1
$\begingroup$

I do not understand fully the proof of Corollary 1.6 from Mumford's Geometric Invariant Theory (§3: Linearization of an invertible sheaf, p 35):

Corollary 1.6
$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\Pic{Pic}$Let $G$ be a connected linear algebraic group $G$ acting on an algebraic variety over $k$, which is proper over $k$. (Mumford's variety is a scheme $X/k$ such that $\overline{X}= X \times \Spec(\overline{k})$ is irreducible and closed; cf. Chap 0.) Let $\mathcal{L}$ be an invertible sheaf on $X$.
Then if $ \overline{X} $ is a normal variety, some tensor power $\mathcal{L}^n$ of $\mathcal{L}$ is always linearizable.

Proof. According to a result of Chevalley [Ch 10], in this case all components of $\Pic(X/k) $ are proper over $k$, hence all reduced components are Abelian varieties. Therefore the connected linear group $G$, being birational to projective space as a variety (?), must act trivially on $\Pic(X/k) $ (Why that's true? comp. 1st question?) […and finally claim follows from previous Prop 1.5 which states that some power $\mathcal{L}^n$ of $\mathcal{L}$ is $G$-linearizable iff some multiple $n \lambda $ of $\lambda \in \Pic(X/k)(k)$ is fixed by $G$, where the $k$-point $\lambda$ of $\Pic(X/k)$ is defined by $\mathcal{L}$.]

Questions:
Why is $G$ as connected linear group birational to projective space?

And why the above leads to the consequence that $G$ acts trivially on $\Pic(X/k) $? We can reduce it to the question why the action of every connected linear group on an Abelian variety is trivial.

A Remark: Mumford nowhere describes this action of $G$ on $ \Pic(X/k)$ in explicit terms, so I assume he means the most natural one, i.e. for $S \to \Spec(k)$ the action of a geometric point $\alpha \in G$ on $X$ induces an action on $X \times_k S$ via $\alpha \times 1_S$ and this one acts on

$$ \Pic(X/k)(S) = \{ \mathcal{M} \text{ invertible sheaf on } X \times_k S \} / \{ \text{ inv. sheaves of the form } p^*_S(\mathcal{K}) \text{ for } \mathcal{K} \text{ invertible on } S \} $$

via taking pullpacks $\mathcal{M} \mapsto (\alpha \times 1_S)^*\mathcal{M}$. Is it correct?

[Ch10] Chevalley, C: Sur la théorie de la variété de Picard.

$\endgroup$
7
  • 1
    $\begingroup$ A linear group is rational (:= birational to affine space) over $\bar{k}$, but this is sufficient here: any morphism of a rational variety onto an abelian variety is trivial. So the orbits of $G$ acting on $\operatorname{Pic}(X/k) $ are trivial. $\endgroup$
    – abx
    Nov 5, 2022 at 18:51
  • 1
    $\begingroup$ ohh, of course I see. Thank you! $\endgroup$
    – user267839
    Nov 5, 2022 at 19:06
  • 1
    $\begingroup$ I think that some Cayley transform trick makes them birational to projective space. $\endgroup$
    – Ben McKay
    Nov 5, 2022 at 19:21
  • 1
    $\begingroup$ in the remark in last paragraphI made a guess as to how I think the action of $G$ on $\text{Pic }(X/k)$ would work in detail. Is it correct or is there a mistake somewhere? $\endgroup$
    – user267839
    Nov 5, 2022 at 22:13
  • $\begingroup$ Yes, it is correct — this is just the definition of the Pic functor. $\endgroup$
    – abx
    Nov 6, 2022 at 4:45

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.