0
$\begingroup$

Let $k$ be an algebraically closed field, $G$ a connected reductive group, $T$ a maximal torus, $W$ the Weyl group and $\chi:G\rightarrow T/W$ the Steinberg morphism.

We know that if $\gamma,\gamma'\in G$ are semisimple and $\chi(\gamma)=\chi(\gamma')$ then they are conjugated.

Is it still true over a local artinian ring $R$ with residue field $R$, or over complete discrete valuation rings with residue field $k$?

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ Define "semisimple" for an $R$-valued point. $\endgroup$
    – Marguax
    Commented Oct 15, 2013 at 22:12
  • $\begingroup$ $g \in G(R)$ is semisimple, if there exists a torus T over R, such that there is a morphism from $T\rightarrow G$ over $R$ which contains $g$ in its image. $\endgroup$
    – prochet
    Commented Oct 16, 2013 at 6:50
  • $\begingroup$ I don't think it is true that infinitesimal multiplicative type subgroups with non-cyclic Cartier dual necessarily lie in a subtorus of $G$ (even over an algebraically closed field), so by your definition the "universal point" of such a subgroup scheme would not be semisimple. Not that there's anything "wrong" with that, as it depends on your motivation (not given), but are you sure that this phenomenon doesn't give you pause about your definition? Do you know if the weaker condition of admitting such a $T$ over a finite flat local extension of $R$ is equivalent to have one over $R$? $\endgroup$
    – Marguax
    Commented Oct 16, 2013 at 6:58
  • $\begingroup$ I don't know for your question on $R$. Probably we need a stronger definition of semisimple. $\endgroup$
    – prochet
    Commented Oct 16, 2013 at 9:31
  • $\begingroup$ There is a stronger notion of semisimplicity regarding to a faithful representation $\rho:G\rightarrow GL_{n}$ that say that $g\in G(R)$ is strongly semisimple if $R[\rho(g)]$ is a finite étale algebra locally a direct factor of $\mathfrak{gl}_{n}(R)$, this definition avoids your example I presume. $\endgroup$
    – prochet
    Commented Oct 16, 2013 at 9:34

0

Reset to default

You must log in to answer this question.