Timeline for Invariant ring of linear algebraic groups
Current License: CC BY-SA 4.0
15 events
when toggle format | what | by | license | comment | |
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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
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Jul 12, 2021 at 10:00 | history | asked | Dr. Evil | CC BY-SA 4.0 |