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, part II

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