Can someone give me a reference for the following or an idea on why it is true? (This is taken from remark 1.5 on page 5 of <http://arxiv.org/abs/0810.0794>.)


Suppose we have an algebraic group $G$ acting on an algebraic variety $X$. If $G$ is a connected unipotent group over $k$ ($\operatorname{char} k > 0$), one can show that the forgetful functor from the equivariant derived category of $\ell$-adic sheaves on $X$ to the derived category of $\ell$-adic sheaves is fully faithful.



Equivariant Derived Category
While the statement is a tautology if one uses Definition 1.3 in the cited paper (the "naive" equivariant derived category), I took the question to be about the usual definition of equivariant derived category (e.g. as in Bernstein-Lunts); as such, it has content.


The proof that the two definitions agree is in Appendix C of the same paper. In particular, this proof shows that the forgetful functor from the usual equivariant derived category is fully faithful. Below is a sketch of a proof of fully faithfullness in case it helps.


To my mind, the key point is that a unipotent group $G$ is isomorphic as a variety over $k$ to an affine space. In particular it has trivial cohomology: $H^j(G;\mathbb Q\_\ell)=0$ for $j>0$. The forgetful functor


$$ F:D\_G(X) \to D(X)$$
has a right adjoint, $E$. To show that $F$ is fully faithful, we just need to check that the unit $$1\_{D\_G(X)} \to EF$$ is an isomorphism (this is easy to see). 


In the language of stacks, this isomorphism is fairly intuitive: there is a morphism $f:X \to X/G$ which is a $G$-torsor - in partiuclar, it is a "fibre bundle with contractible fibres". In this language, the functors $E$ and $F$ are $Rf\_\ast$ and $f^\ast$ respectively. We have 
$$Rf\_\ast f^\ast (\mathcal F) = \mathcal F \otimes Rf\_\ast(\mathbb Q\_{\ell X})$$ by the projection formula. As the fibres of $f$ are contractible $Rf\_\ast (\mathbb Q\_{\ell X}) \simeq \mathbb Q\_{\ell X/G}$. Thus $Rf\_\ast f^\ast \mathcal F \simeq \mathcal F$ as required.


