Does the centralizer of a regular element in a semisimple Lie algebra act by polynomials?

Question

Let $\mathfrak g$ be a semisimple Lie algebra over $\mathbb C$, $\rho : \mathfrak g \to \operatorname{End}(V)$ a finite-dimensional irreducible representation and $x \in \mathfrak g$ regular with centralizer $Z$.

Does $\rho(Z)$ consist of polynomials in $\rho(x)$?

Examples where I know this is true:

• $x$ is semisimple: then $Z$ is a Cartan subalgebra, $\rho(x)$ acts by distinct scalars on the weight spaces $V_\lambda$ and the claim follows from Lagrange interpolation. (This argument doesn't work: the scalars need not be distinct.)
• $\mathfrak g = \mathfrak{sl}_n$ and $\rho = $ standard rep., because $x$ regular implies its minimal polynomial has degree $n$ so the (trace $0$) polynomials in $x$ give all of $Z$.

Answer 1

It is false for the same reason the first example is wrong. Take $\rho$ for example the adjoint representation. As soon as $\mathfrak g$ has rank at least $2$ you can find a regular semisimple $x$ such that some two roots (whose difference is not a root) of $Z$ coincide on $x$, while this is not true for generic $y \in Z$.

The standard representation for $\mathfrak{sl}_n$ is special in that the weights of a Cartan subalgebra differ by roots (rather than linear combinations of such).