Timeline for Representation theory of SU(2) as a discrete group
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2017 at 2:27 | comment | added | Francois Ziegler | @YemonChoi Yes; so it implies that $\mathrm{SU}(2)^\delta$ is not Type I, hence in a sense a “no” answer to question (3) for unitary representations. | |
| Oct 13, 2017 at 2:09 | comment | added | Yemon Choi | For the OP: I have not had time to look closely at arxiv.org/pdf/1611.05801 but does Theorem D answer your questions? | |
| Oct 13, 2017 at 1:59 | comment | added | Yemon Choi | @FrancoisZiegler What is the relevance of Thoma's paper here? AFAIK it deals with discrete Type I groups and shows they are virtually abelian | |
| Oct 12, 2017 at 23:14 | comment | added | Jim Humphreys | @Jens: Probably you trying to ask too many questions here, but for example in (3) the cited paper by Borel-Tits in the algebraic group setting is best compared with a more focused follow-up article by TIts for real Lie groups (if you can track it down): ams.org/mathscinet-getitem?mr=0379749 But you do need to specify finite dimensional representations if that is assumed. | |
| Oct 12, 2017 at 12:19 | comment | added | Francois Ziegler | For (2): mathoverflow.net/questions/191682/…. For (3), allowing infinite-dimensional irreps (you don't say): ams.org/mathscinet-getitem?mr=248288. | |
| Oct 12, 2017 at 12:05 | comment | added | YCor | you should check in the given references, as it should answer your latter question. | |
| Oct 12, 2017 at 11:59 | comment | added | Jens Reinhold | Hmm, okay, that is kind of cheap. I did mean conjugate by an inner automorphism. And what about non-trivial endomorphisms of $SU(2)^{\delta}$. Are there any? | |
| Oct 12, 2017 at 11:57 | comment | added | YCor | Finally (3) is probably tackled in Borel-Tits "homomorphismes abstraits de groupes algébriques simples", see jstor.org/stable/1970833?seq=1#page_scan_tab_contents, or numdam.org/item/SB_1972-1973__15__307_0 for Steinberg's Bourbaki seminar account (the latter is in English). | |
| Oct 12, 2017 at 11:51 | comment | added | YCor | What do you call "conjugate"? if you mean conjugate by an inner automorphism, there are more reps. If you mean by an automorphism of $SL_2(SL_2(\mathbf{C}))$, probably there's none. Additional reps are indeed by "conjugating" the standard rep by an automorphism of the field $\mathbf{C}$ (extended to $SL_2(\mathbf{C})$). For homomorphisms between Lie groups, continuous and smooth are the same; continuous reps of $SU(2)$ into $SL_2(\mathbf{C}$ are indeed all conjugate to the inclusion by an inner automorphism of the target group. | |
| Oct 12, 2017 at 11:31 | history | asked | Jens Reinhold | CC BY-SA 3.0 |