Answer by Marius Buliga for Shortest-path Distances Determining the Metric?
Nice paper, I was not aware of it. There is this paper: I.G. Nikolaev, A metric characterization of riemannian spaces, Siberian Advances in Mathematics, 1999, v. 9, N4, 1-58 see the mathscinet page:...
View ArticleAnswer by Dick Palais for Shortest-path Distances Determining the Metric?
There is an old paper of mine called "On the Differentiability of Isometries" in which I show that if you know a Riemannian manifold $M$ only as a metric space, i.e., you just know its point set and...
View ArticleAnswer by Sergei Ivanov for Shortest-path Distances Determining the Metric?
Concerning the second question. A single closed geodesic is not enough. For example, let $S$ be the equator of the standard sphere. All distances between points of $S$ are realized by paths in $S$, so...
View ArticleAnswer by Paul Siegel for Shortest-path Distances Determining the Metric?
There are other people who frequent MO who are much better equipped than me to answer your question, but I think I can provide a few useful insights.First, you are correct that the Riemannian metric...
View ArticleAnswer by Will Jagy for Shortest-path Distances Determining the Metric?
You are thinking in terms of long geodesics, but your question is local. The metric is defined point by point. To know the metric tensor at a point $A,$ it is only necessary to know the geodesic...
View ArticleShortest-path Distances Determining the Metric?
The metric of a Riemannian manifold determines the shortestdistance between any two points.I assume the reverse holds? That is, if you are given theshortest distance d(x,y) between every pair of...
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