Timeline for Centralizer of a reductive subgroup
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 22, 2022 at 12:15 | comment | added | Windi | @YCor: Thanks for letting me know about it. To be perfectly honest, I simply didn't know there is such thing as "accepting anwsers". I just took care of it. | |
| Dec 22, 2022 at 12:02 | vote | accept | Windi | ||
| Dec 22, 2022 at 1:34 | comment | added | LSpice | @DavidA.Craven, oops, I misread! | |
| Dec 21, 2022 at 21:41 | comment | added | David A. Craven | @LSpice It won't usually. I should have taken $G$ to be simple: if $H<G$ then $L(H)$ is always a submodule of $L(G)$. The question asked for the restriction to be reducible though, not irreducible. I just chose the Lie algebra because proper subgroups are guaranteed to yield reducible restrictions. | |
| Dec 21, 2022 at 13:24 | comment | added | LSpice | @DavidA.Craven, why should your $\rho$ be irreducible on $H$? (A basic example: it isn't if $Z(G)$ has positive dimension.) | |
| Dec 21, 2022 at 6:43 | comment | added | YCor | @Windi as Alex M. suggests, you should accept answers of questions whenever the answers are correct. Accepting correct answers is useful to everybody in this site and shows consideration to people who're trying to help you. Please take this into account. | |
| Dec 20, 2022 at 19:27 | comment | added | David A. Craven | Can't you just take $G$ any reductive group, $H$ any proper reductive subgroup with $C_G(H)=Z(G)$, and then take $\rho$ to be the representation on the Lie algebra of $G$? | |
| Dec 20, 2022 at 17:32 | comment | added | Alex M. | @Windi: Hmm... 24 questions but only 1 accepted answer? You must be very difficult to please. | |
| Dec 20, 2022 at 17:01 | answer | added | groupie | timeline score: 1 | |
| Dec 20, 2022 at 16:58 | comment | added | YCor | And then a variation works here. Take dimension 4, $G=\mathrm{SO}_4$, and $H$ the image of the 3-dim irreducible of $\mathrm{SL}_2$ (i.e., $H$ is the upper left $\mathrm{SO}_3$). The centralizer of $H$ in $\mathrm{GL}_4$ consists of those diagonal$(a,a,a,b)$ matrices, and hence in $\mathrm{SO}(4)$ it consists of $\{\pm 1\}$, and this is precisely the center of $\mathrm{SO}(4)$. | |
| Dec 20, 2022 at 16:47 | comment | added | YCor | Probably it would have been fine to edit your original question, where after a counterexample you now replaced the assumption that $H$ is a proper subgroup, with the assumption that $G$ is irreducible but not $H$ (in the original form $G$ was $\mathrm{SO}_n$ acting on $\mathbf{C}^n$ and the counterexample was the image of an odd-dimensional rep of $\mathrm{SL}_2$). | |
| Dec 20, 2022 at 16:44 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Dec 20, 2022 at 16:33 | history | asked | Windi | CC BY-SA 4.0 |