Timeline for Under what conditions do distances from pivot points uniquely identify a point on a manifold?
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| Aug 10 at 12:08 | comment | added | Ali Taghavi | @RobertBryant Is it a good idea to introduce a generqlized "configuration space" as follows: All n tuples for which the corresponding vector function(squared) is a topological embedding (or in case of empty cut locus case one may consider all n tuple for which the corresponding squared map is isometric or symplectical embedding). In case of negative curvature does it generate a configuration space worth of study? | |
| Aug 8 at 13:19 | comment | added | Robert Bryant | @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. | |
| Aug 8 at 11:49 | comment | added | Ali Taghavi | @RobertBryant Thank you for introducing me the cut locus point. I was not aware of it. Are they same as focal points (or any realtion to it). I think in negative curvature we have more chance for smoothness of the squared distance function yes? | |
| Aug 8 at 11:23 | comment | added | Robert Bryant | 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. | |
| Aug 8 at 11:21 | comment | added | Ali Taghavi | Any way I vaguely remeber the square distance function, distance to a set not to single point, is introduced as a real analytic function(in a book by S. Krantz whose title contains (I thinl) "...real analytic....". I do not remenber the reason this function was introduced in his book but may be his consideration would be helpful in line of your question | |
| Aug 8 at 11:17 | comment | added | Ali Taghavi | +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? | |
| Aug 8 at 10:57 | comment | added | Ali Taghavi | @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? | |
| Aug 8 at 8:46 | comment | added | Robert Bryant | 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 $x_i$. | |
| Aug 8 at 7:46 | history | asked | shuhalo | CC BY-SA 4.0 |