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Let us consider $0\to E\to G\to H\to 0$ an exact sequence of coherent sheaves on a surface $S$ such that $rk G=rk E +1$ and the double dual of $H$ is $\mathcal{O}_S(D)$ for some effective divisor $D$. Is it true that I have the following exact sequence:

$0\to detE\to det G\to i_*\mathcal{O}_D\to 0$,

where $i$ is the inclusion map from $D$ to $S$? In general, which is the behaviour of the determinant with respect to a short exact sequence of sheaves?

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  • $\begingroup$ There are two issues with your question. First of all, there is no natural map of $\det E\to\det G$ given your exact sequence. The only natural map is $\det E\otimes \mathcal{O}(D)\to\det G$. Secondly, if you had an exact sequence of determinants as you mention (possibly unnatural), it forces $\det E=\det G(-D)$. This need not be true in general. For example, let $M$ be an effective divisor and let $G=E(M)\oplus \mathcal{O}$ and the map being $E\to E(M)$. Then the double dual of $H$ is trivial and $\det E\neq \det G$. $\endgroup$
    – Mohan
    Commented Feb 4, 2011 at 16:46

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Provided that $S$ is smooth and $E,G$ are locally free, I find an exact sequence $0\to\det E\to\det G\to i_*\mathcal O_D(\det G\vert_D)\to 0$, as follows. Tensor the exact sequence $0\to\mathcal O_S\to\mathcal O_S(D)\to\mathcal O_D(D)\to 0$ coming from the existence of a section with $\det E$ and use $\det E\otimes \mathcal O_S(D)=\det G$.

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I'm not sure what the purpose of the assumption on the ranks of$E$ and $G$ is (by the way, how are you defining rank? Are you making some assumptions about the sheaves, and is $S$ smooth?). I guess it's so that you can make your assumption about the double dual of $H$.

As YBL mentioned, there is always an isomorphism $det(G) \simeq det(E) \otimes det(H)$ whenever you have a short exact sequence. I'm assuming that your map $0 \to det(E) \to det(G)$ is, under this isomorphism, equivalent to picking out a section of $det(H).$ It is then clear that the cokernel of your map will end up being the same as the cokernel of the map $0 \to \mathcal{O}_S \to det(H).$ This will be true in general, without any of the extra assumptions you've made.

In your particular case, if $det(H) \simeq \mathcal{O}_S(D),$ then this cokernel will be the skyscraper sheaf supported at $D$. Maybe I'm being stupid, but does the assumption on the double dual of $H$ imply this about the determinant?

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  • $\begingroup$ I think so because the determinant of a a coherent sheaf coincides with the determinant of its double dual. In my case $S$ is smooth and so everythink should work as you wrote. Thank you very much! $\endgroup$
    – ginevra86
    Commented Feb 4, 2011 at 15:45
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Your sequence can't be exact as both det(E) and det(G) have rank 1. Det transforms exact sequences into tensor products see this question.

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  • $\begingroup$ What do you mean? The left term has rank $0$, so I do not think that there are any problems with ranks. $\endgroup$
    – ginevra86
    Commented Feb 4, 2011 at 13:53
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    $\begingroup$ sorry..I meant the right term $\endgroup$
    – ginevra86
    Commented Feb 4, 2011 at 14:01

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