The Stacks project

39.12 Equivariant quasi-coherent sheaves

We think of “functions” as dual to “space”. Thus for a morphism of spaces the map on functions goes the other way. Moreover, we think of the sections of a sheaf of modules as “functions”. This leads us naturally to the direction of the arrows chosen in the following definition.

Definition 39.12.1. Let $S$ be a scheme, let $(G, m)$ be a group scheme over $S$, and let $a : G \times _ S X \to X$ be an action of the group scheme $G$ on $X/S$. A $G$-equivariant quasi-coherent $\mathcal{O}_ X$-module, or simply an equivariant quasi-coherent $\mathcal{O}_ X$-module, is a pair $(\mathcal{F}, \alpha )$, where $\mathcal{F}$ is a quasi-coherent $\mathcal{O}_ X$-module, and $\alpha $ is a $\mathcal{O}_{G \times _ S X}$-module map

\[ \alpha : a^*\mathcal{F} \longrightarrow \text{pr}_1^*\mathcal{F} \]

where $\text{pr}_1 : G \times _ S X \to X$ is the projection such that

  1. the diagram

    \[ \xymatrix{ (1_ G \times a)^*\text{pr}_1^*\mathcal{F} \ar[r]_-{\text{pr}_{12}^*\alpha } & \text{pr}_2^*\mathcal{F} \\ (1_ G \times a)^*a^*\mathcal{F} \ar[u]^{(1_ G \times a)^*\alpha } \ar@{=}[r] & (m \times 1_ X)^*a^*\mathcal{F} \ar[u]_{(m \times 1_ X)^*\alpha } } \]

    is a commutative in the category of $\mathcal{O}_{G \times _ S G \times _ S X}$-modules, and

  2. the pullback

    \[ (e \times 1_ X)^*\alpha : \mathcal{F} \longrightarrow \mathcal{F} \]

    is the identity map.

For explanation compare with the relevant diagrams of Equation (

Note that the commutativity of the first diagram guarantees that $(e \times 1_ X)^*\alpha $ is an idempotent operator on $\mathcal{F}$, and hence condition (2) is just the condition that it is an isomorphism.

Lemma 39.12.2. Let $S$ be a scheme. Let $G$ be a group scheme over $S$. Let $f : Y \to X$ be a $G$-equivariant morphism between $S$-schemes endowed with $G$-actions. Then pullback $f^*$ given by $(\mathcal{F}, \alpha ) \mapsto (f^*\mathcal{F}, (1_ G \times f)^*\alpha )$ defines a functor from the category of $G$-equivariant quasi-coherent $\mathcal{O}_ X$-modules to the category of $G$-equivariant quasi-coherent $\mathcal{O}_ Y$-modules.

Proof. Omitted. $\square$

Let us give an example.

Example 39.12.3. Let $A$ be a $\mathbf{Z}$-graded ring, i.e., $A$ comes with a direct sum decomposition $A = \bigoplus _{n \in \mathbf{Z}} A_ n$ and $A_ n \cdot A_ m \subset A_{n + m}$. Set $X = \mathop{\mathrm{Spec}}(A)$. Then we obtain a $\mathbf{G}_ m$-action

\[ a : \mathbf{G}_ m \times X \longrightarrow X \]

by the ring map $\mu : A \to A \otimes \mathbf{Z}[x, x^{-1}]$, $f \mapsto f \otimes x^{\deg (f)}$. Namely, to check this we have to verify that

\[ \xymatrix{ A \ar[r]_\mu \ar[d]_\mu & A \otimes \mathbf{Z}[x, x^{-1}] \ar[d]^{\mu \otimes 1} \\ A \otimes \mathbf{Z}[x, x^{-1}] \ar[r]^-{1 \otimes m} & A \otimes \mathbf{Z}[x, x^{-1}] \otimes \mathbf{Z}[x, x^{-1}] } \]

where $m(x) = x \otimes x$, see Example 39.5.1. This is immediately clear when evaluating on a homogeneous element. Suppose that $M$ is a graded $A$-module. Then we obtain a $\mathbf{G}_ m$-equivariant quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F} = \widetilde{M}$ by using $\alpha $ as in Definition 39.12.1 corresponding to the $A \otimes \mathbf{Z}[x, x^{-1}]$-module map

\[ M \otimes _{A, \mu } (A \otimes \mathbf{Z}[x, x^{-1}]) \longrightarrow M \otimes _{A, \text{id}_ A \otimes 1} (A \otimes \mathbf{Z}[x, x^{-1}]) \]

sending $m \otimes 1 \otimes 1$ to $m \otimes 1 \otimes x^{\deg (m)}$ for $m \in M$ homogeneous.

Lemma 39.12.4. Let $a : \mathbf{G}_ m \times X \to X$ be an action on an affine scheme. Then $X$ is the spectrum of a $\mathbf{Z}$-graded ring and the action is as in Example 39.12.3.

Proof. Let $f \in A = \Gamma (X, \mathcal{O}_ X)$. Then we can write

\[ a^\sharp (f) = \sum \nolimits _{n \in \mathbf{Z}} f_ n \otimes x^ n \quad \text{in}\quad A \otimes \mathbf{Z}[x, x^{-1}] = \Gamma (\mathbf{G}_ m \times X, \mathcal{O}_{\mathbf{G}_ m \times X}) \]

as a finite sum with $f_ n$ in $A$ uniquely determined. Thus we obtain maps $A \to A$, $f \mapsto f_ n$. Since $a$ is an action, if we evaluate at $x = 1$, we see $f = \sum f_ n$. Since $a$ is an action we find that

\[ \sum (f_ n)_ m \otimes x^ m \otimes x^ n = \sum f_ n x^ n \otimes x^ n \]

(compare with computation in Example 39.12.3). Thus $(f_ n)_ m = 0$ if $n \not= m$ and $(f_ n)_ n = f_ n$. Thus if we set

\[ A_ n = \{ f \in A \mid f_ n = f\} \]

then we get $A = \sum A_ n$. On the other hand, the sum has to be direct since $f = 0$ implies $f_ n = 0$ in the situation above. $\square$

Lemma 39.12.5. Let $A$ be a graded ring. Let $X = \mathop{\mathrm{Spec}}(A)$ with action $a : \mathbf{G}_ m \times X \to X$ as in Example 39.12.3. Let $\mathcal{F}$ be a $\mathbf{G}_ m$-equivariant quasi-coherent $\mathcal{O}_ X$-module. Then $M = \Gamma (X, \mathcal{F})$ has a canonical grading such that it is a graded $A$-module and such that the isomorphism $\widetilde{M} \to \mathcal{F}$ (Schemes, Lemma 26.7.4) is an isomorphism of $\mathbf{G}_ m$-equivariant modules where the $\mathbf{G}_ m$-equivariant structure on $\widetilde{M}$ is the one from Example 39.12.3.

Proof. You can either prove this by repeating the arguments of Lemma 39.12.4 for the module $M$. Alternatively, you can consider the scheme $(X', \mathcal{O}_{X'}) = (X, \mathcal{O}_ X \oplus \mathcal{F})$ where $\mathcal{F}$ is viewed as an ideal of square zero. There is a natural action $a' : \mathbf{G}_ m \times X' \to X'$ defined using the action on $X$ and on $\mathcal{F}$. Then apply Lemma 39.12.4 to $X'$ and conclude. (The nice thing about this argument is that it immediately shows that the grading on $A$ and $M$ are compatible, i.e., that $M$ is a graded $A$-module.) Details omitted. $\square$

Comments (12)

Comment #1480 by Li Zhan on

I feel the first row of the diagram (1) should be written as where is the projection to , and is the projection to the second and components.

Comment #1497 by on

Yes and no. See the conventions in Section 39.2. May have been a mistake to have that convention.

Comment #1687 by Mohammed Mammeri on

May be should be replaced by in the first row of diagram (1).

Comment #3382 by Reimundo Heluani on

In Lemma 39.12.2, should be .

Comment #4283 by David Zureick-Brown on

"An" G-equivariant should be "a", and "a" equivariant should be "an".

Comment #5857 by Eli Putterman on

Condition (2) in the definition is not stated in several other places (e.g., Wikipedia; Mumford, GIT). Is this an oversight on other authors' part, or can it be shown to follow from condition (1)?

Comment #5858 by on

If you assume is an isomorphism, then it follows from the other ones, see sentence following the definition. Good exercise. Perhaps we should just assume is an iso and remove (2).

Comment #6011 by Ben church on

Are there interesting examples where is not an isomorphism? All other references I know (Mumford's GIT for example) assume is an isomorphism.

When is a G-cover you might want the notion of a G-equivariant sheaf to agree with descent data in which case better be an isomorphism I think.

Comment #6070 by on

Let me say one more time to make sure everybody is on the same page: given as in Definition 39.12.1 satisfying (1) we have that (2) is equivalent to being an isomorphism. As far as I know nobody ever considers the case where is not an isomorphism.

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