Reading Huybrechts & Lehn's "The geometry of moduli spaces of sheaves" I am stuck with a particular statement that they make in the chapter on GIT without explanation. Namely, they say that for an algebraic $k$-group $G$ acting on a $k$-scheme of finite type $X$, the space of global sections of a $G$-linearized quasi-coherent $\mathcal{O}_X$-module $F$ naturally has the structure of a $G$-representation.

I give here the definition of $G$-linearization, it is essentialy the same one as given by Mumford in GIT: Let $G$ be an algebraic $k$-group, $X$ a $k$-scheme of finite type, and $\sigma : X \times G \rightarrow X$ a group action. Let $F$ be a quasi-coherent $\mathcal{O}_X$-module. Then a $G$-linearization of $F$ is an isomorphism of $\mathcal{O}_{G \times X}$-modules $\Phi : \sigma^* F \rightarrow p_1^*F$, where $p_1$ is the projection $X \times G \rightarrow X$, such that the following cocycle condition is sattisfied: $$ (id_X \times \mu)^*\Phi = p^*_{12}\Phi \circ (\sigma \times id_G)^*\Phi $$ where $p_{12} : X \times G \times G \rightarrow X \times G$ is the projection onto the first two factors.

I don't see how to use the isomorphism $\Phi$ to define a group action of $G$ on $H^0(X,F)$, nor how to use the cocycle condition to show that this action must be linear. Any help would be great, thanks.


{Definition} Let $G$ be an algebraic group and $X$ be a $G$-variety. Let $L\in Pic(X)$ be an invertible sheaf. A $G$-linearization of $L$ consists of an isomorphism $$\phi:\sigma^{\ast}L\simeq p_{2}^{\ast}L$$ that satisfies the following co-cycle condition: There are three maps $$G\times G\times X\rightarrow G \times X$$ given by $1_{G}\times \sigma$, $p_{23}$ and $\mu\times 1_{X}.$ We then require $$(\mu\times 1_{X})^{\ast}\phi= (p_{23})^{\ast}\phi\circ(1_{G}\times \sigma)^{\ast}\phi.$$ \end{definition}

Let's establish G-module struture bit by bit.

Lemma: Let $X$ be a $G$-variety and $\mathscr{L}\in Pic(X)$ be a $G$-linearized invertible sheaf. Then there is a commutative diagram Firure 1 Moreover, $\alpha$ and $\beta$ are isomorphisms. \end{lemma}

Proof Given $a\in \Gamma(G)$ and $\tau\in\Gamma(X,\mathscr{L})$, define $$\alpha(a,\tau)=p_{1}^{\sharp}(a)\cdot p_{2}^{\ast}(\tau)$$ and given $a\otimes b\in \Gamma(G\times G)$ and $\tau^{\prime}\in \Gamma(X,\mathscr{L})$, define $$\beta(a\otimes b\otimes\tau^{\prime})=p_{12}^{\sharp}(a\otimes b)\cdot p_{3}^{\ast}(\tau^{\prime}).$$ Then observe

  1. $p_{3}=p_{2}\circ(\mu\times 1_{X}):G\times G\times X\rightarrow X$
  2. $p_{1}\circ(\mu\times 1_{X})=\mu\circ p_{12}:G\times G\times X\rightarrow G$
  3. Give $f:X\rightarrow Y$ a morphism of schemes and $\mathscr{F}$ as sheaf of $\mathscr{O}_{Y}$-modules and a section $s\in\Gamma(Y,\mathscr{F})$, the pull back $f^{\ast}:\Gamma(Y,\mathscr{F})\rightarrow\Gamma(X,f^{\ast}\mathscr{F})$ is a morphism of $\Gamma(Y,\mathscr{O}_{Y})$-modules. Here the $\Gamma(Y,\mathscr{O}_{Y})$-module structure of $\Gamma(X,f^{\ast}\mathscr{F})$ is induced by the map $\mathscr{O}_{Y}\rightarrow f_{\ast}\mathscr{O}_{X}$. \end{enumerate} The commutativity of the diagram follows immediately from these three observations. That $\alpha$ and $\beta$ are isomorphisms follows from flat base change. \end{proof}


There is a commutative diagram Figure 2

Let me also note that pull back of sheaf is a functor, hence if $$(\mu\times 1_{X})^{\ast}_{p_{2}}:=\Gamma(G\times X,\mathscr{p_{2}}^{\ast}\mathscr{L})\rightarrow \Gamma(G\times G\times X,(\mu\times 1_{X})^{\ast}p_{2}^{\ast}\mathscr{L})$$ and $$(\mu\times 1_{X})^{\ast}_{\sigma}:=\Gamma(G\times X,\sigma^{\ast}\mathscr{L})\rightarrow\Gamma(G\times G\times X,(\mu\times 1_{X})^{\ast}\sigma^{\ast}\mathscr{L}),$$ then $$\phi\circ(\mu\times 1_{X})^{\ast}_{p_{2}}=(\mu\times 1_{X})^{\ast}(\phi)\circ(\mu\times 1_{X})^{\ast}_{\sigma}.$$ Same property holds for $(1_{G}\times\sigma).$ With the notations, we get the Theorem: Let $X$ be a $G$-variety. Then for any $G$-lineraization $\mathscr{L}\in Pic(X)$, $\Gamma(X,\mathscr{L})$ is a $G$-module. \end{theorem}

Proof} Define $\hat{\sigma}:\Gamma(X,\mathscr{L})\rightarrow\Gamma(G)\otimes\Gamma(X,\mathscr{L})$ by the composition $$\hat{\sigma}:=\alpha^{-1}\phi\sigma^{\ast}.$$ Note that the bottom row in Figure 2 is just $(1\otimes\hat{\sigma})$. Then \begin{equation*} \begin{split} (1\otimes \hat{\sigma})\circ\hat{\sigma} =&(\beta^{-1}p_{23}^{\ast}(\phi)(1_{G}\times\sigma)^{\ast}_{p_{2}}\alpha)\circ(\alpha^{-1}\phi\sigma^{\ast})\\ =&\beta^{-1}p_{23}^{\ast}(\phi)(1_{G}\times \sigma)^{\ast}_{p_{2}}\phi\sigma^{\ast}\\ =&\beta^{-1}p_{23}^{\ast}(\phi)(1_{G}\times \sigma)^{\ast}(\phi)(1_{G}\times \sigma)^{\ast}_{\sigma}\phi^{-1}\phi\sigma^{\ast}\\ =&\beta^{-1}(\mu\times 1_{X})^{\ast}(\phi)(1_{G}\times \sigma)^{\ast}_{\sigma}\sigma^{\ast}\\ =&\beta^{-1}(\mu\times 1_{X})^{\ast}(\phi)(\mu\times 1_{X})^{\ast}_{\sigma}\sigma^{\ast} \end{split} \end{equation*}

On the other hand, we have

\begin{equation*} \begin{split} (\mu^{\sharp}\otimes 1)\circ\hat{\sigma}=& (\beta^{-1}(\mu\times 1_{X})^{\ast}_{p_{2}}\alpha)\alpha^{-1}\phi\sigma^{\ast}\\ =&\beta^{-1}(\mu\times 1_{X})^{\ast}_{p_{2}}\phi\sigma^{\ast}\\ =&\beta^{-1}(\mu\times 1_{X})^{\ast}(\phi)(\mu\times 1_{X})^{\ast}_{\sigma}\phi^{-1}\phi\sigma^{\ast}\\ =&\beta^{-1}(\mu\times 1_{X})^{\ast}(\phi)(\mu\times 1_{X})^{\ast}_{\sigma}\sigma^{\ast} \end{split} \end{equation*} We see that $(1\otimes\hat{\sigma})\circ\hat{\sigma}=(\mu^{\sharp}\otimes 1)\circ\hat{\sigma}$. \end{proof}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.