Properties of stabilizers of adjoint action general linear group 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).
 A: 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:

*

*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.

*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.
