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I am reading Peter Topping's notes on Ricci flow: on page 99 a statement is made which is needed for his proof of a version of Perelman's no local volume collapse theorem, but I am not sure why it holds. If you define $\omega_n$ to be the volume of the unit ball in Euclidean $n$-space and $V(p,s)$ to be the volume of the geodesic ball at point $p$ with radius $s$ given a complete Riemannian manifold $(M,g)$, then he argues that the volume ratio always tends to the volume of the Euclidean unit ball. $g(t)$ is a Ricci flow on the manifold for $t \in [0,T]$ and we are working with a smooth metric $g(T)$.

$K(p,s)=\frac{V(p,s)}{s^{n}} \rightarrow \omega_n$

as $s \rightarrow 0$. I am not sure why this would have to hold on an arbitrary complete Riemannian manifold. If you are working on hyperbolic space or something like that with negative curvature, then now surely that relation will not hold as you will have a $\sinh$ term in the denominator meaning that the ratio would tend to $0$, as opposed to the volume of the Euclidean unit ball.

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    $\begingroup$ 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 $s\to0$. $\endgroup$
    – Jeffrey Case
    Commented Mar 12, 2019 at 21:59
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    $\begingroup$ 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 $R$. $\endgroup$
    – Ryan Budney
    Commented Mar 12, 2019 at 23:53
  • $\begingroup$ 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? $\endgroup$
    – Hollis Williams
    Commented Mar 13, 2019 at 23:10

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