1

Suppose we consider $M = (\mathbb{R}^2, d_\infty)$ (here to be clear $d_\infty$ refers to the "maximum metric"). Bounds relating $d_\infty$ to $d_2$ show that this space is homeomorphic to $(\mathbb{R}^2, d_2)$ so $M$ is in some sense "a surface".

Can we find a distance-preserving map $f:M\to\mathbb{R}^m$? I think this proves no: $(0,0), (0,1),(1,0),(1,1)$ are four points mutually unit-distance from each other in $M$ and so would need to map, under $f$, to a regular tetrahedron in $\mathbb{R}^m$ but the lines connecting these points cross (look at the diagonals of this "square tetrahedron") which doesn't happen for any tetrahedra in $\mathbb{R}^m$. Thus I think this proves that there is no distance-preserving map from $M$ to any subset of $R^m$ for $m \in \mathbb{R}^m$.

My first question is: does this proof work? If it doesn't, can we find a different proof of this fact?

My second question is: I have heard in the statement of such things as the Nash embedding theorems and as discussed here, there is a different notion of "isometric embedding" that comes from Riemannian geometry that essentially preserves the lengths of curves in $(\mathbb{R}^2, d_\infty)$ but does not necessarily imply global preservation of the metric. Is there such an isometric embedding from $M$ into $\mathbb{R}^3$? If so, can we describe it?

I understand that this second question requires reinterpreting $M$ as a Riemannian manifold rather than a metric space or something similar but I do not currently have the technical knowledge to read literature to answer or even phrase this question properly, hence me asking on MSE.

Isky Mathews
  • 3,235
  • 11
  • 25
  • Forgive me if I'm missing something, but doesn't $f(x, y) = (x, y, 0, \ldots, 0)$ do the trick (provided $m \geq 2$)? Why is this obvious embedding not isometric? – Joppy Dec 20 '21 at 21:59
  • @Joppy perhaps I am wrong, but say $m=2$ and notice that $d((0,1),(1,0)) = \sqrt{2}$ in $\mathbb{R}^2$ with the Euclidean distance whereas $d((0,1),(1,0))=1$ with the maximum distance. Thus your map doesn't work. – Isky Mathews Dec 20 '21 at 22:02
  • @IskyMathews In the title you have it wrong it should be $\mathbb{R}^m$ with the euclidean metric – MIO Dec 20 '21 at 22:05
  • @aldodecristo: You are correct! That was a very major typo! it made the question completely trivial. – Isky Mathews Dec 20 '21 at 22:08
  • Ah yes if one has the max-norm and the other has the 2-norm then there should be no isometry. – Joppy Dec 20 '21 at 23:25

1 Answers1

2

There is, as you say, no isometry from the plane with the uniform metric into a Euclidean space. I'm not entirely convinced the crossing diagonals are a problem, but here's an alternative proof: In the uniform metric the distance along one edge of the unit square agrees with the Euclidean distance on a number line, so the isometric image of each side of the square is a Euclidean segment. On the other hand, the uniform distance between every pair of boundary points on opposite edges of the unit square is unity, while no two Euclidean segments have this property. (Even more, there is no point in a Euclidean space lying at the same distance from every point of a Euclidean segment.)

As to your second question, the Nash embedding theorem is about Riemannian manifolds, and the plane with the uniform metric is not a Riemannian manifold, i.e., the uniform metric is not induced by a Riemannian metric.

Andrew D. Hwang
  • 78,195
  • 1
    I'm fairly sure your comment about sending segments to Euclidean segments also formalises the argument about crossing diagonals, as it shows the unit square and the diagonals must be sent to a regular tetrahedron as described with these segments as its edges. However, they cross and this cannot happen in $\mathbb{R}^n$. Regardless, great answer. I'll be writing one of my own about Q1 as I forgot that I had another more water-tight proof for that that I think is of some interest to others and then I'll accept yours. – Isky Mathews Dec 21 '21 at 12:19
  • Agreed about the crossing edges: I didn't think the argument was wrong, only that there seem to be details to fill in. (Generally, the simpler the "impossible" configuration, the easier to prove non-existence. :) – Andrew D. Hwang Dec 21 '21 at 12:48