There is an element of every possible length in $[W_{\theta} \backslash W]$

41 Views Asked by At

Let $(W,S)$ be the Weyl group of a root system with base $\Delta$, and let $\theta \subset \Delta$. Let $W_{\theta}$ be the group generated by $\theta$. It is a general result that in every right coset $W_{\theta}w'$ in $W_{\theta} \backslash W$, there is a unique element $w$ of shortest length. It is uniquely characterized by the property that $\ell(xw) = \ell(x) + \ell(w)$ for all $x \in W_{\theta}$, or equivalently by the condition that $w^{-1}(\theta) > 0$.

Let $[W_{\theta} \backslash W]$ be the set of these canonical right coset representatives. The unique longest element $w_0$ of $[W_{\theta} \backslash W]$ is $w_l w_{l,\theta}$, where $w_l$ and $w_{l,\theta}$ are the long elements of $W$ and $W_{\theta}$.

Let $s_{\alpha_1} \cdots s_{\alpha_n}$ be a reduced decomposition of $w_0$, for $\alpha_i \in \Delta$. Is it true that for each $1 \leq i \leq n$, the element $s_{\alpha_1} \cdots s_{\alpha_{i-1}}$ also lies in $[W_{\theta} \backslash W]$? I believe this should be true, but I have not been able to prove this yet.

Attempt: By reverse induction on $i$. The case $i = n$ is clear. For the general case, suppose that $s_{\alpha_1} \cdots s_{\alpha_i}$ lies in $[W_{\theta} \backslash W]$. If $s_{\alpha_1} \cdots s_{\alpha_{i-1}}$ does not lie in $[W_{\theta} \backslash W]$, then there is a $\beta \in \theta$ such that

$$s_{\alpha_{i-1}} \cdots s_{\alpha_1}(\beta) = -\gamma$$ for some positive root $\gamma$. But by induction,

$$s_{\alpha_i} s_{\alpha_{i-1}} \cdots s_{\alpha_1}(\beta) = - s_{\alpha_i}(\gamma) > 0$$ which implies $s_{\alpha_i}(\gamma) < 0$. Since $s_{\alpha_i}$ is a simple reflection, the only way this is possible is if $\gamma = \alpha_i$, with $s_{\alpha_i}(\alpha_i) = -\alpha_i$. So we have

$$s_{\alpha_{i-1}} \cdots s_{\alpha_1}(\beta) = - \alpha_i$$

and

$$s_{\alpha_i}s_{\alpha_{i-1}} \cdots s_{\alpha_1}(\beta) = \alpha_i$$ with both $\beta$ and $\alpha_i$ simple roots. I cannot seem to take the induction argument further than this, but maybe there is something I'm missing.

1

There are 1 best solutions below

0
On BEST ANSWER

Oh. This is a basic argument about root systems. If $w \in W$, and $\alpha \in \Delta$, then $\ell(ws_{\alpha}) = \ell(w) \pm 1$, with $$\ell(ws_{\alpha}) = \ell(w)+1 \iff w(\alpha) > 0$$

Taking $w = s_{\alpha_1} \cdots s_{\alpha_{i-1}}$ and $s = s_{\alpha_i}$, we have $$w(\alpha_i) = s_{\alpha_1} \cdots s_{\alpha_{i-1}}(\alpha_i) = -\beta < 0$$ which implies that $\ell(s_{\alpha_1} \cdots s_{\alpha_i}) = \ell(s_{\alpha_1} \cdots s_{\alpha_{i-1}}) -1$, but this is a contradiction because the decomposition $s_{\alpha_1} \cdots s_{\alpha_i}$ is reduced.