Volume of geodesic balls

Question

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.

Answer 1

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 $|v|\le \epsilon$.