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Let $k$ be a field of characteristic zero. Let $X$ be a smooth DM stack over $k.$ Is the inertia stack $IX$ always smooth over $k$?

I believe this is true, but cannot find a proof in the literature. I would appreciate it if someone can give a proof or a reference of a proof.

PS: We do not assume $X$ is separated.

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Let X be a DM stack over a scheme S. X is called tame if its automorphism groups are of order invertible on S. For example, every DM stack over a field of characteristic zero is tame.

If X is tame and smooth over S, then the (relative) inertia stack $I_{X/S}$ is smooth over S. This is because $I_{X/S}$ can be decomposed as the disjoint union of $Map^{repr}(B\mathbb{Z}/n, X)$, stack of representable maps $B\mathbb{Z}/n \to X$, over integers $n$ invertible on $S$. Each $G=\mathbb{Z}/n$ is linearly reductive over $S$ so these mapping stacks are smooth (the cotangent complex is the $G$-invariants of the cotangent complex $L_X$ pulled back to $BG$).

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  • $\begingroup$ Let $X_n$ be a so-called mapping stack. Is it a DM stack? If yes, why? How does the smoothness of $X_n$ follow from the linear reductivity of $G=\mathbb{Z}/n\mathbb{Z}?$ Can you describe a smooth scheme $Y_n$ with a smooth surjective morphism $Y_n \to X_n?$ $\endgroup$ Jul 13, 2022 at 1:22
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    $\begingroup$ For a wild counterexample, one can take the quotient of $\mathbb A^2$ over a field of characteristic $p$ by the automorphism $(x,y) \to (x, y+x^2)$ of order $p$, whose inertia stack contains as a component (the quotient of) the fixed locus of this automorphism, which is non-reduced. $\endgroup$
    – Will Sawin
    Jul 13, 2022 at 1:22
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    $\begingroup$ @YuhangChen Taking $G$-invariants is an exact functor when $G$ is linearly reductive, so the cotangent complex is still acyclic in negative degrees. Existence of such $Y_n$ follows from smoothness, yes. If you're not comfortable with the mapping stack, you can just describe the cotangent complex at a point of the inertia stack in the same way -- these points correspond to a point of the stack together with an automorphism at that point, and the cotangent complex at this point has the same description as in the answer where $G$ is the cyclic group generated by the automorphism. $\endgroup$ Jul 13, 2022 at 12:16
  • $\begingroup$ @crystalline Thanks for the explanations. I am not familiar with both the mapping stacks and the cotangent complex of a DM stack or an Artin stack. But I think the existence of the decomposition of the inertia stack $IX$ into a disjoint union of mapping stacks requires a proof. You meant this decomposition is not required for a proof like this: the cotangent complex $L_{IX}$ pulled back to each $BG$ is perfect in degree zero, and hence so is $L_{IX}$, which proves that the inertia stack of a smooth tame DM stack is smooth. Did I understand correctly? $\endgroup$ Jul 13, 2022 at 15:57
  • $\begingroup$ @YuhangChen Yes, sounds right. $\endgroup$ Jul 14, 2022 at 10:49

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