We assume that $M$ is a smooth manifold of dimension $m$, and we assume that $g$ is a smooth Riemannian metric $g$ on $M$.

If $M$ is moreover path-connected, then $g$ induces a metric (in the sense of metric spaces) on $M$. The distance between two points $a, b \in M$ is the infimum of the lengths of smooth curves in $M$ connecting $a$ and $b$. The length of a smooth curve $c : [0,1] \rightarrow M$ is the integral

$$ L(c) = \int_0^1 g( \dot c(t), \dot c(t) ) \; dt.$$

The latter expression makes sense if the Riemannian metric $g$ is not smooth but merely continuous. The raises the following question: how low can the regularity of a smooth curve be so that we still have a well-defined metric in terms of path lengths?

For example, in many applications you assume the Riemannian metric to have merely measurable coefficients. This suffices to define the $L^2$ norms on vector fields. The question is whether you still have a distance function in that case.


closed as unclear what you're asking by Nate Eldredge, Ben McKay, Willie Wong, Misha, Anton Petrunin Oct 4 '16 at 17:32

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    $\begingroup$ Since the expression for $L(c)$ involves the derivative of $c$, how does it make sense if $c$ is only continuous but not differentiable? It could even be nowhere differentiable. "Almost everywhere differentiable" is plausible but not correct (see the Cantor staircase function). I suspect the condition you are really looking for is "absolutely continuous". $\endgroup$ – Nate Eldredge Oct 3 '16 at 18:26
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    $\begingroup$ Wait - are you talking about the regularity of the curve $c$ or the metric $g$? It seems you are switching back and forth. $\endgroup$ – Nate Eldredge Oct 3 '16 at 18:27
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    $\begingroup$ Consider the following example: your metric is standard on the plane, but along one line it is twice smaller. Note that $L^2$-norm on the vector fields is the same as in the standard plane, but you can make a shortcut using the spacial line --- depending on the application you have in mind, try to tell what would be right choice of the induced metric in this case? $\endgroup$ – Anton Petrunin Oct 3 '16 at 19:43
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    $\begingroup$ Although tangentially relevant, you might be interested to know that there is a way to define a distance for metrics with very low regularity which doesn't require being able to compute the length of curves : $d_g(x,y)=\sup\{|f(y)-f(x)|,\,f:M\xrightarrow{C^1}\mathbb{R},\, |\nabla f|^2_g\leq 1\}$. This in particular makes sense for measurable metrics. However, it won't see the pathology Anton built. $\endgroup$ – Thomas Richard Oct 4 '16 at 8:24
  • $\begingroup$ @NateEldrege: right, that was just an annoying typo. I corrected it. $\endgroup$ – shuhalo Oct 5 '16 at 2:43