Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Background:

Let $M$ be a Riemannian manifold. Let $p \in M$ and $\epsilon \gt 0$.

For sufficiently small $\epsilon$, the standard definition (correct me if I'm wrong) for the 'geodesic ball of radius $\epsilon$ centered at $p$' is:

$$B = \{ \gamma(1) \mid \gamma \text{ is a geodesic}, \gamma(0)=p, \text{ and } \langle \gamma'(0), \gamma'(0) \rangle \, \lt \epsilon \} .$$

I am wondering: suppose we instead consider the set

$$B_{\text{alt}} = \{ \gamma(\epsilon) \mid \gamma \text{ is a geodesic}, \gamma(0)=p, \text{ and } \langle \gamma'(0), \gamma'(0) \rangle \lt 1 \} .$$

(Roughly speaking, $B$ takes an $\epsilon$-ball in $T_pM$ and runs the geodesics forward by 1, while $B_{\text{alt}}$ takes the unit ball in $T_pM$ and runs the geodesics forward by $\epsilon$. $\langle \cdot , \cdot \rangle$ is the metric inner product.)

Question:

Do we have $B = B_{\text{alt}}$? If not, is there a nice counter-example? Or an intuitive reason why we should not expect the equality to hold?

Note:

I have tried to compute the volume of $B_{\text{alt}}$ (to next-to-leading order in $\epsilon$) in the special case of a two-dimensional manifold with a diagonal metric. Despite many careful checks, my answer does not match the standard answer for the volume of a geodesic ball -- so I suspect that $B \ne B_{\text{alt}}$.

Thanks for any help!

Appendix - calculation of $Vol(B_{alt})$ (added 12/23)

I'm going to redo my calculation of the volume of $B_{alt}$, taking account of joriki's comment below. The problem I was having (which I don't think is resolved by joriki's suggestion to replace $\epsilon$ by $\epsilon^{1/2}$, but will have to check to be sure) is described in the next two paragraphs.

I'm doing the calculation in a particular coordinate chart, assuming the metric is diagonal: $ds^2 = g_{11}(x^1,x^2) (dx^1)^2 + g_{22}(x^1,x^2) (dx^2)^2$. I find that $Vol(B_{alt})$ includes (at next-to-leading order in $\epsilon$) the terms $(\partial_1)^2 g_{11}|_p$ and $(\partial_2)^2 g_{22}|_p$. In a diagonal metric (and using the Levi-Civita connection), the only Christoffel symbol in which $\partial_1 g_{11}$ appears is $\Gamma^1_{11} = \frac{1}{2} g^{11} \partial_1 g_{11}$. However, the formula $R^a_{bcd} = \partial_c \Gamma^a_{db} - \partial_d \Gamma^a_{cb} + \Gamma^a_{ce} \Gamma^e_{db} - \Gamma^a_{de} \Gamma^e_{cb}$ shows that $\partial_1 \Gamma^1_{11}$ doesn't appear in the scalar curvature - the first two terms in the preceeding formula just cancel when all indices are equal to 1.

Now I will explain how these bad terms appear in my calculation. My approach is to calculate $Vol(B_{alt}) = \int_{B_{alt}} dx^1 dx^2 \sqrt{g_{11}(x^1,x^2) g_{22}(x^1,x^2)}$ by Taylor-expanding the integrand about the point $p \leftrightarrow (x^1_0,x^2_0)$. Given $a\ \partial_1 |_p + b\ \partial_2 |_p$ in the unit ball in $T_pM$, the corresponding point in $B_{alt}$ is $(x^1_0 + a\ \epsilon + C(a,b)\ \epsilon^2 + O(\epsilon^3), x^2_0 + b\ \epsilon + D(a,b)\ \epsilon^2 + O(\epsilon^3))$, where $C,D$ are constants we can get from the geodesic equation.(**) There's a Jacobian factor to change variables from $(x^1, x^2)$ to $(a,b)$. But the only way (it seems to me) to get the second derivatives of the metric that appear in $R$ is to Taylor-expand $\sqrt{det\quad g}$ to second order. That's exactly the order at which the bad terms $(\partial_1)^2 g_{11}|_p$ and $(\partial_2)^2 g_{22}|_p$ appear (even if we replace $\epsilon$ by $\epsilon^{1/2}$).

(**) $C(a,b) = -\frac{1}{2}(a^2 \Gamma^1_{11} + b^2 \Gamma^1_{22} + a\ b\ \Gamma^1_{12})|_p$.

Thank you Srivatsan for your TeX edits... I am still fairly new to TeX.

share|improve this question

1 Answer 1

You need to either use $\epsilon^2$ in the first definition or use $\sqrt\epsilon$ in the second. You can see from the geodesic equation that rescaling the parameter of a geodesic yields another geodesic, with the tangent vector at $0$ correspondingly scaled. Since the first definition requires the square of the tangent vector to be $\lt\epsilon$, you'd need to rescale by $\sqrt\epsilon$ to transform this into $\lt1$.

share|improve this answer
    
Thanks for your reply. With this correction, do you think $B = B_{alt}$? I'm still not sure, for the following reason. A standard result I have seen is that the volume of a geodesic ball in two dimensions is $\pi \epsilon^2 [1 - (R / 48) \epsilon ^2 + O(\epsilon^4) ]$, where $R$ is the scalar curvature. I calculated the volume using the second definition ($B_{alt}$) in the special case of a diagonal metric and found terms that could not possibly occur in the scalar curvature. Changing $\epsilon$ to $\epsilon^{1/2}$ in my calculation does not make it match the standard result. –  marlow Dec 23 '11 at 2:09
    
@marlow: Wikipedia has the factor $12$ where you have $48$, and this is also what I get for the surface of a small spherical cap (which is also given here). In case this still doesn't resolve the discrepancy, perhaps you could write out the problematic result that you're getting? –  joriki Dec 23 '11 at 2:43
    
I will look into the numerical factor. Unfortunately that doesn't account for the discrepancy. I added an appendix to my original post to explain the problematic terms I am finding in my calculation of $Vol(B_{alt})$. –  marlow Dec 23 '11 at 6:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.