1
$\begingroup$

Let $G$ be a reductive algebraic group over some algebraically closed field $k$. Recall that, given an algebraic variety $X$ with an action of $G$, it is said that this action is linearizable if there exists a line bundle $\pi: L \to X$ with a fiberwise linear action of $G$ on $L$ for which $\pi$ is equivariant.

As a particular case, we can take $X=G$ and consider the action of $G$ on itself by conjugation. Then, my questions are:

  1. Is this conjugation action linearizable?
  2. If the conjugation action is linearlizable, under what conditions can the line bundle $L$ be taken ample?

In particular, I am concerned with the case $G = SL(2,\mathbb{C})$. In that case, we know that $X // G = \mathbb{A}^1$. As mentioned in this paper of Doebeli, if the GIT quotient is zero-dimensional, the existence of linearlization follows from Luna's slice theorem. However, if the quotient is $1$-dimensional, as in the case of $G=SL(2,\mathbb{C})$, the problem is harder and, in general, the result is false.

Anycase, it seems like, using the results and definitions of that paper, it could be possible to define a linear model for this action satisfying conditions of proposition 2, and thus obtaining that this conjugation action is linearlizable. However, I belive that this procedure too complicated and there must exists a general argument to show that these conjugation actions are always linearizable, but I can't find it. Furthermore, even using that paper, we can't assure that such line bundle is ample.

Thank you so much in advance!

$\endgroup$

1 Answer 1

6
$\begingroup$

Sumihiro proved that every normal quasi-projective $k$-scheme with an action of a smooth affine connected group $G$ has an ample $G$-equivariant line bundle. This applies in particular to any action of $G$ on itself. (Sumihiro's theorem was generalized a few days ago by Brion.)

$\endgroup$
1
  • $\begingroup$ You are completely right. I think that, actually, it can be done directly. Embed $G \subset GL(n, k) \subset k^2$. Now, seen as matrices, given any $g \in G$, the action of $g$, $g \cdot: G \to G$ extends to a linear map $g \cdot: k^{n^2} \to k^{n^2}$ giving us an action of $G$ on $k^{n^2}$ that restricts to the conjugation action on $G \subset k^{n^2}$. $\endgroup$
    – a_g
    Apr 25, 2017 at 8:16

Your Answer

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

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