The image of a Riemannian manifold under a distance-based mapping to $\mathbb{R}^n_{\geq 0}$

Question

Consider $n$ points, $p_i\in M,\, i=1,\ldots , n$, in a Riemannian manifold $M$, and define the map \begin{array} &f: M&\to& \mathbb{R}^n_{\geq 0}\\ \qquad q &\mapsto& \left(\begin{matrix}d(q,p_1)\\ \vdots\\ d(q, p_n)\end{matrix}\right) \end{array} where $d(\cdot,\cdot)$ is the distance function in $M$. What can we say about the topology of $f(M)$? Is $f(M)$ homeomorphic to $M$ for sufficiently large $n$? Is there some literature about this construction?

Answer 1

This is nowhere near as strong as what you asked but here's a funny and easier related result. First, the question can be stated and makes sense for an arbitrary metric space. If $M$ is any metric space we can consider the map

$$Y : M \ni q \mapsto \left( p \mapsto d(p, q) \right) \in \mathbb{R}_{\ge 0}^M$$

where we consider the distances from a point to every other point in $M$. Then we have the following amusing result:

The Yoneda lemma for metric spaces: If $\mathbb{R}_{\ge 0}^M$ is given the $\sup$ metric then $Y$ is an isometric embedding.

(Strictly speaking if $M$ is unbounded the $\sup$ "metric" here can take the value $\infty$. This is a pretty harmless generalization of a metric called an extended metric. Also we don't have to use the full $\mathbb{R}_{\ge 0}^M$, this argument really takes place in the subspace of maps of Lipschitz norm $\le 1$.)

Proof. By definition we have

$$d(Y(q_1), Y(q_2)) = \sup_{p \in M} \left| d(p, q_1) - d(p, q_2) \right|$$

and by the triangle inequality we have $|d(p, q_1) - d(p, q_2) \le d(q_1, q_2)$ with equality if $p = q_1$ or $p = q_2$. So the value of this $\sup$ is exactly $d(q_1, q_2)$. $\Box$

For more on why I called this result the "Yoneda lemma" see here.

This gives a kind of "Cayley's theorem for metric spaces": every metric space is isometric to a subspace of an $\ell^{\infty}$ space. This gives a version of the desired result for $M$ a finite metric space, although note that it's only interesting in this case that the map is an isometry since the fact that it's a homeomorphism is not very interesting here.

If $M$ is an infinite metric space but we only use a finite set of points then we need to know something about when a point in $M$ is determined by its distance to a finite set of points. I think this is true for subspaces of $\mathbb{R}^n$ with the induced metric but you would already need to know that Riemannian manifolds can be isometrically embedded into $\mathbb{R}^n$ to apply this.

Answer 2

Let $(M,d)$ be a metric space satisfying the following: There exists a number $N\in \mathbb N$ such that for every $x\in M$ there is $r=r_x>0$ and points $x_1,...,x_N\in B(x,r)$ for which the map $$ B(x,r)\to \mathbb R_+^N, y\mapsto (d(y,x_1),...,d(y,x_N)) $$ is injective. This is true for all Riemannian manifolds $M$ and the corresponding Riemannian distance functions $d$ (see P.Petersen's "Riemannian Geometry", section 10.3). Suppose $(M,d)$ is also compact. Then we can find a finite subset $Z=\{z_1,...,z_n\}\subset M$ such that every point $x\in M$ is within distance $\epsilon=r/4$ from one of the points of $Z$, where $$ r=\min\{r_{z_1},...,r_{z_n}\}. $$ For every $z_j\in Z$ let $\{x_{1j},..., x_{Nj}\}\subset B(z_j, r)$ be the points as above. Define $$ X=\{ x_{ij}: 1\le i\le N, 1\le j\le n\}\cup Z. $$ Label elements of $X$ arbitrarily, $x_k, 1\le k\le K=N(n+1)$. I claim that the map $$ f: y\mapsto (d(y,x_1),...,d(y,x_K)), y\in M, $$ is injective. Indeed, the restriction of $f$ to each ball $B(x_i, r)$ is injective since $X$ contains the points $x_{1i},...,x_{Ni}$. If $B(z_i, \epsilon)\cap B(z_j, \epsilon)\ne \emptyset$ then $$ B(z_i, \epsilon)\cup B(z_j, \epsilon)\subset B(z_i, 3\epsilon)\subset B(z_i, r)$$ and, hence, $f$ restricts to an injective map on $B(z_i, \epsilon)\cup B(z_j, \epsilon)$. Suppose that $B(z_i, \epsilon)\cap B(z_j, \epsilon)=\emptyset$ and $x\in B(z_i, \epsilon), y\in B(z_j, \epsilon)$. Then $$ d(x, z_i)< d(y, z_i), $$ which implies that $f(x)\ne f(y)$. Since the balls $B(z_i,\epsilon)$ cover the entire $M$, it follows that $f: M\to \mathbb R^K$ is injective. Thus:

Theorem. For every compact connected Riemannian manifold $M$ with Riemannian metric $d$ there exists a finite subset $X=\{x_1,...,x_K\}\subset M$ such that the map $$ f: y\mapsto (d(y,x_1),...,d(y,x_K)), y\in M $$ is an embedding $M\to \mathbb R^K$.