# Representing elements of $H^1(X, K_X)$ on a curve

This is a question about exercise 2 (c), pp.113-114, from Chapter 4 of Voisin's "Hodge Theory and Complex Algebraic Geometry, I".

Let $$X$$ be a compact complex curve with a divisor $$D = \sum x_i$$ on it (all $$x_i$$ are different so there are no multiplicities). Consider the exact sequence $$0 \to K_X \to K_X(D) \stackrel{Res}{\to} \oplus_i \ \mathcal{O}_{x_i} \to 0,$$ where $$Res_i(w) = \int_{\partial D_i} \frac{1}{2\pi i} w$$is the residue of the meromorphic form around $$x_i$$.

We further have the associated long exact sequence $$0 \to H^0 (K_X) \to H^0 (K_X(D)) \to \oplus_i \ H^0 (\mathcal{O}_{x_i}) \stackrel{\delta}{\to} H^1(K_X) \to H^1 (K_X(D)) \to 0.$$

The exercise asks to show that $$\delta(1_{x_i})$$ is the class in $$H^1(K_X)$$ of the form $$\bar \partial \mu_i$$, where $$\mu_i$$ is a differential form of type $$(1,0)$$, which is $$C^{\infty}$$ away from $$x_i$$, and equal to $$\frac{dz_i}{z_i}$$ in a neighborhood of $$x_i$$.

I am assuming that we have to work with the Dolbeault representation of cohomology classes, so $$H^1(K_X) = A^{1,1}(X)/ \bar \partial \left( A^{1,0}(X) \right).$$

1. Comment by @Blazej from this thread says that $$\bar \partial$$ removes the singularity from $$\frac{dz_i}{z_i}$$. In what sense should I interpret this statement? Is it that since away from $$0$$ this form is holomorphic then we just postulate $$\bar \partial \left( \frac{dz_i}{z_i} \right) = 0$$ on that neighborhood?

2. If the previous interpretation is correct then how do I go about the exercise? The only way to compute the connecting homomorphism that I know of is to functorially pick (injective/acyclic/Cech?) resolutions for all the sheaves and chase the resulting diagram. For $$K_X$$ and $$K_X(D)$$ I can choose the Dolbeault, but what do I choose for the skyscraper sheaves?

• The $\mathcal O_{x_i}$ are acyclic, so they resolve themselves. This is also the reason why there is no $H^1(\bigoplus_i \mathcal O_{x_i})$ appearing in the long exact sequence you wrote. Jul 15 '21 at 8:08
• Thanks! The acyclicity didn't occur to me. And for 1. do you agree with the interpretation in the post as it is? Jul 16 '21 at 10:26

This can be done by diagram chase, just as you suggest. For the sheaves $$K_X$$ and $$K_X(D)$$ you choose the Dolbeault resolutions. The sheaf $$\bigoplus_i \mathcal O_{x_i}$$ is acyclic, and hence $$0 \to \bigoplus_i \mathcal O_{x_i} \to \bigoplus_i \mathcal O_{x_i} \to 0$$ is an acyclic resolution. For the diagram chase, we first have to pick a preimage of the section $$1_{x_i}$$ under the map $$A^{0,1}(K_X(D)) \to \bigoplus_j \mathcal O_{x_j}.$$ Now choose a local coordinate $$z_i$$ in a neighbourhood $$U$$ of $$x_i$$ and take $$\mu_i|_U = \frac{d z_i}{z_i}$$. This is a section of $$K_X(D)$$ over $$U$$. Then we may extend $$\mu_i|_U$$ to a global section $$\mu_i$$ by multiplying with a $$\mathcal C^\infty$$-bump-function $$f$$ that is identically $$1$$ in a neighbourhood of $$x_i$$ and has compact support in $$U$$. So $$\mu_i = f \cdot \mu_i|_U$$ is a well-defined global section of $$K_X(D)$$, which has a single pole at $$x_i$$, and hence it is a preimage of $$1_{x_i}$$.
Now by the definition of the connecting homomorphism $$\delta$$ we get $$\bar \partial \mu_i = \delta(1_{x_i})$$.
• Upon further reflection, I don't see how we have a short exact sequence of resolutions: in degree zero indeed we have $0 \to A^{0,0}(K_X) \to A^{0,0}(K_X(D)) \to \oplus_j \mathcal{O}_{x_j} \to 0$, but in degree $1$ we will have $0 \to A^{0,1}(K_X) \to A^{0,1}(K_X(D)) \to 0 \to 0$ which is not exact, right? Jul 19 '21 at 0:07
• @Bananeen Note that even though $\mathcal O_{x_j}$ is acyclic, it's resolution $0 \to \mathcal O_{x_j} \to \mathcal O_{x_j} \to 0$ is not zero in degree 1. So in degree 1 you get an exact sequence $0 \to A^{0,1}(K) \to A^{0,1}(K(D)) \to \bigoplus_j \mathcal O_{x_j} \to 0$. Jul 19 '21 at 7:00
• Something is not right here. I adopt the degree convention like this: a resolution of $\mathcal{F}$ is an exact complex $0 \to \mathcal{F} \to \mathcal{R}^0 \to \mathcal{R}^1 \to ...$. So for the three sheaves in play we have $0 \to K_X \to A^{0,0}(K_X) \to A^{0,1}(K_X) \to 0$, $0 \to K_X(D) \to A^{0,0}(K_X(D)) \to A^{0,1}(K_X(D)) \to 0$, $0 \to \oplus_i \mathcal{O}_{x_i} \to \oplus_i \mathcal{O}_{x_i} \to 0 \to 0$. I cannot see a way to reconcile this with what you are writing at the level of $A^{0,1}$'s. Jul 19 '21 at 23:27