# Equivalence of two distance function on a Riemannian manifold

Let $(M,g)$ be a closed connected $m$ dimensional smooth Riemannian manifold and assume that it is isometrically embedded in a Euclidean space $\mathbb{R}^q$ by $\iota:M\to\mathbb{R}^q$.

$|\ast|$ denotes the Euclidean norm in $\mathbb{R}^q$ and $d(x,y)$ denotes the geodesic distance between $x,y\in M$.

Then, I think that it holds that \begin{eqnarray} \forall\varepsilon>0\ \exists\delta>0\ s.t.\ \forall x,y\in M\ \ \ 0<|y-x|<\delta\Rightarrow\ \dfrac{d(x,y)}{|y-x|}<1+\varepsilon. \end{eqnarray}

But I'm not able to varify the above statement. Can anyone help me?

It is also glad if you show me the following weaker statement: \begin{eqnarray} \exists C>0\ \forall x,y\in M\ \ \ d(x,y)\leq C|y-x|. \end{eqnarray}

Thank you.

• Offhand this looks plausible if $M$ is compact, but it's false in general. Embed $(0, 2\pi)$ in $\mathbf{R}^{2}$ by $\iota(t) = (\cos t, \sin t)$ and consider $x \approx 0$ and $y \approx 2\pi$. – Andrew D. Hwang Feb 2 '15 at 14:13
• "Isometric embedding" means $d(x,y)=|x-y|$; so there is nothing to prove. Or have I missed something? – Christian Blatter Feb 10 '15 at 16:20
• @ChristianBlatter "$\iota : (M, g) \to (N, \bar{g})$ is an isometric embedding" means that the Riemannian metric $g$ on $M$ is the pullback via $\iota$ of the Riemannian metric $\bar{g}$ on $N$. It follows e.g. that $d_M(x,y)$ is the infimum of the $N$-lengths of curves from $x$ to $y$ which are contained entirely in $M$, but in general this is larger than $d_N(x,y)$. – mollyerin Feb 10 '15 at 22:23
• Is this even true for plane curves? For instance $\gamma(t) = [t, t^2\sin(1/t)]?$ – user7530 Feb 11 '15 at 23:36

[Edit: Errors in the original proof (forgot a factor of $||\partial_t \Gamma(1,t)||$ in one of my estimates) have, I believe, been fixed.]

I'm sorry this is so long; I couldn't find a slick proof so I just sat down and did horrendous analysis. I believe the following gives a proof of your claim, but it is horrendous analysis so the probability I've made a proof-breaking mistake is quite nontrivial.

Theorem. Let $M \hookrightarrow \mathbb{R}^n$ be an isometric embedding. Given $\epsilon > 0$ and $p \in M$, there exists $D > 0$ such that for $x, y \in M$ such that $|x - p|, |y - p| < D$, we have $$d_M(x, y) \leq (1+\epsilon)d_{\mathbb{R}^n}(x,y).$$

I'd believe that the following are simple corollaries:

Corollary. By the Nash Embedding Theorem, the same holds replacing $\mathbb{R}^n$ by an arbitrary Riemannian manifold $N$.

Corollary. If $M$ is closed, then given $\epsilon > 0$ there exists $D$ in $M$ such that for $x,y \in M$ with $|x - y| < D$, the same inequality holds. (This follows from a compactness argument, e.g. the argument in frog's answer.)

I think this last answers your question. Onto the discussion of the theorem itself.

Outline of Proof. The proof (that I came up with, anyway) is long and requires some tricky analysis, so I'll try to say a word about the intuition. For $x, y \in M$ we can draw the straight line between $x$ and $y$; the length of this line is the distance between $x$ and $y$ in $M$. If $x$ and $y$ are sufficiently close, the line between them will be contained in a nice tubular normal neighborhood of $M$, and so we can deform it onto $M$ to get an $M$-path between $x$ and $y$. This path is not a geodesic, but it suffices to estimate its length. We do this by computing the first variation of the energy of the paths in the deformation. It turns out that the first variation of the energy of these paths can be estimated in terms of the shape operator of $M$, the length of the deformed path in $M$, and a term that describes how far we had to deform our path to get it onto $M$. The shape operator will satisfy some uniform bound on small (precompact) subsets of $M$, so the goal is to show that the deformation itself can be made to be quite small. For this we use the fact that the tangent space of $M$ approximates $M$ to first order, which roughly shows that for points a distance $\eta$ apart the distance to $M$ should be $O(\eta^2)$. Combining these estimates will give a proof.

Proof. The question is local, so we may assume that $M$ is the image of $f : U \hookrightarrow \mathbb{R}^n$, where $U \subseteq \mathbb{R}^m$ is an precompact open neighborhood of $0$ with the property that there is some smooth extension of $f$ to a larger open set $V \supset \overline{U}$ (this last part allows us to get estimates on sup norms of $f$ and its derivatives).

We let $L$ be an upper bound for the second partial derivatives of $f$, and let $C$ be a constant so that for $x, y \in U$ we have $|y - x| \leq C |f(y) - f(x)|$. (The existence of such a $C$ follows from the fact that $f$ is a smooth embedding, so there's some locally defined diffeomorphism $g$ so that $g \circ f$ is the standard embedding $U \subseteq \mathbb{R}^m \subseteq \mathbb{R}^n$; then $C$ is a Lipschitz constant for $g^{-1}$.)

Sublemma. Let $x, y \in U$ satisfy $|f(x) - f(y)| = \eta$. Then for any point $z$ on the line segment connecting $f(x)$ and $f(y)$, the distance from $z$ to $f(U)$ satisfies $d_{\mathbb{R}^n}(z, f(U)) \leq 2nLC^2\eta^2$.

Proof of Sublemma. By Taylor's theorem, one can write $$f(y) = f(x) + Df_x(y - x) + R(y-x),$$ where the remainder satisfies $$|R(y-x)| \leq nL|y-x|^2 \leq nLC^2 |f(y) - f(x)|^2 = nLC^2 \eta^2.$$ Thus the distance from $f(y)$ to the point $f(x) + Df_x(y-x)$, which lives in the tangent space to $x$, is bounded by $nLC^2 \eta^2$. It follows that for $0 \leq \lambda \leq 1$, the distance between $f(x) + \lambda(f(y) - f(x))$ and the point $f(x) + Df_x(\lambda(y-x))$ is bounded by $\lambda nLC^2 \eta^2$. (This is just looking at the relevant triangle.)

By the same Taylor estimate, applied now to $x + \lambda(y - x)$ in place of $y$, it follows that the point $f(x) + Df_x(\lambda(y-x))$, which lives on the tangent space of $x$, is within a distance of $\lambda^2 nLC^2\eta^2$ of the point $f(x + \lambda(y-x)) \subseteq f(U)$. Adding these two estimates (and ignoring the irrelevant factors of $\lambda$) gives the claim of the sublemma.

Proof of Theorem (ct'd). There is some open $W \subseteq \mathbb{R}^n$ containing $f(U)$ so that for $w \in W$, there is a unique closest point to $w$ in $f(U)$ ($W$ is a small tubular normal neighborhood of $f(U)$). We let $U' \subseteq U$ be small enough so that straight lines in $\mathbb{R}^n$ between points of $f(U')$ remain in $W$.

Now consider $x, y \in U$. Choose a variation field $\Gamma(s,t)$, for $0 \leq s,t \leq 1$, so that $t \mapsto \Gamma(0,t)$ is the constant speed line segment in $\mathbb{R}^n$ with $\Gamma(0,0) = f(x)$ and $\Gamma(0,1) = f(y)$, and so that for each fixed $t$, the curve $s \mapsto \Gamma(s,t)$ is the constant speed straight-line path from $\Gamma(0,t)$ to the point on $f(U)$ closest to $\Gamma(0,t)$. Note that $$\Gamma(s,t) = s\Gamma(1,t) + (1-s) \Gamma(0,t).\tag{*}$$ We sometimes write $\gamma_s(t)$ instead of $\Gamma(s,t)$.

Now, $\gamma_1(t)$ is a curve in $f(U)$ connecting $f(x)$ and $f(y)$, and we will estimate its length. Recall that the energy of a curve $\gamma$ is $E(\gamma) = \int ||\gamma'||^2$. We consider the first variation of the energy of $\Gamma$: \begin{align} \frac{\partial}{\partial s} E(\gamma_s(t)) &= \frac{\partial}{\partial s} \int_0^1 \langle \partial_t \Gamma(s,t), \partial_t \Gamma(s,t) \rangle dt \\ &= 2 \int_0^1 \langle D_s \partial_t \Gamma(s,t), \partial_t \Gamma(s,t) \rangle dt \\ &= 2 \int_0^1 \langle D_t \partial_s \Gamma(s,t), \partial_t \Gamma(s,t) \rangle dt \\ &= -2 \int_0^1 \langle \partial_s \Gamma(s,t), D_t \partial_t \Gamma(s,t) \rangle dt. \end{align} We wish to estimate this last term. By $(*)$, we have $$D_t \partial_t \Gamma(s,t) = s D_t \partial_t \Gamma(1,t).$$ Let $B$ be the second fundamental form of the embedding $f(U) \subseteq \mathbb{R}^n$, and let $B_0$ be a bound on its norm within $U$. Then we can estimate our inner product using the fact that $\partial_s \Gamma(s,t)$ is orthogonal to $f(U)$: \begin{align} \left| \langle \partial_s \Gamma(s,t), D_t \partial_t \Gamma(s,t) \rangle \right| &= s \left| \langle \partial_s \Gamma(s,t), D_t \partial_t \Gamma(1,t) \rangle \right| \\ &= s \left| \Big\langle \partial_s \Gamma(s,t), B\big(\partial_t \Gamma(1,t), \partial_t \Gamma(1,t)\big) \Big\rangle \right| \\ &\leq s B_0 ||\partial_s \Gamma(s,t)|| \cdot ||\partial_t \Gamma(1,t)||^2 \end{align}

By the construction of $\Gamma$, and the sublemma, we have $$||\partial_s \Gamma(s,t)|| = d_{\mathbb{R}^n}(\Gamma(0,t), \Gamma(1,t)) \leq 2nLC^2 |f(y)-f(x)|^2.$$ Thus we have the estimate \begin{align} \left| \frac{\partial}{\partial s} E(\gamma_s(t)) \right| &\leq 4nLC^2 \; |f(y) - f(x)|^2 \int_0^1 ||\partial_t \Gamma(1,t)||^2 dt \\ &= 4nLC^2 E(\gamma_0) E(\gamma_1). \end{align} Integrating, it follows that $$E(\gamma_1) - E(\gamma_0) \leq 4nLC^2 E(\gamma_0) E(\gamma_1),$$ which becomes $$E(\gamma_1) \leq \frac{E(\gamma_0)}{1 - 4nLC^2 E(\gamma_0)}$$ (so long as $E(\gamma_0)$ is small enough that the term on the right is positive).

Since $E(\gamma_0)$ is the square of the $\mathbb{R}^n$-distance from $x$ to $y$, by choosing $U$ small enough (depending only on $\epsilon$) we can ensure that $$\frac{1}{\sqrt{1 - 4nLC^2 E(\gamma_0)}} \leq 1 + \epsilon.$$

Let $\ell_1$ be the length of $\gamma_1$, and $\ell_0$ the length of $\gamma_0$. Since $\gamma_0$ is a geodesic, we have $\ell_0^2 = E(\gamma_0)$; we also have the estimate $\ell_1^2 \leq E(\gamma_1)$, which follows from the CBS inequality. Then the above becomes $$\ell_1 \leq \sqrt{E(\gamma_1)} \leq \frac{\ell_0}{\sqrt{1-4nLC^2E(\gamma_0)} } \leq (1 + \epsilon)\ell_0.$$ Since $d_{f(U)} (f(x),f(y)) \leq \ell_1$ and $d_{\mathbb{R}^n}(f(x),f(y)) = \ell_0$, this proves the theorem.

• $||B(\partial_t\Gamma(1,t),\partial_t\Gamma(1,t))||\leq B_0||\partial_t\Gamma(1,t)||^2$. – stb2084 Feb 11 '15 at 18:16
• @stb2084 Yes, of course you're right. Unfortunately this appears to break the proof as far as I can tell. Morally it still feels like the right approach to me, and simple examples seem to indicate that the thing you want to be true should be true. But I don't see how to fix this, so maybe my intuition is just off. Thanks for reading through all my garbage -- sorry about the mistake! – mollyerin Feb 11 '15 at 22:14
• @stb2084 I believe I've fixed the error you pointed out. – mollyerin Feb 11 '15 at 22:37

EDIT: Thanks to mollyerin for pointing out my wrong statement. I will leave it nevertheless so you can see what mollyerin is referring to. I apologize:

Wrong statement: The statement is wrong in general: Think of a 2-sphere centered at the origin in $\mathbb R^3$ and push the north and south pole ($N$ and $S$) in a smooth way towards the origin. In this way you will be able to make $|N-S|$ arbitrarily small but $d(N,S)$ will be bounded away from zero as much as you want (by stretching your sphere). However, your statement becomes true if you ask for $d(x,y)<\delta$ instead of $|x-y|<\delta$. I can give you a proof of this statement if you wish.

With this you can prove the second statement: The function $$(x,y)\mapsto\frac{|x-y|}{d(x,y)}$$ takes a minimum on $(M\times M)\setminus \{(x,y)\in M\times M, d(x,y)\geqslant \delta\}$ for any $\delta>0$ simply by continuity. For $d(x,y)<\delta$, the boundedness follows from the (altered) statement above.

EDIT 2: @mollyerin: What I thought of was this: Translate the embedding so that $x=0$ and consider for any unit tangent vector $v$ in $T_xM$ the curve $\gamma_v(t):=\exp_p(tv)$. Then for any $y$ sufficiently close to $x$ one has $y=\exp_p(tv)$ for some $v$ and $t$. Moreover one has $d_M(x,\gamma_v(t))=t$ and using the taylor expansion of $|\gamma_v(t)-0|$ one finds

EDIT: This Theorem is in error, but it can be corrected (second line)

Wrong: $$|\gamma_v(t)|=t-t^3\frac{|\ddot{\gamma}(0)|^2}{3!}+O(t^4)$$ Right: $$|\gamma_v(t)|=t-t^3\frac{|\ddot{\gamma}(0)|^2}{4!}+O(t^4)$$ hence the desired quotient satisfies $$1+O(t^2)<1+\varepsilon$$ provided $t$ is small enough. Is there something obvious I'm missing? Thanks for any comments.

• I don't believe your counterexample: For whatever embedding you choose, I can choose $\delta$ much smaller than the $\mathbb{R}^3$-distance between the north and south poles of the sphere, no? – mollyerin Feb 10 '15 at 15:43
• Also, if you have a nice proof of this statement (even for just $d(x,y) < \delta$) I'd be interested to see it, since the proof I worked out is a mess and it's hard for me to be sure it's right. – mollyerin Feb 10 '15 at 15:52
• Thanks for your comments. My "proof" would use the answer you actually gave to one of my questions (math.stackexchange.com/questions/436027/comparison-statement). If $d(x,y)$ is small enough, take a curve (parametrized by arc length) in $M$ that relies $x$ and $y$, say $\gamma:[a,b]\to M$. Then use taylor expansions of $d(\gamma(0),\gamma(t))$ and $|\gamma(0)-\gamma(t)|$ to conclude. I didn't work out the details but I guess it should work this way. If needed I can write it out or think about it more carefully. – frog Feb 10 '15 at 16:08
• I couldn't figure out how to do anything like this -- unfortunately the $M$-geodesic from $x$ to $y$ doesn't have to agree with the line segment from $x$ to $y$ even to first order. I am still interested in more details if you (a) feel like thinking about it and (b) can convince yourself that it should work. – mollyerin Feb 10 '15 at 16:28
• I guess there must be a flaw in my idea but i wrote it as an edit to my answer and I'm grateful for any comments. – frog Feb 11 '15 at 9:56