Almost volume preserving charts for a riemannian manifold

Question

Let $M$ be riemannian manifold. Is it true that for all $p\in M$ and $\varepsilon>0$ there is a chart $\phi:U\rightarrow\mathbb{R^n}$ arround $p$ such that for all nonempty open $V\subseteq U$ $\frac{vol(V)}{vol(\phi (V))}\in (1-\varepsilon,1+\varepsilon)$ ?

Answer 1

Yes, this is true (in fact, such an estimate holds for all Borel measurable $V$).

Fix $p$ and let $\phi: U \to \mathbb{R}^n$ give Riemannian normal coordinates at $p$. For $V \subseteq U$ Borel measurable, the volume $\text{vol}_M(V)$ is given by $$ \text{vol}_M(V) = \int_{\phi(V)} \sqrt{\det(g(x))} d \mu(x), $$ where $d\mu$ is Lebesgue measure, and $g(x)$ is the matrix representing the metric tensor in $\phi$-coordnates (i.e., $g_{ij}(x) = \langle \psi^*(\partial_i), \psi^*(\partial_j) \rangle_{\psi^{-1}(x)}$).

At $0 = \phi(p)$, the matrix $g_{ij}$ is just the identity matrix (because of our choice of $\phi$), and the functions $g_{ij}$ vary smoothly on $\phi(V)$, so, by shrinking $U$ if necessary, we can assume that, for all $x \in U$, we have $$ 1 - \epsilon \leq \sqrt{\det(g(x))} \leq 1 + \epsilon. $$ Your estimate follows immediately.

In this context it seems reasonable also to mention the formula given on this Wikipedia page relating the Riemannian and Lebesgue volume forms in normal coordinates -- the error is quadratic in the coordinate $x$ and determined by the Ricci tensor.