Let $M$ be a smooth Riemannian manifold (without boundary). Let $X\subset M$ be a smooth compact submanifold with boundary, $\dim X=\dim M$.

Under what conditions $X$, equipped with the induced intrinsic metric, is an Alexandrov space with curvature bounded below?

A simple sufficient condition is some convexity of $X$: any point of $X$ has a neighborhood such that any two points of $X$ in it are connected (in $M$) by a unique shortest geodesic, and this geodesic is contained in $X$. Is this condition also necessary?


1 Answer 1


Yes, this condition also necessary.

Assume $X$ is an Alexandrov space. At any point of $\partial_MX$ (the relative boundary of $X$ in $M$) the space of directions is a half-sphere; therefore $\partial_MX$ is also boundary of $X$ as it is defined for Alexandrov spaces.

If curvature $\ge 0$ then the distance function $f$ to the boundary is concave. It follows that the hypersurface $\partial_MX$ can be approximated by level sets of convex functions and therefore locally convex.

If curvature $\ge -1$, we get that $h=\mathrm{sh}\circ f$ satisfies $h''\le h$ in the barier sence and the same conclusion holds.


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