dualizing sheaf of a nodal curve

Question

I'm trying to understand the dualizing sheaf $\omega_C$ on a nodal curve $C$, in particular why is $H^1(C,\omega_C)=k$, where $k$ is the algebraically closed ground field. I know this sheaf is defined as the push-forward of the sheaf of rational differentials on the normalization $\tilde{C}$ of $C$ with at most simple poles at the points lying over the nodal points of $C$ and such that the sum of residues at the two points lying over the node will be zero. I can show that this is indeed an invertible sheaf on $C$, but I have no clue, despite my many attempts, how to show that $H^1(C,\omega_C)=k$. I've been able to show it in some very simple cases using Cech cohomology, but can someone explain to me how to do it in general?

Answer 1

If $\tilde{C}$ is the normalization, with two points $x$ and $y$ being identified under the map $\pi: \tilde{C} \to C$ to the node $z$ of $C$, then we have an exact sequence $$0 \to \Omega^1_{\tilde C} \to \Omega^1_{\tilde C}(x + y) \to k_x \oplus k_y \to 0,$$ where $k_x$ and $k_y$ are the skyscraper sheaves at the points $x$ and $y$. Pushing forward (which is exact because the map $\pi$ is finite, and so in particular affine) we get an exact sequence $$0 \to \pi_* \Omega^1_{\tilde C} \to \pi_*\Omega^1_{\tilde C}(x+y) \to k_z^{\oplus 2} \to 0.$$ Now there is a short exact sequence $0 \to k_z \to k_z^{\oplus 2} \to k_z \to 0$, where the third arrow is just given by adding the two components, and $\omega_C$ is the preimage of (the first copy of) $k_z$ under the surjection $\pi_* \Omega^1_{\tilde C}(x+y) \to k_z^{\oplus 2}$. In conclusion, we have an exact sequence $$0 \to \pi_* \Omega^1_{\tilde C} \to \omega_{C} \to k_z \to 0.$$

Now taking cohomology (and recalling that $H^i(C,\pi_*\mathcal F) = H^i(\tilde{C},\mathcal F)$ for a coherent sheaf on $\tilde{C}$), we obtain $$0 \to H^0(\tilde{C},\Omega^1_{\tilde C}) \to H^0(C,\omega_C) \to H^0(C,k_z) \to H^1(\tilde{C},\Omega^1_{\tilde C}) \to H^1(C,\omega_C) \to 0.$$ (The point here being that $H^1$ of a skyscraper sheaf such as $k_z$ vanishes.)

I claim that in this exact sequence the map $H^1(\tilde{C},\Omega^1_{\tilde C}) \to H^1(C,\omega_C)$ is an isomorphism, and hence that the latter is one-dimensional, since the former is.

For this, it is equivalent to show that the map $H^0(C,\omega_C) \to H^0(C,k_Z) = k$ is surjective.

Now $H^0(C,\omega_C) \subset H^0(C,\pi_*\Omega^1_C(x+y)) = H^0(\tilde{C},\Omega^1(x+y)).$ The residue theorem shows that we may find a differential $\omega \in H^0(\tilde{C},\Omega^1(x+y))$ whose residues at $x$ and $y$ are non-zero. (These residues are then negative to one another.) Thought of as a section of $H^0(C,\pi_*\Omega^1_C(x+y))$, this differential $\omega$ clearly lies in $H^0(C,\omega_C)$. Its image under the map $H^0(C,\omega_C)$ is non-zero (equal to the residue at either $x$ or at $y$, depending on a choice that was implicitly made above), and so indeed $H^0(C,\omega_C) \to k$ is surjective.

Summary: The residue theorem guarantees the existence of sections of $H^0(C,\omega_C)$ which have non-zero residues at $x$ and $y$ when pulled back to $\tilde{C}$, and this in turn shows that $H^1(C,\omega_C)$ is isomorphic to $H^1(\tilde{C},\Omega^1_C)$, and hence is one-dimensional.