# Isometric imbedding of finite metric space into standards spaces [duplicate]

Is it true that any metric space consisting of $n$ points can be isometrically imbedded into $n-1$ dimensional Euclidean space? Hyperbolic space? (For $n=3$ this is true.) If not, what are necessary/sufficient conditions? The case $n=4$ is the first unknown to me case.

What happens with imbeddings into the unit sphere, where certainly one needs some extra conditions on the metric space, like perimeter of each triangle is at most $2\pi$.

## marked as duplicate by Dylan Thurston, Benjamin Steinberg, Joonas Ilmavirta, Yoav Kallus, Ryan BudneyMay 19 '15 at 6:05

• The tripod graph (Y-shaped metric space with 4 elements) cannot be embedded isometrically in any Euclidean space or any rescaled hyperbolic space, or in a sphere of any radius. – YCor May 13 '15 at 13:20

The answer to the first question is definitely "no". A cycle on four vertices with each edge of length one yields a four point metric space which cannot be isometrically embedded to any $\mathbb{R}^n$ or $\mathbb{H}^n$. (Once two opposite points of the cycle have been determined, the remaining two points must map to midpoints of geodesics between them, but there is only one such geodesic.)

The condition in the Euclidean case is that the Gram matrix is positive semi-definite (put one vertex at the origin, then the entries of the matrix are $\|v_i\|$ on the diagonal, and $\langle v_i, v_j\rangle$ elsewhere. These can be expressed in terms of lengths via the parallelogram law (see my arxiv preprint "some observations on the simplex"). For $H^n$ and $S^n$ it's the same, except now the inner products are the cosines or coshes of the distance, and the signature should be $(n, 1)$ in the hyperbolic case, and positive definite in the spherical case.

In the Euclidean case, this is usually expressed via the Cayley-Menger matrix (which google), but this is just a more symmetric way of expressing the Gram matrix condition). In the Euclidean case, the determinant is a multiple of the volume of the simplex, in other space forms, there is no clear meaning to the determinant (that I know of).