Timeline for A question about $R$-points of an complex reductive group.
Current License: CC BY-SA 3.0
8 events
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| Apr 17, 2013 at 12:38 | comment | added | Jim Humphreys | @ayanta: Sorry, my comment on centers was too offhand. It wasn't immediately clear to me how natural the question is in this generality. Certainly one can construct examples of the type you mention. | |
| Apr 17, 2013 at 4:44 | comment | added | user30180 | @Jim Humphreys: it is not generally true that the center of a connected reductive (non-ss) group is connected, so it is unclear what you meant in the first part of the second sentence of your comment. For example, if $G = {\rm{SL}}_n$ and $d|n$ with $1 \le d < n$ then the central pushout of $G$ along $\mu_d \hookrightarrow {\rm{GL}}_1$ is non-ss connected reductive with disconnected center. (Direct product if $d = 1$.) Likewise, $G := {\rm{Spin}}_{2n}$ for $n > 2$ has center $\mu_2 \times \mu_2$ or $\mu_4$ and we can form a non-ss connected reductive central pushout along a $\mu_2$. | |
| Apr 16, 2013 at 23:22 | comment | added | Jim Humphreys | It's not clear to me which motivating examples occur for you of disconnected centers in (disconnected?) reductive groups which are not semisimple. For a connected reductive group, its center is already connected; while for a connected semisimple group, its center is finite and might cause trouble. A standard model for what you are looking at is GL_n, with the semisimple quotient by its center being PGL_n. Points of the latter over various fields can get complicated compared to PSL_n. And does the characteristic matter? Jantzen's book deals especially with prime characteristic. | |
| Apr 16, 2013 at 16:00 | vote | accept | Oliver Straser | ||
| Apr 16, 2013 at 14:33 | history | edited | Oliver Straser | CC BY-SA 3.0 |
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| Apr 16, 2013 at 14:22 | answer | added | Jason Starr | timeline score: 2 | |
| Apr 16, 2013 at 14:16 | comment | added | user30379 | Since $\mathbf{C}$ is algebraically closed, $Z_e$ is a split group of multiplicative type, so $Z_e$ is a split torus. Thus, for any $\mathbf{C}$-algebra $R$, the obstruction to $G(R((t)))/Z_e(R((t))) \rightarrow (G/Z_e)(R((t)))$ being bijective is a class in the etale cohomology set $H^1(R((t)),Z_e)$, which is a power of ${\rm{Pic}}(R((t)))$. For $R$ a field or even artinian local ring, this Pic is trivial and so bijectivity holds. Thus, you have bijectivity on infinitesimal points over $\mathbf{C}$, which probably implies an isomorphism as smooth ind-schemes, yes? | |
| Apr 16, 2013 at 14:07 | history | asked | Oliver Straser | CC BY-SA 3.0 |