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 kintersecting, 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.)

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