# Does the Volume Ratio in a Complete Riemannian Manifold always tend to the Volume of the Euclidean Unit Ball?

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.

Let $$B^n$$ denote the unit ball in $$T_pM$$. For $$s>0$$ small, let $$\varphi_s:B^n\hookrightarrow M$$ be given by $$v\mapsto\exp_p(sv).$$ Let $$g_s$$ be the pull-back metric on $$B^n$$ with respect to $$\varphi_s$$, that is, $$g_s=\varphi_s^*g.$$ Then, by definition, the volume $$V(p,s)$$ is equal to the volume of $$B^n$$ with respect to the metric $$g_s$$. It follows that the volume ratio, $$K(p,s)$$, is equal to the volume of $$B^n$$ with respect to the metric $$h_s:=\frac{g_s}{s^2}.$$ A short computation shows that, as $$s$$ goes to $$0$$, the metric $$h_s$$ converges (uniformly) to the constant metric $$(h_0)_q(u,v)=g_p(u,v),\qquad q\in B^n,u,v\in T_pM.$$ Hence, the corresponding Riemannian volume forms converge to the standard volume form and the claim follows.

Edit: Let us write the metric $$g_s$$ explicitly: for $$q\in B^n$$ and $$u,v\in T_pM$$ we have \begin{align}(g_s)_q(u,v)&=g_{\exp_p(sq)}(d(\exp_p)_{sq}(su),d(\exp_p)_{sq}(sv))\\&=s^2g_{\exp_p(sq)}(d(\exp_p)_{sq}(u),d(\exp_p)_{sq}(v)).\end{align} The metric $$h_s$$ is thus given by $$(h_s)_q(u,v)=g_{\exp_p(sq)}(d(\exp_p)_{sq}(u),d(\exp_p)_{sq}(v)).$$By continuity of $$g$$, we have \begin{align}\lim_{s\to0}(h_s)_q(u,v)&=g_{\exp_p(0)}(d(\exp_p)_0(u),d(\exp_p)_0(v))\\&=g_p(u,v),\end{align} where the convergence is uniform.

• Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
– Tom
Mar 13, 2019 at 23:45
• Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
– Tom
Mar 14, 2019 at 0:16
• @Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $\alpha$, the Riemannian volume form is multiplied by $\alpha^{n/2}$. Mar 14, 2019 at 6:22
• @Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $\varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice. Mar 14, 2019 at 7:21
• @Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out. Mar 14, 2019 at 8:22