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?

  • $\begingroup$ 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. $\endgroup$ Jul 15 '21 at 8:08
  • $\begingroup$ Thanks! The acyclicity didn't occur to me. And for 1. do you agree with the interpretation in the post as it is? $\endgroup$
    – Bananeen
    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 $\bar \partial \mu_i = \delta(1_{x_i})$.

  • $\begingroup$ 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? $\endgroup$
    – Bananeen
    Jul 19 '21 at 0:07
  • $\begingroup$ @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 $0 \to A^{0,1}(K) \to A^{0,1}(K(D)) \to \bigoplus_j \mathcal O_{x_j} \to 0$. $\endgroup$ Jul 19 '21 at 7:00
  • $\begingroup$ 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. $\endgroup$
    – Bananeen
    Jul 19 '21 at 23:27

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