$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$I have some problems to understand the proof of *Proposition 1.5* from Mumford's [Geometric Invariant Theory](https://link.springer.com/book/9783540569633), p 34:


*Corollary 1.5*  

Let $G$ be a connected linear algebraic group acting on an algebraic
variety $X$, that is proper over $k$ (in the book a variety is a
scheme $X/k$ such that
$\overline{X}= X \times \Spec(\overline{k})$ is irreducible
and reduced). Let $\mathcal{L}$ be an invertible
sheaf on $X$, and let $[\mathcal{L}]$ the class regarded as the $k$-rational point of the
Picard scheme $\Pic(X/k)$ associated to $\mathcal{L}$.  

Then some power $\mathcal{L}^n$ is $G$-linearizable if and only if
some multiple
$[\mathcal{L}]^n$ of $[\mathcal{L}] $ is left fixed by induced $G$-action
on $ \Hom\_k(\Spec(k), \Pic (X/k))$.


(at this point one should remark that in the book this left action
by $G$ on $\mathcal{Pic}\_{X/k}(k)$ is not explained in explicit terms.
In the substantively similar question [Corollary 1.6 in Mumford's Geometric Invariant Theory](https://mathoverflow.net/questions/433970/corollary-1-6-in-mumfords-geometric-invariant-theory) I made a remark how I think this
action might work in detail).


[Second important *remark* before
we dip into the proof:  

Recall that more less by [definition](https://ncatlab.org/nlab/show/Picard+scheme) of the Picard functor is given by


$$ \mathcal{Pic}\_{X/k}(S) := H^0(G, R^1p\_{1\*} (\mathcal{O}\_{S \times X}^\*) = \\
 \{ \mathcal{M} \text{ invertible sheaf 
on } X \times\_k S \} / \{ \text{ inv. sheaves of the form }
p^\*\_S(\mathcal{K}) \text{ for } \mathcal{K} \text{ invertible on }
S \}. $$


The proof uses [cp *Chap. 0, §5, (d)*] the fact that the Picard functor is "almost"
representable, that means precisely there exists a $k$-scheme
$\Pic(X/k)$ representing the associated
functor $\text{Hom}( \ , \Pic(X/k))$ which *contains* the
Picard functor $\mathcal{Pic}\_{X/k}$
in the sense that for any $k$-scheme $S$ there is a functorial
inclusion


$$ \iota\_S: \mathcal{Pic}\_{X/k}(S) \hookrightarrow \Hom\_k(S,\Pic(X/k)). $$


In general that's a proper inclusion.
The equality only holds if $X \times\_k S$ admits a section over $S$.



> 
> **The Proof.** The only if is clear. Conversely, suppose $[\mathcal{L}]^n$
> is left fixed by $G$. Then the claim is first that for some $m$, the
> two pullback sheaves $\sigma^\*(\mathcal{L}^{nm})$ and
> $p\_2^\*(\mathcal{L}^{nm})$ (induced by the action and projection maps
> $\sigma, p\_2: G \times X \to X $ on $ G \times X $ are isomorphic.
> To see this, consider the see-saw exact sequence [Rem.: I never saw the term see-saw sequence. I think that is just the exact part of Leray–Serre spectral sequence for higher image sheaf]:
> 
> 
> 


$$ 0 \to H^1(G, \mathcal{O}\_G^\*) \to 
H^1(G \times X, \mathcal{O}\_{G \times X}^\*) \to
H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*).$$



> 
> Since $H^1(G, \mathcal{O}\_G^\*)$ is a finite group (Seminaire
> Chevalley, [9], 5-21), it is enough to show that the image of
> $\sigma^\*(\mathcal{L}^n) \otimes p\_2^\*(\mathcal{L}^n)^{-1}$
> in $H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*)$ is zero. But,
> by the functorial definition of $\Pic (X/k)$
> (cf Chap. 0, §5, (d), page 23)
> 
> 
> 


$$ \mathcal{Pic}\_{X/k}(G) = H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*) \subset
\Hom\_k(G, \Pic (X/k)). $$



> 
> But, as in the proof of proposition 1.4, it holds
> $H^0(G \times X, \mathcal{O}\_{G \times X}^\*) \cong
> H^0(G, \mathcal{O}\_G^\*)$ and the latter is just $k\* \times M$, where
> $M$ is the set of characters, i.e., $\Hom(G, \mathbb{G}\_m)$.  
> 
> *Choose an isomorphism*
> $\phi: \sigma^\*(\mathcal{L}^{nm}) \to p\_2^\*(\mathcal{L}^{nm})$, which
> is the identity on $\{e\} \times X$.
> [The rest of the proof verifies the cocycle condition
> $p^\*\_{23} \phi \circ (1\_G \times \sigma)^\* = (m \times 1\_x)^\* \phi $, that's fine .]
> 
> 
> 


The **question** is why the assumption that the class
$[\mathcal{L}^n] \in \Hom\_k(\Spec(k), \Pic (X/k))$
is fixed by $G$-action, implies that
the pullback sheaves $\sigma^\*(\mathcal{L}^{nm})$ and
$p\_2^\*(\mathcal{L}^{nm})$ are isomorphic, or as remarked that's equivalent to to the question why
the images of classes $[\sigma^\*(\mathcal{L}^{n})]$ and
$[p\_2^\*(\mathcal{L}^{n})]$ in
$ H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*)) \subset
\Hom\_k(G, \Pic (X/k)) $
are identical?


To rephrase it in other terms, the maps $\sigma, p\_2: G \times X \to X $, which are given on geometric points by $(g,x) \mapsto g \cdot x$, respectively $(g,x) \mapsto x$, map the classes $[\mathcal{M}] \in \mathcal{Pic}\_{X/k}(k) $ to classes in $\mathcal{Pic}\_{X/k}(G)$ via taking $[\mathcal{M}]$ to the pullback $[\sigma^\*\mathcal{M}]$, respectively $[p\_2^\*\mathcal{M}]$.  

How do these operations by $\sigma, p\_2$ look like in *explicit* terms as maps between $ \Hom(\Spec(k), \Pic(X/k))$ and $\Hom(G, \Pic(X/k))$? Especially how to construct explicitly
from the pullback of $[\mathcal{L}]^n$ by $\sigma$ and $p\_2$
elements in $\Hom\_k(G, \Pic (X/k)) $ representing
the classes of the images of
$\sigma^\*(\mathcal{L}^{n})$ and
$p\_2^\*(\mathcal{L}^{n})$?


Pictorally, the action and projection morphisms $\sigma, \pr\_X$ should induce following diagram


$$
\require{AMScd}
\begin{CD}
\mathcal{Pic}\_{X/k}(k) @>{\iota\_k} >> \Hom(\Spec(k), \Pic(X/k)) \\
@VV\sigma^\*, p\_2^\*V @VVf\_{\sigma^\*}, f\_{\pr\_X^\*}V \\
\mathcal{Pic}\_{X/k}(G) @>{\iota\_G}>> \Hom(G, \Pic(X/k))
\end{CD}
$$


and I'm interested in the explicit structure of the right vertical maps $f\_{\sigma^\*}, f\_{\pr\_X^\*}:
\Hom(\Spec(k), \Pic(X/k)) \to 
\Hom(G, \Pic(X/k))$ making the diagram commutative
with respect $\sigma^\*, \pr\_X^\*$ on the left and what they do with $[\mathcal{L}^n] \in \Hom\_k(\Spec(k), \Pic (X/k))$.


My conjecture is that the image of
$p\_2^\*(\mathcal{L}^{n})$ in
$\Hom\_k(G, \Pic (X/k)) $ should represent a constant map with image be the $k$-point
$[\mathcal{L}^n]$, while $\sigma^\*(\mathcal{L}^{n})$ the
orbit map of $[\mathcal{L}^n]$ induced by the action of $G$ on $k$-valued points of $ \Pic (X/k)$. This would suggest that $f\_{\sigma^\*}$ and $f\_{\pr\_X^\*}$ should be explicitly given by


$$ [x] \mapsto f\_{\sigma^\*}([x]) := (g \mapsto g \cdot [x]) $$


and respectively


$$ [x] \mapsto f\_{\pr\_X^\*}([x]) := (g \mapsto [x]) $$


i.e. the constant map, where $[x]: \Spec(k) \to \Pic(X/k)$
is any geometric $k$-point of $\Pic(X/k)$ and $ g \cdot [x]:= [g^\*x]$ the induced action on Picard group via pullback. Having this, we assumed $G$ to fix $[\mathcal{L}^n]$, therefore these the images of $[\mathcal{L}^n]$ by these maps would coinside as elements in
$\Hom\_k(G, \Pic (X/k)) $ and should give isomorphic
line bundles over $G \times X$.


Therefore if the $f\_{\sigma^\*}$, $f\_{\pr\_X^\*}$ would be given like I conjecture, this would be consistent with the tacitly used claim in the proof that $[\sigma^\*(\mathcal{L}^{n})]$ and $ [p\_2^\*(\mathcal{L}^{n})]$ are identical as elements in $\mathcal{Pic}\_{X/k}(G) \subset
\Hom\_k(G, \Pic (X/k))$. But I not see how to verify that $f\_{\sigma^\*}$, $f\_{\pr\_X^\*}$ have this form.



Proposition 1.5 in Mumford's Geometric Invariant Theory
I think the fist part of the proof of 1.5 may be rephrased as follows. The hypothesis that $[\mathcal{L}^n]$ is fixed by the $G$-action exactly means that for all $g \in G$, there is an isomorphism:
$$ \sigma^\* \mathcal{L}^n \big|\_{\{g\} \times X} \simeq p\_2^\* \mathcal{L}^n|\_{\{g\} \times X}.$$ Put differently, for all $g \in G$, there is an isomorphism:


$$\sigma^\* \mathcal{L}^n \otimes \left(p\_2^\* \mathcal{L}^n \right)^{-1} \big|\_{\{g\} \times X} \simeq \mathcal{O}\_X.$$
By the Seesaw Theorem, this means that $\sigma^\* \mathcal{L}^n \otimes \left(p\_2^\* \mathcal{L}^n\right)^{-1}$ is the pull-back of a line bundle on $G$. But a Theorem of Chevalley implies $H^{1}(G, \mathcal{O}\_{G}^\*)$ is finite, hence $\sigma^\* \mathcal{L}^n \otimes \left(p\_2^\* \mathcal{L}^n \right)^{-1}$ is torsion and we are done.


