I'm trying to work out the following problem: Find the first two terms of the power series expansion for the volume of a ball of radius r centered at p in a Riemannian Manifold, M with dimension n. We are given that(adsbygoogle = window.adsbygoogle || []).push({});

[tex] Vol(B_r(p)) = \int_S \int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t \mathrm{d}v[/tex]

(I'm ignoring the cut locus distance because we are only interested in small r anyway.) I assume that [itex] S = S^{n-1} = \{v \in T_pM : |v| = 1 \} [/itex]. Clearly the constant term is 0, because a ball of radius 0 has no area in any manifold, so I think that they are really asking for the first two non-zero terms. To find the next term we need to calculate the derivative of [itex] Vol(B_r(p)) [/itex] wrt r.

[tex] \frac{\mathrm{d}}{\mathrm{d}r} Vol(B_r(p)) =

\frac{\mathrm{d}}{\mathrm{d}r}\int_S \int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t \mathrm{d}v = \int_S \frac{\mathrm{d}}{\mathrm{d}r} \left(\int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t\right) \mathrm{d}v = \int_S \det(d(exp_p)_{rv})r^{n-1} \mathrm{d}v[/tex]

I can pass the derivative under the integral sign because S is compact and the inner integral is a continuous function of v. (Right?) So the second term is 0 also, unless n=1. Here's where I run into trouble. I'm not sure how to differentiate [itex]\det(\mathrm{d}(exp_p)_{rv})[/itex]. I do know that [itex]\mathrm{d}(exp_p)_{tv} \cdot tw = J(t)[/itex] where [itex]J(t)[/itex] is the Jacobi Field along [itex]\gamma_v[/itex] defined by [itex]J(0)=0[/itex] and [itex]J'(0)=w[/itex]. My guess is that the solution uses this along with some linear algebra trick to find the derivative.

Thanks in advance for your help!

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# Power Series for Volume of Balls in Riemannian Manifold

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