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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\ad{ad}\DeclareMathOperator\Ad{Ad}$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.

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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. : $$ \begin{pmatrix}a & b & c & d\\ & a &b & c \\ & & a &b \\ & & & a \end{pmatrix} $$

So the centraliser of a matrix in Jordan normal form comprises block diagonal matrices with these Toeplitz blocks where the generalised eigenspaces are and generic blocks where the honest eigenspaces are.

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.

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  • $\begingroup$ 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. $\endgroup$
    – user837898
    May 19, 2022 at 0:56
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    $\begingroup$ 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 $G$ $\endgroup$
    – Callum
    May 19, 2022 at 9:08

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