Timeline for Is the set of rational points of an (almost) simple algebraic group simple?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 1, 2023 at 23:06 | comment | added | YCor | @LSpice yes, sorry I indeed meant "non-simply-connected (semisimple) groups" | |
| Mar 1, 2023 at 18:55 | comment | added | LSpice | @YCor, re, I am confused by "non-semisimple groups (as in most adjoint cases)". Should it be "non-simply connected groups"? | |
| Mar 1, 2023 at 9:05 | comment | added | YCor | @Linus I meant exactly what I wrote. As algebraic groups, $\mathrm{PGL}_n$ and $\mathrm{PSL}_n$ are equal. The consequence is an ambiguity i the notation $\mathrm{PSL}_n(K)$: if it is interpreted as the group of $K$-points of the algebraic group $\mathrm{PSL}_n$, then it is in general larger than the group-theoretic $\mathrm{PSL}(n,K)=\mathrm{SL}_n(K)/$center. (To be clear, in this statement $\mathrm{PGL}_n$ is meant as an algebraic group, i.e. identified to the functor $L\mathrm{PGL}_n(L)$. Quotients of algebraic groups need not be surjective when passing to $K$-points.) | |
| Mar 1, 2023 at 7:16 | comment | added | Linus | @YCor I think it should read $PGL_2=GL_2/Z(GL_2)$ and $PGL_2(K)/PSL_2(K)=K^*/{K^}^2$. | |
| Feb 28, 2023 at 15:29 | answer | added | David E Speyer | timeline score: 4 | |
| Feb 28, 2023 at 7:57 | answer | added | Linus | timeline score: 12 | |
| Feb 28, 2023 at 0:25 | history | became hot network question | |||
| Feb 27, 2023 at 21:42 | comment | added | YCor | For non-semisimple groups (as in most adjoint cases) I think one useful keyword is "Whitehead group". I think I remember they are described by some exact sequences (possibly this is evoked in Serre's "Galois Cohomology")? | |
| Feb 27, 2023 at 21:27 | answer | added | LSpice | timeline score: 12 | |
| Feb 27, 2023 at 21:23 | comment | added | LSpice | I think Steinberg shows that, for a simply connected group over a finite field, there are only a very few small cases where this fails. However, "somewhere in Steinberg" is vast literature to search. | |
| Feb 27, 2023 at 19:18 | comment | added | Gro-Tsen | Maybe the question should be: “are there not-too-restrictive conditions that one can add on the situation that will ensure that $H(K)$ is simple?” | |
| Feb 27, 2023 at 18:19 | answer | added | Dave Benson | timeline score: 13 | |
| Feb 27, 2023 at 16:43 | comment | added | H A Helfgott | Ah, right. Thanks. | |
| Feb 27, 2023 at 16:40 | comment | added | YCor | The abelianization of $\mathrm{PGL}_2(K)$ is $K^*/{K^*}^2$. And, as an algebraic group, $\mathrm{PGL}_2=\mathrm{SL}_2/Z(\mathrm{SL}_2)$. | |
| Feb 27, 2023 at 16:36 | comment | added | H A Helfgott | Not sure I understand this. What equals $K^*/{K^*}^2$? | |
| Feb 27, 2023 at 16:28 | comment | added | YCor | No, e.g., for $G=\mathrm{PSL}_2=\mathrm{PGL}_2$ one gets the abelianization $K^*/{K^*}^2$. | |
| Feb 27, 2023 at 16:24 | history | asked | H A Helfgott | CC BY-SA 4.0 |