0
$\begingroup$

Let $(W,S)$ be a finite irreducible Coxeter-System of rank $n$ and $E$ be a real reflection representation of $W$. Let $x\in E$ and suppose that the isotropy group of $x$ is generated by one element in $S$. Now which are the subgroups of rank $n-1$ that do not stabilize $x$?

$\endgroup$
6
  • 1
    $\begingroup$ When you say "do not stabilize" do you mean that the stabilizer of $x$ in the subgroup is trivial? $\endgroup$
    – S. Carnahan
    Commented Nov 2, 2013 at 0:42
  • $\begingroup$ yes, that is what I mean $\endgroup$
    – Tom
    Commented Nov 4, 2013 at 9:01
  • $\begingroup$ If $s_1$ to $s_n$ are the roots and $s_1$ is the root which fixes $x$ then the group generated by $s_2$ to $s_n$ would be what I need. However, is this all that can happen? $\endgroup$
    – Tom
    Commented Nov 4, 2013 at 12:38
  • $\begingroup$ Tom: There are other possibilities too. Consider do instance what happens in rank 2 cases. $\endgroup$
    – Misha
    Commented Nov 4, 2013 at 19:39
  • $\begingroup$ No I am a bit confused specially by the "answer" of Ben. $\endgroup$
    – Tom
    Commented Nov 5, 2013 at 8:22

1 Answer 1

0
$\begingroup$

You have to consider the Coxeter Graph: The subgroups you are looking for are those that you get by removing one edge...

$\endgroup$
3
  • 1
    $\begingroup$ This is incorrect and does not even attempt to answer the question. $\endgroup$
    – Misha
    Commented Nov 4, 2013 at 19:41
  • $\begingroup$ why? isnt that all clear from the Coxeter Graph?! $\endgroup$
    – Ben
    Commented Nov 6, 2013 at 17:20
  • $\begingroup$ Ben: you also have to think about conjugated of standard parabolics. If you think your "solution" is correct, try to write the details. $\endgroup$
    – Misha
    Commented Nov 6, 2013 at 17:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .