2
$\begingroup$

Let $(M,g)$ be a compact Riemannian manifold. To this Riemannian manifold, we associate a natural number $K(M,g)$ as follows:

$K(M,g)$ is the maximum of all $n\in \mathbb{N}$ such that we have at least $n$ distinct points $x_1,x_2,\ldots,x_n\in M$ such that $d(x_i,x_j)$ is independent of $i\neq j$.

What is this number for the round sphere $S^n$?

Does $K(M,g)$ depends on the Riemannian metric $g$?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ You can fit at most n+1 equidistant points in R^n, and so there are at most n+1 euclidean-equidistant points in the round sphere in R^n. Since geodesic distance is determined by euclidean distance, this solves the problem for the round sphere. mathoverflow.net/questions/30270/… $\endgroup$
    – Dustin G. Mixon
    Commented Mar 8, 2022 at 22:11

1 Answer 1

Reset to default
3
$\begingroup$

$K(M,g)$ depends on the metric, as shown by this question, which implies that we can change the metric of $\mathbb{R}^3$ so it has as many points pairwise at distance $1$ as we want.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ thank you very much for your answer and your attention to my question $\endgroup$
    – Ali Taghavi
    Commented Jun 3, 2022 at 19:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .