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.)


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?


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.


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 $\lt1$.

  • $\begingroup$ 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. $\endgroup$
    – marlow
    Commented Dec 23, 2011 at 2:09
  • $\begingroup$ @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? $\endgroup$
    – joriki
    Commented Dec 23, 2011 at 2:43
  • $\begingroup$ 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 $Vol(B_{alt})$. $\endgroup$
    – marlow
    Commented Dec 23, 2011 at 6:19

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