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Let $G$ be a connected linear algebraic group. This question concerns Hilbert's 14th Problem for the adjoint action of $G$ on itself. Let $k[G]^G$ denote the algebra of regular functions on $G$ invariant under the conjugation action. It is well-known that if $G$ is reductive, then $k[G]^G$ is finitely generated and that it is in fact a polynomial algebra on $r$ generators, where $r$ is the reductive rank of $G$. The main question concerns non-reductive groups:

Question: Is $k[G]^G$ always finitely generated? If not, can we give a necessary/sufficient criteria for when it is finitely generated? Can one "describe" $k[G]^G$ even if it is not finitely generated?

For instance, what does the invariant ring for the group $U_n$ of $n\times n$ unipotent upper triangular matrices look like?

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  • $\begingroup$ The following statement is a particular case of arXiv:1605.00800. (Everything is over an algbraically closed field $k$ of characteristic zero.) Let $U \subseteq {\rm GL}_n$ be the unipotent radical of a Borel. Let $\mathfrak{n}$ be the Lie algebra of $U$, so that $U$ acts on $\mathfrak{n}$ via the adjoint action. Then $k[\mathfrak{n}]^U$ is finitely generated. As far as I understand, in some other paper of the same author the ring of invariants is described more explicitly. This seems kind of similar/related to your last question. $\endgroup$
    – AlexIvanov
    Commented Jul 12, 2021 at 19:25
  • $\begingroup$ Thanks Alexlvanov...It seems in Theorem 4.3 of this paper, it is proved that $k[\mathfrak{n}]^U$ is actually a free polynomial algebra. I cannot figure out if this paper is published. $\endgroup$
    – Dr. Evil
    Commented Jul 13, 2021 at 3:09
  • $\begingroup$ I think for $U_n$ one can check that the ring of invariants is generated by the entries that are just above the diagonal, by inductively checking that invariant functions must be independent of the top-right entry, then the next two entries, then.... $\endgroup$
    – Will Sawin
    Commented Jul 13, 2021 at 16:20
  • $\begingroup$ @WillSawin So are you saying that we should have $k[\mathfrak{u_n}]^{U_n}=k[E_{1,2}^*, ..., E_{n-1,n}^*]$? $\endgroup$
    – Dr. Evil
    Commented Jul 14, 2021 at 3:44
  • $\begingroup$ For $U_n$ an argument is also given in Section 2 of arxiv.org/pdf/2001.00447.pdf. $\endgroup$
    – Dr. Evil
    Commented Jul 14, 2021 at 4:10

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