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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?

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  • $\begingroup$ Doesn't Serre duality imply that group is dual to $H^0(C,\mathcal{O}_C)$? $\endgroup$ – S. Carnahan Mar 15 '11 at 19:04
  • $\begingroup$ Dear Scott, Yes, but my impression was that the OP wanted a direct proof, so to speak. Regards, Matt $\endgroup$ – Emerton Mar 15 '11 at 19:32
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    $\begingroup$ How does Serre duality imply this? In order to prove Serre Duality for a singular curve one first needs the dualizing sheaf, no? $\endgroup$ – HNuer Mar 15 '11 at 21:22
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    $\begingroup$ HNuer: It's a question of terminology. A dualizing sheaf is usually understood as the thing which satisfies Grothendieck-Serre duality. So your question is really: why does the thing defined in your second sentence behave (to some extent) ike a dualizing sheaf? $\endgroup$ – Donu Arapura Mar 15 '11 at 21:47
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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.

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  • $\begingroup$ Dear Matt, in the middle of the displayed short exact sequence in the tenth line, I think you meant $\omega_C$ rather than $\omega_{\tilde C}$. Let me use the occasion to thank you for the recurring pleasure of reading your always lucid and beautifully written posts. $\endgroup$ – Georges Elencwajg Mar 15 '11 at 21:06
  • $\begingroup$ How does the residue theorem imply the existence of differentials with non-zero residues? I thought it just implied that the sum of the residues of any meromorphic differential was zero. $\endgroup$ – HNuer Mar 15 '11 at 21:58
  • $\begingroup$ Dear Georges, Thank you for the correction and for the kind words. Best wishes, Matt $\endgroup$ – Emerton Mar 16 '11 at 0:12
  • $\begingroup$ Dear HNuer, I am referring to the stronger result which says that summing to zero is the only obstruction for finding a differential on a smooth projective curve with at worst simple poles with prescribed residues at some finite set of points (and holomorphic everywhere else). If you aren't already familiar with this statement, I'll leave it as an exercise. Best wishes, Matt $\endgroup$ – Emerton Mar 16 '11 at 0:15
  • $\begingroup$ Dear Matt, I am indeed unfamiliar with that statement, and only know of the result I quoted from Hartshorne's brief discussion in III.7 of his book. Do know of a reference for the stronger statement? Also, in you answer above, isn't $\Omega^1_{\tilde{C}}(x+y)$ the differentials with zeroes at x and y? Wouldn't $\Omega^1_{\tilde{C}}(-x-y)$ be the sheaf of differentials with poles there? This may just be a stupid question, but I want to make sure I understand. Thanks for all the help. $\endgroup$ – HNuer Mar 16 '11 at 5:24

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