## 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: