RIEMANN SURFACES: TALK V: DOLBEAULT COHOMOLOGY NICK MCCLEEREY 0. Complex Differential Forms Consider a complex manifold X n (of complex dimension n) 1, and consider its complexified tangent bundle T C X. We have seen that this splits as T C X = T X 1,0 T X 0,1, refered to respectively as the holomorphic and anti-holomorphic tangent bundles each is a complex vector bundle of complex dimension n, and the former is in fact a holomorphic vector bundle (the other is of course anti-holomorphic). In terms of local (real) coordinates {x 1, y 1,..., x n, y n }, (which we can reform as complex coordinates {z 1, z 1,..., z n, z n } by setting z j = x j + iy j and z j = x j iy j ) we have that the tangent vectors { } z,..., 1 z form a basis for T X 1,0 and n { z 1,..., } z form a basis for T X 0,1, where we define: n z j = 1 [ 2 x j i ] y j, z j = 1 2 [ x j + i y j This splitting also induces a splitting of the (complexified) cotangent bundle, TC X = T X 1,0 T X 0,1 in coordinates, we have that the differential forms {dz 1,..., dz n } are a basis for T X 1,0 and the forms {dz 1,..., dz n } are a basis for T X 0,1, where we define: dz j = dx j + idy j, dz j = dx j idy j. Observe that these are just the dual basis associated to our bases for T X 1,0 and T X 0,1, i.e. dz ( ) j z = δ j k k, ( ) ( ) dzj z = 0, dz j k z = δ j k k, and ( ) dzj z = 0. k As is standard, we are also interested in exterior powers of the cotangent bundle with itself, m TC X, 1 m 2n. Observe that the splitting of T C X also induces a splitting of m TC X, into forms of mixed type : m T C X = T X p,q, where T X p,q := ( p T X 1,0) ( q T X 1,0), and has sections that are locally of the form: dz j1... dz jp dz k1... dz kq, j 1,..., j p, k 1,..., k q {1,..., n} or more succinctly dz I dz J, where I and J are multi-indicies inside{1,..., n}. Such forms are said to be (p, q)-forms, or forms of bidegree (p, q). Note that they are also complex-valued forms of degree m specifying the bidegree just carries more information. Also observe that, as these are totally antisymmetric tensors, there are no forms of type (p, q) with either p or q larger than n in particular, all top forms are of pure type (n, n). 1 For the troubled, all of our manifolds will always be assumed connected and without boundary 1 ].
2 NICK MCCLEEREY We also see that the splitting of the cotangent bundle induces a splitting of the exterior derivative d in the following way: suppose f C (X), and consider the decomposition of the 1-form df into its holomorphic and antiholomorphic components df = π 1,0 df + π 0,1 df. It is rather simple to compute in coordinates that π 1,0 df = f z j dzj and π 0,1 df = f dz j dzj, where the summation convention is understood to be in effect. We are thus led to define two natural operators: and we see immediately that: := π 1,0 d, and := π 0,1 d, d = +. Moreover, from the fact that d 2 = 0, we see that: 2 = 2 = 0, and =. We can generalize these operators to forms of higher degree in the obvious manner if α m T C X is given locally as α = fdzi dz J, we define: α := f z j dzj dz I dz J and α := f z j dzj dz I dz J and extend by linearity to the rest of m TC X. One easily checks that the above properties of and still hold. As an aside, observe that a function f is holomorphic = 0 for all 1 j n. We can rephrase this more succinctly in terms of then f is holomorphic iff f = 0. Similarly, f is anti-holomorphic iff f = 0. We shall extend this to forms of higher degree as well, with the only wrinkle being we will want the form to be of appropriate type we define an m-form α to be a holomorphic m-form if α is in an (m, 0)-form and α = 0. Similarly, α is said to be anti-holomorphic m-form if α is a (0, m)-form such that α = 0. The typing is so the form behaves in all components in a manner similar to an (anti-) holomorphic function. Note that by dimension reasons, there are no holomorphic (or anti-holomorphic) forms of degree larger than n. Thus, the determinate bundle of the holomorphic cotangent bundle is T X n,0 this line bundle is called the canonical bundle, and is often denoted by K X instead of T X n,0 (n.b. T X 0,n is not the anti-canonical bundle the dual of K X is the anti-canonical bundle. It is isomophic to the determinate bundle of the holomorphic tangent bundle, though one often does not take this point of view). iff f z j 1. Dolbeault Cohomology Observe now that, in terms of the previous decompostion, sends T X p,q to T X p+1,q and sends T X p,q to T X p,q+1, and so in particular these operators respect the decomposition. Since and square to 0, this gives us a bi-graded complex, that is mostly symmetric in terms of the information it gives. We will thus more or less ignore and work exclusively with, a move motivated in part by the fact that we can extend to forms with coefficents in an arbitrary holomorphic vector
RIEMANN SURFACES: TALK V: DOLBEAULT COHOMOLOGY 3 bundle, while we cannot do the same with. We thus consider complexes of the form: 0 Γ ( T X p,0) Γ ( T X p,1)... Γ ( T X p,n 1) Γ (T X p,n ) 0 and the associated cohomology groups: H p,q ker (X, C) = Im. These are called the Doulbeault cohomology groups of X, and carry information not only about the topology of X, but also its complex structure. Note that H p,0 (X, C) is exactly the space of holomorphic p-forms. If one is familiar with sheaf cohomology, it is fairly easy to show that: Proposition 1.1. If X is a complex manifold and Ω p X is the sheaf of holomorphic p-forms on X, then H p,q (X, C) = H q (X, Ω p X ). We are interested in extending this construction to differential forms with coefficents in an arbitrary holomorphic vector bundle, i.e. twisted Dolbeault cohomology. First, let us review breifly twisted differential forms. If E is a complex vector bundle over X, we can consider the tensor product TC X E =: T E X. Bilinearity of the tensor product shows this splits as: T EX = T EX 1,0 T EX 0,1. Sections of this vector bundle are refered to as twisted differential forms or E- valued differnetial forms. Locally they are generated by products α s, where α is a differential form and s is a section of E. We would like to define a Doulbeault cohomology on these, so first we need to define a -type operator. The obvious approach is of course just to define (α s) = α s. The problem with this is that if E is not assumed to be holomorphic, for any smooth function u : U C, we would have to have: (α s) = (uα u 1 s) = uα u 1 s + u α u 1 s = (α s) + u α u 1 s. This is of course ludicrous the last term need not vanish as u is only assumed smooth (smoothness is to ensure that u 1 s is still (locally) a section of the smooth bundle E). If however, u where holomorphic, our operator would be well defined, as then u = 0. An approriate assumtion to guarantee this is to assume E to be a holomorphic vector bundle then in order for u 1 s to still be a (holomorphic) section of E we would have to have u holomorphic (as an aside, this shows why we choose to work with over. It also shows that in order to have a derhamtype cohomology theory for twisted forms, extra assumptions on E are needed in particular, one needs for E to be flat). Thus, we can define a -like operator, called E, on differential forms with values in a holomorphic vector bundle E and consider the resulting Doulbeault cohomology, as E obviously squares to 0: m T E X ( p = TEX p,q, TEX p,q := T E X 1,0) ( q T E X 0,1) 0 Γ ( TEX p,0) E ( Γ T E X p,1) E... E ( Γ T E X p,n 1) E Γ (T E X p,n ) 0 H p,q (X, E) := ker E. Im E
4 NICK MCCLEEREY A (trivial) example is to let E be the trivial holomorphic line bundle over X one should immediately see this recovers the previous untwisted Dolbeault cohomology. Also, as before, it is easy to show using sheaf cohomology that: Proposition 1.2. If X is a complex manifold, then: H p,q (X, E) = H q (X, Ω p E). While the analogous statement for regular Dolbeault cohomology looks good and all, this generalization is where the real money is at. As we shall see later on, when E is a line bundle associated to a divisor D, Ω p E is naturally identified with the sheaf of meromorphic p-forms with zeros (and poles) along D with the appropriate order. Working with meromorphic forms (and functions) will prove to be very important later on, as holomorphic functions (and forms) will often prove to be too rigid for example, non-trivial holomorphic forms rarely exist in profusion on compact complex manifolds, and even when they do, they are often difficult to construct. 2. Serre Duality and the Hodge Theorem From here on we shall turn our attention to compact complex manifolds, where much more can be said. Proofs in this section will be surpressed, as they would lead us too far astray from our stated goals for the course specifically, the proof techniques all come from Hodge Theory. The intersted reader is directed the books of Griffiths and Harris, Principles in Algebraic Geometry, and Huybrechts, Complex Geometry. We shall say here however that compactness is absolutely crucial, as the proof techniques rely heavily on PDE methods. Our first result is might appear obvious, but is fairly non-trivial to prove: Theorem 2.1. If X is a compact complex manifold, then H p,q (X, E) is a finitedimensional C-vector space for all 0 p, q n. Our next result is a bit more surprising, and is due to Serre. First, note that in the case for E = C, the trivial line bundle, we have a natural pairing between H p,q (X, C) and Hn p,n q (X, C) given by: (α, β) α β, α H p,q (X, C), β Hn p,n q (X, C). X If we again let E be an arbitray holomorphic vector bundle, we see we can extend this pairing to one between H p,q (X, E) and Hn p,n q (X, E ) by defining: (α s) (β s ) := s (s)α β. Here E is the dual vector bundle to E. Using this, the above pairing now makes sense for H p,q (X, E) and Hn p,n q (X, E ), and we have the following: Theorem 2.2. (Serre Duality) Let X be a compact complex manifold. Then the natural pairing: (α, β) α β, α H p,q (X, C), β Hn p,n q (X, C) X is non-degenerate. In particular, it induces an isomorphism between H p,q (X, E) and H n p,n q (X, E ), the dual of H n p,n q (X, E ).
RIEMANN SURFACES: TALK V: DOLBEAULT COHOMOLOGY 5 We call particular attention to the case of p = 0, which sometimes also goes by the name Serre Duality: Corollary 2.3. If X is a compact complex manifold, then: H q (X, E) = H n q (X, K X E ). Proof. We have simply used Prop. 1.2, the fact that Ω n is the sheaf of sections of K X, and the fact that Ω 0 = C on a compact manifold the last fact is a direct application of the maximum principle, which we shall not prove here. Thrm. 2.1 also gives us another immediately usefull corrollary: Corollary 2.4. If X is a compact complex manifold, then: H p,q (X, E) = H n p,n q (X, E ). Proof. Any finite dimensional vector space is isomorphic to its dual. Already these results cut down pretty drastically the possibilities for the Dolbeault cohomology groups of a compact complex manifold X. More can be said, but it requires more background than we currently have available. We shall thus turn out attention back to the case when E is the trival line bundle. Note then that Corollary 2.4 becomes: Corollary 2.5. If X is a compact complex manifold, then: H p,q (X, C) = H n p,n q (X, C). Proof. The dual of the trivial line bundle is itself. In order to say more, we will need to introduce one more restriction on our manifold, in particular that it be Kähler. The Kähler condition is at its heart a metric condition, so we briefly recall what a Hermitian metric and form are. A Hermitian metric on a complex manifold X is locally given by a positive Hermitian matrix (g ij ) 1 i,j n. The associated Hermitian form is the (1, 1)-form given (locally) by ω := g ij dz i dz j. The Kähler condition is that dω = 0, i.e. ω is closed. A manifold that admits a Kähler form will be a called a Kähler manifold. While the condition is a metric one, it also carries ramifications for the complex structure, and even the topology, on X. For example, a Kähler form is a 2-form, and is also nondegenerate, so ω k will form a non-trivial element in H 2k (X, C) for all 1 k n, where we are talking abou the derham cohomology here. We will not get into any further Kähler geometry here, but suffice it to note that any Riemann surface is trivially Kähler. Our next result may answer a few questions the reader may have been wondering. Let H m (X, C) be the m th -derham cohomology group of X with complex coefficents. Then: Theorem 2.6. (Hodge Theorem) Let X be a compact Kähler manifold. Then: H m (X, C) = H p,q (X, C). Moreover, we have that H p,q (X, C) = H q,p (X, C).
6 NICK MCCLEEREY Pay heed! This theorem does not hold in general for complex manifolds which are not Kähler! It also doesn t usually hold if the manifold is not compact! This result is tempting to just assume to be true trivially, based on the splitting of the groups in the chain complex and just the general nicety of the second statement, and many beginners fall into this trap. The proof, of the splitting in particular, is however very hard, and the Kähler condition is absolutely essentiall, providing necessary relations between,, and d that do not hold on a non-kähler manifold. One such relation that is easy to state and good to know is the following: 2 Lemma 2.7. (-Lemma) If X is a compact Kähler manifold, and α a differential form on X, then the following are equivalent: i) α is d-exact. ii) α is -exact. iii) α is -exact. iv) α is -exact. We shall denote by h p,q := dim C H p,q (X, C). This is called the (p, q)th Hodge number of X. By the Hodge Theorem. hp,q = b m, where b m is the m th Betti number of X, i.e. the complex dimension of H m (X, C). We also see that h p,q = h q,p, and recalling Serre Duality we see that h p,q = h n p,n q also. We can also now combine these two results to deliver a one-line proof of Poincare duality on a Kähler manifold: 3 Theorem 2.8. Let X be a compact Kähler manifold. Then: Proof. H m (X, C) = H m (X, C) = H 2n m (X, C). H p,q (X, C) = 3. Riemann Surfaces H n p,n q (X, C) = H 2n m (X, C). The above two theorems are quite difficult to prove in general. However, we can prove them (fairly) easily on a Riemann surface. We start with Serre Duality for the trivial line bundle (we will not prove it for a general vector bundle): Proof. We clearly only need consider the pairing between H 1,0 (X, C) and H0,1(X, C). Consider a non-zero class [α] H 1,0 (X, C). We wish to find a class [β] H0,1(X, C) so that α β 0. We claim the choice [β] = [iα] works. The class is in X H 0,1 (X, C) as it is -closed for dimensional reasons. Locally, we have: α β = fdz ifdz = f 2 idz dz = 2 f 2 dx dy so that α β 0 for all U. Since α is non-zero, at least one of these is strictly U positive, and the pairing is thus non-degenerate. Standard linear algebra finishes the proof. 2 This result actually holds for a slightly larger class of manifolds, called -manifolds. In particular, it is a weaker condition than being Kähler. 3 There are of course easier, more general proofs this is simply intended as a demonstration of the power of the stated theorems.
RIEMANN SURFACES: TALK V: DOLBEAULT COHOMOLOGY 7 We now prove the Hodge Theorem. We will need to assume the -Lemma 4 and the following easy application of the maximum principle for (pluri)harmonic functions: if a function f on a compact complex manifold satisfies f = 0, then f is constant. Proof. The second claim is trivial here from Serre Duality. As for the decomposition, we do the 2-forms first. As we mentioned before, every top form on a Riemann surface is a (1, 1)-form, so there is an obvious inclusion from H 1,1 (X, C) into H 2 (X, C) if α [0] H 1,1 (X, C), then α is a (1, 0)-form, and hence automatically -closed, making dα = α. The map is clearly surjective, so we need injectivity but this follows immediatly from the -Lemma. For the 1-forms, we first prove the following claim: H 0,1 (X, C) = {anti-holomorphic 1-forms}. To see this, we will find a unique anti-holomorphic form inside a class [α] (X, C). By the -Lemma, there exists a smooth function f on X so that H 0,1 α = f. Then we see that (α f) = 0, so α f is an anti-holomorphic form in [α]. To see it is unique, suppose α g is also anti-holomorphic. Then (f g) = 0, so f g is a constant, and hence f = g. With this, the decomposition is now easy: as before, we have an obvious inclusion of H 1,0 (X, C) into H 1 (X, C), and the above claim now produces a similar inclusion from H 0,1 (X, C) into H 1 (X, C), both of which are injective. Their images are clearly disjoint, so we will be done if we can show every derham class is expressible as a sum of a holomorphic and anti-holomorphic class. To this end, let α be a d-closed 1- form. Then α splits into a (1, 0) and a (0, 1) part, α 1,0 and α 0,1, and moreover α 1,0 = α 0,1 = f, by the -Lemma. Then we claim α + df is the representative we are after. In particular, α 1,0 + f is holomorphic and α 0,1 + f is anti-holomorphic, proving the decomposition. We can now use these results to (almost) immediately determine the Hodge numbers for any compact Riemann surface. We know any compact Riemann surface is topologically an oriented genus g surface. Thus, basic algebraic topology tells us that: H 0 (X, C) = H 2 (X, C) = C 2 and H 2 (X, C) = C 2g. Thus, we immediately conclude that h 0,0 = h 1,1 = 1 and h 1,0 = h 0,1 = g, so that the dimension of H 1,0 (X, C) encodes the genus of the surface, purely topological information. However, we can recover more than topological information from the Dolbeult cohomology. Consider in particular the case of a compex torus. As we will (hopefully) see later, the complex structure on a torus can be recovered from its periods. To do this, we pick an aribtrary non-zero holomorphic 1-form α H 1,0 (X, C) (which recall is 1-dimensional over C), and then consider the group of complex numbers: { } { } Λ := α γ H 1 (X, Z) = α β β H 0,1 (X, Z), γ X 4 This is one such tool we might use. Another, essentially equivalent one, would be the solvability of the Poisson equation, which is the approach taken in Donaldson s book.
8 NICK MCCLEEREY where we define H 0,1 (X, Z) = (π0,1 ı)(h 1 (X, Z)). Here ı is the composition of the inclusion of H 1 (X, Z) into H 1 (X, Z) by Poincare duality and the inclusion of H 1 (X, Z) into H 1 (X, C), and π 0,1 is the linear projection from H 1 (X, C) onto H 0,1 (X, C). Λ is called the period group of α, and is a Z2 -lattice in C. It is relatively straightforward to check that Λ is indeed a lattice, and that moreover X is biholomorphic to C/Λ, but we would need some more basic facts about compact Riemann surfaces. If one is familiar with the Uniformization Theorem, this is nigh trivial.