Can one describe the multiplication of two Bruhat cells?

Question

For $G$ a simple linear algebraic group and $B$ a fixed Borel subgroup, we have the Bruhat decomposition $G = \coprod_{w \in W} B\dot{w}B$, where $W$ is the Weyl group and $\dot{w}$ is any representative of $w$ in $N_G(T)$.

If $s_\alpha \in W$ is a simple reflection, then the following rule is known:

• $(B\dot{s_\alpha} B)(B\dot{w}B) = B\dot{s_\alpha}\dot{ w} B,\hspace{5mm}$ if $\hspace{3mm}l(s_\alpha w) = l(w) + 1$,
• $(B\dot{s_\alpha} B)(B\dot{w}B) = B\dot{w}B \cup B\dot{s_\alpha}\dot{ w} B,\hspace{5mm}$ if $\hspace{3mm}l(s_\alpha w) = l(w) - 1$,

where $l(w)$ is the length of $w \in W$.

This rule for simple reflections implies that for any $w_1, w_2 \in W$

$(B\dot{w}_1 B)(B\dot{w}_2B) = \coprod_{w \in U} B\dot{w}B$,

where $U \subset W$. However, knowing what this subset $U$ is seems quite complicated (I think one can put some partial conditions on $U$ involving Bruhat order).

My question is if anyone has seen a description of which cells appear in $(B\dot{w}_1 B)(B\dot{w}_2B)$ (i.e., what does $U$ look like)? Or, does anyone know of some nice connections between any aspects of this problem and some other theory? It seems likely to me that the decomposition of $(B\dot{w}_1 B)(B\dot{w}_2B)$ is much to complicated to have any single rule for arbitrary $w_1, w_2$. But, I thought I'd pose this question to M.O. just in case. Thanks!

Answer 1

This is to complement Ben's answer. The multiplication you are looking at is closely related to the 0-Hecke monoid. This is the deformation of the Weyl group $W$ where you keep all the braid relations but you replace the involution relation $s^2=1$ with the idempotent relation $s^2=s$. This monoid $H(W)$ is in bijection with $W$ via $w\mapsto h(w)$ (where $w\in W$ and $h(w)\in H(W)$) and can be identified with the set of principal Bruhat ideals with set wise multiplication. Richardson and Springer observed that if $w,w'\in W$, then the unique dense open $B\times B$ orbit in $BwB\cdot Bw'B$ is the $Bw''B$ where $h(w)h(w')=h(w'')$.

Answer 2

If you pick a reduced expression $w_1=s_1\cdots s_m$, then you have $B\dot w_1 B=(B\dot s_1B)\cdots (B\dot s_m B)$. Thus, you can calculate $(B\dot w_1 B)(B\dot w_2 B)=(B\dot s_1B)\cdots (B\dot s_m B)(B\dot w_2 B)$. Now apply the rule you cited for simple reflections a whole bunch of times. The result you'll get is that you get the products of subwords of the expression times $w_2$ with a combinatorial property on the subwords.