Comparison between riemannian distance of a manifold embedded in $\mathbf R^N$ and euclidean distance

Question

Let $M$ be a closed Riemannian manifold isometrically embedded in some $\mathbf R^N$, and moreover let $d_M$ be the Riemannian distance on $M$. It is clear that for $x,y \in M$ : $$ |x-y| \leq d_M(x,y). $$ My question is about the inverse inequality, i.e. do we have a constant $k$ such that $$ d_M(x,y) \leq k |x-y|\;? $$ I think the answer is yes and the idea was to take

\begin{equation*} k = \underset{x,y \in M}{\sup} \frac{d_M(x,y)}{|x-y|} \end{equation*}

We just have to verify that the function defined just above is continue, so typically we have problems when $x$ and $y$ are close. Do we have some asymptotics or some ways to prove it ? I searched some results in books but I couldn't find anything.

Thank you !

Answer 1

As you observed yourself, the remaining problem is a local one. Hence it's fine to assume that $0\in M$ and only consider a small $0$-neighbourhood, where we write $x=(x',x'')\in \mathbb R^{\mathrm{dim}M}\times \mathbb R^{n-\mathrm{dim} M}$ and assume that $x'|_{M}$ consitutes a local coordinate system for $M$. Also write $y=(y',y'')$.

The Taylor series of $y'\mapsto d_M(x',y')^2$ near $y'=x'$ only starts at the second order, and using the integral representation of the remainder we find that $$d_M(x',y')^2=(x'-y')^TG(x',y')(x'-y')$$ for a smooth matrix valued function $G(x',y')$, defined near $0$. (This way $G(x',x')$ is just the matrix representation of the metric tensor $g$ in the $x'$-coordinate.)

Using this, we have $$ \left|\frac{d_M(x,y)}{|x-y|}\right|^2 = |G(x',y')|\times \frac{|x'-y'|^2}{|x-y|^2} \le |G(x',y')|, $$ and the right hand side, being smooth across the diagonal, is locally bounded.