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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!

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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.)

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  • $\begingroup$ 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
    Commented Apr 25, 2017 at 8:16

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