Suppose we have a group $G(q)$ over a finite field $\mathbb{F}_q$. How can the Bruhat decomposition be used in order to calculate the order of $G(q)$? Are there any examples for some particular groups?
2026-03-29 23:59:45.1774828785
On
Bruhat decomposition and the order of the group.
192 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
2
There are 2 best solutions below
0
On
Yes, consider for example the finite group $G=GL(n,q)$. The order $o(n,q)$ of this group and the number $f(n,q)=(1+q)(1+q+q^2)\cdots (1+q+q^2+\cdots q^{n-1})$ of complete flags in $\mathbb{F}_q^n$ is related by $$ o(n,q)=q^{\binom{n}{2}}(q-1)^nf(n,q), $$ and the combinatorial explanation uses the Bruhat decomposition $$ GL(n,q)=\bigcup_{w\in S_n}(\psi^{-1}\phi^{-1})(w) $$ with straightforward maps $\psi\colon GL(n,q) \rightarrow \mathcal{F}(n,q)$ and $\phi\colon \mathcal{F}(n,q) \rightarrow S_n$, where $\mathcal{F}(n,q)$ is the space of complete flags. In this sense, the order of $GL(n,q)$ can be computed using the Bruhat decomposition.
I don't remember the exact reference, but this question is dealt in Meinolf Geck's book An Introduction to Algebraic Geometry and Algebraic Groups.
The problem is that in the Bruhat decomposition $\displaystyle G = \coprod_{w \in W} BwB$, if you write $g = b_1wb_2$ for some element $g \in G$, the elements $b_1$ and $b_2$ are not unique in general. Thus, you have to use a refinement of the decomposition, which is stated in the book and requires several results about the Frobenius map. But I am afraid of not being able to be much more precise (I don't have the book here).
However, the formula is not that easy to use, since it involves some combinatorial computations.