## Intersections of Borel subgroups

There’s an amazing answer on Mathoverflow where Emerton gives a complete and very intuitive proof of the Bruhat decomposition

$\mathrm{GL}(V) = \bigcup_{w \in S_n} BwB$

where $V$ is an $n$-dimensional vector space, $B$ is the subgroup of upper triangular matrices, and $S_n \le \mathrm{GL}(V)$ is the subgroup of permutation matrices.

An an exercise, the reader is left to show that any two Borel subgroups contain a common torus. Since this step is the first point where one sees why the symmetric group is relevant, I’d like to record a proof here.

Proof. Let $U_0 \subseteq U_1 \subseteq \ldots \subseteq U_n$ and $V_0 \subseteq V_1 \subseteq \ldots \subseteq V_n$ be two chains of subspaces of $V$ such that $\mathrm{dim}\ U_j = \mathrm{dim}\ V_j = j$ for each $j$. Say, just for this proof, that $U_j$ is $k$intersecting, if $k$ is minimal such that $U_j \backslash U_{j-1}$ meets $V_k$ (or equivalently, $V_k \backslash V_{k-1}$). Let $f : \{1,\ldots, n\} \rightarrow \{1,\ldots, n\}$ be the function defined by $f(j) = k$ if and only if $U_j$ is $k$-intersecting.

Claim: $f$ is injective. Proof of claim. Suppose that $U_j$ and $U_{j'}$ are $k$-intersecting, where $j \le j'$. By hypothesis, there exists $u \in U_j \backslash U_{j-1}$ and $u' \in U_{j'} \backslash U_{j'-1}$ such that $u, u' \in V_k$. Since $V_k / V_{k-1}$ is $1$-dimensional, there exist scalars $\alpha$, $\alpha'$, not both zero, such that $\alpha u + \alpha' u' \in V_{k-1}$. If $\alpha' = 0$ then $u \in U_j \cap V_{k-1}$, a contradiction. And otherwise, since $U_j \subseteq U_{j'-1}$ we have $\alpha u + \alpha' u' \in (U_{j'} \backslash U_{j'-1}) \cap V_{k-1}$, again a contradiction.

Thus $f$ is a permutation. For each $j \in \{1,\ldots, n\}$ choose $u_j \in (U_j \backslash U_{j-1}) \cap V_{f(j)}$. (Necessarily $u_j \not\in V_{f(j)-1}$.) The Borel subgroups corresponding to the two chains of subspaces clearly both contain the torus having $u_1, \ldots, u_n$ as eigenvectors. $\Box$

Edit. This proof gives a constructive algorithm to find vectors $w_1,\ldots, w_n \in V$ and a permutation $f$ of $\{1,\ldots, n\}$ such that $w_j \in U_j \backslash U_{j-1}$ and $w_{f(j)} \in V_{f(j)} \backslash V_{f(j)-1}$ for each $j$. It is easy to see that $f$ is unique: suppose $g$ has the same property and $g \not= f$. There exists $j$ such that $g(j) < f(j)$. Then $U_j \backslash U_{j-1}$ meets $V_{g(j)} \backslash V_{g(j)-1}$, contradicting the minimality of $f(j)$.

Algebraically the uniqueness of $f$ corresponds to the disjointness of the Bruhat decomposition. Emerton's post has a nice topological proof of this.

Whenever one has a matching problem it's natural to see if there's an easy reason why the condition in Hall's Marriage Theorem holds. This paper by R. Brualdi gives a fairly short proof of this fact. The key observation is the following: if $u_j \in U_j \backslash U_{j-1}$ for each $j$, then given any $j_1 < j_2 < \ldots < j_r$, the set $\cup_{i=1}^r U_{j_i} \backslash U_{j_i-1}$ contains the subspace of $V$ spanned by $u_{j_1}, \ldots, u_{j_r}$. Of course a shorter proof follows from the proof in this post, since Hall’s condition is obviously necessary for a matching to exist.

Shameless plug: this joint paper with John Britnell gives an application of Hall’s Marriage Theorem to group theory where the use of Hall’s theorem appears to be essential. (At least, we know of no other way to prove the result.)