Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
What do you know about the singularities of $C_{\text{red}}$ or of $C$? –  Karl Schwede Oct 23 '12 at 15:53
1  
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
add comment

2 Answers 2

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.

share|improve this answer
    
@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 '12 at 9:54
add comment

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.

share|improve this answer
    
@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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.