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 (39.10.1.1).

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. $\square$

Comment #1480 by Li Zhan on

I feel the first row of the diagram (1) should be written as where $pr_2: G \times G \times X \to X$ is the projection to $X$, and $pr_{23}: G \times G \times X \to G \times X$ is the projection to the second $G$ and $X$ 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 $(1_G\times a)^{\ast} pr_2^{\ast} \mathcal{F}$ should be replaced by $(1_G\times a)^{\ast} pr_1^{\ast} \mathcal{F}$ in the first row of diagram (1).

Comment #3382 by Reimundo Heluani on

In Lemma 39.12.2, $f$ should be $f:Y\rightarrow X$.

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

Comment #6011 by Ben church on

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

When $X/S$ is a G-cover you might want the notion of a G-equivariant sheaf to agree with descent data in which case $\alpha$ 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 $\alpha$ as in Definition 39.12.1 satisfying (1) we have that (2) is equivalent to $\alpha$ being an isomorphism. As far as I know nobody ever considers the case where $\alpha$ is not an isomorphism.

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).