1
$\begingroup$

I'm trying to understand better the following sentence (see link to the reference at the end):

To represent the metric of $n$ points in the Euclidean space, one clearly needs no more than $n−1$ dimensions.

What "representing a metric" means?

Reference: "Dimensionality Reductions in $l_2$ that Preserve Volumes and Distance to Affine Spaces", by Avner Magen, section 1 ("Introduction").

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ It just means to isometrically embed. Think about embedding a 3 point metric space as a triangle, using the triangle inequality. A 4th point can sit in space above those 3 to make a tetrahedron. $\endgroup$
    – Ben McKay
    Commented Apr 5, 2020 at 15:04
  • $\begingroup$ @BenMcKay, thanks for your fast reply. I'm not familiar with a few definitions you mentioned. Would you mind to elaborate a little bit more about "isometrically embedding" and the example you presented? I understood that we have 3 points in an Euclidean space that create a triangle; and a 4th point which isn't collinear with them creates a tetrahedron. How's that related to the question above? Best regards $\endgroup$
    – keyboardAnt
    Commented Apr 5, 2020 at 15:11
  • $\begingroup$ Isometric embedding means an isometry of metric spaces to its image. So you represent a metric space $X$ in Euclidean space $\mathbb{E}^n$ by mapping $\phi \colon X \to \mathbb{E}^n$ with a map $\phi$ which is an isometry to its image, in the induced metric on its image. $\endgroup$
    – Ben McKay
    Commented Apr 5, 2020 at 15:21
  • 1
    $\begingroup$ I hope that this means that any $n$ points of a Euclidean space are contained in an $(n-1)$ affine subspace (which is isometric to the $(n-1)$-dimensional Euclidean space), so we can think of these $n$ points as points of the Euclidean space $\mathbb R^{n-1}$. $\endgroup$
    – Taras Banakh
    Commented Apr 5, 2020 at 17:18

0

Reset to default

Browse other questions tagged .