# Characterization of all $w$ in the Weyl group satisfying $w \geq w_l w_{l, \theta}$

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 $$x \in W_{\theta}$$?

• 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. – Jim Humphreys Sep 16 '19 at 19:07

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 $$w_{l,\theta}w_lw$$.