For an algebraic group $G$ acting on a scheme $X$, $H^0(X,F)$ of a $G$-linearized $\mathcal{O}_X$-module $F$ has the structure of a $G$-representation 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.
 A: {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


*

*$p_{3}=p_{2}\circ(\mu\times 1_{X}):G\times G\times X\rightarrow X$

*$p_{1}\circ(\mu\times 1_{X})=\mu\circ p_{12}:G\times G\times X\rightarrow G$

*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}


Similarly, 
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}
