## 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

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

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.
$\square$

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