I want to prove that the semisimple regular elements of $\mathfrak{sl}_n(\mathbb{C})$ are the traceless elements with $n$ distinct eigenvalues.
Here, I take 'regular' to mean that $\dim Z_\mathfrak{g}(x) = \operatorname{rk} \mathfrak{g}$.
In one direction, $n$ distinct eigenvalues implies $x$ is semisimple and thus so too is $\operatorname{ad}_x$, and it also necessarily implies that the geometric multiplicity of 0 for $x$ is as small as possible, i.e. $x$ is regular.
In the other direction, I am lost.
The wikipedia entry deals with regular elements of Lie groups, not Lie algebras, as indicated by the phrase
However if there are equal eigenvalues, then the centralizer is the product of the general linear groups of the eigenspaces of $M$, and has strictly larger dimension, so that $M$ is not regular.
This is not true for the centraliser $Z_\mathfrak{g}(x) = \{ y \in \mathfrak{g} : \operatorname{ad}_x(y) = 0\}$: the centraliser has no relation (as far as I can tell) to the eigenspaces of $x$ itself.
Where am I going wrong?
Best Answer
A semisimple element $x$ in $\mathfrak{sl}_n(\mathbb C)$ is of the form $x=G x' G^{-1}$ for some $G\in{\rm GL}_n(\mathbb C)$ and $x'=diag(\lambda_1,...,\lambda_n)$. The centralizer of $x$ is the set of matrices $y\in\mathfrak{sl}_n(\mathbb C)$ with $[x,y]=0$, i.e. writing $y=G y' G^{-1}$, $[x',y']=0$. The matrix entries of this equation are $$ [x',y']_{ij}=y'_{ij}(\lambda_i-\lambda_j)=0. $$ Thus the element is regular (meaning that the centralizer has dimension $n-1=rank(\mathfrak{sl}_n(\mathbb C))$) if and only if the $\lambda_i$'s are all distinct (otherwise the centralizer is larger).