Today’s post is about group theory. Well, to state the result, one needs to know just group theory but proving it requires some knowledge of root systems. And one really understands its application when studying algebraic groups. In my opinion, one can rarely say while learning mathematics that s/he will not need a particular result, or something is not useful in their research. So it’s always good to know something and connect to it later when learning something related.

The Bruhat decomposition is a decomposition of a group having a “Tits system” into double cosets. It is used to decompose $GL(n,F)$, the group of $n \times n$ invertible matrices over a field $F$, or any reductive connected closed subgroup of this matrix group.

For a group $G$, a Tits system (named after Jacques Tits) is a triple $(B, N, I), \, B, N$ subgroups of $G$ called a $BN-$pair and $I$ is a set such that the following axioms hold:

1. $T := B \cap N \unlhd N$,
2. $I$ is a set of generators of $W := N / T$ (called the Weyl group) and $s^2 = 1$ for all $s \in I$,
3. If $w \in W$ and $s \in I$, then $wBs \subseteq BwsB \cup BwB$,
4. If $s \in I$, $sBs^{-1} \neq B$,
5. $B, N$ generate $G$.

Remarks:

• (1) says that $W$ is a group.
• In (3), $W$ is not necessarily a subgroup of $G$. Yet, define $wB$ as $wB := nB$, where $w=nT$ and observe the well-definedness.
• We’ll be concerned with “Bruhat cells”, C(w) := BwB. With this notation, (3) is equivalent to $C(w) C(s) \subseteq C(ws) \cup C(w)$.
• From (4), $1 \not\in I$.
• From (2), each $s \in I$ has order 2.

Theorem (Bruhat decomposition).

Let $(B, N, I)$ be a Tits system and $W = N/T$ be the Weyl group. Then,

$G = \coprod_{w\in W} C(w)$.

A (rough) sketch of the proof:

Let $G' = \cap_{w\in W} C(w)$. Prove that $G’$ is a group, and in fact a subgroup of $G$.

$G'$ contains $N$ and $B$ which generate $G$ so $G' = G$.

For disjointness, it suffices to show $w_1 = w_2$ whenever $C(w_1) = C(w_2)$.

Observe that two double cosets are either disjoint or equal.

The canonical example:

$F$ is any field and $GL(n,F)$ is the set of invertible matrices over $F$. The Borel subgroup $B$ consists of upper triangular matrices and take $N$ to be the subgroup of monomial matrices, i.e., each row and each column has a unique nonzero entry. Then $T = B \cap N \cong (F^*)^n$, which is a torus. One does however need to show that this forms a Tits system.

Recently while browsing in the Purdue Math library, I found notes of Armand Borel‘s talk titled `Algebraic groups and Arithmetic groups’. It gives an overview of algebraic groups and its applications. They are very sleek (but mathematically dense) so I decided to type them out. These notes discuss Tits systems, and their deep connections to Geometry, group theory and number theory. You can read them here.