How to resolve a wedge product of vector bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:06:51Z http://mathoverflow.net/feeds/question/41970 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41970/how-to-resolve-a-wedge-product-of-vector-bundles How to resolve a wedge product of vector bundles Fei YE 2010-10-13T03:58:49Z 2010-10-13T07:26:47Z <p>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 $A$) $$0\to S^kA\to S^{k-1}A\otimes B\to S^{k-2}A\otimes \wedge^2B\to \cdots\to \wedge^kB\to \wedge^k C\to 0.$$ </p> <p>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?</p> http://mathoverflow.net/questions/41970/how-to-resolve-a-wedge-product-of-vector-bundles/41990#41990 Answer by Chris Brav for How to resolve a wedge product of vector bundles Chris Brav 2010-10-13T07:26:47Z 2010-10-13T07:26:47Z <p>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.) </p> <p>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).</p> <p>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 $S^{k-i-1}A \otimes \bigwedge^{i+1}B$</p> <p>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).</p> <p>While the above construction is certainly useful in geometry, I'm afraid the above is not geometric in the way you were looking for. </p>