# 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 $(q-1)^3q^3$
$(1,2)$ $s_1$ $\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\\end{pmatrix}$ $q$ 1 $(q-1)^3q^4$
$(2,3)$ $s_2$ $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\\end{pmatrix}$ $q$ 1 $(q-1)^3q^4$
$(1,2,3)$ $s_1s_2$ $\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\\end{pmatrix}$ $q^2$ 2 $(q - 1)^3q^5$
$(1,3,2)$ $s_2s_1$ $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\\end{pmatrix}$ $q^2$ 2 $(q - 1)^3q^5$
$(1,3)$ $s_1s_2s_1$ $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\\end{pmatrix}$ $q^3$ 3 $(q - 1)^3q^6$
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 $G$