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Let $C$ be a cartier divisor on a smooth projective surface in $\mathbb{P}^3$. Then we get the short exact sequence $$0 \to \mathcal{O}_X(-C) \to \mathcal{O}_X(-C_{red}) \to F \to 0$$ for some sheaf $F$. We see that $F$ is supported on $C$. Assuming $C \not= C_{red}$ when is it possible to say that $Ext^2_X(F,\mathcal{O}_X)=0$?

edited Oct 23 at 8:38

Francesco Polizzi
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asked Oct 23 at 8:33

Naga Venkata
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What do you know about the singularities of $C_{\text{red}}$ or of $C$? – Karl Schwede Oct 23 at 15:53

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We can analyze a spectral sequence computing: $$Ext^2(F, O_X)$$ In particular, if we can show that $$H^0(X, \mathcal{E}xt^2(F, O_X))) = 0, H^1(X, \mathcal{E}xt^1(F, O_X))) = 0, H^2(X, \mathcal{E}xt^0(F, O_X))) = 0$$ then we are done. The two terms on the ends are easily seen to be zero and in fact it's easy to see that $$ Ext^2(F, O_X) = H^1(X, \mathcal{E}xt^1(F, O_X))) = H^1(X, O_X(C)/O_X(C_{red}) ). $$ Not sure if this is any help. – Karl Schwede Oct 23 at 16:20

@Schwede: I know this using the spectral sequence on $\mathcal{E}xt$. So not helpful but thanks for the attempt. – Naga Venkata Oct 23 at 16:29

Dear Naga Venkata, no problem. Do you know anything else about the singularities of $C$ or $C_{red}$ or the genus of $C_{red}$? The self intersection of $C$? Anything like that might be useful. – Karl Schwede Oct 23 at 17:26

@Schwede: The self intersection of $C$ and $C_{red}$ is negative and it can be shown that the last map that sasha talks of below is infact injective. I do not have much information about the singularity. However, it is local complete intersection (since Cartier divisor). You can assume that the degree of the surface is $d \ge 5$. This bounds the genus of the curve contained to $\binom{d-1}{3}$. You are welcome to state partial results/ideas by assuming criterion on genus and sigularity. However, you should assume that the curve is not smooth or irreducible. – Naga Venkata Oct 24 at 8:07

Sasha · Answer 1 · 2012-10-23 09:23:21Z UTC

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There is a long exact sequence $$ H^1(X,O_X(C_{red})) \to H^1(X,O_X(C)) \to Ext^2(F,O_X)\to H^2(X,O_X(C_{red})) \to H^2(X,O_X(C)), $$ so $Ext^2(F,O_X) = 0$ if and only if the first map is surjective and the last map is injective.

answered Oct 23 at 9:23

Sasha
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@sasha:I understand that. The reason for the question was to determine surjectivity of the first map. So could you suggest some other (non-equivalent) condition. – Naga Venkata Oct 23 at 9:54

unknown (google) · Answer 2 · 2012-10-23 10:43:54Z UTC

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Maybe something which can help : use Serre duality on X in order to obtain a $H^{0}$. As F is supported on C, one can compute the $H^{0}$ on C and use the adjunction formula to express the dualizing sheaf of X in terms of the dualizing sheaf of C. The conclusion is that the Ext group is zero if and only if $H^{0}(C, F(C) \otimes \omega_{C}^{-1})$ is zero.

answered Oct 23 at 10:43

unknown (google)
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	@unknown: As far as I understand we need $F$ to be locally free to apply Serre duality which is not the case. – Naga Venkata Oct 23 at 12:53
	what I call Serre duality should be called Grothendieck-Serre duality. $Ext^{n-i}(F, \omega_{X}) = H^{i}(F)^{*}$, X projective smooth of dimension n, is true for any coherent sheaf. – unknown (google) Oct 24 at 8:57

Vanishing of Ext group

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