Equivariant derived category and invariant divisor I'm looking for a reference of the following (folklore?) result.
Let $X$ be a smooth projective variety equipped with a $G=\mathbb{Z}/2\mathbb{Z}$ action (we consider the simplest case, everything can be generalized to an arbitrary finite group). Suppose that the invariant subset is a smooth divisor $Z$. Put $Y=X/G$ and let $p:X\to Y$ and $i:Z\to X$ denote the projection and inclusion maps correspondingly. Then on has a semiorthogonal decomposition
$D^G(X)=\left<p^*D(Y), i_*D(Z)\right>$,
where $D(Y)$ stands for the bounded derived category of coherent sheaves, $D^G(X)$ denotes the equivariant counterpart and $i_*D(Z)$ is equipped with a non-trivial $G$-action. (I might be confusing the order of components.)
 A: I know you wanted a reference, but let me try to sketch a proof anyway.
In general, suppose we have an adjoint pair of functors between triangulated categories (or better, cocomplete pretriangulated dg categories or similar(?)). 
$$ R: \mathcal C \leftrightarrows \mathcal D: L$$
where $RL \simeq id_D$ (so $L$ is fully faithful).
Claim (well known, I guess): we have a semiorthogonal decomposition $ (ker(R), im(L))$.
Proof: First, it is clear that $ker(R)$ is contained in the right orthogonal to $im(L)$ by adjunction. On the other hand, if $M$ is contained in the right orthogonal to $ker(R)$, then consider the map $f:LR(M) \to M$ given by the counit of the adjunction. The map $R(f)$ is an isomorphism as $RL\simeq id$, thus the cone of $f$ is in the kernel of $R$. As $M$ is right orthogonal to $ker(R)$, the cone is zero, so $f$ must be an isomorphism. Thus $M$ is contained in the image of $L$. $\square$
Using the notation from the question, we take $\mathcal C=D(X)^G$, $\mathcal D = D(Y)$, $L=p^\ast$ and $R$ is the functor of invariants. Your question is then reduced to: 
Claim: $i_\ast D(Z) = ker(R)$.
Proof: Everything is local, so we may assume $X=Spec(A)$ is affine, then $Y=Spec(A^G)$, and $Z=Spec(A/I)$, where $I$ is the ideal generated by $g(f)-f$, where $f\in A$, and $g$ is the involution. The functor $L= A\otimes_{A^G} (-)$, and $R = (-)^G$.
It is clear that $i_\ast D(Z)$ is contained in the kernel of invariants. To see the other containment, we must prove that if $M$ is in the kernel of invariants then it is annihilated by the ideal $I$. Note that $g(m)=-m$ for any $m\in M$. Consider
$$ (g(f) - f)(g(m)-m) = 2(g(f)-f)m$$
for $f\in A$ and nonzero $m\in M$. This is a $G$-invariant element of $M$, so must be zero, thus $m$ is annihilated by $g(f)-f$ as required. 
A: It would be nice to have that general statement, which looks very similar to the blowup formula.
I don't know if it holds. For curves (essentially over C) this was studied here http://arxiv.org/pdf/math/0410283.pdf (page 5)
You might also find some information here http://arxiv.org/pdf/math/0206144v1.pdf.
