A possible characterization of Euclidean geometry via the curvature of the Median-submanifold

Is there a Riemannian metric $$g$$ on $$\mathbb{R}^2$$ with inducing distance $$d$$ which is not isometric to the standard metric but satisfy the property quoted bellow?

For every two distinct points $$p,q\in \mathbb{R}^2$$, the locus of all points $$z$$ with $$d(z,p)=d(z,q)$$ is a geodesic.

As a higher dimensional version:

Is there a Riemannian metric on $$\mathbb{R}^3$$ which is not isometric to the standard metric but satisfy the property quoted bellow?

For every two distinct points $$p,q\in \mathbb{R}^3$$, the locus of all points $$z$$ with $$d(z,p)=d(z,q)$$ is a 2 dimensional submanifold with constant sectional curvature?

Edit:

According to the comment of Willie Wong, I realized that the previous version of the question should be reconstructed. So I present the question as follows:

Assume that we have a complete Riemannian metric on $$\mathbb{R}^n$$ which satisfies the following property: For every $$2$$ points $$p,q$$, the locus of all points $$\{z\mid d(z,p)=d(z,q)\}$$ is a codimension- $$1$$ smooth submanifold which is a totally geodesic submanifold. Does this imply that the metric is isometric to either Hyperbolic space or the Euclidean space?

• Aren't the hyperbolic metric and the sphere metric examples? – Willie Wong Mar 3 '18 at 1:00
• Also, I don't get how you related the two questions. The higher dimensional generalization of geodesic curvature should not be sectional curvature, but either mean curvature or the entire second fundamental form (in which case the submanifold is totally geodesic). – Willie Wong Mar 3 '18 at 1:02
• @WillieWong yes thank you. These geometries gives us models not isometric to the Euclidean metric. But does the property under first question implies that the curvature is constant?Or in higher domension, assuming that the median submanifold is totally geodesic, can we conclude that the curvature is constant?Thanks for your revision of the second question. – Ali Taghavi Mar 3 '18 at 19:53

The answer to your question is in the affirmative, even in the more general setting of general (pseudo)-Riemannian manifolds.

In the Riemannian case this is a theorem due to Busemann. Consider a Riemannian manifold $$(M,g)$$. The set $$\Sigma_{p,q}$$ consisting of all points equidistant from $$p$$ and $$q$$ on a Riemannian manifold is called the bisector of $$\{p,q\}$$. Busemann's theorem states that if $$\Sigma_{p,q}$$ are totally geodesic for any $$\{p,q\}$$, then $$(M,g)$$ has constant sectional curvature.

A generalization was proven by Beem for pseudo-Riemannian manifolds. The definition of "distance" and "bisector" requires a bit of care. Both are only defined within "simple convex neighborhoods", and the bisector is only a submanifold when $$p$$ and $$q$$ is not joined by a null geodesic. It turns out that this "local" distance function and "local" bisector, for pairs of points that are time-like or space-like separated, is enough to imply Busemann's theorem.

References

Busemann, H., The geometry of geodesics, Pure and applied Mathematics, 6. New York: Academic Press, Inc. X, 422 p. (1955). ZBL0112.37002.

Beem, John K., Pseudo-Riemannian manifolds with totally geodesic bisectors, Proc. Am. Math. Soc. 49, 212-215 (1975). ZBL0301.53026.

• I really thank you Willie for your very perfect and interesting answer. – Ali Taghavi Mar 6 '18 at 7:19