is not a manifold since there is no open
neighbourhood of 1 homeomorphic to an open interval in R.
Example. The sphere S 2 , the torus S 1 × S 1 , the open cylinder (0, 1) × S 1 as well as any
open subset of R2 are two-dimensional manifolds.
An open cover of a topological space X is a collection of open subspaces Uα such that
∪α Uα = X. A manifold M is compact if every open cover of M has a finite subcover or
equivalently if every sequence of points {pi } ⊂ M contains a subsequence which converges
to a point p∞ ∈ M .
By definition, an n-dimensional manifold M is covered by open sets Uα which are homeomorphic via maps fα : Uα → Rn to an open ball in Rn . Denote Uαβ = Uα ∩ Uβ and
fαβ = fβ ◦ fα−1 : fα (Uαβ ) → fβ (Uαβ ). One should think of M as being built from these balls
by 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 are
called transitionfunctions. By construction, the transition functions satisfy:
−1
• fαβ = fβα
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 functions fαβ satisfying these two conditions.
Example. Consider S 1 = {(x, y) : x2 + y 2 = 1} ⊂ R2 . We can use as an open cover the
sets U1 = S 1 − (0, 1) and U2 = S 1 − (0, −1). Then we have homeomorphisms fi : Ui → R
−x
x
given by f1 (x, y) = y−1
and f2 (x, y) = y+1
. One can check that f1−1 : R → U1 is given by
2 −1
. Then f21 = f2 ◦ f1−1 : R∗ → R∗ is given by t 7→ 1t . Thus S 1 is obtained by
t 7→ t22t+1 , tt2 +1
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 the
transition functions) are not only continuous but also C ∞ (infinitely differentiable). In
dimensions at most three, these two notions coincide – in other words, a topological manifold
of dimension one, two or three has a unique differentiable structure. This is not true in higher
dimensions. For example, by a result of Milnor, the seven sphere S 7 has exactly 28 different
differentiable structures.
2
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 S 1 .
• There are two types of surfaces: orientable and non-orientable. The orientable surfaces are classified by their genus g. In other words, for each g ∈ Z≥0 there is exactly
one surface of that genus. The sphere has genus zero, the torus T 2 = S 1 × S 1 has
genus 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 (although
the classification of differentiable 4-manifolds remains open).
• Manifolds of dimension at least 5 (both topological as well as differentiable) have in
principle 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 manifolds
of (complex) dimension n (real dimension 2n). Locally, a complex manifold will have charts
Uα 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 S 2 has the structure of a complex manifold since S 2 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 S 1 by glueing together
two copies of R along R∗ = R − 0.
Given two complex manifolds M1 and M2 a morphism f : M1 → M2 is a continuous map
such that its restriction to charts is holomorphic. We say f is an isomorphism if it has a
holomorphic inverse.
Before continuing we mention a few useful results about holomorphic maps:
• (Open Mapping Theorem) A nonconstant holomorphic map C → C maps open sets
to open sets.
• A holomorphic map which is injective and surjective is biholomorphic (i.e. it has an
inverse 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 incredibly
beautiful mathematics.
It turns out a complex manifold is automatically orientable (see section 3.5). Thus Riemann surfaces are topologically classified by their genus.
The sphere (genus g = 0) has only one complex structure. In other words, two Riemann
surfaces which are homeomorphic to the 2-sphere then are actually isomorphic as Riemann
surfaces (i.e. there is a biholomorphic map between them). This result follows from the
following more general theorem.
Theorem 2.1 (Uniformization Theorem). Any simply connected complex manifold of dimension one is isomorphic to precisely one of the following spaces:
3
• the Riemann sphere Ĉ
• the complex plane C
• the hyperbolic plane H = {z ∈ C : Im(z) > 0}
Intuitively, we think of Ĉ as C with a point added at infinity (thus Ĉ compactifies C). The
uniformization theorem above is remarkable since it says that any simply connected domain
in C (which is not all of C) is isomorphic to H.
Exercise 3. Show directly that the open unit disk ∆ = {z ∈ C : |z| < 1} is isomorphic
z−i
).
to H (hint: use the map z 7→ 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). By
composing with z 7→ az + b for appropriate a and b we can assume f (0) = 0 and f ′ (0) = 1.
Thus f (z) = z + a2 z 2 + a3 z 3 + . . . 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 has
infinitely 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 solution
meaning f is not injective again. Whence f (z) = z.
Exercise 4. Use this to show that
az + b
Aut(Ĉ) = {z 7→
: ad − bc 6= 0}/{dI : d ∈ C} ∼
= P GL2 (C)
cz + d
(recall that we can think of Ĉ as C with a point added at infinity).
a b
Proposition 2.3. Aut(H) = P SL2 (R) where
acts by z =
c d
az+b
.
cz+d
2.1. Riemann Surfaces of Genus One. What are the complex structures one can put
on 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 simply
connected and hence the univeral cover of T 2 , inherits a complex structure from T 2 . By the
Uniformization 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 ball
B around p such that G · p ∩ B = p. Note that the action of Z ⊕ Z on R2 described above is
discrete. On the other hand, the lemma below shows that there is no discrete Z ⊕ Z action
on H. Thus the universal cover of a genus one Riemann surface cannot be H and must be
C.
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
1 s
1 t
can conjugate it to the form
. Since a and b commute it means b =
for
0 1
0 1
4
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 are simultaneously
they commute
λ 0
µ 0
diagonalizable. But then a =
and b =
so again they must generate a
0 λ
0 µ
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 fixedpoint free subgroup Z ⊕ Z ⊂ Aut(C). The fixed-point free automorphisms of C are the
translations tb : z 7→ z + b. Thus any genus one Riemann surface is obtained by quotienting
C by two automorphisms tb and tb′ for some b, b′ ∈ C. Here tb and tb′ correspond to the
generators (1, 0) and (0, 1) of Z ⊕ Z. Note that b and b′ must be pointing in different
directions (i.e. they must be linearly independent over R) or else the quotient is not a genus
one Riemann surface. Rotating and scaling C by an appropriate factor allows one to assume
that b′ = 1 and b ∈ H = {z ∈ C : Im(z) > 0}. Thus we get a map from H to the space of
Riemann surfaces of genus one given by τ 7→ C/h1, τ i. We will denote the Riemann surface
corresponding to τ ∈ H by Eτ (E stands for elliptic curve which is another name for a genus
one Riemann surface).
Proposition 2.6. The map from H to the set of Riemann surfaces is onto. Two points give
the 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
a b
: a, b, c, d ∈ Z, ad − bc = 1}
SL2 (Z) = {
c d
on H is by z 7→ az+b
. To understand where this action comes from recall that tb and tb′
cz+d
corresponded to the basis elements (1, 0) and (0, 1) of Z ⊕ Z. Of course, picking different
basis 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 to
different genus one Riemann surfaces let’s suppose that C/h1, τ i and C/h1, τ ′ i are isomorphic
via some isomorphism f . Then this map lifts to an isomorphism f˜ : C → C such that the
lattice h1, τ i maps to the lattice h1, τ ′ i. Thus f˜(1) = a + bτ ′ and f˜(τ ) = c + dτ ′ where
a, b, c, d ∈ Z satisfy ad − bc = 1 (since the map must be invertible). Hence τ and τ ′ differ
the action of an element of SL2 (Z).
Corollary 2.7. The moduli space M1 (i.e. the parameter space) of genus one Riemann
surfaces is H/SL2 (Z) (which has complex dimension one and happens to be isomorphic to
C).
What are Aut(Eτ )? An automorphism of Eτ is the same as giving an automorphism of C
which 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 the
following extra automorphisms:
• τ = i: z 7→ iz
5
• τ=
√
1+i 3
:
2
z 7→ eiπ/3 z
Aside: Just as a Riemann surface of genus one is the quotient of C by Z ⊕ Z, it turns out
that a Riemann surface C of genus g ≥ 2 is the quotient of H by the fixed point free action of
a group. As before, this group is the fundamental group π1 (C) of C. However, describing all
the possible actions of π1 (C) on H is quite difficult. The moduli space of genus g Riemann
surfaces is denoted Mg and turns out to have have a complex structure of complex dimension
3g −3. In other words, there is a 3g −3 dimensional space parametrizing Riemann surfaces of
genus 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 S 2 .
2.2. Constructing Riemann Surfaces as Curves in P2 . One natural way of describing
manifolds is as the zero set of a polynomial. For instance, the n-dimensional sphere is the
zero set of p = x20 + · · · + 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 effective
way to construct complex manifolds is as the zero set of polynomials in Cn . For example,
z02 + z12 − 1 = 0 is isomorphic (as a complex manifold) to C. Unfortunately, since Cn is not
compact all complex manifolds constructed this way will not be compact. To remedy this
we compactify Cn to get the projective space Pn .
2.2.1. Projective Spaces. The projective space Pn (C) = Pn is defined as the quotient of
Cn+1 − 0 by C∗ = C − 0 where the action is given by λ · (z0 , . . . , zn ) = (λz0 , . . . , λzn ). Just
as a point in Cn is given in the usual coordinates as an n-tuple (z1 , . . . , zn ), a point in Pn is
encoded in homogeneous coordinates as an n + 1-tuple [z0 , . . . , zn ] where one keeps in mind
that [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 point
has can be uniquely represented as [1, t] where t = z1 /z0 . Thus the locus of points where
z0 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 point
added at infinity, so if we believe that P1 should be a complex manifold then it ought to be
the Riemann sphere Ĉ.
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 is
obtained by glueing C to C along C∗ by using the map t = zz01 7→ zz10 = 1t . Since t 7→ 1/t is
a holomorphic map over C∗ this gives P1 a complex structure. Notice that this is the same
glueing map as the one used in section 2.
One can similarly study P2 . A typical point of P2 is [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 is
obtained by glueing to C2 a copy of P1 .
Exercise 5. Check that Pn has a stratification ∐0≤i≤n Cn .
Proposition 2.9. Aut(Pn ) ∼
= P GLn+1 (C) where the action is by left multiplication on
[z0 , . . . , zn ].
6
Proof. Since Pn is Cn+1 − 0 modulo C∗ , any automorphism of Pn lifts to an automorphism of
Cn+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 up
to 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 of
polynomials give us (non-compact) complex manifolds. The equivalent of a polynomial in
the setting of Pn is a homogeneous polynomial.
Definition 2.10. Check that a polynomial f (z0 , . . . , zn ) is homogeneous if for any λ ∈ C − 0
we 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 quotient
Cn+1 /C∗ = Pn .
Exercise 6. A polynomial f (z0 , . . . , zn ) is homogeneous if and only if every monomial
term has the same degree n (which is by definition the degree of f ).
Example. z0 + z12 is not homogenous whereas z03 + z13 is homogeneous of degree 3.
If f (z0 , . . . , zn ) is a general homogeneous polynomial then its zero set in Pn carves out a
complex submanifold of dimension n − 1. For instance, if f (z0 , z1 , z2 ) = z2 then the zero
set are the points in P2 of the form [z0 , z1 , 0] which as we saw is the Riemann sphere S 2 .
Since Pn is compact, the manifold we get will also be compact (since it is a closed subset of
a compact space).
Exercise 7. Show that the zero set of f (z0 , z1 , z2 ) = a0 z0 + a1 z1 + a2 z2 in P2 (where the
constants ai ’s are not all zero) is isomorphic to the Riemann sphere Ĉ (hint: use the action
of P GL3 (C) on P2 ).
Recall that our goal is to construct examples of Riemann surfaces. To do this we can
consider the zero set of a homogeneous polynomial f (z0 , z1 , z2 ) of degree d. For d ≥ 3 the
Riemann surface we get will depend on f but it’s genus will not. So our next question is: if
f 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. Let M be a (topological) manifold and consider an arbitrary
triangulation of M . The Euler characteristic χ(M ) of M is defined as the number of vertices
in the triangulation minus the number of edges plus the number of faces and so on.
Example. Viewing the sphere S 2 as a tetrahedron gives us a triangulation of S 2 with
four vertices, six edges and four faces. Thus χ(S 2 ) = 4 − 6 + 4 = 2. Viewing S 2 as the cube
gives us eight vertices, twelve edges and six faces. Again, we get χ(S 2 ) = 8 − 12 + 6 = 2
(the latter is not technically a triangulation so we should not use it but we still get the right
answer since each face is a polygon).
Theorem 2.11. The Euler characteristic χ(M ) does not depend on the specific triangulation
used.
Proof. Given two triangulations of M one can subdivide them further to obtain a common
subtriangulation. The result follows from the following result:
Exercise 8. Show that χ(M ) remains unchanged after taking subtriangulations.
7
Thus the Euler characteristic χ(M ) is an invariant of M . It provides a way to distinguish
between manifolds: namely, if χ(M1 ) 6= χ(M2 ) then they are not homeomorphic. In the case
of real (orientable) manifolds of dimension two this invariant is enough to distinguish them:
Exercise 9. Show using your favourite triangulation that a Riemann surface of genus g
has Euler characteristic 2 − 2g.
2.2.3. Riemann-Hurwitz Theorem. Suppose f : C1 → C2 is a holomorphic map between two
Riemann surfaces. Locally around each point p ∈ C1 the map p 7→ f (p) can be written as
z 7→ an z n + an+1 z n+1 + . . . . Changing variables one can rewrite it as z → z n . The integer n
is called the ramification index of f at p and denoted ep . If n = 1 then we say f is unramified
at 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→ z n from ∆ to ∆ look like? (identify the branching
points, the ramification points and their indices)
For any holomorphic map f : C1 → C2 the target C2 contains only finitely many branching
points. 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 the
ramification indices as follows:
P
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→ z d . The preimage of a point
w ∈ C ⊂ P1 are the points z ∈ C satisfying z d = w. Thus for w 6= 0 the preimage contains
d points and hence deg(f ) = d. The branching points are 0, ∞ ∈ P1 . The preimage of 0 is 0
where the ramification index is d. Similarly, the preimage of ∞ is ∞ where the ramification
index 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 Euler
characteristics χ(C1 ) and χ(C2 ) in terms of the degree deg(f ) and the ramification indices.
P
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 a
vertex. With this added condition, the preimage of this triangulation gives a triangulation
of 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
P faces in the triangulation of C1 is de
and df respectively. The number of vertices is dv − p∈C1 (ep − 1). Thus
X
X
χ(C2 ) = dv −
(ep − 1) − de + df = d(v − e + f ) −
(ep − 1).
p∈C1
p∈C1
Notice that the sum on the right side is finite since all but a finite number of points have
ramification index one. One can use this result to compute the genus of C1 given the genus
of C2 and knowledge about the map f : C1 → C2 .
d
Example. Consider again the map f : P1 → P1 given by
P z 7→ z . We saw that there are
two 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 hence
χ(P1 ) = 2 (reaffirming our earlier calculation).
Example. In algebraic geometry, Riemann surfaces are also called curves. Consider the
Fermat curve C in P2 given by xn + y n + z n = 0 for some n ∈ N. The map P2 → P1 given
by [x, y, z] → [x, z] is not well defined at [0, 0, 1] (why?) but since [0, 0, 1] is not a point on
C 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 equation
y n = −xn0 − z0n . 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 − z0n = 0 or equivalently xn0 = −z0n . The points
in P1 which satisfy this equation are precisely the points of the form [ζ, 1] where ζ n = −1.
Thus there are precisely n points in P1 over which the fiber of π is ramified and each time
this 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.
since χ(C) = 2 − 2g.
Thus, the genus of C is g = n−1
2
Proposition 2.14. The genus of a general curve of degree n in P2 is
n−1
2
.
Proof. The proof is along the same lines as above except that counting the ramification
points in general is a little harder.
A good way to visually justify this result is as follows. Suppose the curve is given by a
degree n homogeneous polynomial which factors completely into different monomials (for
example, for n = 4 we could have f (x, y, z) = xyz(x − y + z)). Each factor corresponds to a
copy of P1 and thus the product describes the union of n such P1 ’s. Moreover, every two P1 ’s
meet in precisely one point. Thus we have n spheres every pair of which meet in exactly one
point. If one deforms the equation f slightly then one gets a nice (smooth) Riemann surface
and the points in the original curve correspond to n2 small loops in this Riemann surface
which have been pinched to a point. It is then easy to see that in fact the genus of this
. For example, if one takes a torus (genus one Riemann surface)
Riemann surface is n−1
2
and pinches three non-intersecting loops, one gets three spheres each two of which intersect
at a point.
Even though this gives us an effective way of constructing Riemann surfaces as complex
submanifolds in P2 not all Riemann surfaces can arise this way. In particular, since n−1
=2
2
has no solution, we cannot construct genus two Riemann surface. On the other hand we can
generalize this construction by considering the zero set in P3 of two polynomials f and g. If
f and g are chosen to be general this construction yields a Riemann surface. This way one
can construct examples of Riemann surfaces of arbitrary genera.
2.3. Constructing Riemann Surfaces as Covers. Given a curve C one can try to construct a new curve as a cover of C. To describe such a cover it is enough to identify the
degree d of the cover, the branching points p1 , . . . , pm of C over which there is ramification
and a final piece of data known as monodromy. This data tells one how different sheets of
the cover are glued together around each ramification point. We explain further using an
example.
9
Example. The simplest case is when C = P1 and the degree is d = 2. In this case there
are two sheets and each ramification point interchanges these two sheets. In other words,
locally around each ramification point the map looks like z 7→ z 2 . Notice that there is no
extra monodromy data necessary. The only thing we need to specify are the branching points
p1 , . . . , 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 the
cover will have genus 2.
Since the automorphism group of P1 (namely P GL2 (C)) acts triply transitive we can take
p1 = 0, p2 = 1 and p3 = ∞. Each choice of p4 , p5 , p6 gives us a curve of genus two. Since there
are no automorphisms of P1 fixing 0, 1, ∞ these curves turn out to be non-isomorphic to each
other. Thus we get a three dimensional family of genus two curves. Conversely, any genus
two curve can be written as a double cover (a two-to-one cover) of P1 . This explains why the
moduli space of curves of genus two is 3-dimensional. As a consequence of this description,
any Riemann surface of genus two has an involution (namely, the map interchanging the two
sheets). This map is called the hyperelliptic involution.
Example. Consider the triple cover of P1 branched over three points p1 , p2 , p3 . This time
we do need to give the monodromy data. There are three sheets which we will number 1, 2, 3.
We take the monodromy around p1 to be the permutation (123). In other words, over p1
there is a ramification point of order 3 which interchanges the sheets 1 → 2 → 3 → 1. The
monodromy around p2 we take to be (12). In other words, there are two points lying above
p2 . One of the points lies on sheet 3 and is unramified while the other is ramified of order
two and interchanges sheets 1 → 2 → 1. Finally, the monodromy around p3 is (13). Notice
that (123)(12)(13) = id. This is a necessary condition since the monodromy around all three
points should be trivial.
Exercise 12. What is the genus of the resulting triple cover in the example above?
Project: The problem of counting all the possible covers with fixed branching points
p1 , . . . , pm ∈ P1 is known as Hurwitz’s problem. Since describing such covers is the same as
giving the monodromy data this is apriori a combinatorial problem involving the symmetric
group Sn . More recently, this problem has been shown to have amazing and surprising
connexions with the geometry of the moduli space of curves Mg . A potential project may
be to learn a little about the classical Hurwitz problem and explain (in broad terms) this
recent connexion.
2.4. Constructing Riemann Surfaces by Glueing. This method can be used to obtain
very concrete examples of interesting Riemann surfaces with lots of automorphisms. Note
that in general, Riemann surfaces of genera g ≥ 2 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 translating) as shown in figure 1 to get a surface S.
P
1
0
0
1
P
P
1
0
0
1
1
0
0
1
P
P0011
1
0
1
0
1
0
0
1
P
1
0
0
1
1
0
0
1
P
P
Figure 1. The identification shown yields a Riemann surface of genus 2 with
automorphism 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· 10 = 4π accumulating at P . Thus
hood U around P . Note that there is a total angle of 2π
5
√
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 α
of S. One can also exchange the pentagons by using a rotation by π to get an automorphism
of order two. This involution is the hyperelliptic involution from earlier. It turns out these
are all the automorphisms of S so Aut(S) ∼
= Z/10Z.
One can also obtain S as a double cover of P1 branched over the points 0, 1, ζ, ζ 2 , ζ 3 , ζ 4
where ζ is a fifth root of unity. The map P1 → P1 given by z 7→ ζz fixes (as a set) these six
points and lifts to the cover to give the automorphism α ∈ Aut(S) we saw above. One can
apply the same glueing procedure to other shapes to obtain explicit constructions of other
Riemann surfaces.
Exercise 15. What is the genus of the Riemann surface obtained by identifying opposite
sides of an L-shaped figure? Curt McMullen studies such Riemann surfaces in order to
understand the dynamics of ball trajectories on L-shaped billiard tables.
3. Topological Vector Bundles
Let M be a topological manifold. A real vector bundle V over M of dimension n is a family
of real vector spaces of dimension n parametrized by M . More precisely, V is a manifold
with a map π : V → M such that around each point p ∈ M there is an open neighbourhood
p ∈ 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 → M
which we denote In .
11
By definition, a vector bundle is locally trivial (i.e. over small enough open sets U ⊂ M , it
is isomorphic to Rn ×U ). Thus, to describe a vector bundle V over M it is enough to give local
charts 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, fαβ v) in Uβ × Rn ). As before, these transition functions must satisfy:
−1
• fαβ = fβα
on Uα ∩ Uβ
• fβγ ◦ fαβ = fαγ on Uα ∩ Uβ ∩ Uγ
Example. Suppose M = S 1 = {(x, y) : x2 + y 2 = 1}. Along with the trivial line bundle
I1 = M × R there’s also the Möbius line bundle M1 which we describe using the two charts
U0 = S 1 −(0, 1) and U1 = S 1 −(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 like
an (infinite) Möbius strip. As this example illustrates, vector bundles are boring locally (since
they are trivial) but become interesting globally because of the possibility for “twisting”.
π1
π2
A morphism between two vector bundles V1 −→
M and V2 −→
M is a continuous map
g : 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 vector
spaces and not just topological spaces. V1 and V2 are isomorphic if there is an invertible
morphism 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 by
p 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. On
the other hand, show that every section of the Möbius bundle M1 must intersect the zero
section. Conclude that I1 and M1 are not isomorphic.
All the definitions from vector spaces carry over naturally to vector bundles. We summarize them below:
• subbundles: V ⊂ W is a subbundle if there is an injective morphism V ֒→ W . This
means that in each fiber Wp over p we can pick a vector subspace Vp ⊂ Wp such that
the choices vary continuously with p.
1
• direct sums: if V1 and V2 are described by transition functions fαβ
: Uα ∩ Uβ →
2
GLm (R) and fαβ : Uα ∩ Uβ → GLn (R) then the direct sum V1 ⊕ V2 has transition
functions
1
fαβ 0
: Uα ∩ Uβ → GLm+n (R).
2
0 fαβ
1
2
• tensor product: similarly, V1 ⊗ V2 has transition functions fαβ
⊗ fαβ
: (Uα ∩ Uβ ) →
2
1
GLmn (R) where this is short hand for the action v1 ⊗ v2 7→ fαβ v1 ⊗ fαβ v2 .
Exercise 17. Check that the operations ⊕ and ⊗ give the set of vector bundles over M
the 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 ∼
= M1 (the Möbius line bundle). (Hint: draw a picture).
= V2 ∼
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 bundle
′
V 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 locally
the trivialization is π −1 (U ) ∼
= U × Cn . The theory of real and complex vector bundles is very
similar (just like the theory of real and complex vector spaces is very similar). On the other
hand, there are some differences. For example, consider the complex Möbius bundle over S 1
defined as above using the same transition functions
f01 : U0 ∩ U1 → GL1 (R) ֒→ GL1 (C) = C∗
Exercise 20. Show that the complex Möbius line bundle is in fact trivial (hint: find a
section 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 a
nonvanishing 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 has
a dual vector bundle V ∗ . If V is given by transition functions fαβ ∈ GLn (R) then V ∗ has
t −1
transition functions (fαβ
) ∈ GLn (R). Similarly, one defines the dual of a complex vector
bundle.
Fact: We will show later that if V is a real vector bundle then V ∼
= V ∗ but if V is a
complex 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 n
comes equipped with an intrinsic n dimensional vector bundle TM called the tangent bundle.
∗
The dual of this bundle is the cotangent bundle TM
. Strangely, it is more natural to describe
the cotangent bundle so we shall do that first.
∗
The fibers of (TM
)p over a point p ∈ M should correspond to the vector space mp /m2p
∞
where mp ⊂ C is the ideal consisting of functions vanishing at p. It is easy to check that
mp /m2p 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 . TM
will
be trivial over U0 and we choose as basis n vectors which we call dx1 , . . . , dxn . Now take
another chart U1 around p with coordinates y1 , . . . , yn . Inside U0 ∩ U1 the two charts give
us 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 TM
over U1 given by dy1 , . . . , dyn . Then the transition
information for the cotangent bundle over U0 ∩ U1 is given by:
n
X
∂yi
dyi =
dxj .
∂x
j
j=1
∂yi (x1 ,...,xn )
∈ GLn (R). The fact that the
In other words, the transition function is f01 =
∂xj
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 TM
. The local basis
∂
elements for TM are denoted ∂xi . The transition functions are then
n
X ∂yj ∂
∂
=
.
∂xi
∂x
∂y
i
j
j=1
Exercise 22. A somewhat tedious but important exercise everyone should do at least
once in their lifetime is to check that these transition functions do in fact satisfy the glueing
conditions.
3.2. Interlude: Categories, Complexes and Exact Sequences. A category C consists
of a class of objects Ob(C) and a class of maps mor(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 satisfies
1b ◦ 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 spaces
over k and the morphisms are linear maps between vector spaces.
Example. The category of abelian groups. The objects are abelian groups and the morphisms are group homomorphisms between them.
Example. The category of topological spaces. The objects are topological spaces and the
morphisms are continuous maps between them.
A category does not necessarily have kernels and cokernels. For example, one cannot
define the kernel of a map in the category of topological spaces. Since we want to be able
to take kernels, cokernels and direct sums we will work in an enriched category known as
an abelian category. Clearly, the categories of vector spaces and abelian groups are abelian
categories.
Consider a sequence of maps A· :
f1
f2
fn
A0 −
→ A1 −
→ A2 → · · · → An−1 −→ An .
A· is a complex if for each i we have Im(fi−1 ) ⊂ Ker(fi ). In this case we can talk of the
homology Hi (A· ) = Ker(fi )/Im(fi−1 ). If Im(fi−1 ) = Ker(fi ) then the sequence is called
exact. 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:
p
i
0→A−
→B−
→ 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 → 0
14
where the first map is a 7→ 2a and the second map is the natural projection. It is easy to
check 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 exact
sequence splits (hint: define the orthogonal complement).
i
p
Proposition 3.2. A short exact sequence 0 → A −
→B−
→ C → 0 splits if and only if there
exists a map f : B → A such that f ◦ i = idA (or equivalently, a map g : C → B such that
p ◦ g = idC ).
Proof. Exercise 24. .
A map F between two categories C and D is called a functor. It takes objects to objects
and morphisms to morphism such that if 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,
P nonnegative functions
ψα : 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 every
open cover has a locally finite open refinement).
Just as you have inner products h·, ·i for real vector spaces (or Hermitian metrics for
complex vector spaces) there are analogous concepts for vector bundles. An inner product
h·, ·iV on a vector bundle V → M is a map V ×M V → R+ which restricted to any fiber Vp
is an inner product h·, ·iVp : Vp × Vp → R+ .
We know how to construct an inner product on a fixed vector space. The following theorem
tells 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 V
admits an inner (Hermitian) product.
Proof. Choose an open cover Uα of M over which V is trivial. For each α construct an
inner product h·, ·iα for VUP
By the theorem above we can find a partition of unity ψα
α.
subordinate to Uα . Then α ψα h·, ·iα gives an inner product on V (we use that the sum
of two inner products on a vector space is also an inner product since inner products are
positive definite).
This result is a useful tool. We use it next:
i
p
Proposition 3.5. Any short exact sequence of vector bundles 0 → E1 −
→E−
→ E2 → 0 over
M 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
i
→ E is an injection E1 ⊂ E is a subbundle. Pick an inner product
f ◦ i = idE1 . Since E1 −
h·, ·iE on E and take f to be the orthogonal projection E → E1 with respect to h·, ·iE . Corollary 3.6. Any short exact sequence of vector bundles 0 → E1 → E → E2 → 0 splits
as E ∼
= E1⊥ where ⊥ is defined with respect to any inner product on E.
= E1 ⊕ E1⊥ with E2 ∼
A good picture to have in mind is that of the Möbius bundle M1 over S 1 and the associated
short 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 surface
C. Consider a general section σ : C → L. We can produce such a section by giving it locally
and then glueing it together using a partition of unity. Locally, the line bundle L is trivial
so it looks like C × ∆ → ∆ where ∆ is the open unit disk. In this local picture, σ is just a
map ∆ → C.
By perturbing σ we can insist that it is transverse to the zero section. Then locally, the
inverse image of 0 ∈ C under the map σ : ∆ → C is a finite number of points. Each point
p ∈ C where σ intersects the zero section is called a zero of σ. Around each such point
p the section σ is a map σ : ∆ → C where p = 0 ∈ ∆ and σ(0) = 0. The differential
dσ : T0 ∆ → T0 C is a nonsingular two-by-two matrix. We denote by sgn(p) the sign of the
determinant of this matrix. Notice that there was an ambiguity since the map σ : ∆ → C
is defined up to post-multiplication by C∗ . Fortunately, multiplying by a complex number
does not change the sign of det dσ. Of course, this would not be true if we were dealing with
a real vector bundle.
P
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 give
the 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. The
section on the right has no zeroes.
Warning: If we just count points then we get a well defined element of Z/2Z. This is
what we do with real line bundles. Fortunately, in the case of complex line bundles we are
able 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 section
which is nonzero everywhere.
Example. The tangent line bundle on P1 has degree 2. This corresponds to the famous
saying that you cannot comb the hair on a coconut (i.e. the tangent bundle of S 2 is not
trivial). The more precise saying ought to be that you can comb the hairs on a coconut
everywhere except at two points (the north pole and the south pole). At these points the
matrix dσ describes a rotation and has determinant one. Whence both points contribute +1
to the degree.
Exercise 25. Show that the degree of the tangent bundle on a Riemann surface of genus
g is 2 − 2g.
Exercise 26. Show that deg(L1 ⊗ L2 ) = deg(L1 ) + deg(L2 ) (hint: if σi (i = 1, 2) are
sections 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. Then
L1 ∼
= 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 do
this 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) is
trivial.
Example. Let’s construct all the complex line bundles on S 2 . Think of S 2 as two disks
D1 and D2 glued along their circumferences. By the above result, the restriction of any line
bundle L on S 2 to D1 and D2 is trivial. Thus to describe L we simply need indicate how to
glue the trivial line bundles over D1 to the trivial line bundle over D2 along the boundaries
∂D1 = ∂D2 = S 1 . This is equivalent to giving the clutching map f : S 1 → GL1 (C) = C∗ . If
you have such a map you can consider its winding number. It turns out this number is in
fact the degree of the line bundle L. The theorem below tells you that if you have two line
bundles with the same winding number then they are isomorphic (since you can interpolate
between the two clutching maps to obtain a family of line bundles).
Theorem 3.9. If V → X × [0, 1] is a family of vector bundles of vector bundles over X then
V0 ∼
= 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 Chern
classes c1 (V ), . . . , cn (V ). These classes are not integers but rather cohomology classes ci (V ) ∈
H 2i (X, Z) where H 2i (X, Z) is the 2i-th cohomology group of X. In the case X has dimension
two (Riemann surfaces) we find that H 2 (X, Z) = Z so for a line bundle L we get deg(L) =
c1 (L) ∈ H 2 (X) ∈ Z.
3.5. The Determinantal Line Bundle. Suppose V → C is a vector bundle of dimension n
over a Riemann surfaces C with transition functions fαβ : Uα ∩Uβ → GLn (C). The associated
determinantal line bundle det V 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 bundles
then det(E) ∼
= det(E1 ) ⊗ det(E2 ).
Proof. Since we know the sequence splits we have E ∼
= E1 ⊕ E2 . In particular, this means
that the transition functions for E are just f1 ⊕f2 where fi (i = 1, 2) is the transition function
of Ei . Then det(f1 ⊕ f2 ) = det(f1 ) det(f2 ) which (by definition) are the transition functions
for 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 then
deg(E) = deg(E1 ) + deg(E2 ).
17
Proof. This follows immediately from the proposition above and the fact that for line bundles
L1 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 bundle
of dimension 2n. This is achieved by embedding GLn(C) ֒→ GL2n (R). This embedding is
r cos θ r sin θ
done componentwise via z = reiθ 7→
. A complex vector bundle V is said
−r sin θ r cos θ
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 diagram commutes:
/ GL2n (R)
GLn (C)
det
HH
HH det
HH
HH
HH
$
det
/ R∗
/ GL2 (R)
C∗
This means that it is enough to show any complex line bundle L is orientable. If the transition
functions of L are fαβ = reiθ (where r, θ are functions) then the transition
functions for the
r cos θ r sin θ
real
corresponding two dimensional real vector bundle Lreal are fαβ
=
. Then
−r sin θ r cos θ
real
the determinantal (real) line bundle det Lreal has transition functions det fαβ
= r 2 ∈ R∗ .
The key point is that r2 > 0 so these maps can be deformed to constant maps to R+ and
hence 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 ′ ⊕ I1
where 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 a
non-vanishing section of this subbundle it follows the subbundle is the trivial line bundle I1 .
Putting a metric on V we take V ′ = I1⊥ . Then V = V ′ ⊕ I1 .
Proposition 3.15. Let V → M be a vector bundle over a manifold M of dimension
dimR (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 dimension
of the base, one can take a general section and perturb it locally around vanishing points so
that it no longer vanishes.
Corollary 3.16. If V is a complex vector bundle on a Riemann surface then V ∼
= det(V )⊕Im
where m = dim(V ) − 1.
18
Corollary 3.17. The ring R of vector bundles on a Riemann surface is isomorphic to
Z[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 ′.
= V ′ ⊕ In then V ∼
3.7. Holomorphic Vector Bundles. A holomorphic vector bundle is a complex vector
bundle where the transition functions Uαβ → GLn (C) are not just smooth but also holomorphic. It is important to realize that the concept of a holomorphic vector bundle only makes
sense over a complex manifold (for example, Riemann surface). The total space V → M of
a holomorphic vector bundle is a complex manifold (though it is not compact). Two holoπ
π′
morphic vector bundles V −
→ M and V ′ −
→ M are isomorphic if there exists a holomorphic
′
map f : V → V such that f ◦ π ′ = π (i.e. maps the fiber Vp over p ∈ M to the fiber Vp′ also
over p ∈ M ) and which acts linearly on the fibers (i.e. the restricted map fp : Vp → Vp′ is
linear).
Example. The trivial vector bundle In is holomorphic.
Example. If M is a complex manifold of (complex) dimension n then the tangent bundle
TM is a holomorphic vector bundle of dimension n. This is because the transition functions
describing M are holomorphic (by definition) and hence their derivatives, which are used to
used to describe the transition functions for TM , are also holomorphic.
Exercise 29. Show that V is holomorphic if and only if V ∗ is holomorphic.
3.7.1. Holomorphic Line Bundles over P1 . Recall that the complex line bundles over P1 are
in bijection with Z via the degree map. Let us give the complex line bundle Ln of degree
n ∈ Z a holomorphic structure. As a complex manifold we view P1 as two copies U0 and
U1 of C glued together along C∗ via the map f01 : z 7→ 1/z. To build Ln as a holomorphic
line bundle we take it to be trivial over both copies of C and hence we need only give the
glueing map gn : C∗ → GL1 (C) ∼
= C∗ . Since we want Ln to be holomorphic gn must also be
holomorphic. 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 trivial
line bundle since the transition function g0 is the map z 7→ 1. Moreover, the transition
function for Lm ⊗ Ln is gm · gn = gm+n and thus Lm ⊗ Ln ∼
= Lm+n . This is expected since
deg(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 over
U0 and U1 such that over U0 ∩ U1 = C∗ the glueing is made via w = 1/z. By definition,
19
∂
∂
T |U0 ∼
i and T |U1 ∼
i. The glueing over U0 ∩ U1 is then
= U0 × Ch ∂z
= U1 × Ch ∂w
d 1 ∂
1 ∂
∂
∼
=− 2
.
∂z
dz z ∂w
z ∂w
This means the transition function for T is z 7→ −1/z 2 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/z 2 is
in fact isomorphic to L2 where the transition function is 1/z 2 .
Since L∗n ∼
= L−n it follows that the holomorphic cotangent line bundle TP∗1 of P1 is isomorphic to L−2 .
Warning: Though every complex line bundle over a Riemann surface C can be given a
holomorphic structure it is not true that this structure is unique. The picture one should
keep in mind is the sequence of maps
deg
holomorphic vector bundles over C → complex vector bundles over C −−→ Z
By what we said earlier, the second map is an isomorphism. The first map is surjective but
not necessarily injective. If C = P1 the first map is in fact an isomorphism but if C has genus
g ≥ 1 then it is not injective. Finally, be aware that over arbitrary complex manifolds, the
first map may be neither injective nor surjective. For example, if C is a genus one Riemann
surface then there exist complex line bundles on C × C which does not have a holomorphic
structure (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 of
genus g ≥ 1 splits as a direct sum of line bundles. The earlier proof (proposition 3.5) which
worked for complex vector bundles breaks down in the case of holomorphic vector bundles
because the partition of unity and subsequently the inner product are not holomorphic (they
are only smooth).
3.8. Sections of Holomorphic Vector Bundles. Let π : V → M be a holomorphic
vector 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 bundle
then 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 as
a (topological) complex line bundle then Γ(C, I1top ) is the space of all C ∞ maps C → C and
hence has infinite dimension. On the other hand, if we view I1 as a holomorphic line bundle
then Γ(C, I1hol ) is the space of holomorphic maps C → C. Since C is compact we know any
such holomorphic map must be constant. Thus Γ(C, I1hol ) ∼
= 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 z
and w. Over U0 the bundle L2 is trivial so a holomorphic section is the same as a holomorphic
map C → C given by z 7→ f (z). Similarly, over U1 a section is the same as a holomorphic
map w 7→ g(w). We need these two local sections to agree on the overlap U0 ∩ U1 ∼
= C∗ . We
20
know z = 1/w and the glueing of L2 over U0 ∩ U1 is done by 1/z 2 . Thus, in the trivialization
over U1 , the section f (z) is z12 f (z) = w2 f (1/w). This must equal the holomorphic function
g(w) over U0 ∩ U1 . Whence Γ(P1 , L2 ) is isomorphic to the space of holomorphic functions
f (z) such that w2 f (1/w) is also a holomorphic function. These functions are precisely the
polynomials of degree at most two (i.e. of the form a+bz +cz 2 ). Whence dim(Γ(P1 , L2 )) = 3.
Exercise 31. Check that Γ(P1 , Ln ) corresponds to vector space of polynomials a0 + a1 z +
· · · + an z 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) < 0
then 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 L
vanishes. 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
a a
only if det x y > 0. But since f is holomorphic we know ax = by and ay = −bx so that
bx by
the determinant is a2x + b2x > 0.
Exercise 32. In the proof of this proposition we saw that if a holomorphic section of L
vanishes at n points then deg(L) = n. Use this to show that deg(Ln ) = n (hint: use the
earlier 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 ρU V : F(U ) → F(V ) which satisfy ρU U = id and
ρU W = ρV W ◦ ρU V whenever W ⊂ V ⊂ U .
Usually we will write the restriction ρU V (s) of s ∈ F(U ) to F(V ) as s|V . A sheaf is a presheaf
with the following added condition. Let {Ui }i∈I be an open cover of U and suppose that
si ∈ F(Ui ) satisfy si |Ui ∩Uj = sj |Ui ∩Uj for 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 good
exercise to check these are indeed sheaves:
Example. Locally constant sheaf Z is defined by assigning Z to each connected component
of U . In other words, Z(U ) is isomorphic to Z⊕c where c is the number of connected
components 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).
hol
Example. Let X be a complex manifold. The sheaf OX
= OX associates to an open U
the group of holomorphic functions f : U → C. Notice that if X is compact this means
∗
by associating to U the group of holomorphic
OX (X) ∼
= C. Similarly one can define OX
21
maps f : U → C∗ .
Example. If E is a holomorphic vector bundle then one can define the associated sheaf
sh(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 between
viewing 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 (by
definition) if E1 and E2 are vector bundles on X then sh(E1 ⊕ E2 ) = sh(E1 ) ⊕ sh(E2 ). The
same 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 )).
C if p ∈ U
Example. Let p ∈ X be a point. The skyscraper sheaf Cp is defined by
0 if p 6∈ U
Example. Let C be a Riemann surface. Then OC (−p) associates to U the group of
holomorphic maps U → C which vanish at p. If C is compact then OC (−p)(C) = 0. Similarly
one defines OC (p) by associating to U the group of holomorphic maps U → C which have at
most a pole of order one at p. This definition generalizes to give OC (kp) for any k ∈ Z. For
2
2
instance, suppose C = P1 and p = 0 ∈ P1 . Then Γ(P1 , O(2p)) = { ax +bxy+cy
: a, b, c ∈ C}
x2
1
and in general dim(Γ(P , 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 to
U the group of holomorphic sections of L|U which vanish at p. Notice that if p 6∈ U then
L(−p)(U ) = L(U ).
C if p ∈ U, U 6= X
. Then F is a presheaf but
Example. Consider F given by F(U ) =
0 otherwise
not a sheaf. However, if you remove the condition that F(U ) = C only if U 6= X then you
get the skyscraper sheaf Cp which is a sheaf. We mention the following fact without proof:
Fact: There exists a unique way to turn a presheaf into a sheaf. The process is called
sheafification. Recall that a presheaf is not a sheaf if there is no unique lift s satisfying
s|Ui = si whenever si |Ui ∩Uj = sj |Ui ∩Uj . Sheafification essentially throws in such s whenever
necessary 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 of
F. 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
/
F2 (U )
ρU V
F1 (V )
ρU V
φV
/
F2 (V )
Example. If f : X → R is a C ∞ map then we have a morphism C∞ (R) → C∞ (R) given by
g 7→ f g.
22
Example. Let C be a Riemann surface and f : C → C a meromorphic map with a pole at
p ∈ C of order k ∈ Z+ . Then we get a map φf : OC (−kp) → OC given by g 7→ f g. More
generally, 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 group
kerφU : F1 (U ) → F2 (U ). On the other hand, U 7→ imφU : F1 (U ) → F2 (U ) only defines a
presheaf. The sheaf im(φ) is defined to be the sheafification of this presheaf. As before, we
φi−1
φi
say that a sequence of sheaves · · · Fi−1 −−→ Fi −
→ Fi+1 → · · · is exact if ker(φi ) = im(φi−1 )
for all i.
Example. Fix a point p ∈ P1 . The sequence
is exact. Similarly,
0 → OP1 (−p) → OP1 → Cp → 0
0 → OP1 (−2p) → OP1 (−p) → Cp → 0
is exact. On the other hand, even though the map OP1 (−p) → Cp is surjective the map on
global 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 precise
phenomenon that allows us to develop the theory of sheaf cohomology.
4.1. Cech Cohomology. Cech cohomology is a cohomology theory for sheaves. Chapter
III section 4 of [Ha] contains a short note on the subject. The rest of chapter III of [Ha]
as well as [I] provide an introduction to the derived functor approach to sheaf cohomology.
Although we will not talk about derived functors this is a very important concept and a
viable subject for a project.
Fix a complex manifold X and a sheaf F on X. The Cech complex
d
d
d
0
1
2
C · (F, Uα ) = C 0 −
→
C1 −
→
C2 −
→
···
Q
with respect to a cover {Uα }α∈I of X is defined as follows. We let C 0Q
= α∈I F(Uα ) so
that an element of C 0 has the form {fα } with fα ∈ F(Uα ). Next C 1 = α,β∈I F(Uα ∩ Uβ )
Q
where an element looks like {fα,β }. Similarly C 2 = α,β,γ∈I F(Uα ∩ Uβ ∩ Uγ ) and so on. The
differentials are:
• d0 : C 0 → C 1 given by {fα } 7→ {fα |Uα ∩Uβ − fβ |Uα ∩Uβ ∈ F(Uα ∩ Uβ )}
• d1 : C 1 → C 2 given by {fα,β } 7→ {fβ,γ − fα,γ + fα,β ∈ F(Uα ∩ Uβ ∩ Uγ )}
• dn : C n → C n+1 is defined similarly as an alternating sum.
Thanks to the alternating signs di+1 ◦ di = 0 so that we get a complex. The cohomology
of this complex gives us the Cech cohomology groups Ȟ n (X, F) = ker(dn )/im(dn−1 ). The
last 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 fine
enough. 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
Ȟ n (Uα1 ∩ · · · ∩ Uαi , F) = 0 = Ȟ n (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 U
the higher Cech cohomology is zero.
Thus, in order to calculate Cech cohomology, it is enough to choose an open cover of X
where all the open sets Uα1 ∩ · · · ∩ Uαi are convex domains.
Exercise 33. Show that Ȟ 0 (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 Ȟ 1 (X, OX
) does have a nice interpretation – it is the group
parametrizing holomorphic line bundles on X. But first, let’s calculate the Cech cohomology
of some simple spaces and sheaves.
Example. Let’s compute the Cech cohomology of the constant sheaf Z on S 1 . We cover
S 1 using arcs U1 , U2 , U3 which overlap pairwise but where U1 ∩ U2 ∩ U3 = ∅. Since the arcs as
well as their intersections are convex domains this cover is good enough to use for computing
cohomology. Now C 0 = Z3 = (a, b, c) and C 1 = Z9 where the differential d0 : C 0 → C 1 is
given by
d(a, b, c) = (0, a − b, a − c, b − a, 0, b − c, c − a, c − b, 0).
This means that Ȟ 0 (X, Z) = h(a, a, a)i ∼
= Z (as expected since Γ(S 1 , Z) = Z). The kernel of
d1 contains elements of the form (0, x, y, −x, 0, z, −y, −z, 0) ∈ C 1 and thus Ȟ 1 (S 1 , Z) ∼
= Z.
i
1
Similarly one can check that Ȟ (S , Z) = 0 for i > 1.
Exercise 34. Calculate Ȟ i (S 1 , Z) using a cover of S 1 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 Ȟ i (C∗ , Z) ∼
= Ȟ i (S 1 , 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 exists
some 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 −0
extends to a holomorphic function on C2 .
Theorem 4.4. If U is a domain of holomorphy then for any holomorphic vector bundle V
on U the higher Cech cohomology is zero.
Thus, in order to calculate Cech cohomology of holomorphic vector bundles, it is enough
to choose an open cover of X where all the open sets are domains of holomorphy.
Aside: This whole issue of choosing an open cover might seem a little strange. The right
way to think about it is that you want to use open sets which have zero higher cohomology.
This approach suffers from the chicken and egg problem in that we cannot define Cech
cohomology until we choose a cover and yet we cannot choose a cover until we know its
cohomology. 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 most
useful method to use in order to decide on a cover.
Example. Consider C = P1 with the trivial holomorphic line bundle. Recall that the
corresponding 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 Ȟ 0 (C, OC ) = C. To compute Ȟ 1 (C, OC )
we need to understand holomorphic maps U0 ∩ U1 = C∗ → C. Any such map looks like
z 7→ · · · + a−n z −n + · · · + a−1 z −1 + a0 + a1 z 1 + · · · + an z n + . . .
and hence can be written as f − g where f = a0 + a1 z 1 + . . . is holomorphic on U0 and
−g = a−1 z −1 + a−2 z −2 + . . . is holomorphic on U1 . This means that Ȟ 1 (P1 , OP1 ) = 0. The
higher 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:


Z i=0


 C i=0
 2g
Z
i=1
i
i
Cg i = 1
Ȟ (C, OC ) =
Ȟ (C, Z) =
Z
i
=
2



0 i≥2
 0 i≥3
Proposition 4.5. We have the following vanishing results for cohomology:
• if X is a topological manifold of real dimension n then Ȟ i (X, Z) = 0 for i > n
• (Grothendieck) if X is a complex manifold of complex dimension n and V any holomorphic vector bundle on X then Ȟ i (X, V ) = 0 for i > n.
4.1.1. Fine Sheaves. Fine sheaves are a class of sheaves which have the useful property that
they 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 open
set U containing x one can find an open x ∈ V ⊂ U such that F(V ) 6= 0. Similarly, the
support supp(φ) of a morphism of sheaves φ : F1 → F2 consists of points x ∈ X such
that for any open set U containing x one can find an open subset x ∈ V ⊂ U such that
φV : F1 (V ) → F2 (V ) is nonzero.
Example. The skyscraper sheaf Cp for p ∈ X has support supp(Cp ) = {p}. On the other
hand, supp(OX ) = X.
Definition 4.6. A sheaf F over X is fine if for every locally finite cover {Uα } of X by open
sets there exist morphisms φα : F → F such that
• supp(φ
α ) ⊂ Uα
P
• α φα = id
Example. The main examples of fine sheaves for us are C ∞ (R) and C ∞ (C). To define
φα one takes a partition of unity {ψα } with respect to the open cover {Uα }. Then φα is
multiplication by ψα (i.e. takes a function f to ψα f ).
Example. Z, OX and C ∞ (C∗ ) are not (generally) fine sheaves. For instance, if X is
a compact complex manifold then the only morphisms OX → OX is multiplication by a
constant. All of these have support all of X so if we pick any nontrivial cover {Uα } of X we
cannot find the required morphisms φα with support in Uα .
Theorem 4.7. If F is a fine sheaf on X then Ȟ 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 withPrespect to {Uα }. Since F is fine we can find
φα : F → F such that supp(φα ) ⊂ Uα and α φα = id. Since each φα is a sheaf morphism
it
maps of C · giving us a complex φα (C · ) for each α. Since
Pcommutes with the differential
i
·
i
·
α φα = id we have that H (C ) = ⊕α H (φα (C )). In other words, the morphisms φα break
up the cohomology H i (C · ) = Ȟ 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 Ȟ i (X, φα (F)) =
Ȟ i (Uα , φα (F)|Uα ). Since we chose the Uα ’s small enough these groups are zero for i > 0
which means that Ȟ i (X, F) = ⊕α H i (φα (F)) = 0 for i > 0.
4.1.2. The Exponential Sequence. A map of complexes E · → F · consists of maps fi : E i → F i
such that d◦fi = fi+1 ◦d (i.e. the maps fi commute with the differential). A sequence of maps
0 → E · → F · → G· → 0 is short exact if for each i the sequence 0 → E i → F i → Gi → 0 is
exact.
Lemma 4.8 (snake lemma). A short exact sequence of complexes 0 → E · → F · → G· → 0
leads 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 onto
we can find an element u′ ∈ F i which maps onto u. Now consider v = d(u′ ). Since du = 0
the image of v under the map F i+1 → Gi+1 is zero (all squares are commuting). Thus there
must be some v ′ ∈ E i+1 which maps to v ∈ F i+1 . We map u to v ′ . It now remains to check
that this map is well defined (i.e. that 0 ∈ Gi maps to zero) and that the rest of the maps
are 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 topological
space X then the following sequence is exact
· · · → Ȟ i (X, E) → Ȟ i (X, F) → Ȟ i (X, G) → Ȟ i+1 (X, E) → . . .
Proof. Choose an open cover of X which we can use to compute the Cech cohomology groups
of 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 exact
sequence of cohomology groups.
Corollary 4.10. If 0 → E → F P
→ G → 0 is a short exact sequence of sheaves then χ(E) −
χ(F) + χ(G) = 0 where χ(A) = i (−1)i rank(Ȟ i (X, A)) denotes the Euler characteristic of
the sheaf A.
Proof. A general fact from algebra
if 0 → A1 → · · · → An → 0 is an exact
P says that
i
(−1)
rank(A
sequence of abelian groups then
i ) = 0. The result follows since by the
i
previous corollary
· · · → Ȟ i (X, E) → Ȟ i (X, F) → Ȟ i (X, G) → Ȟ i+1 (X, E) → . . .
is an exact sequence.
26
Example. Fix a complex manifold X and consider the exponential sequence
exp
∗
0 → Z → OX −−→ OX
→0
The first map takes a ∈ Z to the constant map in OX . The second map is the exponential
f 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 map
f : U → C∗ we can take the logarithm to obtain a holomorphic map log f : U → C. This
∗
shows exp : OX → OX
is onto. Finally, we have exactness in the middle since the elements of
∗
OX which are mapped by exp to one (the identity in OX
) are precisely the integers (because
exp(2πit) = 1 exactly when t ∈ Z).
∗
Since Ȟ 0 (X, Z) = Z, Ȟ 0 (X, OX ) = C and Ȟ 0 (X, OX
) = C∗ we get a long exact sequence
of abelian groups:
exp
exp
∗
0 → Z → C −−→ C∗ → Ȟ 1 (X, Z) → Ȟ 1 (X, OX ) −−→ Ȟ 1 (OX
) → Ȟ 2 (X, Z) → . . .
exp
exp
Since the map C −−→ C∗ is onto the sequence 0 → Z → C −−→ C∗ → 0 is in fact exact. So
the sequence above splits to give us the long exact sequence:
exp
∗
0 → Ȟ 1 (X, Z) → Ȟ 1 (X, OX ) −−→ Ȟ 1 (OX
) → Ȟ 2 (X, Z) → Ȟ 2 (X, OX ) → . . .
If X is a Riemann surface of genus g we saw that Ȟ 1 (X, Z) = Z2g , Ȟ 1 (X, OX ) = Cg ,
Ȟ 2 (X, Z) = Z and Ȟ 2 (X, OX ) = 0. Thus we get the long exact sequence:
∗
0 → Z2g → Cg → Ȟ 1 (X, OX
)→Z→0
Example. Similarly we can consider the exponential sequence
exp
0 → Z → C ∞ (C) −−→ C ∞ (C∗ ) → 0
Since C ∞ (C) is a fine sheaf (see section 4.1.1) we know Ȟ i (X, C ∞ (C)) = 0 for i > 0. Then
the corresponding long exact sequence implies Ȟ 1 (X, C ∞ (C∗ )) ∼
= Ȟ 2 (X, Z). So if X is a
Riemann surface we get Ȟ 1 (X, C ∞ (C∗ )) ∼
= Z.
4.2. Line Bundles and Cech Cohomology.
∗
Theorem 4.12. If X is a complex manifold then Ȟ 1 (X, OX
) parametrizes holomorphic line
bundles on X.
d
d
0
1
Proof. Pick a cover {Uα }nof X and denote by C 0 −
→
C1 −
→
C 2 → . . . the Cech complex
o
Q
∗
−1
of OX
. Then ker(d1 ) =
where fαβ : Uα ∩ Uβ → C∗ . Also
α,β fαβ : fαβ · fαγ · fβγ = 1
o
nQ
−1
for some gα : Uα → C∗ .
im(d0 ) =
α,β fαβ : fαβ = gα gβ
Choose {Uα } fine enough so that any line bundle is trivial over any Uα1 ∩ · · · ∩ Uαn . If
L is a line bundle choose a trivialization tα : L|Uα → C × Uα for each α. Denote the
glueing functions of L with respect to this trivialization by fαβ . Then the glueing condition
implies fαβ · fβγ = fαγ which means {fαβ } ∈ ker(d1 ). However, this element depends on the
trivialization 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 fαβ
= gα−1 fαβ gβ which differ from the old
glueing functions by an element of im(d0 ).
∗
Thus, given a line bundle L we get a well defined element of ker(d1 )/im(d0 ) = Ȟ 1 (X, OX
).
1
∗
Conversely, given an element of Ȟ (X, OX ) 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 surface
C of genus g is isomorphic to an n-dimensional torus Cg /Λ where Λ ⊂ Cg is an integral
lattice of rank 2g.
Proof. Let’s go back to the short exact sequence:
0 → Z2g → Cg → Ȟ 1 (C, OC∗ ) → 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 Ȟ 1 (C, OC∗ ) → 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 and
not in terms of counting zeros of a general section (as we did in section 3.4).
∗
The total space Ȟ 1 (X, OX
) of holomorphic line bundles on X is called the Picard group
and denoted P ic(X). The group operation is the tensor ⊗ and the inverse of a line bundle L
is the dual line bundle L∗ (recall that L ⊗ L∗ ∼
= I1 ). P icd (X) ⊂ P ic(X) denotes the subspace
of line bundles of degree d. P ic0 (X) is called the Jacobian and denoted J(X). Notice that
P icd (X) ⊗ P ice (X) → P icd+e (X) so that J(X) is a subgroup of P ic(X).
If X = C is a Riemann surface of genus g then the corollary above implies P ic(C) ∼
=
J(C) × Z and also that J(C) is a compact complex manifold Cg /Λ of dimension g. As the
complex structure of C varies, the lattice Λ ∈ Cg moves and we get different tori. Even
though these tori are topologically the same (they are homeomorphic to (S 1 )2g ) as complex
manifolds they are different. A classical theorem of Torelli (essentially) says that one can
recover the Riemann surface C from its Jacobian J(C).
∗
The complex and real topological analogues of the sheaf OX
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 Ȟ 1 (X, C ∞ (C∗ )) parametrizes complex
(topological) line bundles and Ȟ 1 (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. Ȟ 1 (X, C ∞ (C∗ )) ∼
= Ȟ 2 (X, Z) and Ȟ 1 (X, C ∞ (R∗ )) ∼
= Ȟ 1 (X, Z/2).
exp
Proof. The exponential sequence for C ∞ (C∗ ) is 0 → Z → C ∞ (C) −−→ C ∞ (C∗ ) → 0 leading
to the long exact sequence
deg
· · · → Ȟ 1 (X, C ∞ (C)) → Ȟ 1 (X, C ∞ (C∗ )) −−→ Ȟ 2 (X, Z) → Ȟ 2 (X, C ∞ (C)) → . . .
Since C ∞ (C) is a fine sheaf we have Ȟ i (X, C ∞ (C)) = 0 for i > 0 implying that the map
deg
Ȟ 1 (X, C ∞ (C∗ )) −−→ Ȟ 2 (X, Z) is an isomorphism.
28
exp
The exponential map for C ∞ (R) is 0 → 0 → C ∞ (R) −−→ C ∞ (R>0 ) → 0. This shows that
∞
C (R) ∼
= C ∞ (R>0 ). On the other hand, we also have the short exact sequence
m
0 → Z/2 → C ∞ (R∗ ) −
→ C ∞ (R>0 ) → 0
where m(f ) = f 2 . Since C ∞ (R>0 ) ∼
= C ∞ (R) is fine the higher cohomology vanishes and the
long exact sequence becomes
m
0 → Z/2 → Γ(C ∞ (R∗ )) −
→ Γ(C ∞ (R>0 )) → Ȟ 1 (X, Z/2) → Ȟ 1 (X, C ∞ (R∗ )) → 0
√
But for any map g : X → R>0 we have a well defined square root g. This means
m
Γ(C ∞ (R∗ )) −
→ Γ(C ∞ (R>0 )) is surjective and hence Ȟ 1 (X, Z/2) ∼
= Ȟ 1 (X, C ∞ (R∗ )).
Corollary 4.16. If C is a Riemann surface of genus g then complex and real line bundles
are parametrized via the degree map by Z and (Z/2)2g respectively.
Proof. This follows immediately from the previous proposition and the fact that Ȟ 2 (C, Z) =
Z and Ȟ 1 (C, Z/2) = (Z/2)2g .
Exercise 36. Show that Ȟ 1 (S 1 , C ∞ (C∗ )) = 0 and conclude that all complex line bundles
on S 1 are trivial.
Exercise 37. Show that up to isomorphism there are exactly two real line bundles on
1
S , namely the trivial and the Möbius line bundle.
4.3. Riemann-Roch and Serre Duality. Let L be a holomorphic line bundle on a Rie0
mann surface C of genus g. What is the dimension of the space of global sections HP
(C, L) =
Γ(L)? To answer this question we first calculate the Euler characteristic χ(C, L) = i (−1)i hi (L)
where hi (L) = rankH i (C, L). Since dimC (C) = 1 Grothendieck’s theorem (4.5) tells us
hi (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 some
n ∈ Z (later we will show this sheaf corresponds to a line bundle of degree n). To calculate
χ(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 on
a point. Since a point is a zero dimensional complex manifold Grothendieck’s theorem
tells 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 get
χ(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 surface
C of genus g then χ(C, V ) = deg(V ) + (1 − g)rank(V ). In particular, if L is a line bundle of
degree d then χ(C, L) = d − g + 1.
We will give a proof in the next section. What is remarkable about this result is that
χ(C, L) only depends on the topological isomorphism class of L. For example, if C is a curve
1 if p = q
Thus h0
of genus g > 0 and p, q ∈ C are two points then h0 (OC (p − q)) =
0 if p 6= q
29
(and consequently also h1 ) depends not only on the topological type (deg(OC (p − q)) = 0
for any p, q ∈ C) but also on the holomorphic type. On the other hand, the difference
χ(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 problem
has 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 the
cotangent bundle of C (this is also known as the canonical line bundle). Then H 1 (C, V ) =
H 0 (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 ≥ 0
then 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 ) = 0
and h1 (Ln ) = −n − 1.
Example. If C is a Riemann surface of genus g then h1 (KC ) = h0 (KC∗ ⊗ KC ) = h0 (OC ) =
1. But χ(KC ) = (2g − 2) − g + 1 = g − 1 which means h0 (KC ) = g. Note that showing
directly from definition that the canonical line bundle has a g dimensional space of sections
is 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 degree
zero then h0 (L) = 0 unless L ∼
= OC ).
Example. Suppose L is a line bundle on a Riemann surface C of genus g such that
deg(L) > 2g − 2. Then deg(L∗ ⊗ KC ) < 2 − 2g + 2g − 2 = 0 so that h1 (L) = 0 and
h0 (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 if
deg(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 that
H 0 (V ) is finite dimensional. Then we can consider among all sections of V the one which
vanishes to highest possible order d. But then H 0 (V ⊗OC (−np)) = 0 for any n > d and point
p ∈ 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 meromorphic
section. Although rather believable this is not a trivial statement to prove and requires some
analysis. It would seem like we gave a proof above but in the process we used Riemann-Roch
whose proof will turn out to rely on this proposition. In fact, if one were to trace through the
arguments carefully, one would find that proving the existence of meromorphic sections of
line 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 bundle
V 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 this
holds 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 an
OX (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 commutes 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 contains
too many morphisms. So from now on we will work only in the category of OX -modules
where the objects are OX -modules and the morphisms are OX -module maps between them.
Consequently, a few of our old 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 defined
by F ∗ (U ) = HomOX (F(U ), OX (U )) where we consider just OX -module morphisms rather
than all sheaf morphisms. Also, the tensor product of F1 and F2 is F1 ⊗OX F2 rather than
F1 ⊗Z F2 .
Lemma 4.22. If F is a sheaf of OX -modules then HomOX (OX , F) = H 0 (X, F) = Γ(F).
More generally, if L is a line bundle then HomOX (L, F) = H 0 (X, F ⊗OX L∗ ) = Γ(F ⊗OX L∗ ).
Proof. Suppose the constant section 1 ∈ Γ(OX ) maps to s ∈ Γ(F). If a ∈ OX (U ) then since
the 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 ⊗OX L∗ ∼
= OX →
∗
F ⊗OX L (and vice versa). Then by the previous argument such maps correspond to elements
of Γ(F ⊗OX L∗ ).
Recall that an R-module M is free of rank n if M ∼
= R⊕n . Since any rank n vector bundle
⊕n
V is locally trivial we can find open sets U where V |U ∼
. So over such sets V is
= OX
isomorphic 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 contained
in 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 free
sheaf of rank n and conversely. Moreover, any morphism of vector bundles is a morphism
of OX -modules (althought the converse is not true).
Proof. We saw already that if V is a vector bundle of rank n then the corresponding sheaf is
locally free of rank n. Conversely, suppose V is a locally free sheaf of rank n. For simplicity
of notation let’s assume n = 1. By definition we have open sets Uα and isomorphisms
fα : V |Uα → OX (Uα ). So over Uα ∩ Uβ we can consider fα ◦ fβ−1 to get a map OX (Uα ∩ Uβ ) →
OX (Uα ∩ Uβ ). Since this map is a morphism of OX -modules, following the lemma above we
just need to give the map 1 7→ a ∈ OX (Uα ∩ Uβ ). Since this morphism is invertible we must
31
∗
have a ∈ OX
(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 linear
and a linear map Cm → Cn commutes with the C action. The reason why the converse is
not true is that the fiberwise map may drop in rank. For example, consider the map between
the trivial line bundles over C given by multiplication by z. This map is an isomorphism of
fibers over z 6= 0 but drops rank at z = 0. The quotient is then the skyscraper sheaf C0 at
z = 0 which is not a vector bundle. Hence we do not allow this in the case of vector bundles
but do allow this map when working with sheaves because the skyscraper sheaf is a perfectly
good sheaf (even though it’s not locally free).
Exercise 40. Show that the tensor product F1 ⊗OX F2 of two locally free sheaves F1 and
F2 of ranks n1 and n2 is a locally free sheaf of rank n1 n2 (hint: this is a local question so
you 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. line
bundles) in terms of divisors.
Definition P
4.25. A divisor D on a Riemann surface C is a (formal) finite Z-linear
sum of
P
n
points D = i=1 ai pi 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 taking
OC (kp)(U ) to be sections of U which vanish to order k at p (if k ≤ 0) or have poles of order
at most
P k at p (if k > 0). We extend this definition to give us sheaves OC (D) for any divisor
D = i ai pi .
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→ f1 f2 is an isomorphism) and conclude that
OC (D)∗ ∼
= OC (−D).
Lemma 4.26. Let D be a divisor on a Riemann surface C. The sheaf OC (D) is a locally
free sheaf of rank one and degree deg(D).
P
Proof. Since OC ( i ai pi ) ∼
= ⊗i OC (ai pi ) it is enough to prove the assertion for D = np where
n ∈ Z and p ∈ C. Away from p this sheaf is isomorphic to OC so we just need to prove that
in a small enough neighbourhood U around p the sheaf is isomorphic to OC (U ). If z is a
local 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 that
n ≥ 0. Thus we have the global section 1 ∈ Γ(OC (np)). Let U be a small neighbourhood of
p. Then OC (np) is trivialized over U by g 7→ (z − p)n · g. So the section 1 around p locally
looks like (z − p)n . Away from p the line bundle OC (np) is trivial so the section 1 does not
intersect the zero section. Hence 1 vanishes only at p where it vanishes to order n. Thus, by
our 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 some
divisor D.
32
P
Proof. Suppose L has a global section s which vanishes along some divisor D = i ai pi where
ai > 0 (i.e. it vanishes to order ai at pi ). Then this gives a section of L ⊗OC OC (−D) which
does not vanish. Thus L ⊗OC OC (−D) ∼
= OC (D). More generally, if
= OC which means L ∼
L has no section then by proposition 4.19 we know that L ⊗OC OC (np) has a section and so
L ⊗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 of
principal 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 a
principal 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 for
each d ∈ Z (up to isomorphism). So if p, q ∈ P1 then we must have OP1 (dp) ∼
= OP1 (dq) for
every d ∈ Z. This means that dp ≡ dq so by the theorem above there must be some map
f : P1 → P1 such that f −1 (0) − f −1 (∞) = dp − dq. If we take p = 0 and q = ∞ then we can
write down this map explicitly as z 7→ z d .
Example. If C is a Riemann surface of genus g > 0 then OC (p) 6∼
= OC (q) for any
p 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 appreciate
the lemma recall that if E ֒→ F is an inclusion of locally free sheaves then the quotient need
not 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 fiber
∼
over z is multiplication by z − p (because the isomorphism OC (p)(U ) −
→ OC (U ) is given by
g 7→ (z − p)g). Thus at p the rank of the map drops from one to zero and that is why at p
we get a cokernel which shows up as the skyscraper sheaf Cp .
In our case we have a vector bundle V and we want to find a line subbundle L ֒→ V so
that 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 sections
s ∈ Γ(V ⊗OC L∗ ). As in the example above, the quotient sheaf will be locally free if and only
if 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 of
highest possible degree such that there exists a non-zero map L → V then quotient sheaf is
locally free.
Proof. A map L → V corresponds to a section in s ∈ H 0 (C, V ⊗OC L∗ ). If deg(L) is large
enough then there are no global sections and if deg(L) is really negative then there must be
33
some global sections (4.19). ThusPsuch an L of highest degree exists. Suppose that s vanishes
along some nonzero divisor D = i ai pi with ai ∈ Z>0 . Then H 0 (C, V ⊗OC L∗ ⊗OC OC (−D))
contains a section which means there is a map L ⊗OC OC (D) → V . This is a contradiction
since deg(L ⊗OC OC (D)) > deg(L).
Theorem 4.30 (Riemann-Roch). If V is a holomorphic vector bundle on a Riemann surface
C 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 short
exact 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 RiemannRoch 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 remainsPto prove Riemann-Roch for line bundles
P L. We know L = OC (D) for some
divisor D = i ai pi so we proceed by induction on i |ai |. Suppose without loss of generality
that p1 > 0 so that have the short exact sequence 0 → OC (D − p1 ) → OC (D) → Cp . By
induction 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 which
follows 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 ∈ Z
there 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 uniquely
isomorphic to a direct sum of line bundles ⊕ni=1 OP1 (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 line
bundles. Using lemma (4.29) we take a line bundle L = OP1 (d) of highest degree so that we
have a short exact sequence 0 → L → V → V ′ → 0 where V ′ is also a vector bundle. By
n−1
OP1 (di ).
induction we can assume V ′ splits as ⊕i=1
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) →
n−1
OP1 (di − d − 1) → 0. Since H 0 (OP1 (−1)) = H 1 (OP1 (−1)) = 0
V ⊗OP1 OP1 (−d − 1) → ⊕i=1
∼
if we look at the long exact sequence in cohomology we get H 0 (V ⊗OP1 OP1 (−d − 1)) −
→
n−1
OP1 (di − d − 1)) which is nonzero since di − d − 1 ≥ 0. But this means V has a
H 0 (⊕i=1
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
H 1 (G ∗ ⊗OP1 E) = H 1 (⊕i OP1 (d−di )) = H 0 (⊕i OP1 (d−di )∗ ⊗OP1 KP1 ) = H 0 (⊕i OP1 (di −d−2)) = 0
since KP1 = OP1 (−2) and di − d − 2 < 0. Thus the sequence splits and hence V splits as the
direct sum of line bundles.
Finally, suppose ⊕i OP1 (ai ) ∼
= ⊕i OP1 (bi ) for some ai , bi ∈ Z with a1 ≤ · · · ≤ an and
b1 ≤ · · · ≤ bn . If an < bn then tensoring both sides by OP1 (−bn ) we get ⊕i OP1 (ai − bn ) ∼
=
⊕i OP1 (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, as
before, that an−1 = bn−1 . Continuing this way we show ai = bi for all i thus completing the
proof.
i
p
Proposition 5.2. If 0 → E −
→F −
→ G → 0 is a short exact sequence of vector bundles on a
complex manifold X and H 1 (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
p
can be done if the following map induced by p is surjective: HomOX (G, F) −
→ HomOX (G, G)
since we can take as g any preimage of idG ∈ HomOX (G, G). But
HomOX (G, F) = HomOX (OX , G ∗ ⊗OX F) = Γ(G ∗ ⊗OX F).
Similarly, HomOX (G, G) = Γ(G ∗ ⊗OX G). So we need to show that Γ(G ∗ ⊗OX F) → 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 ∗ ⊗OX E) → Γ(G ∗ ⊗OX F) → Γ(G ∗ ⊗OX G) → H 1 (G ∗ ⊗OX E) → · · ·
Since H 1 (G ∗ ⊗OX E) = 0 this implies Γ(G ∗ ⊗OX F) → Γ(G ∗ ⊗OX G) is surjective and the
result follows.
Lemma 5.3. If 0 → E → F → G → 0 is a short exact sequence on a complex manifold X
and V a vector bundle then 0 → E ⊗OX V → F ⊗OX V → 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. Suppose
V has rank one. We can pick such an open cover where V |Uα is trivial and hence isomorphic
to OX (Uα ). Then over Uα we have E(Uα ) ⊗OX (Uα ) V (Uα ) ∼
= E(Uα ) and similarly with F
and G so the sequence remains exact. The same thing happens if V has rank greater than
one.
Of course, in general a short exact sequence does not necessarily remain exact after ten×2
soring. For example, consider the sequence of abelian groups 0 → Z −→ Z → Z/2 → 0. If
0
we tensor (over Z) with Z/2 then we get 0 → Z/2 −
→ 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 Riemann
surface of genus one. First we take a closer look at line bundles on C. We know that the
space J(C) of degree zero holomorphic line bundles on C is a torus of dimension one. This
means that J(C) is also a genus one Riemann surface.
Proposition 5.4. If C is a genus one Riemann surface then J(C) ∼
= P icd (C) for any
=C∼
d ∈ Z.
Proof. Recall that P icd denotes the space of line bundles on C of degree d. If E is a line
bundle of degree d then we have an isomorphism J(C) → P icd (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 ∗ because
E ⊗OC E ∗ ∼
= OC . Thus all P icd (C) are isomorphic (as d varies).
35
A proof that J(C) ∼
= C can be obtained directly from the description of J(C) as the
1
quotient of H (C, OC ) ∼
= C by a lattice Λ. However, we give a more direct proof by showing
that the map C → P ic1 (C) given by p 7→ OC (p) is an isomorphism. To see that the map
is injective we need to show OC (p) 6∼
= OC (q) unless p = q. This we saw to hold true for
any 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) for
some point p ∈ C. By Riemann-Roch, if deg(L) = 1 then h0 (L) = 1 since by Serre duality
h1 (L) = h0 (L∗ ) = 0 (KC ∼
= OC because C has genus one). So L has a global section which
vanishes at exactly one point p (since deg(L) = 1). Consequently, as we saw in the proof of
theorem (4.27) this means L ∼
= OC (p).
Aside: One can check that the group structure on J(C) coming from ⊗ is the same as
the natural group structure on C coming from the fact that C ∼
= C/Λ.
For C a Riemann surface of arbitrary genus, the map C → P ic1 (C) used in the proof
above 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-Jacobi
map C → P ic1 (C) given by p 7→ OC (p) is injective. Moreover, if C has genus one then it is
an isomorphism.
Definition 5.6. A vector bundle V is called indecomposable if it cannot be written as a
direct sum V ′ ⊕V ′′ of nonzero subbundles. V is irreducible if it does not contain any nonzero
subbundles 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 ′′ . Of
course, in the case of complex (or real) vector bundles the two notions coincide (proposition
3.5).
Denote by Bun(r, d) the space of indecomposable vector bundles of rank r and degree d on
C. 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 degree
2
h 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 reference
is 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.
36