Bruhat decomposition for general linear group of degree three over a field

This article describes the details of the Bruhat decomposition for the general linear group of degree three over a field $K$. Let $G=GL(3,K)$ and $B$ denote the Borel subgroup of upper-triangular matrices. We have that:

$G=\bigcup _{w\in W}BwB$

where $W$ is the Weyl group, which in this case can be identified with symmetric group:S3. In other words, there is a set map $G\to W$ whose fibers are the double cosets of $B$, and whose restriction to the subgroup $W$ of $G$ is the identity map. The map is well defined because every double coset of $B$ intersects $W$ at a unique point.

Note that the set map is not a homomorphism of groups.

Another way of putting this is that there is a set map from the left coset space $G/B$ to $W$ that sends a left coset containing an element of $W$ to that element of $W$, and that is invariant under the left action of $B$ by multiplication.

Interpretation in terms of flags

The mapping:

$G/B\to W$

can be interpreted as follows: an element of $G/B$ is a complete flag of subspaces for the three-dimensional space $K^{3}$, and the mapping to $W$ describes its relative position with respect to the standard flag (the one stabilized by $B$). If the flag is equal to the standard flag, then the map sends it to the identity element of $W$, otherwise it is sent to one of the non-identity element of $W$. The generic flag gets sent to the anti-diagonal permutation, corresponding to $(1,3)$.

Finite field case

In the finite field case, for a finite field with $q$ elements, the fibers for the Bruhat map:

$G/B\to W$

can be computed explicitly.

Element of $W$	Expression in Bruhat terms	Matrix	Size of fiber in $G/B$ (equals $q$ to the power of the Bruhat word length)	Degree of polynomial (= word length of Bruhat word)	Size of fiber in $G$	Explanation
$()$	empty word	${\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\\\end{pmatrix}}$	1	0	$(q-1)^{3}q^{3}$
$(1,2)$	$s_{1}$	${\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\\\end{pmatrix}}$	$q$	1	$(q-1)^{3}q^{4}$
$(2,3)$	$s_{2}$	${\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\\\end{pmatrix}}$	$q$	1	$(q-1)^{3}q^{4}$
$(1,2,3)$	$s_{1}s_{2}$	${\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\\\end{pmatrix}}$	$q^{2}$	2	$(q-1)^{3}q^{5}$
$(1,3,2)$	$s_{2}s_{1}$	${\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\\\end{pmatrix}}$	$q^{2}$	2	$(q-1)^{3}q^{5}$
$(1,3)$	$s_{1}s_{2}s_{1}$	${\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\\\end{pmatrix}}$	$q^{3}$	3	$(q-1)^{3}q^{6}$
Total (6 rows)	--	--	$q^{3}+2q^{2}+2q+1=(q+1)(q^{2}+q+1)$ equals size of $G/B$	--	$q^{3}(q-1)^{3}(q+1)(q^{2}+q+1)$ equals order of $G$