Let $f: \mathcal{X}\to X$ be a morphism from a smooth DMstack $\mathcal{X}$ to its coarse moduli space $X$. Assume that $X$ is also smooth. Is it true that $Lf^*$ is fully faithful and induces an equivalence of $D^b(coh(X))$ with an admissible subcategory of $D^b(coh(\mathcal{X}))$?
1 Answer
If both $\mathcal{X}$ and $X$ are locally Noetherian and regular, then $f$ is flat. Then $Lf^*$ is the usual pullback $f^*$. If $\mathcal{X}$ is tame, then the natural transformation $$\theta:\text{Id} \Rightarrow f_*f^*,$$ is a natural isomorphism. However, when $\mathcal{X}$ is not tame, this can fail. For instance, let $k$ be a field of characteristic $p>0$, let $(\mathbb{Z}/p\mathbb{Z})_k$ denote the usual finite, étale $k$group scheme whose underlying group of $k$points is $\mathbb{Z}/p\mathbb{Z}$, and consider the ArtinSchreier action,
$$
\mu:(\mathbb{Z}/p\mathbb{Z})_k\times_k \mathbb{P}^1_k \to \mathbb{P}^1_k, \ a\cdot[s,t] = [s+at,t].
$$
Let $q:\mathbb{P}^1_k \to \mathcal{X}$ be the associated quotient stack, which is a smooth DeligneMumford stack.
Consider the $(\mathbb{Z}/p\mathbb{Z})_k$invariant $k$morphism,
$$
F:\mathbb{P}^1_k \to \mathbb{P}^1_k, \ F([s,t]) = [s^pst^{p1},t^p].
$$
This is the uniform categorical quotient in the category of $k$schemes. Denote the target by $X$.
Thus $F$ factors through a $1$morphism of stacks,
$$
f:\mathcal{X} \to X,
$$
and this is a coarse moduli space of $\mathcal{X}$. But now consider the fiber over the closed point $\infty = [1,0]$ in $X$. The schemetheoretic fiber of $F$ is the closed subscheme $$Z(t^p) \cong \text{Spec}\ k[(t/s)]/\langle (t/s)^p \rangle.$$ Moreover, the induced action of $(\mathbb{Z}/p\mathbb{Z})_k$ on $Z(t^p)$ by $k$morphisms is
$$
\mu_\infty: (\mathbb{Z}/p\mathbb{Z})_k \times_k \text{Spec}\ k[(t/s)]/\langle (t/s)^p \rangle \to \text{Spec}\ k[(t/s)]/\langle (t/s)^p \rangle,$$
$$ \mu_\infty(a)^*(t/s) = (t/s)/(1+a(t/s)) = (t/s)(1a(t/s)+a^2(t/s)^2 + \dots + (a)^{p2}(t/s)^{p2}),
$$
for every $a\in \mathbb{Z}/p\mathbb{Z} \subset k$.
In particular, the $\mathbb{Z}/p\mathbb{Z}$invariant $k$subspace of $k[(t/s)]/\langle (t/s)^p \rangle$ is spanned by $1$ and $(t/s)^{p1}$. Since the invariant subspace is not $1$dimensional, for the skyscraper sheaf $\kappa(\infty)$ on $X$ supported at $\infty$, the natural transformation
$$
\theta_{\kappa(\infty)} : \kappa(\infty) \to f_*f^*\kappa(\infty),
$$
is not an isomorphism.

$\begingroup$ Thanks for the answer, Jason, but could you explain, please, why $f$ is flat and $\theta: Id\to Rf_{*}f^*$ is isomorphism in the tame case? $\endgroup$ Aug 26, 2015 at 11:56

$\begingroup$ First of all, in the tame case, $Rf_*$ equals $f_*$. In the nontame case, $Rf_*$ does not even map $D^b$ to $D^b$. $\endgroup$ Aug 26, 2015 at 12:13

$\begingroup$ Flatness of $f$ is a corollary of the local flatness criterion, cf. Theorem 23.1, p. 179 of H. Matsumura, Commutative Ring Theory, Cambridge U. Press. However, via the ChevalleyShephardTodd theorem, your hypotheses are extremely strong. $\endgroup$ Aug 26, 2015 at 12:17

3$\begingroup$ Both exactness of $f_*$ and $\theta$ being isomorphic can be checked etale locally on $X$. Thus, assume that $\mathcal{X}$ is $[Y/G]$ with induced map $F:Y\to X$ and $G$ prime to the characteristic. Then $f_*f^*$ is the same as $(F_*F^*())^G$. Since $F$ is finite (hence affine), $F_*$ is exact. Since $G$ is prime to the characteristic $()^G$ is exact. Thus $f_*$ is exact. Because $f_*$ is exact, to check $\theta$ is isomorphic, it suffices to consider $\theta_{\mathcal{O}}$. This case follows from the definition of the categorical quotient of $Y$ by $G$. $\endgroup$ Aug 26, 2015 at 12:43