2
$\begingroup$

Let $G=GL(n,\mathbb{C})$ and let us consider $x \in GL(n,\mathbb{C})$. I'd like to know whether the following is true: the stabilizer for the conjugation action $C(x)$ is special in the sense that every $C(x)$ principal bundle in the etale topology is also a Zariski principal bundle. Everything should be considered here in the algebro-geometric setting.

For regular semisimple and regular nilpotent elements that should be true, but for general $x$ things get a little bit messy and I do not know how to end up the proof(even understand whether that should be true or not).

Improve this question
$\endgroup$
4
  • $\begingroup$ You might as well take $x$ to be any $n\times n$-matrix, and then using the Jordan decomposition you can reduce the question to looking at centralizers of nilpotent elements in Levi subgroups of $GL_n(\mathbb{C})$, i.e. nilpotent elements in $GL_{n_1} \times \dots \times GL_{n_k}$. It therefore suffices to consider nilpotent elements in $GL_n$, by looking at each factor individually. $\endgroup$
    – Jef
    Commented Apr 15, 2021 at 19:41
  • $\begingroup$ I've tried to do that but the centralizer of a nilpotent matrix seems not to easy to compute actually $\endgroup$
    – Tommaso Scognamiglio
    Commented Apr 16, 2021 at 10:16
  • $\begingroup$ I agree, but at least you only need to understand their reductive quotients. This is because an extension of special groups is special and a unipotent group over $\mathbb{C}$ is special. So if $x$ is nilpotent, the unipotent radical $R_u(C(x))$ is special and we just need to know that $C(x)/R_u(C(x))$ is special. $\endgroup$
    – Jef
    Commented Apr 16, 2021 at 10:23
  • $\begingroup$ Actually if each Jordan block has different size the group seems more pleasant. In the situation where one has different Jordan blocks of the same size,I wasn't really able to find relevant normal subgroups even in the case of a $4 \times 4$ matrix with two blocks of size $2$. Do you have any idea whether the guess is true or not? $\endgroup$
    – Tommaso Scognamiglio
    Commented Apr 16, 2021 at 11:05

1 Answer 1

Reset to default
5
$\begingroup$

This is true and follows from:

Claim: Let $x$ be a $n\times n$ matrix with $\mathbb{C}$-coefficients. Then the centralizer $C(x)$ of $x$ in $GL_n(\mathbb{C})$ fits into a short exact sequence $1\rightarrow U\rightarrow C(x) \rightarrow \prod_{n_i} GL_{n_i}\rightarrow 1$, where $U$ is a unipotent group and $n_i$ a sequence of integers.

Since an extension of special groups is special and since $\mathbb{G}_a$ and $GL_k$ are special, $C(x)$ is special too, answering your question. For the proof of the claim:

  1. It suffices to prove the claim for $x$ nilpotent. Indeed, let $x = x_s+x_n$ be the Jordan decomposition into semisimple and nilpotent parts. Then $C(x)$ is the centralizer of $x_n$ in $C(x_s)$. Since $C(x_s)$ is a product of general linear groups (easy to see by taking $x_s$ a diagonal matrix) and $x_n$ is a nilpotent element of $Lie(C(x_s))$ (a product of matrix Lie algebras), we are done by looking at each factor.
  2. For $x$ nilpotent, $C(x)$ is an extension of a unipotent group and a product of general linear groups. This follows from the explicit computation in section IV.1.7-1.8 in the paper below.

Springer, T. A.; Steinberg, R., Conjugacy classes, Sem. algebr. Groups related finite Groups Princeton 1968/69, Lect. Notes Math. 131, E1-E100 (1970). ZBL0249.20024.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .