Let $f: \mathcal{X}\to X$ be a morphism from a smooth DM-stack $\mathcal{X}$ 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 1


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 Artin-Schreier 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 Deligne-Mumford stack. Consider the $(\mathbb{Z}/p\mathbb{Z})_k$-invariant $k$-morphism, $$ F:\mathbb{P}^1_k \to \mathbb{P}^1_k, \ F([s,t]) = [s^p-st^{p-1},t^p]. $$ 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 scheme-theoretic 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 $$ \mu_\infty: (\mathbb{Z}/p\mathbb{Z})_k \times_k \text{Spec}\ k[(t/s)]/\langle (t/s)^p \rangle \to \text{Spec}\ k[(t/s)]/\langle (t/s)^p \rangle,$$ $$ \mu_\infty(a)^*(t/s) = (t/s)/(1+a(t/s)) = (t/s)(1-a(t/s)+a^2(t/s)^2 + \dots + (-a)^{p-2}(t/s)^{p-2}), $$ 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 $(t/s)^{p-1}$. 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 non-tame 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 Chevalley-Shephard-Todd 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.