## 100.19 Automorphism groups

Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ correspond to $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$. In this situation we often use the phrase “let $G_ x/k$ be the automorphism group algebraic space of $x$”. This just means that

\[ G_ x = \mathit{Isom}_\mathcal {X}(x, x) = \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathcal{I}_\mathcal {X} \]

is the group algebraic space of automorphism of $x$. This is a group algebraic space over $\mathop{\mathrm{Spec}}(k)$. If $k'/k$ is an extension of fields then the automorphism group algebraic space of the induced morphism $x' : \mathop{\mathrm{Spec}}(k') \to \mathcal{X}$ is the base change of $G_ x$ to $\mathop{\mathrm{Spec}}(k')$.

Lemma 100.19.1. In the situation above $G_ x$ is a scheme if one of the following holds

$\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is quasi-separated

$\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is locally separated,

$\mathcal{X}$ is quasi-DM,

$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-separated,

$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is locally separated, or

$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is locally quasi-finite.

**Proof.**
Observe that (1) $\Rightarrow $ (4), (2) $\Rightarrow $ (5), and (3) $\Rightarrow $ (6) by Lemma 100.6.1. In case (4) we see that $G_ x$ is a quasi-separated algebraic space and in case (5) we see that $G_ x$ is a locally separated algebraic space. In both cases $G_ x$ is a decent algebraic space (Decent Spaces, Section 67.6 and Lemma 67.15.2). Then $G_ x$ is separated by More on Groupoids in Spaces, Lemma 78.9.4 whereupon we conclude that $G_ x$ is a scheme by More on Groupoids in Spaces, Proposition 78.10.3. In case (6) we see that $G_ x \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite and hence $G_ x$ is a scheme by Spaces over Fields, Lemma 71.10.8.
$\square$

Lemma 100.19.2. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ be a point. Let $P$ be a property of algebraic spaces over fields which is invariant under ground field extensions; for example $P(X/k) = X \to \mathop{\mathrm{Spec}}(k)\text{ is finite}$. The following are equivalent

for some morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the class of $x$ the automorphism group algebraic space $G_ x/k$ has $P$, and

for any morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the class of $x$ the automorphism group algebraic space $G_ x/k$ has $P$.

**Proof.**
Omitted.
$\square$

Lemma 100.19.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ be a point. The following are equivalent

for some morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the class of $x$ setting $y = f \circ x$ the map $G_ x \to G_ y$ of automorphism group algebraic spaces is an isomorphism, and

for any morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the class of $x$ setting $y = f \circ x$ the map $G_ x \to G_ y$ of automorphism group algebraic spaces is an isomorphism.

**Proof.**
This comes down to the fact that being an isomorphism is fpqc local on the target, see Descent on Spaces, Lemma 73.11.15. Namely, suppose that $k'/k$ is an extension of fields and denote $x' : \mathop{\mathrm{Spec}}(k') \to \mathcal{X}$ the composition and set $y' = f \circ x'$. Then the morphism $G_{x'} \to G_{y'}$ is the base change of $G_ x \to G_ y$ by $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. Hence $G_ x \to G_ y$ is an isomorphism if and only if $G_{x'} \to G_{y'}$ is an isomorphism. Thus we see that the property propagates through the equivalence class if it holds for one.
$\square$

## Comments (2)

Comment #4347 by Chris Birkbeck on

Comment #4497 by Johan on