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Let $M$ be a smooth Riemannian manifold of dimension $n$ and scalar curvature $R$ (with respect to the Levi-Civita connection). Let $m \in M$ and let $B$ be the geodesic ball of radius $r$ centered at $m$. That is,

$B = \{ Exp_m(v)\ |\ \ v\in T_mM,\ ||v||< r\ \} $

Then for small $r$, the volume of $B$ is:

\begin{equation} Vol(B) = (constant)\ r^n\ [1 - \frac{R}{6(n+2)}r^2 + O(r^4)\ ] \end{equation}

where the constant depends only on $n$, and $R$ is evaluated at $m$. So $R$ basically tells us the difference between the volume of a small ball in $M$ and the volume of a small ball in Euclidean $\mathbb{R}^n$ (to leading order in the radius $r$).


1) Is there any generalization of this to the case of a semi-Riemannian manifold?

2) If not, is there an analogous result for the case of a Lorentzian manifold (that is, a semi-Riemannian manifold whose metric signature has $n-1$ pluses and one minus)?

3) If not, is there some other result that gives a nice way to interpret the scalar curvature $R$ in a semi-Riemannian (or Lorenztian) manifold?


The problem I see is this: to define a geodesic ball, we want to start with a ball "of radius $r$" in $T_mM$. But the metric is indefinite, so there is no norm. I looked in Barrett O'Niell's book "Semi-Riemannian Geometry" but did not find the answer there.

Thanks in advance for your help!

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In the $(k+1)$ hyperbolic space $H^{k+1}$ we have that $\mathrm{vol}(B_r)=\pi^{k/2}\Gamma(k/2)e^{kr}/k!+O(re^{(k-2)r})$. That helps? – emiliocba Jan 22 '12 at 6:02
@emiliocba I'm afraid I don't see it. A hyperbolic space is an example of a Riemannian manifold, isn't it? The scalar curvature may be negative, but the metric is still positive definite. – marlow Jan 22 '12 at 16:36

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