Lemma 101.29.3. Let $\mathcal{X}$ be an algebraic stack such that $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact. Then there exists a well-ordered index set $I$ and for every $i \in I$ a reduced locally closed substack $\mathcal{U}_ i \subset \mathcal{X}$ such that

each $\mathcal{U}_ i$ is a gerbe,

we have $|\mathcal{X}| = \bigcup _{i \in I} |\mathcal{U}_ i|$,

$T_ i = |\mathcal{X}| \setminus \bigcup _{i' < i} |\mathcal{U}_{i'}|$ is closed in $|\mathcal{X}|$ for all $i \in I$, and

$|\mathcal{U}_ i|$ is open in $T_ i$.

We can moreover arrange it so that either (a) $|\mathcal{U}_ i| \subset T_ i$ is dense, or (b) $\mathcal{U}_ i$ is quasi-compact. In case (a), if we choose $\mathcal{U}_ i$ as large as possible (see proof for details), then the stratification is canonical.

**Proof.**
Let $T \subset |\mathcal{X}|$ be a nonempty closed subset. We are going to find (resp. choose) for every such $T$ a reduced locally closed substack $\mathcal{U}(T) \subset \mathcal{X}$ with $|\mathcal{U}(T)| \subset T$ open dense (resp. nonempty quasi-compact). Namely, by Properties of Stacks, Lemma 100.10.1 there exists a unique reduced closed substack $\mathcal{X}' \subset \mathcal{X}$ such that $T = |\mathcal{X}'|$. Note that $\mathcal{I}_{\mathcal{X}'} = \mathcal{I}_\mathcal {X} \times _\mathcal {X} \mathcal{X}'$ by Lemma 101.5.6. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is quasi-compact as a base change, see Lemma 101.7.3. Therefore Proposition 101.29.1 implies there exists a dense maximal (see proof proposition) open substack $\mathcal{U} \subset \mathcal{X}'$ which is a gerbe. In case (a) we set $\mathcal{U}(T) = \mathcal{U}$ (this is canonical) and in case (b) we simply choose a nonempty quasi-compact open $\mathcal{U}(T) \subset \mathcal{U}$, see Properties of Stacks, Lemma 100.4.9 (we can do this for all $T$ simultaneously by the axiom of choice).

Using transfinite recursion we construct for every ordinal $\alpha $ a closed subset $T_\alpha \subset |\mathcal{X}|$. For $\alpha = 0$ we set $T_0 = |\mathcal{X}|$. Given $T_\alpha $ set

\[ T_{\alpha + 1} = T_\alpha \setminus |\mathcal{U}(T_\alpha )|. \]

If $\beta $ is a limit ordinal we set

\[ T_\beta = \bigcap \nolimits _{\alpha < \beta } T_\alpha . \]

We claim that $T_\alpha = \emptyset $ for all $\alpha $ large enough. Namely, assume that $T_\alpha \not= \emptyset $ for all $\alpha $. Then we obtain an injective map from the class of ordinals into the set of subsets of $|\mathcal{X}|$ which is a contradiction.

The claim implies the lemma. Namely, let

\[ I = \{ \alpha \mid \mathcal{U}_\alpha \not= \emptyset \} . \]

This is a well-ordered set by the claim. For $i = \alpha \in I$ we set $\mathcal{U}_ i = \mathcal{U}_\alpha $. So $\mathcal{U}_ i$ is a reduced locally closed substack and a gerbe, i.e., (1) holds. By construction $T_ i = T\alpha $ if $i = \alpha \in I$, hence (3) holds. Also, (4) and (a) or (b) hold by our choice of $\mathcal{U}(T)$ as well. Finally, to see (2) let $x \in |\mathcal{X}|$. There exists a smallest ordinal $\beta $ with $x \not\in T_\beta $ (because the ordinals are well-ordered). In this case $\beta $ has to be a successor ordinal by the definition of $T_\beta $ for limit ordinals. Hence $\beta = \alpha + 1$ and $x \in |\mathcal{U}(T_\alpha )|$ and we win.
$\square$

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