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 an 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.

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.

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 an 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{zg}(𝐴))$ where $\mathfrak{zg}(𝐴)$ 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.

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 an 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.