# Centralizer of an element in a matrix Lie group whose Jordan form is given

$$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\ad{ad}\DeclareMathOperator\Ad{Ad}$$Let $$G\subset \GL_n(\mathbb{C})$$ be a complex matrix Lie group, (e.g., $$\SO_n$$, $$\Sp_{n}$$), and let $$x\in G$$ with Jordan form $$u$$ ($$u$$ may not belong to $$G$$) given, I wonder how to compute the centralizer $$Z_{G}(x)$$.

I know the Lie algebra of $$Z_{G}(x)$$ is $$\{Y\in \mathfrak{g}\mid \Ad(x)Y=Y\}$$. But it's hard for me to find a matrix $$t\in \GL_n(\mathbb{C})$$ such that $$tut^{-1}\in G$$. I tried to find $$U\in \mathfrak{gl}_n(\mathbb{C})$$ such that $$\exp(U)=u$$ and then find a $$T\in \mathfrak{gl}_n(\mathbb{C})$$ such that $$\ad(T)(U)\in \mathfrak{g}$$, but I don't think that means $$\exp(T)u(\exp(T))^{-1}\in G$$.

Any help will be appreciated.

You’re question talks abut the centraliser but you then seem instead to be unsure how to relate an arbitrary Jordan normal to an element of $$G$$. I’ll try to answer both.
Firstly, the centraliser. A matrix of the form: $$\begin{pmatrix}\lambda & 1 & & \\ & \ddots &\ddots \\ & & \ddots &1 \\ & & & \lambda \end{pmatrix}$$ has centraliser given by upper triangular Toeplitz matrices e.g. : $$\begin{pmatrix}a & b & c & d\\ & a &b & c \\ & & a &b \\ & & & a \end{pmatrix}$$
Now if we have a $$v$$ conjugate to $$u$$, their centralisers are conjugate as well by the same element so we just need fo find a way to relate $$u$$ to an element of $$G$$. In practice this is a bit tedious so I'll be brief. We just need to pick $$v$$ with the right eigenvalues, dimension of eigenspaces and generalised eigenspaces, etc. We then find a basis of eigenvectors and generalised eigenvectors (arranged in Jordan chains) and then our conjugacy matrix has these as columns.
Of course there will be many such $$v$$ in general so it would usually make more sense to start with $$v$$ find its Jordan normal form and use that to compute the centraliser.
• I know for conjugate matrices, their centralizers are conjugate; but here the Jordan form u may not belong to G, so the centralizer in G of the element it represents may not be conjugate to the centralizer of u. Also, I think what you did is for $G=GL_n$, but general matrix Lie group can be more complicated. May 19, 2022 at 0:56
• Sorry, I didn't make that part clear. The centraliser of $x \in G$ is just the intersection of its centraliser in $GL_n$ with $G$. So take the conjugation of the centraliser of $u$ and then intersect it with $G$ May 19, 2022 at 9:08