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Let $X$ be an algebraic variety. Consider an exact sequence $$0\to A\to B\to C\to 0$$ of vector bundles on $X$. I have seen in different papers the following type resolution of wedge product of $C$ (or $A$) $$0\to S^kA\to S^{k-1}A\otimes B\to S^{k-2}A\otimes \wedge^2B\to \cdots\to \wedge^kB\to \wedge^k C\to 0.$$

Question: does this resolution come from certain geometric context? Is there a proof which involves certain geometric aspects, for example, using projective bundles associated to the vector bundles?

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up vote 6 down vote accepted

The resolution in question is the $k$th graded piece of the symmetric algebra of the complex $A \rightarrow B$, where $A$ is considered to lie in even degrees (say degree $2$) and $B$ in odd degrees (say degree $1$). (Also, a kind of Koszul complex.)

The way I think of this: forgetting the differential $A \rightarrow B$ for a minute, we just have a graded vector bundle $A \oplus B$ with $A$ in degree $2$, $B$ in degree $1$. Since the symmetric algebra construction $\mathcal{S}$ takes sums to tensor products, we have $\mathcal{S}(A \oplus B) \simeq \mathcal{S}(A) \otimes \mathcal{S}(B)$. Since $A$ is in degree $2$, $S(A)=S(A)$ (commutative) and since $B$ is in degree $1$, $\mathcal{S}(B) \simeq \bigwedge(A)$ (the Koszul sign rule for graded tensor products makes this super-commutative).

Now you have to bring in the differential. To do this, it is useful to use the bialgera structure on the symmetric algebra construction, where the coproduct comes from applying $\mathcal{S}$ to the diagonal map $X \rightarrow X \oplus X$. To see how to get the differential out of this: given $S^{k-i}A \otimes \bigwedge^{i} B$, comultiply in the first factor to end up in $S^{k-i-1}A \otimes A \otimes \bigwedge^{i}B$, now apply the map $A \rightarrow B$ to the middle factor of $A$ to end up in $S^{k-i-1}A \otimes B \otimes \bigwedge^{i}B$, and finally, multiply the last two factors together to end up in $S^{k-i-1}A \otimes \bigwedge^{i+1}B$

This is a standard construction and can be found for instance in the paper of Akin, Buchsbaum, Weyman on Schur complexes, or in the book of Weyman on Cohomology of vector bundles and syzygies. If people have other references, I'd be glad to know about them, since the above mentioned ones are from the point of view commutative algebra (which is for me non-optimal).

While the above construction is certainly useful in geometry, I'm afraid the above is not geometric in the way you were looking for.

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Thanks a lot. The construction you given is still new to me. I had tried to use properties of projective bundles to construct the sequence, eventually I can get a resolution of $S^kC$ in stead of $\wedge^kC$. – Fei YE Oct 13 '10 at 13:30
If you switch the gradings in the description that I gave, then you should get a resolution of $S^{k}C$. Can you say more about what your construction using projective bundles? – Chris Brav Oct 13 '10 at 18:32
Here is a rough idea. Let $\pi : \mathbb{P}(B)\to X$ be the projection morphism. The ideal of $\mathbb{P}(C)$ in $\mathcal{O}_{\mathbb{P}(B)}$ is the image of the composition $\pi^*A(-1)\to \pi^*B(-1)\to \mathcal{O}_{\mathbb{P}(B)}$. So $\pi^*A(-1)\to \mathcal{O}_{\mathbb{P}(B)}$ gives a Koszul resolution of $\mathcal{O}_{\mathbb{P}(C)}$. Twist by $\mathcal{O}(r)$ and then apply $\pi_*$, one should get something. – Fei YE Oct 13 '10 at 22:32
Fei, you should accept the answer... – diverietti Jan 25 '11 at 23:56

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