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My question is somewhat related to Existence of a Riemannian metric inducing a given distance, but in a much simpler setting (as I am just starting to learn about Riemannian geometry). I hope it makes sense!

Let $S \subset \mathbb{R}^n$ be a compact, connected, non-convex subset of $\mathbb{R}^n$ (say with smooth boundary). Assume $S$ is also of dimension $n$. (In particular, I have in mind a non-convex two-dimensional subset of $\mathbb{R}^2$.)

Equip $S$ with the distance $d$ defined (in words) by \begin{align} \text{For } x,y \in S, \quad d(x,y) &= \text{infimum (over all paths) of the Euclidean length of a path from $x$ to $y$}\\ &\quad \text{ which stays entirely within $S$} \end{align}

Assuming there is a Riemannian metric $g(\cdot,\cdot)$ which is induced from this distance ($g$ will not be the Euclidean inner-product?...), can we say say anything about the Ricci curvature of the Riemannian manifold $(S,g)$ (non-negative...) ?

Obviously when $S$ is convex then $d$ is the regular Euclidean distance and $g$ is just the inner-product inherited from $\mathbb{R}^n$. If we lose convexity, it seems the curvature is no longer zero, but I don't really have a good sense for what happens.

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Lost in a Maze
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1 Answers1

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The interior of $S$ is open, hence locally convex. In an arbitrary Euclidean ball $B$ contained in $S$, the inf-of-path-lengths metric $d$ restricts to the Euclidean metric on $B$. Now, a Riemannian metric $g$ is "determined locally" by its inf-of-path-lengths metric. Since $(B, d)$ is Euclidean for every ball $B \subset S$, it follows that if a Riemannian metric $g$ on $S$ induces $d$, then $(S, g)$ is locally Euclidean, hence flat.

Andrew D. Hwang
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