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:



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> 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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Characterization of all $w$ in the Weyl group satisfying $w \geq w_l w_{l, \theta}$
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 $w\_{l,\theta}w\_lw$.


