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.

$\begingroup$ 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. $\endgroup$– YCorApr 2, 2019 at 9:24

$\begingroup$ Just out of curiosity, is there a connected Riemannian manifold with bounded sectional curvature, for which the convergence ($\#$) fails? $\endgroup$– YCorApr 2, 2019 at 9:27

$\begingroup$ Probably not. But I didn't want to be too daring. $\endgroup$– user130903Apr 2, 2019 at 9:37

$\begingroup$ What if $X$ is compact  say, a sphere? You obviously need some more conditions (e.g. negative curvature) for the question to make sense. $\endgroup$– Alex GavrilovApr 2, 2019 at 12:22

1$\begingroup$ @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. $\endgroup$– quaragueApr 2, 2019 at 12:41
1 Answer
The idea somehow being that volume growth is largest with constant negative curvature
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 $(n1)k$, 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 $\epsilon \to 0$.

$\begingroup$ Concerning your last statement, and related to it: how does the area of metric spheres grow in a complete manifold with bounded (in my case constant) curvature? Is it guaranteed to be at most exponential? Could you please give a reference? $\endgroup$– Alex M.Jun 23 at 5:36

$\begingroup$ A reference for the growth of the volume would be equally good, of course, if you know any. Thank you. $\endgroup$– Alex M.Jun 23 at 6:07

$\begingroup$ @AlexM.: Yes, it is at most exponential. In a space of constant negative curvature, see here for the volume, which you can easily check is a function that grows exponentially. $\endgroup$ Jun 24 at 5:38