# 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: