scalar curvature

Question

I am studying scalar curvature. It is the trace of the Ricci operator. I read that its geometric meaning follows from this formula $\frac{Vol_M(B(p,t))}{Vol_{\mathbb{R}^n}(B(o,t))}=1-\frac{1}{6(n+2)}k(p)t^2+o(t^2)$ where $k$ is the scalar curvature. From this formula one knows that if the scalar curvature is positive the geodesic balls with small radius have smaller volume than that of the same ball in $\mathbb{R}^n$. Unfortunately, I don't know the meaning of "volume of the geodesic ball" (i.e. $Vol_M(B(p,t))$) and I cannot find it. Can someone help? Thanks in advance for the help and I apology for my English.

Answer 1

Remember that on a complete Riemannian manifold $M$, geodesic length defines a distance metric: $$d(x,y) = \min\{\mbox{length of geodesics connecting $x$ and $y$}\}.$$ So the geodesic ball $B(p,t)$ is just the ball about $p$ of radius $t$ in this distance metric. It is the same as $\exp(B(0,t))$, where $B(0,t)\subset T_pM$.

The volume of the geodesic ball is its volume as a Borel subset of $M$, i.e., $$Vol_M(B(p,t)) = \int_{B(p,t)}dVol_M,$$ where $dVol_M$ is the volume form on $M$ induced by the Riemannian metric. (In local coordinates, if the metric is $g_{ij}$, $dVol_M = \sqrt{\det(g_{ij})} dx_1\wedge\cdots\wedge dx_n$.)