Proof that $[X/G]$ is an Algebraic Stack

We prove that if $G$ is a smooth algebraic group acting on an algebraic space $X$ , then the quotient stack $[X/G]$ is an algebraic stack. This is one of the most fundamental and frequently-used results in the theory of algebraic stacks. The proof consists of two main parts: verifying that $[X/G]$ is a stack, that its diagonal is representable, and that it admits a smooth atlas.

Statement

Theorem4.12Quotient stacks are algebraic

Let $S$ be a scheme, $G$ a smooth algebraic group scheme over $S$ , and $X$ an algebraic space over $S$ with a $G$ -action $\sigma : G \times_S X \to X$ . Then:

$[X/G]$ is a stack over $(\mathrm{Sch}/S)_{\mathrm{fppf}}$ .
The diagonal $\Delta : [X/G] \to [X/G] \times_S [X/G]$ is representable by algebraic spaces.
The natural morphism $\pi : X \to [X/G]$ is a smooth surjection (and is a $G$ -torsor).

In particular, $[X/G]$ is an algebraic (Artin) stack over $S$ , with $\pi$ as a smooth atlas.

Part 1: $[X/G]$ is a stack

ProofPart 1 -- Stack axioms

Recall that $[X/G]$ is the category fibered in groupoids over $(\mathrm{Sch}/S)$ whose objects over a scheme $T$ are pairs $(P \to T, \phi : P \to X)$ where $P$ is a $G$ -torsor over $T$ and $\phi$ is $G$ -equivariant.

We verify the two stack axioms: descent for morphisms and descent for objects.

Axiom 1 (Descent for morphisms): Let $(P_1, \phi_1)$ and $(P_2, \phi_2)$ be two objects over $T$ . We need to show that the presheaf $\operatorname{Isom}_T((P_1, \phi_1), (P_2, \phi_2))$ is a sheaf in the fppf topology on $(\mathrm{Sch}/T)$ .

An isomorphism $(P_1, \phi_1) \xrightarrow{\sim} (P_2, \phi_2)$ is a $G$ -equivariant isomorphism $\psi : P_1 \xrightarrow{\sim} P_2$ over $T$ with $\phi_2 \circ \psi = \phi_1$ . Consider the algebraic space $I = \operatorname{Isom}_T^G(P_1, P_2) \times_{\operatorname{Mor}_T(P_1, X)} \{\phi_1\}$ where $\operatorname{Isom}_T^G(P_1, P_2)$ is the sheaf of $G$ -equivariant isomorphisms from $P_1$ to $P_2$ (which is an algebraic space since $G$ -torsors are algebraic spaces and isomorphisms between them form algebraic spaces) and the fiber product imposes $\phi_2 \circ \psi = \phi_1$ .

Since $I$ is represented by an algebraic space, its functor of points is an fppf sheaf. Hence $\operatorname{Isom}_T$ is a sheaf.

Axiom 2 (Descent for objects): Let $\{T_i \to T\}$ be an fppf cover and let $(P_i, \phi_i) \in [X/G](T_i)$ be objects with descent data: isomorphisms $\alpha_{ij} : (P_i, \phi_i)|_{T_{ij}} \xrightarrow{\sim} (P_j, \phi_j)|_{T_{ij}}$ satisfying the cocycle condition on triple overlaps $T_{ijk}$ .

Since $G$ -torsors satisfy fppf descent (a fundamental result in descent theory -- the stack $\mathrm{B}G$ is a stack for the fppf topology), the torsors $P_i$ with the descent data $\alpha_{ij}$ glue to a $G$ -torsor $P$ over $T$ .

The equivariant maps $\phi_i : P_i \to X$ are compatible (since the $\alpha_{ij}$ respect the maps to $X$ ), so by descent for morphisms of algebraic spaces, they glue to a $G$ -equivariant map $\phi : P \to X$ .

Thus $(P, \phi) \in [X/G](T)$ is the descended object. This proves $[X/G]$ is a stack. $\square$

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Part 2: Representability of the diagonal

ProofPart 2 -- Representable diagonal

We must show: for any scheme $T$ and any two objects $(P_1, \phi_1), (P_2, \phi_2) \in [X/G](T)$ , the sheaf $\operatorname{Isom}_T((P_1, \phi_1), (P_2, \phi_2))$ is represented by an algebraic space over $T$ .

Step 2a: Reduce to trivial torsors. An isomorphism from $(P_1, \phi_1)$ to $(P_2, \phi_2)$ is a $G$ -equivariant isomorphism $\psi : P_1 \xrightarrow{\sim} P_2$ over $T$ with $\phi_2 \circ \psi = \phi_1$ .

First, consider the sheaf of all $G$ -equivariant isomorphisms $\psi : P_1 \to P_2$ (without the condition on $\phi$ ). This is represented by the contracted product: $\operatorname{Isom}_T^G(P_1, P_2) \cong P_1 \times^G P_2^{\mathrm{op}}$ where $P_2^{\mathrm{op}}$ is $P_2$ with the opposite $G$ -action ( $g$ acts by $g^{-1}$ ). Explicitly, this is $(P_1 \times_T P_2)/G$ where $G$ acts diagonally by $g \cdot (p_1, p_2) = (p_1 \cdot g^{-1}, g \cdot p_2)$ .

Since $P_1$ and $P_2$ are $G$ -torsors (hence smooth over $T$ ), and $G$ acts freely on $P_1 \times_T P_2$ , the quotient $P_1 \times^G P_2^{\mathrm{op}}$ is an algebraic space, smooth over $T$ (in fact, it is a $G$ -torsor twisted by the "difference" of $P_1$ and $P_2$ ).

Step 2b: Impose the compatibility with $\phi$ . The condition $\phi_2 \circ \psi = \phi_1$ cuts out a closed subspace. More precisely, consider the two morphisms: $\operatorname{Isom}_T^G(P_1, P_2) \xrightarrow{\psi \mapsto \phi_2 \circ \psi} \operatorname{Mor}_T^G(P_1, X) \xleftarrow{\phi_1} T$ The isomorphism sheaf $\operatorname{Isom}_T((P_1,\phi_1), (P_2,\phi_2))$ is the fiber product: $\operatorname{Isom}_T((P_1,\phi_1), (P_2,\phi_2)) = T \times_{\operatorname{Mor}_T^G(P_1, X)} \operatorname{Isom}_T^G(P_1, P_2).$

Now, $\operatorname{Mor}_T^G(P_1, X) \cong X \times^G P_1^{\mathrm{op}}$ (equivariant maps from a torsor to $X$ correspond to sections of the associated $X$ -bundle). This is an algebraic space over $T$ . The map $\phi_1 : T \to \operatorname{Mor}_T^G(P_1, X)$ is a section of the structure map, hence a locally closed immersion.

Therefore, the fiber product is an algebraic space over $T$ (as a base change of algebraic spaces). In fact, if $X$ is separated, the map $\phi_1 : T \to \operatorname{Mor}_T^G(P_1, X)$ is a closed immersion, so $\operatorname{Isom}_T$ is a closed subspace of $\operatorname{Isom}_T^G(P_1, P_2)$ .

Explicit description: Locally trivializing $P_1$ and $P_2$ , the isomorphism sheaf becomes: $\operatorname{Isom}_T((P_1,\phi_1), (P_2,\phi_2))(T') = \{g \in G(T') : g \cdot \phi_1|_{T'} = \phi_2|_{T'}\}$ which is the "transporter" from $\phi_1$ to $\phi_2$ in $G$ . This is a closed subscheme of $G_{T'}$ (the fiber product $G \times_{X \times X, (\sigma, \phi_2) \times \phi_1} T$ ), hence an algebraic space. $\square$

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Part 3: The atlas is smooth and surjective

ProofPart 3 -- Smooth atlas

We prove that the natural morphism $\pi : X \to [X/G]$ is representable, smooth, and surjective.

Definition of $\pi$ : The morphism $\pi$ sends a point $x \in X(T)$ (i.e., a $T$ -valued point of $X$ ) to the object $(G \times T \xrightarrow{\mathrm{pr}_2} T, \; G \times T \xrightarrow{(g,t) \mapsto g \cdot x(t)} X)$ of $[X/G](T)$ . That is, $\pi(x)$ is the trivial $G$ -torsor with the equivariant map determined by the orbit of $x$ .

Representability of $\pi$ : We need to show that for any scheme $T$ and any object $(P, \phi) \in [X/G](T)$ , the fiber product $X \times_{[X/G]} T$ (via $\pi$ and the classifying map $T \to [X/G]$ ) is an algebraic space.

The fiber product $X \times_{[X/G]} T$ parametrizes pairs: a point $x \in X(T')$ and an isomorphism between $\pi(x)$ and $(P, \phi)|_{T'}$ in $[X/G](T')$ . Such an isomorphism is a $G$ -equivariant isomorphism $\psi : G \times T' \xrightarrow{\sim} P|_{T'}$ with $\phi \circ \psi(1, t') = x(t')$ .

This is equivalent to: a section $s : T' \to P|_{T'}$ (given by $s(t') = \psi(1, t')$ ), which is the same as a trivialization of $P$ over $T'$ . The condition $\phi \circ \psi(1, \cdot) = x$ means $x = \phi \circ s$ .

So: $X \times_{[X/G]} T \cong P$ (the total space of the $G$ -torsor $P$ ). Indeed, a point of $P$ over $T'$ is a pair (base point in $T'$ , point in the fiber of $P$ ), and the map to $X$ sends it to $\phi$ of that point.

$\pi$ is a $G$ -torsor: The above shows $X \times_{[X/G]} T \cong P$ , which is a $G$ -torsor over $T$ by assumption. Since $P \to T$ is smooth (as $G$ is smooth), the morphism $\pi : X \to [X/G]$ is smooth.

$\pi$ is surjective: For any algebraically closed field $k$ and any object $(P, \phi) \in [X/G](k)$ , the $G$ -torsor $P$ over $\operatorname{Spec} k$ is trivial (by Hilbert 90, since $k$ is algebraically closed and $G$ is smooth). A trivialization $s : \operatorname{Spec} k \to P$ gives a point $x = \phi \circ s \in X(k)$ with $\pi(x) \cong (P, \phi)$ . Hence $\pi$ is surjective on geometric points.

By the definition of surjective morphisms of stacks (surjective on geometric points), $\pi$ is surjective.

Conclusion: $\pi : X \to [X/G]$ is a smooth, surjective, representable morphism from an algebraic space $X$ . This is a smooth atlas for $[X/G]$ , proving that $[X/G]$ is an algebraic (Artin) stack. $\square$

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Verification of the groupoid structure

RemarkThe groupoid of the atlas

The smooth atlas $\pi : X \to [X/G]$ gives rise to the groupoid: $R = X \times_{[X/G]} X \cong G \times X$ with:

Source: $s(g, x) = x$ .
Target: $t(g, x) = g \cdot x$ .
Composition: $c((h, g \cdot x), (g, x)) = (hg, x)$ .
Identity: $e(x) = (1, x)$ .
Inverse: $i(g, x) = (g^{-1}, g \cdot x)$ .

This is the action groupoid of the $G$ -action on $X$ . Both $s$ and $t$ are smooth (since $G$ is smooth and $\sigma$ is a morphism of algebraic spaces), confirming that $R \rightrightarrows X$ is a smooth groupoid and $[X/G] \cong [X / (G \times X)]$ in the groupoid notation.

Special cases and refinements

ExampleFree action: quotient is an algebraic space

If $G$ acts freely on $X$ (i.e., the stabilizer $G_x$ is trivial for all geometric points $x$ ), then the diagonal of $[X/G]$ is a monomorphism (since $\operatorname{Isom}$ is at most a singleton). This means $[X/G]$ is an algebraic space, not just a stack.

The proof: the transporter $\{g \in G : g \cdot x_1 = x_2\}$ is either empty (if $x_1, x_2$ are in different orbits) or a single point (if they are in the same orbit, by freeness). So $\operatorname{Isom}$ is represented by a monomorphism, and $[X/G]$ is a sheaf of sets.

When the action is free and proper, $[X/G]$ is a separated algebraic space, and in fact agrees with the geometric quotient $X/G$ .

ExampleFinite group: DM stack

If $G$ is a finite group scheme (e.g., a constant group $G$ ), then $G$ is etale (in characteristic zero or when $|G|$ is invertible), and the atlas $\pi : X \to [X/G]$ is etale (since $G$ is etale implies $G$ -torsors are etale covers). Hence $[X/G]$ is a Deligne--Mumford stack.

The groupoid $R = G \times X \rightrightarrows X$ has etale source and target maps, confirming the etale groupoid structure.

ExampleG_m-action: Artin but not DM

For $G = \mathbb{G}_m$ acting on $X = \mathbb{A}^1$ by scaling, the atlas $\pi : \mathbb{A}^1 \to [\mathbb{A}^1/\mathbb{G}_m]$ is smooth (but not etale, since $\mathbb{G}_m$ has dimension 1). The stabilizer at the origin is $\mathbb{G}_m$ (positive-dimensional), so $[\mathbb{A}^1/\mathbb{G}_m]$ is an Artin stack but not DM.

The groupoid is $R = \mathbb{G}_m \times \mathbb{A}^1 \rightrightarrows \mathbb{A}^1$ with $s(t, x) = x$ and $t(t, x) = tx$ . Both source and target are smooth.

ExampleProper action: separated stack

If the action map $\Psi : G \times X \to X \times X$ defined by $(g, x) \mapsto (x, g \cdot x)$ is proper, then the diagonal of $[X/G]$ is proper (since the diagonal fibers are the stabilizers, which are the fibers of $\Psi$ along the diagonal). In particular, $[X/G]$ is a separated algebraic stack.

For $G$ reductive acting on a proper variety $X$ : the action is always proper (since $G$ is affine and $X$ is proper), so $[X/G]$ is separated.

ExampleNon-separated quotient stack

Let $G = \mathbb{G}_a$ (the additive group) act on $\mathbb{A}^2$ by $t \cdot (x, y) = (x, y + tx)$ . The orbits in $\{x \neq 0\}$ are the horizontal lines $\{x = c, c \neq 0\}$ , while $\{x = 0\}$ consists of infinitely many fixed points (each $(0, y)$ is fixed). The action is not proper (the map $\mathbb{G}_a \times \mathbb{A}^2 \to \mathbb{A}^2 \times \mathbb{A}^2$ is not proper along $\{x = 0\}$ ).

The quotient stack $[\mathbb{A}^2/\mathbb{G}_a]$ is an algebraic (Artin) stack, but it is not separated: the stabilizer at points $(0, y)$ is $\mathbb{G}_a$ (which is not proper).

The role of smoothness of $G$

RemarkWhy G must be smooth

The proof crucially uses that $G$ is smooth over $S$ :

$G$ -torsors are smooth: A $G$ -torsor $P \to T$ is a smooth morphism (locally on $T$ , $P \cong G \times T$ , and $G$ is smooth).
The atlas is smooth: $X \times_{[X/G]} T \cong P$ is smooth over $T$ since $P$ is a $G$ -torsor.

If $G$ is not smooth (e.g., $G = \mu_p$ in characteristic $p$ , which is not smooth), then $G$ -torsors are not smooth morphisms -- they are only finite flat. In this case, $\pi : X \to [X/G]$ is flat and surjective but not smooth, and $[X/G]$ may fail to be an Artin stack (in the sense requiring a smooth atlas).

However, $[X/G]$ is still a stack, and for finite flat group schemes, it is an algebraic stack under a generalized definition (where the atlas is allowed to be fppf rather than smooth). The theory of algebraic stacks is sometimes extended to accommodate this.

Summary

The proof that $[X/G]$ is an algebraic stack consists of three independent parts:

Stack axioms: Verified using descent for $G$ -torsors and equivariant morphisms.
Representable diagonal: The isomorphism sheaf $\operatorname{Isom}_T((P_1,\phi_1), (P_2,\phi_2))$ is an algebraic space, constructed as a fiber product involving the contracted product $P_1 \times^G P_2^{\mathrm{op}}$ .
Smooth atlas: The map $\pi : X \to [X/G]$ is smooth and surjective because its base change along any test morphism $T \to [X/G]$ is a $G$ -torsor (hence smooth).

The smoothness of $G$ is essential for the atlas to be smooth. The result applies to all smooth group actions and produces the most important class of algebraic stacks in practice.

Proof that [X/G][X/G][X/G] is an Algebraic Stack