Divergence theorem on Hyperbolic space

Question

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?

Answer 1

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.

Answer 2

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.