## 95.19 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 95.19.1. In the situation above $G_ x$ is a scheme if one of the following holds

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

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

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

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

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

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

Proof. Observe that (1) $\Rightarrow$ (4), (2) $\Rightarrow$ (5), and (3) $\Rightarrow$ (6) by Lemma 95.6.1. 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 62.6 and Lemma 62.15.2). Then $G_ x$ is separated by More on Groupoids in Spaces, Lemma 73.9.4 whereupon we conclude that $G_ x$ is a scheme by More on Groupoids in Spaces, Proposition 73.10.3. 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 66.10.8. $\square$

Lemma 95.19.2. 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

1. 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

2. 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. $\square$

Remark 95.19.3. Let $P$ be a property of algebraic spaces over fields which is invariant under ground field extensions. Given an algebraic stack $\mathcal{X}$ and $x \in |\mathcal{X}|$, we say the automorphism group of $\mathcal{X}$ at $x$ has $P$ if the equivalent conditions of Lemma 95.19.2 are satisfied. For example, we say the automorphism group of $\mathcal{X}$ at $x$ is finite, if $G_ x \to \mathop{\mathrm{Spec}}(k)$ is finite whenever $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is a representative of $x$. Similarly for smooth, proper, etc. (There is clearly an abuse of language going on here, but we believe it will not cause confusion or imprecision.)

Lemma 95.19.4. 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

1. 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

2. 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 68.10.15. 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 isomorpism 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. $\square$

Remark 95.19.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ be a point. To indicate the equivalent conditions of Lemma 95.19.4 are satisfied for $f$ and $x$ in the literature the terminology $f$ is stabilizer preserving at $x$ or $f$ is fixed-point reflecting at $x$ is used. We prefer to say $f$ induces an isomorphism between automorphism groups at $x$ and $f(x)$.

Comment #4347 by on

Typo in the proof of Lemma 95.19.4 : In the second to last sentence it should say "Hence $G_x \to G_y$ is an isomorphism"

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