Intersection of affine subspaces of finite codimension in Hilbert space

Question

I'm wondering whether the following assertion is true:

Any two affine subspaces of the same finite codimension in a ($\infty$-dimensional) Hilbert space either are parallel or have nonempty intersection.

If no condition on codimensions is given, I can figure out some counter-exemples: take $L$ any affine subspace and take $M$ to be the affine space generated by a translation of $L$ with any line.

Answer 1

I finally got to find the complete answer: two affine subspaces of finite codimension in a Hilbert space intersect themselves if their sum is the whole space. And they do not generically intersect otherwise.

Proof

Let $L=p+E$ and $M=q+F$ be two affine subspaces of finite codimension $r$ and $s$ in the Hilbert space $\mathcal{H}$, $E$ and $F$ are linear spaces.

Let us denote by $\pi$ the orthogonal projector on $E^\perp$. We compute some preliminary relations on codimensions.

1. $\mathrm{codim}_{E^\perp}\pi(F)=\mathrm{codim}_\mathcal{H}(E+F)=\dim E^\perp -\dim (\pi(F))=\mathrm{codim}_\mathcal{H}E-\dim\pi(F)$
2. $\mathrm{codim}_\mathcal{H}(E\cap F)=\mathrm{codim}_F(\ker \pi\big\vert_F)+\mathrm{codim}_\mathcal{H}F=\dim\pi(F)+\mathrm{codim}_\mathcal{H}F$
3. $\dim \pi(F)=\mathrm{codim}_\mathcal{H}E-\mathrm{codim}_\mathcal{H}(E+F)$
4. $\dim \pi(F)=\mathrm{codim}_\mathcal{H}(E\cap F)-\mathrm{codim}_HF$

We define now the (finite dimensional) Hilbert space $\mathcal{H}':=\mathcal{H}/(E\cap F)$. The reason to choose this quotient is that $p+E\cap F$ and $q+E\cap F$ cannot intersect themselves if $p\ne q$. So $L$ and $M$ intersect if and only if their images in $\mathcal{H}'$ intersect.

Let us denote by $E'$ and $F'$ the quotient image of $E$ and $F$ in $\mathcal{H}'$. We compute their dimensions.

1. $\dim \mathcal{H}'=\mathrm{codim}_\mathcal{H}(E\cap F)=\mathrm{codim}_\mathcal{H}E+\mathrm{codim}_\mathcal{H}F-\mathrm{codim}_\mathcal{H}(E+F)=\mathrm{codim}_\mathcal{H}E-\dim E'$
2. $\dim E'=\mathrm{codim}_E(E\cap F)=\mathrm{codim}_\mathcal{H}(E\cap F)-\mathrm{codim}_\mathcal{H}E=\mathrm{codim}_\mathcal{H}F-\mathrm{codim}_\mathcal{H}(E+F)$
3. $\dim F'=\mathrm{codim}_F(E\cap F)=\mathrm{codim}_\mathcal{H}(E\cap F)-\mathrm{codim}_\mathcal{H}F=\dim \pi(F)$

We resume this:

1. $\dim \mathcal{H}'=r+s-\mathrm{codim}(E+F)$
2. $\dim E'=r-\mathrm{codim}(E+F)$
3. $\dim F'=s-\mathrm{codim}(E+F)$

The sufficient condition we are looking for is $$\dim E'+\dim F'\ge \dim \mathcal{H}',$$ which corresponds $\mathrm{codim}_\mathcal{H}(E+F)=0$, so we want $$E+F=\mathcal{H}.$$

If this condition is satisfied, then the two affine subspaces do intersect themselves (generically into a single point). Otherwise, any two generic affine subspaces not verifying this condition, do not intersect. And it is easy to construct counter-exemples...