# Divergence theorem on Hyperbolic space

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Given a vector field, say $F$, defined on a manifold $U$, the divergence theorem states that: $$\int_U\nabla \cdot F dV=\int_{\partial U} F d \Sigma .$$

Well if the manifold is $\mathbb R ^n$ and $F$ only depends on the (geodesic) distance of two given points in $\mathbb R^n$ one can apply the divergence theorem considering the volume of the sphere centered in one of the points, whose radius, $R$, is the distance between the considered points, and $\partial U$ is just the sphere's surface. Then the divergence theorem can be writen explicitly as

$$\int_{S^n}\nabla \cdot F \space \space R^n\sin^{n-1}(\theta)d\theta d\Omega=F \space R^n \int_{ S^{n-1}} \sin^{n-1}(\theta)d\theta d\Omega .$$

But what if the considered manifold is the hyperbolic space? What would be the volume element? $R^n\sinh^{n-1}(\theta)d\theta d\Omega$, as the wiki article suggests? If so, what would be interpretation of $R$ and the angles?

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user PML
asked Feb 11, 2013
There's a non-metrical version of the divergence theorem using the language of differential forms. Once you apply the Hodge star and duality between covectors and vectors, you have a vector-language divergence theorem in any Riemann manifold.

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user Ryan Budney
Thank you. I shall look into that =)

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user PML

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OK, so the easiest way to get at this is to consider the unit sphere--this can be embedded in 3-space with equation:

$$x^{2} + y^{2} + z^{2} = r^{2}$$

Or alternately, with $r = {\rm cons}$. Using the normal Jacobian business, we can find that the measure of this space is $\int dV = \int r^{2}\sin\theta d\theta d\phi$

Now, consider the hyperbolic plane. this is given by the equation

$$x^{2} - y^{2} - z^{2} = r^{2}$$

Now, let's explicitly do our jacobian business after choosing $r= {\rm cons}$. And to save ourselves some work, we're going to do our best to make it nice and simple. First, we make the substitution $y = \rho \sin \theta$ and $z = \rho \cos \theta$, which means that $y^{2} + z^{2} = \rho^{2}$, and our constraint is

$$x^{2} - \rho^{2} = r^{2}$$

Now, to make our lives yet simpler, we make the substitution $x = r\cosh \phi$ and $\rho = r \sinh \phi$. Using the identity $\cosh^{2} \phi- sinh^{2}\phi = 1$, we see that our constraint is automatically satisfied, and simple algebra can convince you that any point $x,y,z$ for $x > 0$ can be solved for some $r, \phi, \theta$. All in all we have:

$$x = r \cosh \phi$$ $$y = r \sinh\phi \sin \theta$$ $$z = r \sinh\phi \cos\theta$$

Now, remembering that our constraint is $r={\rm cons}$, we don't have to take $r$ derivatives when taking differentials and substituting back into our Euclidean distance formula:

$$ds^{2} = dx^{2} + dy^{2} + dz^{2}$$

After all of this, we'll find a new metric tensor where

$$ds^{2} = g_{ab}dx^{a}dx^{b}$$

Finally, our volume element will be:

$$dV = \sqrt{{\rm det}\,g}d\theta d\phi$$

I'll leave it as an excersise for you to do all of the differentials and matrix algebra required to show that it comes out in the form you've given above.

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user Jerry Schirmer
answered Feb 11, 2013 by (130 points)
Now this is what I was looking for. Thank you. I get that you basicaly embedded the $\mathbb H^2$ in the 3 dimensional minkowski space. I've seen that in a [wikipedia] (en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator) article but I just couldn't find the nice expression for the surface formed by the points that were at the same distance from a given point (center). (After your answer I just feel dumb =)) Thank you, really, for the very elucidative answer.

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user PML
hum, when you say Euclidean distance formula shouldn't it be the Minkowski one?I've done the calculations to get the metric and the volume element and it seems that it should be the minkowski metric...

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user PML
@PML: it works in both cases. The minkowski case is cleaner, since the $r,\phi,\theta$ system is orthonormal in the minkowski case. If you embed in Euclidean space, though, you will still find that $\sqrt{\rm det g} = r^{2}\sinh\phi d\theta d\phi$, you'll just have to content with $g_{r\phi}$ components to your metric tensor.

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user Jerry Schirmer
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What wiki suggest is right, this happens because the metric tensor of hyperbolic manifold includes $\sinh$, I doubt that it's possible to give simple geometrical interpretation of angels, but regarding $R$, it's circle radios on this manifold, in other words, if you will take the Circumference of a circle on this manifold and divide it by $2\pi$ you will get $R$, here I still supposing that this hyperbolic manifold is 2-dimentional, visualizing this on higher dimensional hyperbolic manifolds is almost impossible, I can imagine it by "analogy" only.

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user TMS
answered Feb 11, 2013 by (40 points)
hummm, would then be correct to say that: $\int_{ S^n}\nabla \cdot F \space \space R^n\sinh^{n-1}(\theta)d\theta d\Omega=F \space R^n \int_{ S^{n-1}} \sinh^{n-1}(\theta)d\theta d\Omega$ ? I mean would the submanifold still be a n-sphere and it's boundary a sphere-(n-1)?

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user PML
You integrating according to some simply connected, closed surface inside a hyperbolic manifold, there is no usual spheres anymore, you still able to integrate according to a sphere "inside" the hyperbolic manifold, but then you will need to find the m-Sphere equation in n-hyperbolic coordinates, not the spherical one!

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user TMS

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