Isometric embedding of $(\mathbb{R}^2, d_\infty)$ into $(\mathbb{R}^m,d_2)$?

Solution 1:

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.