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Let $X$ be a finite set with a (true) metric $d$ and $|X| = n$. Does there exist a set $Y$ of $n$ points in $R^n$ with a pseudo-Riemannian metric with signature $(n - k, k, 0)$ for some integer $k$ such that the matrix of distances between points is the same?

Background: I've been trying to understand Multidimensional Scaling (MDS) and what happens when there are negative eigenvalues in the inner-product matrix. I know from this post that a true metric does not guarantee a positive-definite inner-product matrix. In other words, there are finite metric spaces that are not representable in Euclidean space.

However, every example I've seen can produce a non-distorted embedding so long as some of the dimensions are allowed to have imaginary values. In these examples, the imaginary dimensions correspond to negative eigenvalues of the inner-product matrix when performing MDS. Is this always the case? If we allow some dimensions of the embedding to be imaginary, will MDS always produce an exact embedding?

I'm also a bit unsure on the terminology, so I apologize if this question has already been asked with more precise language.

Michael Hardy
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Steve Riley
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  • If asked whether every finite metric space is isometric to a subset of Euclidean space with the usual metric, the first thing I think of is the set ${1,\ldots,n} \cup{p}$ where the distance between two points in ${1,\ldots,n}$ is just the absolute difference, and the distance from $p$ to any other point is $1.$ But if $p$ is the north pole and $1,\ldots, n$ are on the equator and distance is measured along the surface of the earth, then that's it. – Michael Hardy Nov 19 '23 at 00:39
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    @MichaelHardy Does that satisfy the triangle inequality when $n>3$? – James Hanson Nov 19 '23 at 15:24
  • REVISION OF A HASTY COMMENT: If asked whether every finite metric space is isometric to a subset of Euclidean space with the usual metric, the first thing I think of is the set ${1,\ldots,n} \cup {p}$ where the distance between two points in ${ 1, \ldots, n}$ is just the absolute difference, and the distance from $p$ to any other point is bigger than any distance between two of those $n$ points. But if $p$ is the north pole and $1,\ldots,n$ are on the equator and distance is measured along the surface of the earth, then that's it. @JamesHanson – Michael Hardy Nov 19 '23 at 18:02

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