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