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

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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