$\mathbb{R}^{n}$ is not quasi isometric to $\mathbb{R}^{m}$ for $m \ne n$

Question

Currently, I am reading geometric group theory. I came across a problem that says that $\mathbb{R}^{n}$ is not quasi-isometric to $\mathbb{R}^{m}$ for $m \ne n$. We have to use Borsuk-Ulam theorem for that. But I have no idea how to do this.

Also, I know that $\mathbb{R}^{2}$ is not quasi-isometric to $\mathbb{R}$ using Borsuk-Ulam theorem.

Any help will be highly appreciated.

Answer 1

Since $\Bbb{R}^n$ is quasi-isometric to $\Bbb{Z}^n$ it is enough to show for the question that $\Bbb{Z}^m$ is quasi-isometric to $\Bbb{Z}^n$ if and only if $m=n$. A proof of this can be easily found in lecture notes on geometric group theory. For example, see these notes by Kaisala, Proposition $5.7$, page $18$. For the above result see Corollary $5.8$.

Edit: Also solutions using Borsuk-Ulam can be found in such lecture notes, e.g. here, page $5$, or here, page $7$ and $8$, etc.

Answer 2

Assume that $f: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ is a $(\lambda,K)$-quasi-isometry and $n > m$. We will approximate it by a continuous map and then show that a precomposition with $i : S^m \longrightarrow \mathbb{R}^n$ gives a map violating Borsuk-Ulam theorem.

First, let's find a continuous map approximating $f$. Consider cube grid in $\mathbb{R}^n$ with vertices at $\mathbb{Z}^n$. Subdivide each $n$-cube into $n$-simplices, this gives a triangulation of $\mathbb{R}^n$ (with uniformly small simplices, which will become important later). Let $g :\mathbb{R}^n \longrightarrow \mathbb{R}^m$ be a map, which agrees with $f$ on points in integer coordinates, and elsewhere is given by a linear interpolation with respect to the triangulation.

If $x \in \mathbb{R}^n$, there exists $y \in \mathbb{Z}^n$ with $d(x,y) \leq \sqrt n/2$. Let $z \in \mathbb{Z}^n$ be a point in an $n$-simplex containing $x$ furthest from $y$.

\begin{equation} \begin{split} d(f(x),g(x)) \leq d(f(x),f(y)) + d(f(y),g(y)) + d(g(y),g(x)) \\ \leq (\lambda \sqrt n/2 + K) + 0 + (\lambda d(y,z) +K) \\ \leq \frac{3}{2}\lambda \sqrt n + 2K \end{split} \end{equation}

Let $i:S^m \longrightarrow \mathbb{R}^n$ be an inclusion, which embeds $S^m$ as a sphere of radius $R$. Then $g \circ i$ is continuous map.

If $x$ and $-x$ is a pair of antipodal points on $i(S^m) \subset R^n$, then

\begin{equation} \begin{split} d(g(x), g(-x)) \geq d(f(x), f(-x)) -3 \lambda \sqrt n - 4K \\ \geq \frac{2R}{\lambda} -3 \lambda \sqrt n - 5K \end{split} \end{equation}

If we take $R > \frac{\lambda}{2}(3 \lambda \sqrt n + 5K)$, the right-hand side is positive and $g(x) \ne g(-x)$. This contradicts Borsak-Ulam theorem.

Answer 3

Say $m<n$. Borsuk-Ulam obviously implies that $\mathbf{R}^n$ and $\mathbf{R}^m$ are not homeomorphic, and even that $\mathbf{R}^n$ cannot be mapped continuously injectively into $\mathbf{R}^m$.

Assume by contradiction that there is a quasi-isometric embedding $f:\mathbf{R}^n\to\mathbf{R}^m$. Fix a nonprincipal ultrafilter $\omega$, and define $\tilde{f}(x)=\lim_{k\to\omega}\frac{1}{k}f(kx)$. Then $\tilde{f}$ is a bilipschitz embedding. We get a contradiction.

[If one wants to avoid ultrafilters and just use a classical countable choice, instead, use that $f_k(x)=f(kx)/k$ is bounded for all given $x$, to find a pointwise limit $g$ of the family $f_k|_{\mathbf{Q}^n}$. This pointwise limit is then a bilipschitz embedding of $\mathbf{Q}^n$ into $\mathbf{R}^m$, and hence extends to the completion as a bilipschitz embedding of $\mathbf{R}^n$ into $\mathbf{R}^m$.]