Definition of arc length on manifolds without parametrization

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SUMMARY

The discussion centers on defining arc length on differentiable manifolds without relying on parametrization. It establishes that while the standard definition of arc length involves parametrization, it is possible to conceptualize arc length in a way that is independent of it, particularly in the context of Riemannian geometry. The conversation highlights that defining arc length on a general manifold is complex due to the nature of local coordinate systems and their transition mappings, which may not preserve distance functions. The participants suggest that while a canonical method does not exist, specific cases may allow for parametrization-free definitions.

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  • Differentiable manifolds
  • Riemannian geometry
  • Geodesics and their properties
  • Understanding of parametrization in mathematical contexts
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  • Explore Riemannian metrics and their implications on arc length
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  • Investigate parametrization-free definitions of arc length in specific manifolds
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Mathematicians, particularly those specializing in differential geometry, Riemannian geometry, and anyone interested in the theoretical aspects of arc length 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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