Timeline for Does the Volume Ratio of a Geodesic Ball for a Complete Riemannian Manifold tend to the volume of a Unit Ball in Euclidean $n$-space?

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Mar 13, 2019 at 23:10	comment	added	Hollis Williams		Hi Ryan, I see that you can do a Taylor expansion of the metric to get the linear and quadratic terms, how does that imply the result for convergence to the Euclidean unit ball?
Mar 12, 2019 at 23:53	comment	added	Ryan Budney		Or stated another way, if you pull-back the Riemann metric along the exponential map $T_p M \to M$ you get a metric on $T_p M$. The 2nd order Taylor expansion of that metric, at the origin has only a constant and quadratic term. The constant term is just the Riemann metric of $M$ restricted to $T_p M$. The quadratic term is $\pm (1/3) <R(q,v)w,q>$ where $R$ is the curvature tensor, and $q$ is the point of $T_p$ where you evaluate the function. $v$ and $w$ are input vectors for your metric. The $\pm$ sign depends on your convention for $R$.
Mar 12, 2019 at 21:59	comment	added	Jeffrey Case		Note that he is taking the limit as $s\to0$. The basic idea is that in geodesic normal coordinates, $g_{ij}=\delta_{ij}+O(s^2)$ for $s$ small, and hence $V(p,s)=\omega_ns^n + O(s^{n+2})$. In your example of hyperbolic space, $V(p,s)=\omega_n\int_0^s\sinh^{n-1}t\,dt$ and the claimed limit follows from the fact $s^{-1}\sinh s\to1$ as $s\to0$.
Mar 12, 2019 at 21:29	history	asked	Hollis Williams	CC BY-SA 4.0