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This is a follow-up to this question.

Let $\mathfrak{g}$ be a semisimple complex Lie algebra, $A \in \mathfrak{g}$ a regular nilpotent element (i.e. its centralizer is of dimension equal to the rank of $\mathfrak{g}$). I wish to understand the set of elements which commute with A.

I would be happy if the answer is something like "the polynomials of $A$ which are still in $\mathfrak{g}$". So in order to make sens of polynomials, we need a representation of the Lie algebra. So the precise question is:

Is there a faithful irreducible representation $𝑟$ such that for all regular nilpotent elements $𝐴$ we have the following equality: $ \mathbb{C}[𝑟(𝐴)]∩𝑟(\mathfrak{g})=𝑟(\mathfrak{z_g}(𝐴))$ where $\mathfrak{z_g}(𝐴)$ denotes the centralizer of $𝐴$?

For example take $\mathfrak{sl}_n$. Then the standard representation fulfils the question since $𝐴$ is the matrix with 1 on the over-diagonal (up to conjugacy $A$ is unique here). And a matrix $𝐵$ commutes with $𝐴$ iff $𝐵=𝑃(𝐴)$ with $𝑃$ a polynomial without constant term and of degree smaller than $𝑛$. This is due to the fact that $𝐴$ is cyclic here.

I also analyzed the question for $\mathfrak{so}_n$ where it works for the standard representation. But I have no clue how to do it in general without a case by case treatment of the classical Lie algebras.

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    $\begingroup$ I added "faithful" because the 1-dimensional trivial representation satisfies the condition and is clearly not what you're interested in. Also maybe you know whether the question reduces to the case of $\mathfrak{g}$ simple. Namely, if $\rho_i$ works for $\mathfrak{g}_i$, $i=1,2$, does $\rho_1\otimes\rho_2$ work for $\mathfrak{g}_1\times\mathfrak{g}_2$? $\endgroup$
    – YCor
    Commented Jul 17, 2019 at 14:48
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    $\begingroup$ I don't know the answer off-hand, but I'd like to point out that many properties of nilpotent elements and orbits are established on a case-by-case basis, with proofs in the exceptional cases also involving reduction to Levi components. The question itself appears close to Kostant's work on Toda lattices, which may be worth studying. $\endgroup$
    – Victor Protsak
    Commented Jul 17, 2019 at 18:39
  • $\begingroup$ Does anything in the question mathoverflow.net/q/160891/27465 and/or its answers help here? $\endgroup$
    – Torsten Schoeneberg
    Commented Mar 2 at 18:27

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