Definition of arc length on manifolds without parametrization

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Curves on differentiable manifolds are functions from real intervals to the manifold, and arc length is determined by the curve's image rather than its parametrization. While the standard definition of arc length involves parametrization, it is possible to define arc length in a way that does not rely on it, particularly in Euclidean spaces. However, defining arc length on general manifolds without parametrization is challenging due to the nature of local coordinate systems and their transition mappings, which may not preserve distance functions. Some suggest that a piecewise-geodesic approach could generalize the concept of arc length, but this still requires a fixed metric for practical calculations. Ultimately, while parametrization may seem unnecessary, it plays a crucial role in defining arc length and understanding the geometry of curves on manifolds.
  • #31
and \Vert f\Vert_{\infty} is the measure theory notation for the minimum real number K>0 such that K>|f| almost everywhere, also known as the essential supremum of |f|. Must be what he means.
 
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  • #32
I'm very sorry for you lost your detailed post. Next time don't forget to copy&paste your text somewhere before pressing the preview or submit button. Thank you for repeating it shortly.
 
  • #33
mma said:
It is shown that sup(||grad ax||) = 1 (by local geodesic calculation)

I thought originally that this "local geodesic calculation" is trivial. But now I see that I don't really know how is it.

It is clear that taking a short geodesic \gamma(s) starting from a(x) and parametrized by its arc length, then

d(x,y) = a_x(y) - a_x(x) = \int_0^{d(x,y)}\dot{\gamma}(a_x)|_{\gamma(s)} ds

and from this follows \dot{\gamma}(a_x) = 1.

But this means only that g(\mathrm{grad}(a_x), \dot{\gamma}) =1, and of course we know that \Vert\dot{\gamma}\Vert = 1.

But how follows \Vert\mathrm{grad}(a_x)\Vert = 1 from this?
 
  • #34
I suspect that \mathrm{grad}(a_x) = \dot{\gamma}.

Here a_x(y) := d(x,y), the distance between x and y (I forgot to mention this in my previous post), and \gamma is a geodesic through x, parametrized by its arc length)

So, my question is: how can I prove this?
 
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  • #35
mma said:
It is clear that taking a short geodesic \gamma(s) starting from a(x)

Of course strarting from x and not from a(x). Sorry for the mistyping.
 
  • #36
mma said:
I suspect that \mathrm{grad}(a_x) = \dot{\gamma}.

Here a_x(y) := d(x,y), the distance between x and y (I forgot to mention this in my previous post), and \gamma is a geodesic through x, parametrized by its arc length)

So, my question is: how can I prove this?

Because the level sets of the distance function are perpedicular to the gradient vector of it, and these level sets are n-1-dimensional submanifolds, it would be enough to prove that the geodesics passing through x are always perpedicular to the level sets of the d(x,y) distance function. Could anybody prove this?
 
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  • #37
mma said:
it would be enough to prove that the geodesics passing through x are always perpedicular to the level sets of the d(x,y) distance function.

Bingo! It's http://en.wikipedia.org/wiki/Gauss%27s_lemma_(Riemannian_geometry)"
 
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