Let $W$ be the Weyl group of a root system $\Phi$ with base $\Delta$ and system of positive roots $\Phi^+$. Let $S = \{ w_{\alpha} : \alpha \in \Delta \}$ be the set of simple reflections corresponding to elements of $\Delta$. Let $\theta \subset \Delta$, and let $w_0 = w_l w_{l,\theta}$, where $w_l$ and $w_{l,\theta}$ are the long elements of $W$ and $W_{\theta} = \langle w_{\alpha} : \alpha \in \theta \}$.

The element $w_0$ is characterized uniquely by the following property: in each left coset of $W_{\theta}$ in $W$, there is a unique representative of smallest length, and $w_0$ is the longest of all these special representatives. It has the property that $\ell(w_0x) = \ell(w_0) + \ell(x)$ for all $x \in W_{\theta}$.

I noticed it follows immediately from here that if $x \in W_{\theta}$, then $w_0x \geq w_0$ in the Bruhat order. I was wondering whether the converse is true. So my question is:

If $w \in W$, and $w \geq w_0$ in the Bruhat order, then do we have $w = w_0x$ for some $x \in W_{\theta}$?

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    $\begingroup$ Note that Nathan's answer implies that your assumption about $W$ being a Weyl group is too restrictive; it just needs to be a finite Coxeter group. $\endgroup$ Sep 16, 2019 at 19:07

1 Answer 1


Yes, this follows from the fact that $x \mapsto w_l x$ is an antiautomorphism of the Bruhat order on a finite Coxeter group. (See Björner and Brenti, Proposition 2.3.4, for example, but their $w_0$ is your $w_l$) You also need the fact that $w_l$ is an involution. (For example, Björner and Brenti, Proposition 2.3.2.)

If $w\geq w_0$, then $w_lw\leq w_{l,\theta}$, which further implies that $w_lw\in W_\theta$. Thus $w=w_l(w_lw)=w_0(w_{l,\theta}w_lw)$, so the desired $x$ is $w_{l,\theta}w_lw$.


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