3
$\begingroup$

I imagine the following result is folklore

Theorem. Those $k$-dimensional subspaces $\zeta \subset \mathbb{R}^n$ $(1 \leq k \leq n-1)$ for which the orthogonal projection of the lattice $\mathbb{Z}^n$ onto $\zeta$ is dense in $\zeta$ form a subset of full measure in the Grassmannian $G_k(\mathbb{R}^n)$.

What would be a (good) reference? Is there a "standard" proof?

Below in the answers I'll give a simple proof of my own that just uses a simple result in the geometry of numbers (an immediate corollary of van der Corput's extension of Minkowski's first theorem in the geometry of numbers):

If $C \subset \mathbb{R}^n$ is a $0$-symmetric closed convex set with infinite volume, then it contains an infinite number of integer points.

$\endgroup$

1 Answer 1

1
$\begingroup$

Disclaimer: this is not an answer to the question, but a long comment I wish to keep separate. The proof of the result in the question follows immediately from the two lemmas below.

Definition. Two subspaces $L, K \subset V$ of a finite-dimensional spaces $V$ are said to have minimal intersection if the dimension of their intersection is the least possible compatible with their dimensions and the dimension of $V$.

Lemma 1. If every subspace of $\mathbb{R}^n$ that is spanned by integer vectors has minimal intersection with the orthogonal complement of a $k$-dimensional subspace $\zeta$ $(1 \leq k \leq n-1)$, then the orthogonal projection of $\mathbb{Z}^n$ onto $\zeta$ is dense in $\zeta$.

Proof. Let $D_\epsilon \subset \zeta$ be the closed disc of radius $\epsilon > 0$ centered at the origin, and let $C_\epsilon$ be the cylinder $\pi_\zeta^{-1}(D_\epsilon)$, where $\pi_\zeta : \mathbb{R}^n \longmapsto \zeta$ is the orthogonal projection onto $\zeta$. Note that $C_\epsilon$ is a $0$-symmetric, closed, convex set with infinite volume and hence contains an infinite number of integer vectors.

Claim. The dimension of the subspace $\eta_\epsilon$ spanned by the (infinite) set of all integer vectors in $C_\epsilon$ is greater than $k$, the dimension of the subspace $\zeta$.

Indeed, $\eta_\epsilon$ is spanned by integer vectors and if its dimension were less than or equal to $k$, then by hypothesis it would intersect $\zeta^\bot$ only at the origin. This would imply that the intersection of $\eta_\epsilon$ and the cylinder $C_\epsilon$ is compact and hence it could not contain an infinite number of integer vectors. This would contradict the definition of $\eta_\epsilon$.

Notice now that if the dimension of $\eta_\epsilon$ is greater than $k$, then the minimal intersection hypothesis implies that its projection onto $\zeta$ is surjective. Hence, we can find integer vectors $\mathbf{m}_1$, $\mathbf{m}_2$, $\dots$, $\mathbf{m}_k$ in $C_\epsilon$ such that their projections form a basis of $\zeta$. The norm of these projections is less than or equal to $\epsilon$ and so the lattice spanned by these vectors is a subset of $\pi_\zeta(\mathbb{Z}^n)$ which is $2 \epsilon$ close to every point in $\zeta$. Since $\epsilon > 0$ was arbitrary, we conclude that $\pi_\zeta(\mathbb{Z}^n)$ is dense in $\zeta$.

Lemma 2. The set of $k$-dimensional subspaces of $\mathbb{R}^n$ that have minimal intersection with every subspace spanned by integer vectors has full measure in the Grassmannian $G_k(\mathbb{R}^n)$.

Proof. Fix a dimension $p$ between $1$ and $n-1$ and a $p$-dimensional subspace $\eta$. Consider the set of all $(n-k)$-dimensional subspaces $\zeta^\bot$ that do not intersect $\eta$ minimally. This is a finite union of lower-dimensional submaifolds and hence has measure zero in $G_{n-k}(\mathbb{R^n})$. The union of all these sets as $\eta$ ranges over the countable set of all subspaces spanned by integer vectors has still measure zero and so its complement has full measure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.