$\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Proj{Proj}\DeclareMathOperator\Pic{Pic}$I have a question about an example for a line bundle not admitting a
$G$-linearization from Mumford's GIT, page 33:


We consider the action of $\PGL(n+1)$ on projecive space
$\mathbb{P}^n= \Proj k[X\_0,\dotsc, X\_n]$.
Observe that $\PGL(n+1)$ is given as the open affine
subscheme of


$$ \mathbb{P}^{n^2+2n} = \Proj k[a\_{00},\dotsc, a\_{0n};
a\_{10}, \dotsc , a\_{nn}]$$


complementary to the determinants hypersurface $\det(a\_{ij})=0$. The
action morphism $ \sigma: \PGL(n+1) \times \mathbb{P}^n \to \mathbb{P}^n $ is determined by


\begin{gather\*}
\sigma^\*(\mathcal{O}\_{\mathbb{P}^n}(1)) \cong
 p\_1^\*(\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)) \otimes 
p\_2^\*(\mathcal{O}\_{\mathbb{P}^{n}}(1)) \\
\sigma^\*(X\_i)= \sum\_{j=0}^n p\_1^\*(a\_{ij}) \otimes p\_2^\*(X\_j)
\end{gather\*}


where $p\_1, p\_2$ are
projections canonical projections.


Mumford claims that $\mathcal{O}\_{\mathbb{P}^n}(1) $ admits no
$\PGL(n+1)$-linearization, because
the restriction of $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$
to the open subscheme $\PGL(n+1)$ has order $n+1$ in
$\Pic[\PGL(n+1)]$, and is therefore not trivial.


My question is why the fact that $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$
restricted to the affine open $\PGL(n+1)$ is not trivial, implies that
$\mathcal{O}\_{\mathbb{P}^n}(1) $ admits no
$\PGL(n+1)$-linearization?


[Indeed, $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$ has order $n+1$ in
$V$ because $\Pic(\mathbb{P}^{n^2+2n}) \to
\Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a\_{ij}))=
\Pic(\PGL(n+1))$ induces an isomorphism
$\Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a\_{ij})) \cong
\mathbb{Z}/(\deg(\det(a\_{ij}))\mathbb{Z}$.]


To turn it another way round, why if $\mathcal{O}\_{\mathbb{P}^n}(1) $
would admit a $\PGL(n+1)$-linearization, then
the restriction of $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$
to $\PGL(n+1)$ must be trivial? I conjecture that this argument can somehow reduced to an easy comparison of orders of group elements in groups $\Pic(X)$, $\Pic^G(X)$ but I do not see how it can be directly related.


Maybe it somehow helps to know that we have always a morphism of groups $ \Pic^G(X) \to \Pic(X)$ which is not neccessarily injective.



Example of a line bundle not admitting a $\operatorname{PGL}(n+1)$-linearization in Mumford's GIT
$PGL(n+1)$-linearization of $\mathcal O\_{\mathbb P^n}(1)$ could be used to produce isomorphism $\gamma : p\_2^\*(\mathcal O\_{\mathbb P^n}(1)) \to \sigma^\*(\mathcal O\_{\mathbb P^n}(1))$, hence, taking to attention isomorphism
$$ \sigma^\*(\mathcal{O}\_{\mathbb{P}^n}(1)) \cong
 p\_1^\*(\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)) \otimes 
p\_2^\*(\mathcal{O}\_{\mathbb{P}^{n}}(1)) $$
which you have mentioned, we have equality of Picard classes $([\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)],[\mathcal{O}\_{\mathbb{P}^{n}}(1)])=(0,[\mathcal{O}\_{\mathbb{P}^{n}}(1)])$ in $Pic(PGL(n+1)) 
\times Pic(\mathbb P^n) = Pic(PGL(n+1) \times \mathbb P^n)$, so, $[\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)] = 0 $ in $Pic(PGL(n+1))$


How to construct such $\gamma$? $PGL(n+1)$-linearization of $\mathcal O\_{\mathbb P^n}(1)$ is a choice of morphism $\tilde{\sigma} : PGL(n+1) \times L \to L$ (where $L$ is a total space of the line bundle $\mathcal O\_{\mathbb P^n}(1)$) such that the diagram
$$\require{AMScd}
\begin{CD}
p\_2^\*(\mathcal O\_{\mathbb P^n}(1)) = PGL(n+1) \times L @>\tilde{\sigma}>> L\\
@V{}VV @V{}VV \\
PGL(n+1) \times \mathbb P^n @>{\sigma}>> \mathbb P^n
\end{CD}$$
commutes. Total space of the line bundle $\sigma^\*(\mathcal O\_{\mathbb P^n}(1))$ is by definition a fiber product $(PGL(n+1) \times \mathbb P^n) \times\_{\mathbb P^n} L$, hence, by universal property of fiber products the morphism $\gamma : p\_2^\*(\mathcal O\_{\mathbb P^n}(1)) \to \sigma^\*(\mathcal O\_{\mathbb P^n}(1))$ of $PGL(n+1) \times \mathbb P^n$-varieties exists. $\gamma$ is an isomorphism since for every $g \in PGL(n+1)$ induced morphism $\gamma\_g : O\_{\mathbb P^n}(1) \to g^\* O\_{\mathbb P^n}(1)$ is an isomorphism.


