# Volume of balls in homogeneous manifolds

Let $$X=G/H$$ be a homogeneous manifold, where $$G$$ and $$H$$ are connected Lie groups and assume there is given a $$G$$-invariant Riemannian metric on $$X$$. Let $$B(R)$$ be the closed ball of radius $$R>0$$ around the base point $$eH$$ and let $$b(R)$$ denote its volume. Is it rue that $$\lim_{\varepsilon\to 0}\ \limsup_{R\to\infty}\ \frac{b(R+\varepsilon)}{b(R)}=1?\qquad(\#)$$ The idea somehow being that volume growth is largest with constant negative curvature in which case it is exponential and thus satisfies our claim.

• I think it's known (to be checked, since these kind of statements have several variants, the first being maybe due to Pansu) that in case $G$ (and hence $X$) has polynomial growth, then for some $d$, $b(R)/R^d$ has a limit in $]0,\infty[$. In this case we already have $\lim b(R+\varepsilon)/b(R)=1$ for each fixed $\varepsilon$, which is not a formal consequence of having polynomially bounded growth.
– YCor
Apr 2, 2019 at 9:24
• Just out of curiosity, is there a connected Riemannian manifold with bounded sectional curvature, for which the convergence ($\#$) fails?
– YCor
Apr 2, 2019 at 9:27
• Probably not. But I didn't want to be too daring.
– user130903
Apr 2, 2019 at 9:37
• What if $X$ is compact - say, a sphere? You obviously need some more conditions (e.g. negative curvature) for the question to make sense. Apr 2, 2019 at 12:22
• @AlexGavrilov If $X$ is compact, then $b(R)=vol(X)$ for $R$ large enough, so the limsup is already $1$ for fixed $\varepsilon$. This is not the interesting case but I don't see a problem here. Apr 2, 2019 at 12:41

is essentially the content of the Bishop–Cheeger–Gromov comparison theorem. See for instance Lemma 36 of Peter Petersen's Riemannian Geometry, 2ed. It states, in his notation, that in any complete Riemannian manifold $$(M,g)$$ of dimension $$n$$ with Ricci curvature bounded below by $$(n-1)k$$, the function $$r \mapsto \frac{\operatorname{vol} B(p,r)}{v(n,k,r)}$$ is nonincreasing, where $$B(p,r)$$ is the ball in $$(M,g)$$ centered at $$p$$ with radius $$r$$, and $$v(n,k,r)$$ is the volume of the ball of radius $$r$$ in the space form of dimension $$n$$ and constant sectional curvature $$k$$. A homogeneous space certainly has bounded Ricci curvature, and the interesting case for us is when $$k$$ is negative, so that $$v(n,k,r)$$ is the volume in hyperbolic space, which as you say grows exponentially.
So this result tells us that $$\frac{\operatorname{vol} B(p,r)}{v(n,k,r)} \ge \frac{\operatorname{vol} B(p,r+\epsilon)}{v(n,k,r+\epsilon)}$$ or in other words $$1 \le \frac{\operatorname{vol} B(p,r+\epsilon)}{\operatorname{vol} B(p,r)} \le \frac{v(n,k,r+\epsilon)}{v(n,k,r)}.$$ As $$r \to \infty$$ the right side converges to something of the form $$e^{c \epsilon}$$, which in turn goes to $$1$$ as $$\epsilon \to 0$$.