Question about proof of positive roots under reflection

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.

• 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. – Jim Humphreys Nov 2 '17 at 14:38

1 Answer

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.

• Thank you, now I almost have it! Could you please only give me a little hint how to prove that $B(\alpha, \alpha_s) \leq 0$? – Diglett Nov 2 '17 at 12:22
• You say it yourself: Note that cosines are non-negative since $m(s,t)\geq 2$ for $t\neq s$ :-) – Chris McDaniel Nov 2 '17 at 18:00
• 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? – Diglett Nov 3 '17 at 13:39