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$ \newcommand{dist}{\operatorname{dist}} \newcommand{B}{\mathbb{B}} $

Let $\mathcal {M}$ be a Riemannian manifold, $p \in S \subset \mathcal{M}$ and $r>0$. Denote $S_{r} := S \cap\B(p,r)$.

Question 1: For small values of $r$, Is there a relation similar to the following $$ \dist(\exp_p^{-1}(u);\exp_p^{-1}(S_r)) \leq C(r)\dist(u;S_r) \tag{1} $$ for every $u \in \B(p,r)$. Note that $C(r)$ may be related to the curvature.

I know that the inequality (1) holds with $C(r) = 1$, for a Hadamard manifold (since the exponential function has non-expansion property for a Hadamard manifold).

Question 2: Is there an asymptotic relation between $\dist(u;S_r)$ and LHS of (1) for small values of $r$.

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  • $\begingroup$ Try looking in Isaac Chavel's book, Riemannian Geometry, for Toponogov's Comparison Theorem. I think it will tell you that two sided control on sectional curvature gives two sided control on the expansion and contraction of geodesic distances. $\endgroup$
    – Ben McKay
    Commented May 24, 2018 at 11:22

1 Answer 1

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Lemma 3.24 at page 87 of Alexander Grigor'yan's "Heat Kernel and Analysis on Manifolds" says that (I reformulate it slightly)

For any point $p \in M$ and chart $(U', h)$ around $p$ there exist a $U \subseteq U'$ and a constant $C \ge 1$ such that for all $x,y \in U$ we have

$$\frac 1 C \| h(x) - h(y) \| \le d(x,y) \le C \| h(x) - h(y) \| \ .$$

Taking $h = \exp_p ^{-1}$ and $v \in S_r$ you get

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \le d(u,v) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \ ,$$

whence by taking $\sup _{v \in S_r}$ one gets

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \le d(u,S_r) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \ ,$$

the first inequality being what you are asking for.

By looking at Grigor'yan's proof, it seems to me that for small enough $U$ and $r$ one may choose $C$ independent of $p$ and $r$.

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  • $\begingroup$ Thanks! It is an answer to the Question 1. Do you know that $C \to 1$ when $r \to 1$ necessarily, in the case $h=\exp_p^{-1}$ and for minimal $C$s, $\endgroup$
    – Mahdi - Free Palestine
    Commented May 25, 2018 at 10:47
  • $\begingroup$ @Mahdi: Notice that there is no guarantee that for $r \to 1$ you will remain in the open subset where $\exp$ is a diffeomorphism. I believe that the question that you ask in the above comment is ill posed. $\endgroup$
    – Alex M.
    Commented May 29, 2018 at 14:54
  • $\begingroup$ Oh! actually, I mean $r \to 0$! $\endgroup$
    – Mahdi - Free Palestine
    Commented Jun 6, 2018 at 8:14

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