Let $\mathfrak{g}$ be a semisimple complex Lie algebra, $A \in \mathfrak{g}$ a principal 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.

More specifically, given an irreducible finite-dimensional representation $r$ of $\mathfrak{g}$, the answer seems to be the set of all polynomials $P$ applied to $r(A)$ such that the result is still in the image $r(\mathfrak{g})$. Is that true? Is the space $P \in \mathbb{C}[x]$ with $P(r(A)) \in r(\mathfrak{g})$ of dimension the rank of $\mathfrak{g}$ ?

EDIT: Given $\mathfrak{g}$ as above, is there an irreducible representation $r$ such that for all principal nilpotent elements $A$ we have the following equality: $\mathbb{C}[r(A)] \cap r(\mathfrak{g}) = r(\mathfrak{zg}(A))$ where $\mathfrak{zg}(A)$ denotes the centralizer of $A$?

For example take $\mathfrak{sl}_n$ with standard representation and $A$ the matrix with 1 on the over-diagonal. Then 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 verified the statement for $\mathfrak{so}(2n)$ with the standard representation. But I have no clue how to do it in general.

Let $\mathfrak{g}$ be a semisimple complex Lie algebra, $A \in \mathfrak{g}$ a principal 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.

More specifically, given an irreducible finite-dimensional representation $r$ of $\mathfrak{g}$, the answer seems to be the set of all polynomials $P$ applied to $r(A)$ such that the result is still in the image $r(\mathfrak{g})$. Is that true? Is the space $P \in \mathbb{C}[x]$ with $P(r(A)) \in r(\mathfrak{g})$ of dimension the rank of $\mathfrak{g}$ ?

For example take $\mathfrak{sl}_n$ with standard representation and $A$ the matrix with 1 on the over-diagonal. Then 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 verified the statement for $\mathfrak{so}(2n)$ with the standard representation. But I have no clue how to do it in general.

Edit: follow-up question.

Let $\mathfrak{g}$ be a semisimple complex Lie algebra, $A \in \mathfrak{g}$ a principal 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.

More specifically, given an irreducible finite-dimensional representation $r$ of $\mathfrak{g}$, the answer seems to be the set of all polynomials $P$ applied to $r(A)$ such that the result is still in the image $r(\mathfrak{g})$. Is that true? Is the space $P \in \mathbb{C}[x]$ with $P(r(A)) \in r(\mathfrak{g})$ of dimension the rank of $\mathfrak{g}$ ?

For example take $\mathfrak{sl}_n$ with standard representation and $A$ the matrix with 1 on the over-diagonal. Then 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 verified the statement for $\mathfrak{so}(2n)$. But I have no clue how to do it in general.

Let $\mathfrak{g}$ be a semisimple complex Lie algebra, $A \in \mathfrak{g}$ a principal 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.

More specifically, given an irreducible finite-dimensional representation $r$ of $\mathfrak{g}$, the answer seems to be the set of all polynomials $P$ applied to $r(A)$ such that the result is still in the image $r(\mathfrak{g})$. Is that true? Is the space $P \in \mathbb{C}[x]$ with $P(r(A)) \in r(\mathfrak{g})$ of dimension the rank of $\mathfrak{g}$ ?

EDIT: Given $\mathfrak{g}$ as above, is there an irreducible representation $r$ such that for all principal nilpotent elements $A$ we have the following equality: $\mathbb{C}[r(A)] \cap r(\mathfrak{g}) = r(\mathfrak{zg}(A))$ where $\mathfrak{zg}(A)$ denotes the centralizer of $A$?

For example take $\mathfrak{sl}_n$ with standard representation and $A$ the matrix with 1 on the over-diagonal. Then 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 verified the statement for $\mathfrak{so}(2n)$ with the standard representation. But I have no clue how to do it in general.