Two subspaces, part II

Good news first. Given a 2-dimensional subspace M of \mathbb R^4 (or \mathbb C^4) such that e_j\notin M\cup M^\perp (j=1,2,3,4), we can always find a coordinate subspace N such that M and N are in generic position. (Recall that generic position means (M\cup M^\perp)\cap (N\cup N^\perp)=\{0\}, which is feasible only when the dimension of M and N is half the dimension of the ambient space.)

Indeed, suppose that none of the subspaces \langle e_1,e_2\rangle, \langle e_1,e_3\rangle, \langle e_1,e_4\rangle are in generic position with M. By the pigeonhole principle, either M or M^\perp meets at least two of these subspaces. We may assume M meets \langle e_1,e_2\rangle and \langle e_1,e_3\rangle. Since M is two-dimensional, it follows that M\subset \langle e_1,e_2,e_3\rangle. Hence e_4\in M^\perp, a contradiction.

The story is different in n=6 and higher dimensions. Consider the space

M=\langle e_1-e_2,e_1-e_3,e_4+e_5+e_6\rangle, with M^\perp=\langle e_1+e_2+e_3,e_4-e_5,e_4-e_6\rangle

The condition \forall j\ e_j\notin M\cup M^\perp holds. Yet, for any 3-dimensional coordinate space N, either N or its complement contains at least two of vectors e_1,e_2,e_3, and therefore intersect M. Damn perfidious pigeons.

So it’s not enough to demand that M\cup M^\perp does not contain any basis vectors. Let’s ask it to stay away from the basis vectors, as far as possible. By the Pythagorean theorem, the maximal possible distance of e_j to M\cup M^\perp is 1/\sqrt{2}, attained when e_j is equidistant from M and M^\perp. Let’s call M an equidistant subspace if this holds for all e_j. There are at least two other natural ways to express this property:

  • In the basis \{e_1,\dots,e_{2n}\}, the projection onto M is a matrix with 1/2 on the diagonal
  • Reflection across M sends e_j into a vector orthogonal to e_j. As a matrix, this reflection has 0s on the diagonal.

In two dimensions, the only equidistant subspaces are y=x and y=-x. In higher dimensions they form a connected subset of the Grassmannian \mathrm{Gr} (n/2, n) (self-promotion).

Is every equidistant subspace in generic position with some coordinate subspace?

We already saw that the answer is yes when n=2,4. It is also affirmative when n=6 (G. Weiss and V. Zarikian, “Paving small matrices and the Kadison-Singer extension problem”, 2010). I am 75% sure that the answer is yes in general.

Two subspaces

The title is borrowed from a 1969 paper by Paul Halmos. Two subspaces M and N of a Hilbert space H are said to be in generic position if all four intersections M\cap N, M^\perp\cap N, M\cap N^\perp, M^\perp\cap N^\perp are trivial. It may be easier to visualize the condition by writing it as (M\cup M^\perp)\cap (N\cup N^\perp)=\{0\}. The term “generic position” is due to Halmos, but the concept was considered before: e.g., in 1948 Dixmier called it “position p”.

Let us consider the finite-dimensional case: H is either \mathbb R^n or \mathbb C^n. The dimension count shows that there are no pairs in generic position unless

  1. n=2k, and
  2. \mathrm{dim}\, M=\mathrm{dim}\, N=k.

Assume 1 and 2 from now on.

In the simplest case H=\mathbb R^2 the situation is perfectly clear: two lines are in generic position if the angle between them is different from 0 and \pi/2. Any such pair of lines is equivalent to the pair of graphs \{y=0, y=kx\} up to rotation. Halmos proved that the same holds in general: there exists a decomposition H=H_x\oplus H_y and a linear operator T\colon H_x\to H_y such that the generic pair of subspaces if unitarily equivalent to \{y=0, y=Tx\}.

If we have a preferred orthonormal basis e_1,\dots,e_n in H, it is natural to pay particular attention to coordinate subspaces, which are spanned by some subset of \{e_1,\dots,e_n\}. Given a subspace M, can we find a coordinate subspace N such that M and N are in generic position? The answer is trivially no if M\cup M^\perp contains some basis vector. When n=2 this is the only obstruction, as is easy to see:

Lines in generic position

In higher dimensions… later