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.



Volume of balls in homogeneous manifolds

> 
> 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 $(n-1)k$, 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 $\epsilon \to 0$.


