The Riemann distance function $d(p,q)$ is usually defined as the infimum of the lengths of all piecewise smooth paths between $p$ and $q$.

Does it change if we take the infimum only over smooth paths? (Note that if a smooth manifold is connected, then it is smoothly path connected).

I am quite certain the distance does not change. I think that every piecewise smooth path can be approximated by a smooth path.

Around any singular point of the original path, we can take a coordinate ball, and create smomehow a smoothing of a relevant segment of the path which is not much longer than the original.

An explicit construction such as this can be found here. However, the point there is only to show smooth path connectivity, and we also need some bound on the "added length".

Partial Result (Reduction to the case of Euclidean metric):

I show that the specific Riemannian metric does not matter. That is, if we can create a smoothing with small elongation measured by one metric $g_1$ then we can do the same for any other metric $g_2$.

Hence it is enough to prove the claim for $\mathbb{R}^n$ with the standard metric.


Since the question is local (we focus around some point $p$ of non-smoothness of the original piecewise-smooth path) we can take an orthonormal frame for $g_1$, denoted by $E_i$. write $g_{ij}=g_2(E_i,E_j)$, I want to find $\text{max} \{g_2(v,v)|v\in \mathbb{S}^{n-1}_{g_1}\} = \text{max} \{g_2(v,v)|v=x^iE_i , x=(x^1,...,x^n) \in \mathbb{S}^{n-1}_{Euclidean}\} = \text{max} \{g_{ij}x^ix^j| \sum(x^i)^2=1 \} = \text{max} \{x^T \cdot G \cdot x | \|x\|=1 \} = \text{max}{\lambda(G)}$.

Since the roots of a polynomial are continuous in in terms of its coefficients, and the coefficients of the charactersitic polynomial of a matrix depends continuously on the matrix entries, it follows that the eigenvalues of a matrix depends continuously on the matrix entries. Hence, since the matrix $g_{ij}(q)$ is a continuous function of $q$, it follows that if we restrict to a compact small enough neigbourhood of $p$ we the function $f(q)= \text{max}{\lambda(g_{ij}(q))}$ is continuous and in particularly bounded by some constant $C$. Hence for any path $\gamma$ which is contained in a small enough neighbourhood of $p$ $L_{g_2}(\gamma) \le \sqrt C L_{g_1}(\gamma)$.

In particular we can take $g_1$ to be the pullback metric of the standrad Euclidean metric via some coordinate ball around $p$. Now solving the problem for the Euclidean case (which implies solving it for $g_1$), we obtain a solution for an arbitrary $g_2$ as required.


If you already reduced this problem to the $\mathbb{R}^n$ case, then we should be able to tackle it with the usual analytical methods. The following is probably a bit of technical overkill but should work.

As far as I can see, the only problem is to smoothly connect two pieces with an arbitrarily small loss of length. Assume we have two smooth paths $$p_1: [a,0] \to \mathbb{R}^n $$ and $$p_2: [0,b] \to \mathbb{R}^n $$ such that $p_1(0)=p_2(0)$.

Now for some arbitrarily small $\delta$ we want to construct $\tilde{p}:[a,b]\to \mathbb{R}^n$ such that $\tilde{p}(a)=p_1(a)$, $\tilde{p}(b)=p_2(b)$ and $$\mathop{Length}(\tilde{p}) = \mathop{Length}(p_1)+ \mathop{Length}(p_2)+\varepsilon$$ For this construct a partition of unity: Let $$\varphi_1(t) := \begin{cases} 1 & \text{ for } t\leq -\delta \\ 0 & \text{ for } t\geq \delta \\ \frac{e^{-1/(x-\delta)}}{ e^{-1/(x-\delta)}+e^{-1/(x+\delta)}} & \text{ for }t \in (-\delta,\delta) \end{cases} $$ and $\phi_2(t) = 1-\phi_1(t)$. Then the $\phi_i(t)$ are smooth and $|\dot{\phi(t)}|\leq \frac{4}{\delta}$. (I haven't checked the details here but there definitely is such a function)

I think we can safely assume both parts to be parameterised by arc length (that is $|\dot{p}_i(t)| = 1$ for all $t$) and we can smoothly extend them (for example by their taylor series) a bit, so that they are defined on $[a,\delta]$ and $[-\delta,b]$.

Now we define $$\tilde{p}:[a,b]\to \mathbb{R}^n; t \mapsto \phi_1(t) p_1(t)+\phi_2(t) p_2(t)$$ (technically we need to extend the $p_i$ to all of $[a,b]$, but the corresponding $\phi_i$ is $0$ anyway, so this works as a definition) Then $\tilde{p}$ is smooth since it is built from smooth functions and it is equal to $p_1$ on $[a,-\delta]$ and $p_2$ on $[\delta,b]$, so we need only to care about the middle.

But now for $t\in[-\delta,\delta]$ $$ |\tilde{p}(t)| = |\dot{\phi}_1(t)p_1(t)+\dot{\phi}_2(t)p_2(t)+\phi_1(t)\dot{p_1}(t) + \phi_2(t)\dot{p_2}(t)|$$ so via triangular inequality and noting that $\dot{\phi}_2= -\dot{\phi}_1$ $$ \leq \underbrace{|\dot{\phi}_1(t)|}_{\leq 4/\delta} |p_1(t)-p_2(t)| + \phi_1 (t) \underbrace{|p_1(t)|}_{=1} + \phi_2(t) \underbrace{|p_2(t)|}_{=1}$$ $$ \leq 4/\delta |p_1(t)-p_2(t)| + \phi_1(t)+\phi_2(t)$$ thus since $p_1(0)=p_2(0)$ and $|\dot{p}_i|=1$ implies $|p_1(t)-p_2(t)| \leq 2|t| \leq 2\delta$: $$\leq 4/\delta \cdot 2\delta + 1 = 9$$

So we only need to integrate: $$\mathop{Length}(\tilde{p}) = \int_a^b |\tilde{p}(t)| dt $$ $$= \int_a^{-\delta} |\tilde{p}(t)| dt + \int_\delta^b |\tilde{p}(t)| dt + \int_{-\delta}^{\delta} |\tilde{p}(t)| dt $$ $$ \leq \mathop{Length}(p_1) + \mathop{Length}(p_2) + 2\delta \cdot 9$$ which is what we wanted.

Of course we are actually not on $\mathbb{R}^n$ but on some open subset, but for a small enough $\delta$ this should pose no problem.

edit: Maybe some slight change, if you do not believe in extending $p_1$ and $p_2$, it is also possible to just reparamterize them in such a way that $p_1(\delta) = p_2(-\delta)$, since we only need $p_1(t)$ and $p_2(t)$ to be close for small $t$ and not actually to be equal anywhere.

  • $\begingroup$ Thanks for your solution! About the possibility of extending $p_1$,$p_2$ , I am not sure the argument of using Taylor series works, because real smooth functions are not necessarily analytic. However, some people require the existence of (local) extensions as part of the defintion of smoothness of maps from closed intervals into manifolds, so I accept that. $\endgroup$ – Asaf Shachar Aug 16 '15 at 21:29

No, the distance stays the same. As mentioned in The formula for a distance between two point on Riemannian manifold, the reason to allow piecewise smooth curves is to be able to concatenate them, getting other piecewise smooth curves.

For any piecewise smooth curve from $p$ to $q$ there is a smooth one of almost the same length. For example, suppose $\gamma$ has a point of nonsmoothness at $t=t_0$, with one-sided derivatives not matching. Insert a little loop based at $\gamma(t_0)$ that begins with the velocity vector $\gamma'(t_0-)$ and ends with velocity vector $\gamma'(t_0+)$. The point of nonsmoothness is gone, and the added length can be atbitrarily small.

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    $\begingroup$ @NormI Human: I agree what you suggest is the right idea, I just think it is not completely trivial to explicitly build such a smooth loop. In particular, all the higher order derivatives should also agree with the original ones at $t_0$. $\endgroup$ – Asaf Shachar Aug 1 '15 at 11:40

An additional remark to the answer.

On a Riemannian manifold (without "missing" points, e.g., complete) the minimum length in any homotopy class of path exists and is attained by a geodesic path, which is necessarily smooth. If the manifold is of some reasonably finite topological type (compact is much more than enough), the infimum of the geodesic lengths will in fact be attained by one of the geodesic paths, so that the minimum distance between two points is always realized by a geodesic.

  • $\begingroup$ This is certainly not true. Consider $\mathbb{R}^2$ minus a point with the usual metric. Then no path, smooth or otherwise, attains minimum distance between, say, $(-1, 0)$ and $(1, 0)$, in any fixed homotopy class. $\endgroup$ – Qiaochu Yuan Aug 10 '15 at 16:56
  • $\begingroup$ This is in fact true for every complete Riemannian manifold. (See Hopf-Rinow Theorem). $\endgroup$ – Asaf Shachar Aug 10 '15 at 16:56
  • $\begingroup$ @qiaochu, Sorry, I forgot some adjectives in the answer to separate this issue from the one of attaining the minimum among what might be infinitely many homotopy classes. I guess I want "(geodesically) complete". $\endgroup$ – ASCII Advocate Aug 10 '15 at 16:57
  • $\begingroup$ (for example, from wikipedia: "The theorem does not hold in infinite dimensions: (Atkin 1975) showed that two points in an infinite dimensional complete Hilbert manifold need not be connected by a geodesic.[2] " ). I haven't seen the paper but probably this involves building shorter and shorter handles in additional dimensions, to play the same role as the missing point in the example with the punctured plane. @QiaochuYuan $\endgroup$ – ASCII Advocate Aug 10 '15 at 17:02

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