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Let $X$ be a smooth manifold of dimension $n$ equipped with a Riemannian metric.

Suppose that $x_1, \dots, x_m$ are pivot points on that manifold. We consider the distance functions $$ f_i(x) = d( x_i, x ) $$ and thus assign the distance vector $$ f(x) = \Big( f_i(x) \Big)_{1 \leq i \leq m} $$ to each point $x \in X$.

Over which subsets $S \subseteq X$ and with how many pivot points is $f$ injective? Note that the pivot points are known and ordered.

I presume that at least $n+1$ are needed on an $n$-dimensional manifold, and $S$ is generally not the whole manifold.

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    $\begingroup$ If you only require that there be a non-empty $S\subset X$ on which $f$ is injective, then $n$ points in general position suffice. For example, two distinct points in the Euclidean plane will give you an injection on either half-plane bounding the line joining the points. If you want $S = X$, then, generally, you will need more than $n{+}1$ points, but perhaps no more than $2n{+}1$ points. By the way, you might want to let $f_i$ be the square of the distance to $x_i$ so that it will be smooth on a neighborhood of $x_i$. $\endgroup$
    – Robert Bryant
    Commented Aug 8 at 8:46
  • $\begingroup$ @RobertBryant let's consider the square of distance instead of disrance as you pointed out to. I wonder can one prove the easy or advanced Withney embedding theorem or Nash isometric embedding theorem via such function $f$? More precisely: Assume that $(M,g)$ is a Riemannian manifold are there a finite number of pivot points for which the function $f$ (with square distance) would be an isometric embedding? I did not read the Nash proof but I wonder is his proof based on such a function? $\endgroup$
    – Ali Taghavi
    Commented Aug 8 at 10:57
  • $\begingroup$ +1 for you question I think it generates more questions on some particular subcategory of smooth manifolds: For examples what is a precise example of a symplectic manifold and a finite even number of points for which the square distance vector function f you mentioned is a symplectic embedding? $\endgroup$
    – Ali Taghavi
    Commented Aug 8 at 11:17
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    $\begingroup$ Unfortunately, even the squared distance function is not generally smooth on a compact manifold (look at the sphere), it's just smooth away from the cut locus. $\endgroup$
    – Robert Bryant
    Commented Aug 8 at 11:23
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    $\begingroup$ @AliTaghavi: In the complete, simply connected, negative curvature case, the squared distance function is smooth, yes. However, if the surface is completely but not simply-connected, the cut locus will not be empty. $\endgroup$
    – Robert Bryant
    Commented Aug 8 at 13:19

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