Uniformly quasi-isometric patches over classes of Riemannian manifolds Suppose $(M^d,g)$ is a closed, connected Riemannian manifold.
Is there a constant $R > 0$ such that for all $z \in (M,g)$, for all $x \in B_R(z)$, \begin{equation} \frac{1}{2} \lVert \xi \rVert_{\mathbb{R}^d}^2
\leq \sum_{ij} g^{ij}(x) \xi_i \xi_j 
\leq 2 \lVert \xi \rVert_{\mathbb{R}^d}^2 \end{equation}
 for all $\xi \in \mathbb{R}^d$, where $g^{ij}(\cdot)$ is expressed in exponential coordinates centered at $z$?
My real question is this: can a constant $R>0$ be found satisfying the same inequalities
for all closed, connected $d$-dimensional Riemannian manifolds whose sectional curvatures are bounded both below and above?
 A: I haven't done much with the metric tensor, but you can resolve this problem by compactness. Either the $g^{ij}$ approach $\delta_{ij}$ (actually, just have to approach values between 1/2 and 2 on the diagonal and be sufficiently small off the diagonal) as you approach $z$ or they don't.


*

*If they do, then around each $z$ pick an $R$ sufficiently small that the inequalities hold at $z$ and in the ball of radius $R$, then cover the space by all balls at different $z$'s with radius $R/2$. Choose a finite subcover (centered at $z_1,...,z_n$ with radii $R_1/2,...,R_n/2$), and let $R'$ be the smallest $R_i/2$. Then $R'$ works, since every ball of radius $R'$ lies entirely in one of the balls around $z_i$ with radius $R_i$, so the inequalities hold.

*If the tensor doesn't approach the identity near enough as you approach $z$, then $x=z$ is a counterexample.


Edit: I researched it more; Since you are using exponential coordinates case 1 holds. (The metric tensor approaches the identity, according to https://physics.stackexchange.com/questions/20781/are-there-any-clear-and-expressive-plainword-sense-of-metric-tensor-components).
