Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
As you can clearly see that small $d_{\mathcal{M}}$ implies $d_{}$ small? –  Elias 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

2 Answers 2

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

share|improve this answer

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 $d\le d_{\mathcal M}$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.