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S Jul 30, 2021 at 4:02 history bounty ended CommunityBot
S Jul 30, 2021 at 4:02 history notice removed CommunityBot
S Jul 22, 2021 at 2:27 history bounty started Dr. Evil
S Jul 22, 2021 at 2:27 history notice added Dr. Evil Canonical answer required
Jul 14, 2021 at 19:27 comment added Will Sawin @AlexIvanov In characteristic zero, the log/exp maps give an isomorphism between their formal completions. That's the only relationship I know.
Jul 14, 2021 at 19:24 comment added AlexIvanov @Dr.Evil: Unfortunately, I don't know whether the paper is published or not, I just found it on arxiv. Sorry!
Jul 14, 2021 at 19:24 comment added AlexIvanov @WillSawin and Dr.Evil: by the way, is there some clear relation between $k[Lie G]^G$ and $k[G]^G$ (adjoint actions)? It seems that for G = U_n we even have $k[\mathfrak{u}_n]^{U_n} = k[U_n]^{U_n}$ (probably this should follow using Will Sawin's explicit description, or some log/exp maps, which make sense since the group is unipotent). On the other side, I cannot see why such equality should hold for $G = GL_n$. This seems to me like a reasonably general question.
Jul 14, 2021 at 12:05 comment added Will Sawin Yes, that's what I'm saying.
Jul 14, 2021 at 4:10 comment added Dr. Evil For $U_n$ an argument is also given in Section 2 of arxiv.org/pdf/2001.00447.pdf.
Jul 14, 2021 at 3:44 comment added Dr. Evil @WillSawin So are you saying that we should have $k[\mathfrak{u_n}]^{U_n}=k[E_{1,2}^*, ..., E_{n-1,n}^*]$?
Jul 13, 2021 at 16:20 comment added Will Sawin 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....
Jul 13, 2021 at 3:09 comment added Dr. Evil 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.
Jul 12, 2021 at 19:25 comment added AlexIvanov 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.
Jul 12, 2021 at 10:01 history edited Matthieu Romagny CC BY-SA 4.0
looks --> look
Jul 12, 2021 at 10:00 history asked Dr. Evil CC BY-SA 4.0