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

It's well known that if E is a vector bundle with Chern roots $a_1,\ldots, a_r$, then the Chern roots of the $p$th exterior power of E consist of all sums of $k$ distinct $a_i$'s. I would like to say the same is true if E is just a torsion-free coherent sheaf on $P^n$. It seems non-obvious, though, maybe because an exterior power isn't generally an additive functor.

Presumably this is either false or also well known, but I can't find a reference.

share|improve this question
    
For non-singular varieties, every coherent sheaf has a resolution by vector bundles. Maybe that helps? –  Kevin H. Lin Mar 31 '10 at 4:02
    
Incidentally, I don't know a reference for the statement I just made. –  Kevin H. Lin Mar 31 '10 at 4:03
    
Yes. One can think of the Chern classes of a coherent sheaf formally, by taking a finite resolution by vector bundles and defining Chern classes from their behaviour on exact sequences. It's not clear (at least to me) what to do with multilinear constructions, though. At least on the level of modules, I don't think there is any general construction of a free resolution of an exterior power of a module with a known resolution. –  Graham Denham Mar 31 '10 at 4:43
    
This is Example 6.5.1 in Hartshorne, Kevin. You do not need non-singularity, unless you want finite resolutions. –  Mariano Suárez-Alvarez Mar 31 '10 at 4:45
    
Thanks Mariano. Yeah, I had meant to type "finite". –  Kevin H. Lin Mar 31 '10 at 6:14
show 1 more comment

2 Answers

up vote 3 down vote accepted

My guess would be that the formula you want does not extend to the case of coherent sheaves. As indicated in Mariano and David answers (which has unfortunately been deleted), the best hope to compute is via a resolution $\mathcal F$ of $E$ by vector bundles. In general, for 2 perfect complexes $\mathcal F, \mathcal G$ of vector bundles, there is a formula for the localized chern classes $$ch_{Y\cap Z}(\mathcal F \otimes \mathcal G) = ch_Y(\mathcal F)ch_Z(\mathcal G)$$, with $Y,Z$ being the respective support. Unfortunately, this only gives the right formula for the "derived tensor product".

So to mess up the formula, one can pick $E$ such that $Tor^i(E,E)$ are non-trivial. I think an ideal sheaf of codimension at least 2 would be your best bet for computation purpose.

share|improve this answer
    
For those who are curious, I had a wrong answer, now deleted. I think Hailong's answer should work, but I haven't been able to find a counter-example. –  David Speyer Mar 31 '10 at 17:57
    
Thanks, guys. That clears things up. For my application I can compute with Lebelt's resolution (MR0450253) to resolve the exterior powers -- a special case of Mariano's references -- but I didn't want to do that if I could just argue by general principles. –  Graham Denham Mar 31 '10 at 18:49
add comment

This is really a reference for the problem in Graham's comment, rather than an answer to the question. See

  • Tchernev, Alexandre B. Acyclicity of symmetric and exterior powers of complexes. J. Algebra 184 (1996), no. 3, 1113--1135. MR1407888

  • Weyman, Jerzy. Resolutions of the exterior and symmetric powers of a module. J. Algebra 58 (1979), no. 2, 333--341. MR0540642

where the symmetric and exterior powers of finite free resolutions of modules are considered. Under some conditions on the determinantal ideals determined by the maps in the reslutions, the complexes obtained are resolutions of the symmetric and exterior powers of the modules.

share|improve this answer
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.