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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 $c$ 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.

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.

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 $c$ 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.

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 $c$ 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.