Greatest element of ${}^IW$

Question

Let $\mathfrak{g}$ be a finite dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$. Let $W$ be the associated Weyl group and let $\Phi$ be its root system. We write $\Phi^+$ for the set of positive roots in $\Phi$.

Fix a subset of simple roots $I$. We define $ {}^IW := \{w\in W: w<s_\alpha w \ \text{for all }\alpha\in I\}, $ where $<$ is the Bruhat ordering on $W$.

According to Section 3.2 of KOSTANT MODULES IN BLOCKS OF CATEGORY $\mathcal{O}_S$:

${}^IW$ is an interval (i.e., it has a least and a greatest element).

It is obviously that $e$ is the least element of ${}^IW$. How to show the fact that ${}^IW$ also has a greatest element? I have read Deodhar's paper mentioned in Section 3.2, but I cannot find a proof for that.

Maybe it is a silly question, in my opinion, the statement "${}^IW$ is an interval with least element $u$ and greatest element $v$" means the following: ${}^I W=[u,v]:=\{x\in{}^IW: u\le x \le v\}$, where $\le$ is the Bruhat ordering.
I understand a finite Coxeter group contains a unique longest word (which is a maximal element), but I think the $v$ in my interpretation is the maximum element: $x\le v$ for all $x\in {}^IW$, which may not equal to the maximal element $w_0$: $w_0\le x\implies x=w_0$. Is my interpretation correct or not?

Answer 1

Let $W$ be a Weyl group and $\le $ be the Bruhat ordering on $W$. Recall that a poset $P$ is said to be directed if for any $u, w \in P$, there exists $z \in P$ such that $u, w \le z$.

By Proposition 2.2.9 of Combinatorics of Coxeter Groups (by Anders Björner， Francesco Brenti), we get $(W,\le)$ is a directed poset. Since $W$ is finite, let $|W|=n$. Then by mathematical induction, it holds that for all $u_1,\cdots,u_n\in W$, there exists $w_0\in W$ such that $u_1,\cdots,u_n\le w_0$. i.e. $x\le w_0$ for all $x\in W$.

By darij grinberg's suggestion, let $w_0=w_I{}^Iw$, where $w_I\in W_I$ and ${}^Iw\in {}^IW$. The map $x\mapsto $ ${}^IW$-part of $x$, is a Bruhat ordering preserving map, we get $y\le {}^Iw$ for all $y\in {}^IW$. Hence ${}^IW=[e,{}^Iw]^I$, where $[u,v]^I=\{x\in {}^IW: u\le x\le v\}$. My interpretation is correct and ${}^IW$ has a maximum element ${}^Iw$.

Answer 2

For the symmetric group on three letters, with simple roots $\alpha_1, \alpha_2$ and $I=\{\alpha_1\}$. From the table:

We have

${}^IW:=\{w\in W: w<s_\alpha w\ \text{for all }\alpha\in I\} =\{w\in W: w<s_1 w\} =\{123,132,312\} =\{e,s_2,s_2s_1\} =\{x\in {}^IW :e\le x\le s_2s_1\} =[e,s_2s_1]^I$.

It seems Hugh Thomas' counterexample does not work. Did I miss anything?