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?

  • 3
    $\begingroup$ This follows from Lemma 6.8 in Anders Björner, Michelle Wachs, Generalized quotients in Coxeter groups, Trans. Amer. Math. Soc. 308 (1988), pp. 1--37, since $W$ has a maximum element. (Okay, you'll have to substitute $w^{-1}$ for $w$ everywhere, since they are considering $W^I$ rather than $\left. ^I W \right.$) $\endgroup$ Commented Aug 2, 2019 at 11:45
  • $\begingroup$ Thank you very much!!! But where can I find the proof for the fact $W$ has a maximum element? I can just find the fact that $W$ has a maximal element. $\endgroup$ Commented Aug 2, 2019 at 12:13
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    $\begingroup$ Does "the longest word" mean anything to you in the context of Coxeter groups? If not, see Lemma 9.8 (a) of George Lusztig, Hecke algebras with unequal parameters, arXiv:math/0208154v2 and take $I = \varnothing$. $\endgroup$ Commented Aug 2, 2019 at 12:27
  • $\begingroup$ If I've got it right, then, just in terms of identifying the element, you take, as @darijgrinberg mentions, the longest word in the simple reflections $s_\beta$ with $\beta \not\in I$ (i.e., a word that is not equal to any shorter word, and whose length is maximal among all words with this property). Proving that this word works is, well, some work, but that'll at least identify it. $\endgroup$
    – LSpice
    Commented Aug 2, 2019 at 12:31
  • $\begingroup$ I agree the greatest element should be the longest word, but how to see it is also the maximum element in the sense of my interpretation? Since I only know how to show it is the maximal element with respect to Bruhat ordering. $\endgroup$ Commented Aug 2, 2019 at 14:50

2 Answers 2


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 ${}^Iw$.

  • $\begingroup$ For the symmetric group on three letters, with simple roots $\alpha_1,\alpha_2$ and $I=\{\alpha_1\}$, we have that $^IW$ is not an interval in Bruhat order. $\endgroup$ Commented Aug 3, 2019 at 18:46
  • $\begingroup$ But it seems my proof works. Where does I get wrong? $\endgroup$ Commented Aug 5, 2019 at 7:15
  • $\begingroup$ Nothing in your proof says that the map to the interval is surjective, and in fact it is not. $\endgroup$ Commented Aug 6, 2019 at 20:55

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

enter image description here

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?

  • $\begingroup$ The interval in Bruhat order from the identity to $s_2s_1$ includes both $s_2$ and $s_1$. --- Having reread what you've written, it seems that you are using a nonstandard definition of "interval". $\endgroup$ Commented Aug 6, 2019 at 20:56

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