-1
$\begingroup$

Since I did not receive a lot of responses on Math Stack Exchange I would like to repost this question here.

Let $(W, S)$ be a finite Coxeter system. Furthermore, let $V$ be a real vector space with a (finite) basis $\{ \alpha_s | s \in S \}$.

For every $s \in S$ one can define the reflection $\sigma_s : V \to V, v \mapsto v - 2 B(v, \alpha_s) \alpha_s $ where $B$ is the bilinear form defined by $B(\alpha_s, \alpha_t ) = - \cos(\frac{\pi}{m(s,t)})$.

There also exists a unique group homomorphism$\sigma : W \to \operatorname{GL}(V)$ such that $\sigma(s) = \sigma_s$, the so-called geometric representation of $(W,S)$. From now on we denote $\sigma(w) (v)$ for $v \in V, w \in W$ by just $w(v)$.

Now let $\Phi = \{w(\alpha_s) | w \in W, s \in S\} $ be the set of all roots and $\Pi = \{ v \in \Phi | v = \sum_{t \in S} c_t \alpha_t, c_t \geq 0 \forall t \in S\} $ the system of positive roots. Now consider $\alpha_s \in V$ for $s \in S$, a so-called simple root.

At this moment I am dealing with the proof of $s(\Pi \setminus \{ \alpha_s \}) = \Pi \setminus \{ \alpha_s \}$.

Let $\alpha \in \Pi \setminus \{ \alpha_s \}$. Then one can write $\alpha = \sum_{t \in S} c_t \alpha_t$ where the coefficients are non-negative and there exists a $t_0 \in S$ with $c_{t_0} > 0$.

Applying $s$ leads to $s(\alpha) = \sum_{t \neq s} c_t \alpha_t + \mu \alpha_s$ where I calculated $\mu = - c_s + 2 \sum_{t \neq s} c_t \cos(\frac{\pi}{m(s,t)}) $. Note that the cosines are non-negative since $m(s,t) \geq 2$ for $t \neq s$.

But I don't really know why $\mu \geq 0$ though since the $-c_s \leq 0$ could mess everything up.

It drives me insane that I could not find an answer yet. Could someone tell me why it still works? I really appreciate your help.

$\endgroup$
  • $\begingroup$ This is a very old textbook result, easily proved at an appropriate early place in the development. So it doesn't seem to qualify as research-level at all. $\endgroup$ – Jim Humphreys Nov 2 '17 at 14:38
1
$\begingroup$

Note that $B(\alpha,\alpha_s)\leq 0$ for every $\alpha\in\Pi$. Then if $\alpha=\sum_{t\neq s}c_t\alpha_t+c_s\alpha_s$, we have \begin{align*} B(\alpha,\alpha_s)= & \sum_{t\neq s}c_tB(\alpha_t,\alpha_s)+c_sB(\alpha_s,\alpha_s)\\ =& -\sum_{t\neq s}c_t\cos\left(\frac{\pi}{m(s,t)}\right)+c_s\\ & \leq 0 \end{align*} This should give you what you need.

$\endgroup$
  • $\begingroup$ Thank you, now I almost have it! Could you please only give me a little hint how to prove that $B(\alpha, \alpha_s) \leq 0$? $\endgroup$ – Diglett Nov 2 '17 at 12:22
  • $\begingroup$ You say it yourself: Note that cosines are non-negative since $m(s,t)\geq 2$ for $t\neq s$ :-) $\endgroup$ – Chris McDaniel Nov 2 '17 at 18:00
  • $\begingroup$ I'm really sorry because I really tried to thought about your remark for a while but I still can't see it. What makes it impossible for $c_s$ to be greater than $\sum_{t\neq s}c_t\cos\left(\frac{\pi}{m(s,t)}\right)$ in your inequality? $\endgroup$ – Diglett Nov 3 '17 at 13:39

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.