I have two questions (somewhat related) regarding local geometry on a SMOOTH, COMPACT Riemannian manifold. I still have a hard time getting a "good" understanding of local geometry.


Question 1:


It is true that there exists $\epsilon>0$ such that for all $r < \epsilon$, there exists $c\_g >0$ (indep. of $x\in M$) such that


$$ c\_g r^n \leq Vol\_g (B\_x(r)), \forall x \in M, $$


where $B\_x(r)$ is the geodesic ball centered at $x$. In other words, is it true that small geodesic ball are "comparable" to Euclidean balls. 


Question 2:


I am trying to compute an integral on a "small" geodesic ball, namely the following:


$$ I = \int\_{(exp\_{x\_0}(B\_{0}(R))} \rho(x\_0,x)^{2-n} dV\_g$$


Using normal coordinates at $x\_0$, we should have that


\begin{eqnarray\*}
 I & =& \int\_{(exp\_{x\_0}(B\_{0}(R))} \rho(x\_0,x)^{2-n} dV\_g. \\
& \leq & C\_g \int\_0^{R} \rho^{2-n} \rho^{n-1} (1 + O(\rho^2)) d\rho \\
\end{eqnarray\*}


I really need to get that last inequality but I am very unsure about it. I think it should be true, at least for small enough $R$ (hopefully for $R$ smaller than the $\epsilon$ defined in question 1. 


Is that possible? Any feedback would be appreciated.



Volume of geodesic balls
The answer to both questions is 'yes'.


To see this, just consider the exponential map $\exp:TM\to M$ and look at the pullback of the Riemannian volume form $dV$, say $\Omega = \exp^\*(dV)$ on $TM$. By the usual expansion in normal coordinates, there will be a smooth function $\phi$ on $TM$ that vanishes to order $2$ along the zero section such that, for each $x\in M$, the pullback of $\Omega$ to $T\_xM$, say $\iota\_x^\*\Omega$ satisfies $\iota\_x^\*\Omega = (1+\iota\_x^\*\phi) dV\_x$ where $dV\_x$ is the ordinary Euclidean volume form on $T\_xM$ (considered as a Euclidean vector space). This immediately proves what you want, since you can now use compactness to show that, when $\epsilon$ is sufficiently small, one has $|\phi(v)|\le C|v|^2$ for some constant $C>0$ and all $v\in TM$ with $|v|\le \epsilon$.


