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.

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