It is a wellknown fact that for an Enriques surface $Y$, if $D^b(Y)\cong D^b(X)$ for some smooth projective variety $X$, then $X\cong Y$. In other words, Enriques surfaces have no nontrivial FourierMukai Partners. I was wondering if the answer to this question is known if we consider twisted derived categories instead. In particular, what is known about $(X,\alpha)$ such that $D^b(Y)\cong D^b(X,\alpha)$, where $Y$ is an Enriques surface as above and $\alpha$ is a Brauer class on the smooth projective variety $X$.
There is a natural pair $(X,\alpha)$ that you can construct. Only in this pair $X$ is not a smooth projective variety but is a smooth orbifold surface. If you choose a genus one pencil $Y \to \mathbb{P}^{1}$ on the Enriques surface, then the relative Picard stack $\mathcal{X} = \mathcal{P}ic^{0}(Y/\mathbb{P}^{1})$ of the pencil is a FourierMukai partner. The stack $\mathcal{X}$ is a $\mathbb{G}_{m}$ gerbe over an orbifold surface $X$. The moduli space of $X$ is the relative Jacobian $J$ of the chosen genus one fibration. $J$ is a rational elliptic surface, and $X$ is $J$ equipped with a $\mathbb{Z}/2$ orbifold structure along two smooth fibers of the anticanonical map $J \to \mathbb{P}^{1}$, i.e. along the two fibers dual to the double fibers of $Y \to \mathbb{P}^{1}$. The gerbe $\mathcal{X} \to X$ gives you a nontrivial $2$torsion element $\alpha \in Br(X)$ and the derived category of weight one sheaves on $\mathcal{X}$ is equivalent to the $\alpha$twisted derived category of $X$. So you get an equivalence $D^{b}(Y) \cong D^{b}(X,\alpha)$.

$\begingroup$ Thanks Tony, I agree this is very natural. Is this the only thing that can happen though? Is there something that rules this out when X is a genuine smooth projective surface? $\endgroup$ – HNuer Feb 19 '15 at 15:01

$\begingroup$ My guess is that it will be impossible to do it with a genuine surface. One might be able to check this directly by looking at the classification of surfaces, listing all possible twisted Hodge structures, and then checking that the required lattice $E_{8}(2)\oplus H(2)\oplus H$ can never appear in a twist of a surface. I have not done this carefully but it seems doable. You basically have to rule out K#s and rational elliptic surfaces. $\endgroup$ – Tony Pantev Feb 20 '15 at 13:42