# Vanishing of Ext group

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 $Ext^2_X(F,\mathcal{O}_X)=0$?

• What do you know about the singularities of $C_{\text{red}}$ or of $C$? – Karl Schwede Oct 23 '12 at 15:53
• 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 '12 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 '12 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 '12 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 '12 at 8:07

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.
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.
• @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 '12 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. – user25309 Oct 24 '12 at 8:57