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Is the following statement true for finite reflection groups?

Let $G$ be a finite reflection group acting on $\mathbb{R}^n$, let $x, y\in \mathbb{R}^n$ and let $z$ be in the orbit of $y$. If $\langle x,y\rangle = \langle x, z\rangle$, then there exists $g$ in the stabiliser of $x$ such that $z = gy$.

An important addition: $x$ and $y$ are in the same chamber.

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    $\begingroup$ What is off-topic here? If this is trivial for someone, why not giving a hint where to look for the answer? My co-author and I are genuinely stuck with this, and I thought this is what this web site is for: ask experts in the area for a hint. $\endgroup$
    – Violetta
    Commented Sep 24, 2014 at 12:51
  • $\begingroup$ I didn't vote to close the question but when I read it I'm a little confused. Your condition $<x,y>=<x,z>$ seems too easy to satisfy to have any conclusion. Can't you just choose any $x$ orthogonal to both $y$ and $z$? Or does $<\cdot, \cdot>$ not represent inner product? $\endgroup$
    – Ryan Budney
    Commented Sep 24, 2014 at 15:05
  • $\begingroup$ No, the site is for research level problems, it is definitely not for asking experts to give hints. On the other hand, I am surprised that nobody has given any indication why they have voted to close, and the question does not seem completely trivial to me, so I have voted to reopen. $\endgroup$
    – Derek Holt
    Commented Sep 24, 2014 at 16:32
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    $\begingroup$ No: Take the dihedral group $D_3$ acting on $\mathbb R^2$. You have 6 chambers. Take $y$ in the interior of a chamber, and rotate it to $z$, then $z$ is not in the next chamber, but the second next. Now $x\ne 0$ is perpendicular to $z-y$, so it lies in the interior of a chamber (the one between or its negative), so the stabilizer of $x$ is trivial. $\endgroup$
    – Peter Michor
    Commented Sep 24, 2014 at 18:32
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    $\begingroup$ Thank you for the responses. I am sorry I forgot to add that $x$ and $y$ must be in the same chamber. Otherwise this is indeed a bit trivial. $\endgroup$
    – Violetta
    Commented Sep 24, 2014 at 21:48

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Compilation of my remarks:

The answer to the original question is no: Take the dihedral group $D_3$ acting on $\mathbb ℝ^2$. You have 6 chambers. Take $y$ in the interior of a chamber, and rotate it to $z$, then $z$ is not in the next chamber, but the second next. Now $x\ne 0$ is perpendicular to $z−y$, so it lies in the interior of a chamber (the one between or its negative), so the stabilizer of $x$ is trivial.

If $y$ and $z$ lie in the same chamber: The closed chamber is a fundamental domain and is the orbit space (no further identification on walls). Thus $y=z$ since they are in the same orbit, thus the statement is true.

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