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

Suppose that $G$ is an algebraic group over a field $k$. Then for any $k$-algebra $R$, a fiber functor from $\text{Rep}_k(G)$ to the category of projective modules over $R$ is the same as a $G$-torsor on $\text{Spec}(R)$, by Theorem 3.2 of Deligne-Milne.

Now suppose that instead of a $k$-linear faithful exact tensor functor into the category of projective modules, I have a functor (taking exact sequences of representations to distinguished triangles, and $\otimes$ to $\otimes^L$, say) to the bounded derived category of coherent modules over $R$. Do I get some sort of "derived $G$-bundle" on $\text{Spec}(R)$? What should that mean?

share|improve this question
add comment

1 Answer

I think you want your functor to be a delta-functor from the abelian category of representations to D^b_{Coh}(R) which is moreover compatible with tensor products. By the way. I suggest we replace D^b_{Coh}(R) by the derived category of perfect complexes, just to be in vogue.

A special case is where R = k. Then D^b_{Coh}(k) is the category of finite dimensional graded k-vector spaces endowed with a suitable structure of derived category. Suppose we only look at those functors where delta is always zero. (By the way, if the group scheme is linearly reductive, this will always be the case.) Then we obtain simply a k-linear faithful exact tensor functor into the category of graded vector spaces. If we forget the grading, then we obtain a G-torsor over k as you say. Suppose that this torsor is trivial, e.g., if k is algebraically closed. Then what's left over is a grading on each representation of G, compatible with tensor product. Such a thing is given by a cocharacter, i.e., a group scheme homomorphism G_m ---> G.

Thus, in the very special case just discussed, there seems to be a little bit more structure, than for usual tensor functors. Before trying to answer your very interesting question in full, we should try to answer it in the case where G is the additive group over k. My feeling would be that delta still has to be zero...

share|improve this answer
Is it really true that $G$ linearly reductive implies $\delta=0$? For example, consider a cts homomorphism $\Gamma\rightarrow G(k)$ where $\Gamma$ is a procyclic group ($k$ could be discrete or have an interesting topology). Then the functor $V\mapsto V^\Gamma$ may not be exact, but the group cohomology $H^i(\Gamma,V)$ is likely computed by the complex $V\rightarrow V$, where the map is $\gamma-1$ for a topological generator $\gamma$ of $\Gamma$. If $V\mapsto V^\Gamma$ is exact, it looks to me like I get 2 $G$-torsors (one for $H^0$, one for $H^1$), but if not, I don't think $\delta=0$. –  Rebecca Bellovin Dec 23 '13 at 13:28
add comment

Your Answer


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.