7
$\begingroup$

In the Wikipedia paper on Hadamard spaces, it is written that every flat Hadamard space is isometric to a closed convex subset of a Hilbert space. Looking through references provided by this Wikipedia paper, I found that this fact is proved as Theorem 7.2 in the lecture notes of Jacob Lurie. Since those notes contain no references and no historical information, it is not clear:

To whom this important result should be attributed? And when it was proved for the first time?

Remark. I am asking this question because, the mentioned Theorem 7.2 in Lurie notes implies the following nice characterization of metric spaces, which are isometric to Hilbert spaces:

Theorem. A metric space $(X,d)$ is isometric to a Hilbert space if and only if it is nonempty, complete, and satisfies the following two (first-order) conditions:

  1. $\forall x,y\in X\;\exists z\in X\;\; \big[d(x,z)=d(x,y)+d(y,z)=2d(x,y)\big]$;
  2. $\forall x,y\in X\;\exists m\in X\;\forall z\in X\;\big[d(z,m)^2=\tfrac12d(x,z)^2+\tfrac12d(y,z)^2-\tfrac 14d(x,y)^2\big]$.

I hope that this characterization is known (to the specialists). So the question is to whom should it be attributed?

Is this theorem the simplest (in a reasonable sense) metric characterization of Hilbert spaces?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Another metric characterization of flat Hadamard spaces can be found in exercise 0.0.2 on p.3 of arxiv.org/pdf/1701.03483.pdf. Namely, a complete length space $X$ is isometric to a convex subset of a Hilbert space if and only if every 4-point-quadruple in $X$ admits an isometric embedding into a Euclidean space. Is it simpler? $\endgroup$
    – Igor Belegradek
    Commented Mar 22, 2023 at 16:00
  • 1
    $\begingroup$ @IgorBelegradek Thank you for the comment and link to the book (very interesting) but try to write down this characterization as a (desirably first-order) property of the metric, I am not sure that it will be so short. $\endgroup$
    – Taras Banakh
    Commented Mar 22, 2023 at 16:33

1 Answer 1

Reset to default
1
$\begingroup$

Probably I.G. Nikolaev. See Theorem 10.10.13 in Burago, Dmitri; Burago, Yuri; Ivanov, Sergei "A course in metric geometry" [https://www.ams.org/books/gsm/033/][1]

Hat tip to A. Eskenazis who provided me with the references.

Improve this answer
$\endgroup$