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Say a compact n-manifold $\mathcal{M}$ is embedded in $\mathbb{R}^m$, $m > n$. If $d_{\mathcal{M}}$ is the geodesic distance on $\mathcal{M}$, and $d$ the Euclidean distance in $\mathbb{R}^m$, then clearly small $d_{\mathcal{M}}$ implies small $d$.

It seems that small $d$ should imply small $d_{\mathcal{M}}$ (since $\mathcal{M}$ is compact, it should have positive reach $\sigma > 0$). Is this known to be true?

Thank you.

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As you can clearly see that small $d_{\mathcal{M}}$ implies $d_{}$ small? – MathOverview Nov 12 '13 at 15:09
Detailed answer requires a detailed question. This question does not make any smoothness assumptions on the embedding. Without such assumption the claim is false as you can notice by looking at the graph of $\sqrt{|x|}$. – studiosus Nov 12 '13 at 20:16
I took the smoothness assumption for granted, and also that $\mathcal M$ is given the induced Riemannian metric from $\mathbb R^m$. The inequality $d\leq d_{\mathcal M}$ is clear, I think, for the euclidean distance is the infimum of the length of any piecewise smooth curve in $\mathbb R^m$, whereas $d_{\mathcal M}$ is the infimum over the curves contained in $\mathcal M$. The answer below claims that an inequality of the form $d_{\mathcal M} \leq L d$ (for some $L > 0$) holds, but I'm unable to work out the proof in detail. – Francesco Genovese Nov 13 '13 at 11:11

The proof is similar to the argument I gave here. I will consider the more general case of smooth embeddings $$ i: X\to Y $$ where $X$ is compact and $X, Y$ are Riemannian manifolds. First of all, the same argument I gave here shows that $i$ is $R$-Lipschitz for some $R$ (I used only differentiability of $i$ and nothing else). Therefore, we need to show the bilipschitz property of $i$. Recall that every compact submanifold $Z=i(X)\subset Y$ has positive normal injectivity radius, i.e., a positive constant $r$ such that the normal exponential map $\exp_Z: \nu_Z\to Y$ is a diffeomorphism onto its image when restricted to the subset $$ B_r \nu_Z= \{v\in \nu_Z: |v|<r\}. $$ Here $\nu_Z$ is the normal bundle of $Z$ in $Y$. The proof of this assertion is given in the book by Guillemin and Pollack "Differential Topology" (they do it for embeddings to $R^m$ but that's what you care about and their argument is general). The image $N=\exp(B_r(\nu_Z))$ is a certain open neighborhood of $Z$, a tubular neighborhood. Since $\exp_Z$ is a diffeomorphism, its inverse (let's call it $\log_Z$) is smooth. Composing $\log_Z$ with the projection $\nu_Z\to Z$ is gives a smooth map $p: N\to Z$ which is a retraction (fixes $Z$ pointwise) by the construction. Let $\bar B_{r'} \nu_Z$ denote the closure of $B_{r'} \nu_Z$ in $\nu_Z$, $0<r'<r$. Its image under $\exp_Z$ is a compact submanifold with boundary $K\subset N$. Therefore, the restriction of $h$ to $K$ is $L$-Lipschitz, as I noticed above (where you can use any Riemannian metric on $Z$ you like, e..g, the one comping from $X$). Since $h$ is right-inverse to the inclusion map $i: Z\to Y$, it follows that the map $$ i: Z\to K$$ is $L$-bilipschitz, where I use the restriction of the Riemannian distance function from $Y$ to $K$. Since $K$ contains an open neighborhood of $Z$ in $Y$, it follows that the map $i$ is locally bilipschitz, i.e., it is bilipschitz when restricted to open $\epsilon$-balls in $B$ where $\epsilon>0$ is sufficiently small (less than the minimum or $r'$ and the injectivity radii of $Z$ and $Y$).

Now, we can use the point-set topology. Consider the map $$ \phi(z_1, z_2)= \frac{d_Z(z_1, z_2)}{d_Y(z_1,z_2)}, (z_1,z_2)\in Z^2\setminus D, $$ where $D$ is the diagonal in $Z\times Z$. Now, restrict $\phi$ to the compact subset $C\subset Z^2$ consisting of pairs of points $z_1, z_2$ so that $d_Z(z_1,z_2)\ge \epsilon$. Since the distance functions are continuous, $\phi$ is also continuous on $C$. Since $C$ is compact, the function $\phi$ is bounded above by some constant $R'$. Thus, the $R'$-bilipschitz inequality (for the map $i$) holds for all pairs of points in $Z$ within distance (computed in $Z$) at least $\epsilon$. On the other hand, we already proved the $L$-bilipschitz inequality for points in $Z$ within distance $<\epsilon$. Thus, the map $i$ is $\max(R, L, R')$-bilipschitz. qed

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Welcome to Math.SE! The answer is affirmative. Otherwise you would have two sequences $x_k,y_k$ of points in $\mathcal{M}$ such that $d(x_k,y_k)\to 0$ but $d_\mathcal{M}(x_k,y_k)\ge \epsilon$. Since $\mathcal{M}$ is compact, there is a point $p$ to which these sequences converge in the $d$ metric. The point $p$ has a neighborhood $U$ in $\mathbb R^m$ such that there is a diffeomorphism $\Phi$ of $U$ onto some $V\subset\mathbb R^m$ which straightens $U\cap\mathcal{M}$ into a piece of a hyperplane. Since the geodesic distance between $\Phi(x_k)$ and $\Phi(y_k)$ tends to zero as $k\to\infty$, we have a contradiction.

The argument is nonconstructive, as it has to be. The example of a very flat ellipse $x^2+(y/\epsilon)^2=1$ shows that there is no universal upper estimate on $d_{\mathcal M}$ in terms of $d$. (In contrast to $d\le d_{\mathcal M}$).

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