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Let $k$ an algebraically closed field. Let $O=k[[\pi]]$ and $F=k((\pi))$ and $G\rightarrow GL_{n}$ a faithful representation of a semisimple group.

Let $A, B\in G(O)\cap G(F)^{rs}$ (rs for regular semisimple)) such that there exists $g\in GL_{n}(O)$ such that

$gAg^{-1}=B$,

Does there exist an element $g_{1}\in G(O)$ such that $g_{1}Ag_{1}^{-1}=B$?

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Do any of your conditions prevent $G$ from being a torus? –  S. Carnahan Oct 8 '13 at 9:08
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Even $SL(2)$ inside $GL(2)$ seem to fail. –  Marc Palm Oct 8 '13 at 9:24
    
I would prefer the question to be a comparison of the orbits... my naive idea is that more useful things can be said there. definitely the answer to the question in its current form is no. anyway +1. –  Marc Palm Oct 8 '13 at 21:47
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it doesn't fail for $SL_{2}$ because if you have $g\in GL_{2}(O)$ that conjugates, then $\frac{1}{\lambda}g\in SL_{2}(O)$ where $\lambda=\sqrt\det(g)$ that exists because $k$ is algebraically closed. –  prochet Oct 9 '13 at 20:35
    
Ah okay, I wasn't seeing the algebraically closed field part. I was thinking about local fields. Still S.Carnahan's example applies, doesn't it? –  Marc Palm Oct 10 '13 at 8:08

1 Answer 1

Here is a counterexample: Let $a,b$ be distinct units in $k$ such that $ab \neq 1$, and let $G = SL_2 \times SL_2$ be given the usual block diagonal embedding into $GL_4$. Then the matrices $$A = \left( \begin{smallmatrix} a \\ & 1/a \\ & & b \\ & & & 1/b \end{smallmatrix} \right) \qquad \text{and} \qquad B = \left( \begin{smallmatrix} b \\ & 1/b \\ & & a \\ & & & 1/a \end{smallmatrix} \right)$$ are conjugate in $GL_4(O)$, but not $G(O)$. The underlying problem in this example seems to be that the normalizer of the torus in $G$ is too small.

I suspect Mark Palm's suggestion for $G = SL_2 \hookrightarrow GL_2$ will admit a counterexample when $k$ has characteristic 2, but I have not worked it out.

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