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This question is inspired by Hartshorne's exercise II.5.7 (c-d): the problem reads: Let $0\rightarrow \mathcal{F}'\rightarrow\mathcal{F}\rightarrow\mathcal{F}''\rightarrow0$ be a short exact sequence of locally free sheaves. Then for any $r$ there is a finite filtration $F$ of $S^r(\mathcal{F})$ (the sheaf of symmetric algebras over $\mathcal{O}_X$) such that $F^pS^r(\mathcal{F})/F^{p+1}S^r(\mathcal{F})$ is isomorphic to $S^p(\mathcal{F}')\otimes S^{r-p}(\mathcal{F}'')$ (and similarly for exterior algebras).

So basically, I had this feeling that a spectral sequence might be lurking around somewhere; my thinking is this: if we can write down a spectral sequence of a graded complex that converges to $S^r(\mathcal{F})$, then we would naturally be guaranteed a grading of this object which agrees with the $E_\infty$-page. So is there some sort of way to write down a complex so that we heuristically have something like $E_2^{p,q}\Rightarrow S^{p+q}(\mathcal{F})$ where if we go through and read off the sum of the diagonals of the $E_\infty$-page, we get something like $S^{p}(\mathcal{F}')\otimes S^{r-p}(\mathcal{F}'')$?

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I'm not sure there would be anything exactly like that, but you would have a spectral sequence converging to the cohomology of $S^r(\mathcal{F})$ associated to the above filtration. The $E_1$ term would be the cohomology of the associated graded. – Donu Arapura Oct 21 '11 at 8:31
up vote 4 down vote accepted

Take the symmetric power of the complex $F'' \to F'[1]$. However, this is a very exotic point of view on the question.

ADDITION. A complex is a chain of morphisms with zero compositions. You can consider complexes in the derived category as well. In some sense those can be thought of as bicomplexes in the original abelian category. Now if there is a two-term complex $A \to B$ you can consider its symmetric power: $$ S^n A \to S^{n-1}A\otimes B \to S^{n-2}A \otimes \Lambda^2B \to \dots \to \Lambda^n B. $$ If you apply this to the complex $F'' \to F'[1]$ and note that $\Lambda^k(F'[1]) = S^kF'[k]$, you will get a complex $$ S^nF'' \to S^{n-1}F''\otimes F'[1] \to S^{n-2}F'' \otimes S^2F'[2] \to \dots \to S^nF'[n]. $$ Since the original complex was quasiisomorphic to $F$, this one is quasiisomorphic to $S^nF$. Now the stupid filtration of this complex gives the required filtration of $S^nF$.

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I'm sorry but I'm not exactly sure what you mean by this; if you could elaborate more that would be fantastic. More specifically, I'm not sure what you mean by the chain complex $F''\rightarrow F'[1]$. Also, does taking the symmetric power of a chain complex just mean applying the functor $S(-)$? – Geoffrey Oct 28 '11 at 0:01
I have written an addition to the answer. Hope now it is more clear. – Sasha Oct 29 '11 at 7:15
Ah thank you very much; I guess this is a pretty idiosyncratic way of looking at this example but thanks a lot for your reply. – Geoffrey Oct 31 '11 at 20:23

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