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).



Properties of stabilizers of adjoint action general linear group
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](https://zbmath.org/?q=an:0249.20024).


