Do Berkovich homogenous spaces exist?

Let G be a k-analytic group, and let H be a closed subgroup of G. Then does there exist a k-analytic space, which can be reasonably called the quotient G/H?

Commentary: I realise that I am not being overly precise here. This is partly because I am not sure in exactly what form I want to ask this question. In lieu of this, I will mention an example, that I suspect may produce an interesting quotient space.

Let Spec(A) be an affine group scheme of finite type over ko. Let G be the analytification of the generic fibre of Spec(A). On $A\otimes_{k^\circ} k$, one can define the seminorm where ||a|| is defined to be the infimum of all |t| where $t\in k^\times$ is such that $a\in tA=t(A\otimes_{k^\circ} 1)\subset A\otimes_{k^\circ} k$. Take the completion of this tensor product with respect to this seminorm, and let H be its spectrum (which is then naturally an affinoid subgroup of G).

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I seem to recall that this is discussed in the last chapter of Berkovich's book and the answer is at least sometimes yes. But I'm sure the real experts will be along soon. –  David Speyer May 6 '10 at 4:14
@David: I don't think there's anything in Berkovich's book about quotients for $k$-analytic groups; he has some discussion about analytifying algebraic groups and their quotients, but nothing about an intrinsic analytic theory of quotients. –  BCnrd May 7 '10 at 21:40
@Peter: Is $|k^{\times}|$ discrete? I assume $A$ is flat (and finitely presented) over $k^0$. My impression is that you want $H$ to be the "generic fiber" of the formal completion of special fiber. E.g., if $G = {\rm{SL}}_ n$ then $H$ is "unit ball", which has as $k'$-points precisely ${\rm{SL}}_ n(O_ {k'})$ for any $k'/k$. Are you trying to endow an analytic structure on "affine Grassmannian" type of objects? I don't think this is well-suited to Berkovich's theory; such an $H$ is a $k$-analytic domain in $G$, but not open & not closed for $k$-analytic topology (has "boundary", etc.) –  BCnrd May 7 '10 at 21:48
|k^\times| can be assumed discrete or not, I don't mind which right now. A should be flat (or smooth) and finitely presented. It would be cool if one could define some sort of k-analytic space that was like an affine Grassmannian or affine flag variety, but maybe this is not possible. –  Peter McNamara May 8 '10 at 19:56
@Peter: In the equi-char case, the sort of thing you ask for is an ind-scheme over the residue field, no kind of (interesting) analytic structure since you're sort of quotienting by something "open". Or rather, the complex-algebraic story involves a Laurent series field over $\mathbf{C}$, so maybe you want to work over $k(\!(t)\!)$ with $t$-adic topology? Alas, analytic structure of $k$ now does nothing. That being said, Marty Weissman at UC Santa Cruz was thinking about this sort of thing in the mixed-char case, so perhaps email him directly to ask if he came up with anything on it. –  BCnrd May 8 '10 at 22:58

1 Answer

There is an old and forgotten paper of Z. Bosch on homogeneous spaces for affinoid groups (in German), where he construct rigid analytic quotients for affinoid groups and also proves some properties. The same arguments should work in the case of Berkovich spaces.

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