Residues and Integrals

This post is about a computation every algebraic geometry student should do, but that none of my courses covered. Let X be a smooth, projective curve over k. Then H^1(X, \Omega) is a one dimensional k-vector space. If you’ve read Hartshorne III.7 carefully, you’ll remember that there is a canonical isomorphism: Tr: H^1(X,\Omega) \to k. Explicitly, let p and q be two points of X; consider the open cover (X \setminus \{ p\}) \cup (X \setminus \{ q\}) of X and let \omega be a holomorphic 1-form on X \setminus \{ p, q\}. Let c be the cocycle X \setminus \{ p,q \} \mapsto \omega. Then Tr sends the cocycle c to the residue, at p, of \omega. (A good question which I might ask on a qual one day: Why is it OK that this is asymmetric in p and q?)

On the other hand, suppose that k = \mathbb{C}. Then H^1(X, \Omega) is isomorphic to H^{1,1}(X). An element of H^{1,1}(X) is a \overline{\partial}-closed (1,1) form, modulo \overline{\partial}-exact (1,1)-forms. But, because X only has two real dimensions, this simplifies: Every 2 form is a (1,1)-form and every 2-form is closed because there are no 3-forms. So an element of H^{1,1}(X) is a 2-form modulo \overline{\partial}-exact 2-forms. It turns out that this is the same as a 2-form modulo d-exact two forms. In other words, H^{1,1}(X) is the same as the deRham cohomology group H^2(X, \mathbb{C}) that we learn about in differential geometry. And we know a canonical map H^2(X, \mathbb{C}) \to \mathbb{C}: Take the integral!

The point of this post is to compute the relation between Tr and \int. I invite you to try it yourself, then meet me on the other side to see if we got the same answer.

UPDATE: I claimed earlier that this was easy to show H^{1,1} \cong H^2 for curves. As Akhil Mathew points out, it seems to only be easy to show that there is a well defined surjection H^{1,1} \to H^2. Since I only need the map to exist, I’ll leave it at that for now.

In order to do this computation, we need to remember exactly how the isomorphism H^1(X, \Omega) \cong H^{1,1}(X) is built. From now on, all my sheaf cohomology computations will be with respect to the analytic topology. (Also known as “the picture I am drawing when I draw a Riemmann surface”.) By GAGA, for coherent sheaves on X, sheaf cohomology gives the same answers with respect to the analytic and Zariski topologies.

Let \mathcal{E}^{p,q} be the sheaf of C^{\infty} (p,q)-forms. So a section of \mathcal{E}^{1,0} locally looks like f(z) dz where f is a smooth, complex valued function, while a section of \Omega locally looks like f(z) dz where f is a complex-analytic function. See my earlier post for a review of these notions.

We have a short exact sequence

0 \to \Omega \longrightarrow \mathcal{E}^{1,0} \overset{\overline{\partial}}{\longrightarrow} \mathcal{E}^{1,1} \to 0.

Locally, the first map sends a 1-form of the form f(z) dz, with f analytic, to itself, considered as an example of a 1-form of the form g(z) dz, with g smooth. The second map sends a 1-form of the form g(z) dz, with g smooth, to (\partial g/\partial \overline{z}) d \overline{z} dz . Exactness on the right is by a variant of the Poincare lemma.

Now, \mathcal{E}^{p,q} is a module over the sheaf of C^{\infty} functions, and that sheaf has partitions of unity. A standard argument then shows that H^{i}(X,\mathcal{E}^{p,q}) vanishes for i>0. See Voisin, Hodge Theory and Algebraic Geometry I Proposition 4.36, for the details.

So, we have a long exact sequence of groups:
0 \to H^0(\Omega) \longrightarrow H^0(\mathcal{E}^{1,0}) \longrightarrow H^0(\mathcal{E}^{1,1}) \to H^1(\Omega) \longrightarrow 0.

Our task is to take the class of c in H^1(\Omega) and lift to a volume form \eta in H^0(\mathcal{E}^{1,1}), then compare \int \eta to Tr(\omega).

We first refine the open cover X = (X \setminus \{ p \}) \cup (X \setminus \{ q \}). Let P be a small open discs centred at p and let P' be a smaller closed disc, contained in P and containing p. Choose Q and Q' similarly, with P and Q disjoint. Let U = X \setminus (P' \cup Q'). Our new open cover will be X = P \cup Q \cup U. We want to pull the Cech cocycle c back to a cocycle on the new cover.

A technical note: To pull back c, we need to decide how the new cover is a refinement of the old one. In other words, for each open set in the new cover, we need to choose a specific set in the old cover which contains it. Our choice is that U is regarded as a subset of (X \setminus \{ q \}).

With the above choice, the pullback of c to the new cover is \omega|_{P \setminus P'} on P \cap U and 0 on Q \cap U. Call this cocycle c'.

We now look for a preimage of c' in H^0(\mathcal{E}^{1,1}). Remembering how the boundary map in Cech cohomology is defined, we first find sections of \mathcal{E}^{1,0} on P, Q and U whose difference on the overlaps is c'. We will take 0 on Q, 0 on U and \theta on P. Here \theta must be a smooth (1,0) form, whose restriction to P \setminus P' is \omega. A standard trick with bump functions shows that such a smooth extension exists. (Exercise!)

Let \eta be the (1,1) form which is \overline{\partial} \theta on P and 0 on X \setminus P'. We check that this is well defined: On {P \setminus P'}, we have \theta = \omega so \overline{\partial} \theta = \overline{\partial} \omega which is 0 because \omega is holomorphic. By the definition of the boundary map in Cech cohomology, \eta is a preimage of c. Our job is to compute \int \eta.

At this point, the sheaf cohomology is done, and we switch to complex analysis. Everything is supported on P, so we can work locally. The question is this: Let p \in P' \subset P, in the complex plane, be a point, a closed disc, and an open disc. Let \omega = f(z) dz be a holomorphic 1-form on P \setminus \{ p \}. Let \theta = g(z) dz be a smooth 1-form, which agreees with \omega on P \setminus P'. What is \int_{P} \overline{\partial}(\theta)?

Here is a nonrigourous way I like to think about this process. Let P be the disc of radius 1 and let \omega = dz/z. Suppose that, instead of requiring that \omega and \theta coincided on an annulus, we only had to make them match up on the boundary of P. Then we could use \theta = \overline{z} dz. In general, turning \omega into \theta is like noticing that 1/z equals \overline{z} on the unit circle, but smeared out over an annulus.

The integral \int_{P} \overline{\partial}(\theta) is crying out for an application of Stokes’ theorem. Since \theta is a (1,0)-form, \overline{\partial}(\theta) is d \theta. So, after shrinking P a little to make sure \omega extends to the boundary, we have

\int_{P} \overline{\partial}(\theta) = \int_{P} d \theta = \int_{\partial P} \theta = \int_{\partial P} \omega.

But the last is simply the (2 \pi i) (\mbox{the residue of} \ \omega \ \mbox{at} \ p). So \int \eta= (2 \pi i) Tr(\omega) The computation is done.

6 thoughts on “Residues and Integrals

  1. Maybe it’s not as easy as I suggested. Every 2-form on a Riemmann surface is a (1,1) form, so these are both quotients of the vector space of (1,1)-forms.

    If \alpha is a (1,0)-form, then d(\alpha) = \overline{\partial}(\alpha), so every \overline{\partial}-exact form is also d-exact. So we get a surjective map H^{1,1} \to H^2.

    But I’m not seeing an easy argument that this map is injective. Of course, you can use Serre and Poincare duality to show that both sides are one dimensional. But I’d rather not invoke that.

  2. I don’t think there is a way to show injectivity without the Hodge decomposition in some form. Clearly it’s enough to know it for H^2, but you can do a bit better. If you know that the inclusion of the constant sheaf C into the structure sheaf induces a surjection on H^1, then you’re done by taking the holomorphic Dolbeault resolution of C.

  3. Let \omega be a \partial-exact 2-form. Split it into real and imaginary parts \omega=\omega_r+i\omega_i. These are also \partial-exact so there are (0,1)-forms \theta_r and \theta_i such that \partial\theta_r=\omega_r, \partial\theta_i=\omega_i.

    Construct the (1,0)-form \tilde{\theta}=\bar{\theta}_r+i\bar{\theta}_i. Then \bar{\partial}\tilde{\theta}=\omega. A direct calculation gives

    \bar{\partial}(\bar{\theta}_r+i\bar{\theta}_i)=\bar{\partial}\bar{\theta}_r+i\bar{\partial}\bar{\theta}_i=\overline{\partial\theta_r}+i\overline{\partial\theta_i}=\bar{\omega}_r+i\bar{\omega}_i=\omega_r+i\omega_i=\omega

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