Bruhat Order

Bruhat Order

Let $u,w\in W$. Then (i) $u\stackrel{t}{\to }w$ means that $u^{-1}w=t\in T$ and $\ell (u)<\ell (w)$. (ii)$u\to w$ means that $u\stackrel{t}{\to }w$ for some $t\in T$. (iii) $u\le w$ means that there exist $w_i\in W$ such that $u=u_0\to u_1\to \cdots \to u_{k-1}\to u_k=w.$

Observations

(i) $u<w$ implies $\ell (u)<\ell (w)$. (ii) $u<ut⟺\ell (u)<\ell (ut)$, for all $u\in W$ and $t\in T$. (iii) The identity element $e$ satisfies $e\le w$ for all $w\in W$ (any reduced decomposition $w=s_1\dots s_q$ induces $e\to s_1\to s_1{s}_2\to \cdots \to s_1\dots s_q=w$).

The Bruhat graph is the directed graph whose nodes are the elements of $W$ and whose edges are given by (ii). Bruhat order is the partial order relation on the set $W$ defined by (iii).

Lemma

For $u,w\in W$, $u\ne w$, let $w=s_1{s}_2\dots s_q$ be reduced, and suppose that some reduced expression for $u$ is a subword of $s_1{s}_2\dots s_q$. Then, there exists $v\in W$ such that the following hold: (i) $v>u$. (ii) $\ell (v)=\ell (u)+1$. (iii) Some reduced expression for $v$ is a subword of $s_1{s}_2\dots s_q$.

Subword Property

Let $w=s_1{s}_2\dots s_q$ be a reduced expression. Then, $u\le w⟺$ there exists a reduced expression $u=s_{i_1}{s}_{i_2}\dots s_{i_k}$, $1\le i_1<\cdots <i_k\le q$.

Corollary

For $u,w\in W$, the following are equivalent: (i) $u\le w$. (ii) Every reduced expression for $w$ has a subword that is a reduced expression for $u$. (iii) Some reduced expression for $w$ has a subword that is a reduced expression for $u$.

Corollary

Bruhat intervals $[u,w]$ are finite (even if $S$ is infinite). In fact, $\text{card}[u,w]\le 2^{\ell (w)}$.

Corollary

The mapping $w\mapsto w^{-1}$ is an automorphism of Bruhat order (i.e., $u\le w⟺u^{-1}\le w^{-1}$).

Chain Property

If $u<w$, there exists a chain $u=x_0<x_1<\cdots <x_k=w$ such that $\ell (x_i)=\ell (u)+i$, for $1\le i\le k$.

Lifting Property

Suppose $u<w$ and $s\in D_L(w)\setminus D_L(u)$. Then, $u\le sw$ and $su\le w$.