Fiber functors to derived categories

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?

• 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