The Stacks project

95.15 Quotients by group actions

At this point we have introduced enough notation that we can work out in more detail what the stacks $[X/G]$ of Section 95.13 look like.

Situation 95.15.1. Here

  1. $S$ is a scheme contained in $\mathit{Sch}_{fppf}$,

  2. $B$ is an algebraic space over $S$,

  3. $(G, m)$ is a group algebraic space over $B$,

  4. $\pi : X \to B$ is an algebraic space over $B$, and

  5. $a : G \times _ B X \to X$ is an action of $G$ on $X$ over $B$.

In this situation we construct a category $[[X/G]]$1 as follows:

  1. An object of $[[X/G]]$ consists of a quadruple $(U, b, P, \varphi : P \to X)$ where

    1. $U$ is an object of $(\mathit{Sch}/S)_{fppf}$,

    2. $b : U \to B$ is a morphism over $S$,

    3. $P$ is an fppf $G_ U$-torsor over $U$ where $G_ U = U \times _{b, B} G$, and

    4. $\varphi : P \to X$ is a $G$-equivariant morphism fitting into the commutative diagram

      \[ \xymatrix{ P \ar[d] \ar[r]_{\varphi } & X \ar[d] \\ U \ar[r]^ b & B } \]
  2. A morphism of $[[X/G]]$ is a pair $(f, g) : (U, b, P, \varphi ) \to (U', b', P', \varphi ')$ where $f : U \to U'$ is a morphism of schemes over $B$ and $g : P \to P'$ is a $G$-equivariant morphism over $f$ which induces an isomorphism $P \cong U \times _{f, U'} P'$, and has the property that $\varphi = \varphi ' \circ g$. In other words $(f, g)$ fits into the following commutative diagram

    \[ \xymatrix{ P \ar[d] \ar[rrrd]_\varphi \ar[r]^ g & P' \ar[d] \ar[rrd]^{\varphi '} \\ U \ar[rrrd]_ b \ar[r]^ f & U' \ar[rrd]^{b'} & & X \ar[d] \\ & & & B } \]

Thus $[[X/G]]$ is a category and

\[ p : [[X/G]] \longrightarrow (\mathit{Sch}/S)_{fppf}, \quad (U, b, P, \varphi ) \longmapsto U \]

is a functor. Note that the fibre category of $[[X/G]]$ over $U$ is the disjoint union over $b \in \mathop{\mathrm{Mor}}\nolimits _ S(U, B)$ of fppf $U \times _{b, B} G$-torsors $P$ endowed with a $G$-equivariant morphism to $X$. Hence the fibre categories of $[[X/G]]$ are groupoids.

Note that the functor

\[ [[X/G]] \longrightarrow G\textit{-Torsors}, \quad (U, b, P, \varphi ) \longmapsto (U, b, P) \]

is a $1$-morphism of categories over $(\mathit{Sch}/S)_{fppf}$.

Lemma 95.15.2. Up to a replacement as in Stacks, Remark 8.4.9 the functor

\[ p : [[X/G]] \longrightarrow (\mathit{Sch}/S)_{fppf} \]

defines a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$.

Proof. The most difficult part of the proof is to show that we have descent for objects. Suppose that $\{ U_ i \to U\} _{i \in I}$ is a covering in $(\mathit{Sch}/S)_{fppf}$. Let $\xi _ i = (U_ i, b_ i, P_ i, \varphi _ i)$ be objects of $[[X/G]]$ over $U_ i$, and let $\varphi _{ij} : \text{pr}_0^*\xi _ i \to \text{pr}_1^*\xi _ j$ be a descent datum. This in particular implies that we get a descent datum on the triples $(U_ i, b_ i, P_ i)$ for the stack in groupoids $G\textit{-Torsors}$ by applying the functor $[[X/G]] \to G\textit{-Torsors}$ above. We have seen that $G\textit{-Torsors}$ is a stack in groupoids (Lemma 95.14.9). Hence we may assume that $b_ i = b|_{U_ i}$ for some morphism $b : U \to B$, and that $P_ i = U_ i \times _ U P$ for some fppf $G_ U = U \times _{b, B} G$-torsor $P$ over $U$. The morphisms $\varphi _ i$ are compatible with the canonical descent datum on the restrictions $U_ i \times _ U P$ and hence define a morphism $\varphi : P \to X$. (For example you can use Sites, Lemma 7.26.5 or you can use Descent on Spaces, Lemma 74.7.2 to get $\varphi $.) This proves descent for objects. We omit the proof of axioms (1) and (2) of Stacks, Definition 8.5.1. $\square$

Proposition 95.15.3. In Situation 95.15.1 there exists a canonical equivalence

\[ [X/G] \longrightarrow [[X/G]] \]

of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$.

Proof. We write this out in detail, to make sure that all the definitions work out in exactly the correct manner. Recall that $[X/G]$ is the quotient stack associated to the groupoid in algebraic spaces $(X, G \times _ B X, s, t, c)$, see Groupoids in Spaces, Definition 78.20.1. This means that $[X/G]$ is the stackification of the category fibred in groupoids $[X/_{\! p}G]$ associated to the functor

\[ (\mathit{Sch}/S)_{fppf} \longrightarrow \textit{Groupoids}, \quad U \longmapsto (X(U), G(U) \times _{B(U)} X(U), s, t, c) \]

where $s(g, x) = x$, $t(g, x) = a(g, x)$, and $c((g, x), (g', x')) = (m(g, g'), x')$. By the construction of Categories, Example 4.37.1 an object of $[X/_{\! p}G]$ is a pair $(U, x)$ with $x \in X(U)$ and a morphism $(f, g) : (U, x) \to (U', x')$ of $[X/_{\! p}G]$ is given by a morphism of schemes $f : U \to U'$ and an element $g \in G(U)$ such that $a(g, x) = x' \circ f$. Hence we can define a $1$-morphism of stacks in groupoids

\[ F_ p : [X/_{\! p}G] \longrightarrow [[X/G]] \]

by the following rules: On objects we set

\[ F_ p(U, x) = (U, \pi \circ x, G \times _{B, \pi \circ x} U, a \circ (\text{id}_ G \times x)) \]

This makes sense because the diagram

\[ \xymatrix{ G \times _{B, \pi \circ x} U \ar[d] \ar[r]_{\text{id}_ G \times x} & G \times _{B, \pi } X \ar[r]_-a & X \ar[d]^\pi \\ U \ar[rr]^{\pi \circ x} & & B } \]

commutes, and the two horizontal arrows are $G$-equivariant if we think of the fibre products as trivial $G$-torsors over $U$, resp. $X$. On morphisms $(f, g) : (U, x) \to (U', x')$ we set $F_ p(f, g) = (f, R_{g^{-1}})$ where $R_{g^{-1}}$ denotes right translation by the inverse of $g$. More precisely, the morphism $F_ p(f, g) : F_ p(U, x) \to F_ p(U', x')$ is given by the cartesian diagram

\[ \xymatrix{ G \times _{B, \pi \circ x} U \ar[d] \ar[r]_{R_{g^{-1}}} & G \times _{B, \pi \circ x'} U' \ar[d] \\ U \ar[r]^ f & U' } \]

where $R_{g^{-1}}$ on $T$-valued points is given by

\[ R_{g^{-1}}(g', u) = (m(g', i(g(u))), f(u)) \]

To see that this works we have to verify that

\[ a \circ (\text{id}_ G \times x) = a \circ (\text{id}_ G \times x') \circ R_{g^{-1}} \]

which is true because the right hand side applied to the $T$-valued point $(g', u)$ gives the desired equality

\begin{align*} a((\text{id}_ G \times x')(m(g', i(g(u))), f(u))) & = a(m(g', i(g(u))), x'(f(u))) \\ & = a(g', a(i(g(u)), x'(f(u)))) \\ & = a(g', x(u)) \end{align*}

because $a(g, x) = x' \circ f$ and hence $a(i(g), x' \circ f) = x$.

By the universal property of stackification from Stacks, Lemma 8.9.2 we obtain a canonical extension $F : [X/G] \to [[X/G]]$ of the $1$-morphism $F_ p$ above. We first prove that $F$ is fully faithful. To do this, since both source and target are stacks in groupoids, it suffices to prove that the $\mathit{Isom}$-sheaves are identified under $F$. Pick a scheme $U$ and objects $\xi , \xi '$ of $[X/G]$ over $U$. We want to show that

\[ F : \mathit{Isom}_{[X/G]}(\xi , \xi ') \longrightarrow \mathit{Isom}_{[[X/G]]}(F(\xi ), F(\xi ')) \]

is an isomorphism of sheaves. To do this it suffices to work locally on $U$, and hence we may assume that $\xi , \xi '$ come from objects $(U, x)$, $(U, x')$ of $[X/_{\! p}G]$ over $U$; this follows directly from the construction of the stackification, and it is also worked out in detail in Groupoids in Spaces, Section 78.24. Either by directly using the description of morphisms in $[X/_{\! p}G]$ above, or using Groupoids in Spaces, Lemma 78.22.1 we see that in this case

\[ \mathit{Isom}_{[X/G]}(\xi , \xi ') = U \times _{(x, x'), X \times _ S X, (s, t)} (G \times _ B X) \]

A $T$-valued point of this fibre product corresponds to a pair $(u, g)$ with $u \in U(T)$, and $g \in G(T)$ such that $a(g, x \circ u) = x' \circ u$. (Note that this implies $\pi \circ x \circ u = \pi \circ x' \circ u$.) On the other hand, a $T$-valued point of $\mathit{Isom}_{[[X/G]]}(F(\xi ), F(\xi '))$ by definition corresponds to a morphism $u : T \to U$ such that $\pi \circ x \circ u = \pi \circ x' \circ u : T \to B$ and an isomorphism

\[ R : G \times _{B, \pi \circ x \circ u} T \longrightarrow G \times _{B, \pi \circ x' \circ u} T \]

of trivial $G_ T$-torsors compatible with the given maps to $X$. Since the torsors are trivial we see that $R = R_{g^{-1}}$ (right multiplication) by some $g \in G(T)$. Compatibility with the maps $a \circ (1_ G, x \circ u), a \circ (1_ G, x' \circ u) : G \times _ B T \to X$ is equivalent to the condition that $a(g, x \circ u) = x' \circ u$. Hence we obtain the desired equality of $\mathit{Isom}$-sheaves.

Now that we know that $F$ is fully faithful we see that Stacks, Lemma 8.4.8 applies. Thus to show that $F$ is an equivalence it suffices to show that objects of $[[X/G]]$ are fppf locally in the essential image of $F$. This is clear as fppf torsors are fppf locally trivial, and hence we win. $\square$


Lemma 95.15.4. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $G$ be a group algebraic space over $B$. Then the stacks in groupoids

\[ [B/G],\quad [[B/G]],\quad G\textit{-Torsors},\quad \mathcal{G}/\mathcal{B}\textit{-Torsors} \]

are all canonically equivalent. If $G \to B$ is flat and locally of finite presentation, then these are also equivalent to $G\textit{-Principal}$.

Proof. The equivalence $G\textit{-Torsors} \to \mathcal{G}/\mathcal{B}\textit{-Torsors}$ is given in Lemma 95.14.10. The equivalence $[B/G] \to [[B/G]]$ is given in Proposition 95.15.3. Unwinding the definition of $[[B/G]]$ given in Section 95.15 we see that $[[B//G]] = G\textit{-Torsors}$.

Finally, assume $G \to B$ is flat and locally of finite presentation. To show that the natural functor $G\textit{-Torsors} \to G\textit{-Principal}$ is an equivalence it suffices to show that for a scheme $U$ over $B$ a principal homogeneous $G_ U$-space $X \to U$ is fppf locally trivial. By our definition of principal homogeneous spaces (Groupoids in Spaces, Definition 78.9.3) there exists an fpqc covering $\{ U_ i \to U\} $ such that $U_ i \times _ U X \cong G \times _ B U_ i$ as algebraic spaces over $U_ i$. This implies that $X \to U$ is surjective, flat, and locally of finite presentation, see Descent on Spaces, Lemmas 74.11.6, 74.11.13, and 74.11.10. Choose a scheme $W$ and a surjective ├ętale morphism $W \to X$. Then it follows from what we just said that $\{ W \to U\} $ is an fppf covering such that $X_ W \to W$ has a section. Hence $X$ is an fppf $G_ U$-torsor. $\square$

Remark 95.15.5. Let $S$ be a scheme. Let $G$ be an abstract group. Let $X$ be an algebraic space over $S$. Let $G \to \text{Aut}_ S(X)$ be a group homomorphism. In this setting we can define $[[X/G]]$ similarly to the above as follows:

  1. An object of $[[X/G]]$ consists of a triple $(U, P, \varphi : P \to X)$ where

    1. $U$ is an object of $(\mathit{Sch}/S)_{fppf}$,

    2. $P$ is a sheaf on $(\mathit{Sch}/U)_{fppf}$ which comes with an action of $G$ that turns it into a torsor under the constant sheaf with value $G$, and

    3. $\varphi : P \to X$ is a $G$-equivariant map of sheaves.

  2. A morphism $(f, g) : (U, P, \varphi ) \to (U', P', \varphi ')$ is given by a morphism of schemes $f : T \to T'$ and a $G$-equivariant isomorphism $g : P \to f^{-1}P'$ such that $\varphi = \varphi ' \circ g$.

In exactly the same manner as above we obtain a functor

\[ [[X/G]] \longrightarrow (\mathit{Sch}/S)_{fppf} \]

which turns $[[X/G]]$ into a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$. The constant sheaf $\underline{G}$ is (provided the cardinality of $G$ is not too large) representable by $G_ S$ on $(\mathit{Sch}/S)_{fppf}$ and this version of $[[X/G]]$ is equivalent to the stack $[[X/G_ S]]$ introduced above.

[1] The notation $[[X/G]]$ with double brackets serves to distinguish this category from the stack $[X/G]$ introduced earlier. In Proposition 95.15.3 we show that the two are canonically equivalent. Afterwards we will use the notation $[X/G]$ to indicate either.

Comments (2)

Comment #1427 by yogesh more on

Thank you so much for writing out Prop. 73.14.3 out in great detail! I have always wanted to see this done. On line 1346 of the tex file it says, "because as desired."

But going from 1342 to 1344 you seem to use (the are interchanged). This makes me wonder whether should be right multiplication by , but I am just skimming things really fast so I don't know for sure.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04UV. Beware of the difference between the letter 'O' and the digit '0'.