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

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