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I hope somebody can give me a good reference for the following:

Let $G$ be a complex reductive group $H$ be a closed subgroup. Let further $R$ be any $\mathbb{C}$-algebra. Then the canonical map $$G(R)/H(R)\to (G/H)(R)$$ is known to be injective but in general not surjective. See for example [1].

So now my question:

Let $Z_e:=Z(G)_e$ be the connected component of center of $G$ containing $e$, is then the canonical map $$G(\mathbb{C}((t)))/Z_e(\mathbb{C}((t)))\to (G/Z_e)(\mathbb{C}((t))) $$ a bijection.

Or even more, does the canonical map above induce an isomorphism of ind-varietes $$ G(\mathbb{C}( (t)))/Z_e(\mathbb{C}((t)))\cong (G/Z_e)(\mathbb{C}((t))) $$

Remark:

Note that if we take $ Z$ to be the center of $G$ and $R=\mathbb{C}[[t]]$ then the map above is also surjective (this follows from the fact $\mathbb{C}[[t]]$ is strict henselian, hence by SGA III every map $Spec \ \mathbb{C}[[t]]\to G/Z$ can be lifted to a map $Spec \ \mathbb{C}[[t]]\to G $)

[1] Jantzen, Jens Carsten Representations of algebraic groups. Second edition. Mathematical Surveys and Monographs, 107. American Mathematical Society

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    $\begingroup$ 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? $\endgroup$
    – user30379
    Commented Apr 16, 2013 at 14:16
  • $\begingroup$ 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. $\endgroup$
    – Jim Humphreys
    Commented Apr 16, 2013 at 23:22
  • $\begingroup$ @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 $\mu_2$. $\endgroup$
    – user30180
    Commented Apr 17, 2013 at 4:44
  • $\begingroup$ @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. $\endgroup$
    – Jim Humphreys
    Commented Apr 17, 2013 at 12:38

1 Answer 1

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By Hilbert's Theorem 90, every torsor for a split torus over a field is trivial. Thus, as pranavk has commented, this should give surjectivity.

$\textbf{Edit.}$ The first answer I wrote (now changed) applied to the full center $Z$. I did not realize that the OP is asking about the quotient by $Z_e$, the connected component of the identity. I have corrected my answer.

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  • $\begingroup$ Now i am a little confused, so you say the statement holds also for the full center? $\endgroup$
    – Oliver Straser
    Commented Apr 16, 2013 at 14:32
  • $\begingroup$ @Strasser: No, the argument above does not apply to the full center. My first answer (which was only up for a few minutes) was a counterexample for the full center $Z$ in $\textb{SL}_2$. $\endgroup$
    – Jason Starr
    Commented Apr 16, 2013 at 15:45

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