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Jul 12, 2012 at 0:36 vote accept Oliver Jones
Jul 12, 2012 at 0:36 answer added Oliver Jones timeline score: 0
Jul 9, 2012 at 21:15 history edited Robert Bryant
added the differential geometry tag
Jul 9, 2012 at 12:27 answer added Robert Bryant timeline score: 6
Jul 4, 2012 at 0:02 comment added Oliver Jones @Misha: Correct Misha. The Fubini-Study, Begman, and the Euclidean metrics all have potentials of this form. @Yang: I'm speaking locally here; obviously the cut locus will need to be avoided. However, the base point $p$ is arbitrary; you simply avoid the cut locus of $p$. I've thought about this some more and I realized that the potential is also a function of $|z|^2=\sum_i|z_i|^2$. Here $(z_1,\cdots, z_n)$ are local coordinates. Rotationally symmetric metrics have this property.
Jul 3, 2012 at 22:14 comment added Deane Yang Also, I am under the impression that the statement is not literally true for a compact manifold, since the Hessian of the distance function has to degenerate at the cut locus. It seems to me that the question needs to be stated more carefully.
Jul 3, 2012 at 22:13 comment added Deane Yang For a hermitian symmetric space of constant curvature, this holds for any point $p$ (as defined in Misha's comment). It should not be difficult to give an example where it holds for only one particular choice of $p$ but not for others. So presumably the question is about manifolds where the potential can be written as a function of distance from a point $p$, for any choice of $p$?
Jul 3, 2012 at 21:52 comment added Misha @Igor: He means that a potential for the metric is of the form $f(x)=h(d(p,x))$ where $p$ is a fixed point and $h$ is some function of one variable.
Jul 3, 2012 at 18:47 history edited user9072 CC BY-SA 3.0
correction for umlaut
Jul 3, 2012 at 14:58 comment added Igor Rivin Could you elaborate on this question? How can a metric fail to be a function of geodesic distance?
Jul 3, 2012 at 1:41 history asked Oliver Jones CC BY-SA 3.0