Section 101.19 (0DTR): Automorphism groups

101.19 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 101.19.1. 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 101.6.1. 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 68.6 and Lemma 68.15.2). Then $G_ x$ is separated by More on Groupoids in Spaces, Lemma 79.9.4 whereupon we conclude that $G_ x$ is a scheme by More on Groupoids in Spaces, Proposition 79.10.3. 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 72.10.8. $\square$

Lemma 101.19.2. 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. $\square$

Remark 101.19.3. 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 101.19.2 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 101.19.4. 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 74.11.15. 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. $\square$

Remark 101.19.5. 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 101.19.4 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 $f(x)$ .

Comments (2)

Comment #4347 by Chris Birkbeck on August 13, 2019 at 13:42

Typo in the proof of Lemma 101.19.4 : In the second to last sentence it should say "Hence $G_x \to G_y$ is an isomorphism"

Comment #4497 by Johan on September 02, 2019 at 12:41

Thanks and fixed here.

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The Stacks project

101.19 Automorphism groups

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