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Let $k$ be a field. Let $X$ be a connected tame DM stack over $k.$ Let $IX$ be the inertia stack of $X.$ Then $IX$ is a disjoint union of connected components.

Is this always a finite union? If not, is it always countable?

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    $\begingroup$ What about $B\mathbb{Z}$ over $\mathbb{Q}$? $\endgroup$ Jul 26, 2022 at 20:06
  • $\begingroup$ @AriyanJavanpeykar Thanks for the example. So finiteness is too much to ask for. The next question is then whether $IX$ is a countable union of connected components. $\endgroup$ Jul 26, 2022 at 23:10
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    $\begingroup$ Let $G$ be a uncountable (abstract) group and consider $BG$ over $\mathbb{Q}$? $\endgroup$ Jul 27, 2022 at 7:05
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    $\begingroup$ @YuhangChen What? $G$ is discrete by definition, in particular it is an (uncountable) disjoint union of Spec(k)'s, so obviously etale (but not of finite type). $\endgroup$ Jul 28, 2022 at 3:03
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    $\begingroup$ @YuhangChen What's your definition of tame DM stack? According to arxiv.org/abs/math/0703310 such a stack has, by assumption, a finite diagonal. In that case, the inertia stack is finite over $X$, so that finiteness of connected components follows. $\endgroup$ Jul 28, 2022 at 7:16

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As Ariyan's examples in the comments show, you need to at least impose some separation axioms on $X$, at least quasi-separated. Note that $BG$ for $G$ an infinite discrete group is not quasi-separated. Moreover, to get finiteness you also need to impose some finiteness assumption on $X$ as well. An example statement is the following.

Lemma: Suppose $X$ is a noetherian and quasi-separated Deligne-Mumford stack. Then $IX$ is noetherian and so in particular it has finitely many connected components.

Proof: Note that noetherian = locally noetherian + quasi-compact. Since $X$ is quasi-separated, the diagonal is quasi-compact so $IX \to X$ is quasi-compact and thus $IX$ is quasi-compact. Since $X$ is Deligne-Mumford $IX \to X$ is quasi-finite so $IX$ is also locally noetherian.

For a counterexample that is separated but not noetherian, we can take $R = k[x_1, \ldots, x_n, \ldots]$ with the $\mathbb{G}_m$ action such that $x_i$ has weight $i$. Then $X = [\mathrm{Spec} R \setminus \{(0,\ldots, 0, \ldots)\}/\mathbb{G}_m]$ has finite inertia but the inertia stack has infinitely many components.

For a counterexample that is separated and locally noetherian but not quasi-compact we can take an infinite chain of weighted projective lines $\mathcal{P}(n,n+1)$ where the point at infinity on $\mathcal{P}(n,n+1)$ is glued to the point at $0$ on $\mathcal{P}(n+1, n+2)$.

For a counterexample that is separated but not locally noetherian, we can take the union of the infinitely many coordinate axes in $\mathrm{Spec} k[x_1, \ldots, x_n, \ldots]$ and then put a $\mu_d$ stabilizer at the coordinate $x_d = 1$ on the $d^{th}$ component.

In each of these cases the inertia is in fact finite not just quasi-finite and it has countably many components. In general this phenomena is related to the fact that the stratification of $X$ into gerbes was infinite. Whenever $IX \to X$ is quasi-compact (so in particular if $X$ is quasi-separated) there is a well-ordered stratiication of $X$ into gerbes $U_i$ (Stacks Project Tag 06RF). We can choose the $U_i$ to be quasi-compact and connected, and $IU_i \to U_i$ is quasi-finite and quasi-compact by our assumptions.

If moreover $X$ is locally noetherian, then so is $IU_i$ for each $i$ so $IU_i$ has finitely many components. If we drop the locally noetherian assumption but assume separated, then $IU_i \to U_i$ is finite flat which again implies that $IU_i$ has finitely many connected components. In either case, the cardinality of the index set $I$ for the stratification gives an upper bound on the cardinality of the connected components of inertia.

If you just assume quasi-separated but neither locally noetherian nor separated, then I'm not sure if we can still deduce that $IU_i$ has finitely many components since connected components on quasi-compact but non-locally noetherian spaces can be quite strange.

Finally the question about whether the cardinality of the set $I$ indexing the stratification into gerbes is countable seems to be open, see Tag 06RG.

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  • $\begingroup$ In the Lemma, is $X$ quasi-separated over a field? $\endgroup$ Jul 28, 2022 at 21:28
  • $\begingroup$ I think quasi-separated over an affine scheme is good enough. $\endgroup$ Jul 28, 2022 at 21:56

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