Proposition 101.29.1. 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.

## 101.29 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

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 100.10) in order to study this stratification.

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

**Proof.**
According to Proposition 101.28.9 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 101.5.5.

Choose a presentation $\mathcal{X} = [U/R]$. Let $G \to U$ be the stabilizer group algebraic space of the groupoid $R$. By Lemma 101.5.7 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 101.5.1). 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 67.32.1 we see that $W \subset U$ is dense. Note that $W \subset U$ is $R$-invariant by More on Groupoids in Spaces, Lemma 79.6.2. Hence $W$ corresponds to an open substack $\mathcal{U} \subset \mathcal{X}$ by Properties of Stacks, Lemma 100.9.11. 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 101.25.4 and 101.27.11. $\square$

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 101.31.1. It turns out that there doesn't always exist a finite stratification by gerbes. Here is an example.

Example 101.29.2. 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

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

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

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

and so on, and

$\mathcal{X}_{\infty }$ 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 101.29.1 find possibly infinite stratifications by gerbes...!

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

If $\beta $ is a limit ordinal we set

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

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 101.29.4. We can wonder about the order type of the canonical stratifications which occur as output of the stratifications of type (a) constructed in Lemma 101.29.3. 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 stacks.project@gmail.com.

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