The Stacks project

100.29 Stratification by gerbes

The goal of this section is to show that many algebraic stacks $\mathcal{X}$ have a “stratification” by locally closed substacks $\mathcal{X}_ i \subset \mathcal{X}$ such that each $\mathcal{X}_ i$ is a gerbe. This shows that in some sense gerbes are the building blocks out of which any algebraic stack is constructed. Note that by stratification we only mean that

\[ |\mathcal{X}| = \bigcup \nolimits _ i |\mathcal{X}_ i| \]

is a stratification of the topological space associated to $\mathcal{X}$ and nothing more (in this section). Hence it is harmless to replace $\mathcal{X}$ by its reduction (see Properties of Stacks, Section 99.10) in order to study this stratification.

The following proposition tells us there is (almost always) a dense open substack of the reduction of $\mathcal{X}$

Proposition 100.29.1. Let $\mathcal{X}$ be a reduced algebraic stack such that $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact. Then there exists a dense open substack $\mathcal{U} \subset \mathcal{X}$ which is a gerbe.

Proof. According to Proposition 100.28.9 it is enough to find a dense open substack $\mathcal{U}$ such that $\mathcal{I}_\mathcal {U} \to \mathcal{U}$ is flat and locally of finite presentation. Note that $\mathcal{I}_\mathcal {U} = \mathcal{I}_\mathcal {X} \times _\mathcal {X} \mathcal{U}$, see Lemma 100.5.5.

Choose a presentation $\mathcal{X} = [U/R]$. Let $G \to U$ be the stabilizer group algebraic space of the groupoid $R$. By Lemma 100.5.7 we see that $G \to U$ is the base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ hence quasi-compact (by assumption) and locally of finite type (by Lemma 100.5.1). Let $W \subset U$ be the largest open (possibly empty) subscheme such that the restriction $G_ W \to W$ is flat and locally of finite presentation (we omit the proof that $W$ exists; hint: use that the properties are local). By Morphisms of Spaces, Proposition 66.32.1 we see that $W \subset U$ is dense. Note that $W \subset U$ is $R$-invariant by More on Groupoids in Spaces, Lemma 78.6.2. Hence $W$ corresponds to an open substack $\mathcal{U} \subset \mathcal{X}$ by Properties of Stacks, Lemma 99.9.11. Since $|U| \to |\mathcal{X}|$ is open and $|W| \subset |U|$ is dense we conclude that $\mathcal{U}$ is dense in $\mathcal{X}$. Finally, the morphism $\mathcal{I}_\mathcal {U} \to \mathcal{U}$ is flat and locally of finite presentation because the base change by the surjective smooth morphism $W \to \mathcal{U}$ is the morphism $G_ W \to W$ which is flat and locally of finite presentation by construction. See Lemmas 100.25.4 and 100.27.11. $\square$

The above proposition immediately implies that any point has a residual gerbe on an algebraic stack with quasi-compact inertia, as we will show in Lemma 100.31.1. It turns out that there doesn't always exist a finite stratification by gerbes. Here is an example.

Example 100.29.2. Let $k$ be a field. Take $U = \mathop{\mathrm{Spec}}(k[x_0, x_1, x_2, \ldots ])$ and let $\mathbf{G}_ m$ act by $t(x_0, x_1, x_2, \ldots ) = (tx_0, t^ p x_1, t^{p^2} x_2, \ldots )$ where $p$ is a prime number. Let $\mathcal{X} = [U/\mathbf{G}_ m]$. This is an algebraic stack. There is a stratification of $\mathcal{X}$ by strata

  1. $\mathcal{X}_0$ is where $x_0$ is not zero,

  2. $\mathcal{X}_1$ is where $x_0$ is zero but $x_1$ is not zero,

  3. $\mathcal{X}_2$ is where $x_0, x_1$ are zero, but $x_2$ is not zero,

  4. and so on, and

  5. $\mathcal{X}_{\infty }$ is where all the $x_ i$ are zero.

Each stratum is a gerbe over a scheme with group $\mu _{p^ i}$ for $\mathcal{X}_ i$ and $\mathbf{G}_ m$ for $\mathcal{X}_{\infty }$. The strata are reduced locally closed substacks. There is no coarser stratification with the same properties.

Nonetheless, using transfinite induction we can use Proposition 100.29.1 find possibly infinite stratifications by gerbes...!

Lemma 100.29.3. 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

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

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

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

  4. $|\mathcal{U}_ i|$ 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 99.10.1 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 100.5.6. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is quasi-compact as a base change, see Lemma 100.7.3. Therefore Proposition 100.29.1 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 99.4.9 (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. $\square$

Remark 100.29.4. We can wonder about the order type of the canonical stratifications which occur as output of the stratifications of type (a) constructed in Lemma 100.29.3. A natural guess is that the well-ordered set $I$ has cardinality at most $\aleph _0$. We have no idea if this is true or false. If you do please email

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