In a reflection group, the longest word $w_0$ contains all simple reflections

Question

This is Exercise 2 of section 1.8 in Humphreys' "Reflection groups and Coxeter groups", p.16.

The longest word $w_0$ in a finite reflection group $W$ acting on a Euclidean space (with a specified basis/system of simple roots $\Delta$) is defined as the unique element of $W$ that maps the fundamental chamber to its inverse. (This makes sense because $W$ acts simply transitively on the chambers.) I am trying to show that, in every reduced expression of $w_0$, every simple reflection occurs at least once.

I am trying to proceed by contradiction: suppose that $w_0$ admits the reduced expression $w_0 = s_1\cdots s_r$, where $s_i = s_{\alpha_i}$ for some simple roots $\alpha_i$, and that there exists a missing simple root $\beta \neq \alpha_i \ \forall i$. Intuitively I want to show that $w_0s_\beta$ is a longer word than $w_0$, contradicting the fundamental property of the latter, but I find myself unable to do so. Any hint would be greatly appreciated!

Answer 1

I believe your argument still works. Suppose $w_0=s_{\alpha_1}\cdots s_{\alpha_r}$ is a reduced expression in terms of various simple reflections, and suppose $\beta$ is a simple root distinct from all these $\alpha_i$.

I assume there is some positive definite bilinear form $(-,-)$ on the Euclidean space, so the simple reflections are defined by the formula $$ s_\alpha(\beta)=\beta-2\frac{(\beta,\alpha)}{(\alpha,\alpha)}\alpha. $$

The idea is that when you apply these simple reflections, they just add some multiple of their corresponding simple root to the root you plug in, but they'll never add a multiple of $\beta$, since they're all distinct from $\beta$. So the simple reflections in the reduced expression of $w_0$ can't change the coefficient of $1$ for $\beta$, as the simple roots are linearly independent.

That is, we must have $w_0(\beta)=\beta+\sum_{\alpha_i\in\Delta\setminus\{\beta\}}c_i\alpha_i$ for some coefficients $c_i$. This image is necessarily a root, and since the simple roots are linearly independent, we are ensured a positive coefficient of $\beta$, hence $w_0(\beta)\in\Phi^+$.

Answer 2

I believe I have found an easy way to proceed. We know that $w_0$ maps every positive root to a negative root; in particular, it maps $\beta$ to a negative root. But by Proposition 1.4 of the book, for any simple root $\alpha ­\in \Delta$, the simple reflection $s_\alpha$ permutes all the positive roots that are distinct from $\alpha$. Therefore, if $\beta \neq \alpha_i$ for all $i$, the word $w_0 = s_{\alpha_1}\cdots s_{\alpha_r}$ sends $\beta$ to another positive root, a contradiction. (First $s_{\alpha_r}$ sends $\beta$ to a positive root; then $s_{\alpha_{r-1}}$ sends that positive root to another positive root; and so on.)

EDIT: Wait, this doesn't necessarily work. There is no reason why $s_{\alpha_{r-1}}$ should send $s_{\alpha_r}(\beta)$ to another positive root unless we know that $s_{\alpha_r}(\beta) \neq \alpha_{r-1}$...