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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.

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  • $\begingroup$ Just look at the definition given in 1.3. If you are attentive to the definition being used (rather than try to import another definition taken for elsewhere, for example) then you will see that it is a tautology. $\endgroup$
    – grghxy
    Jun 18, 2015 at 5:54

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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.

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