Let $A=\lim\_{r \rightarrow +\infty} \frac{Vol(B(o,r))}{\omega\_{n} r^{n}}$ for any Riemannian manifold $(\mathbb{M}^{n},g)$ with nonnegative Ricci curvature. Here $\omega\_{n}$ is the volume of unit ball in $\mathbb{R}^n$. We call $A$ the cone angle at infinity or asymptotic volume ratio, and the manifold is cone-like if $A>0$. I got a direct question with this definition. Let $(\mathbb{M}^{n},g)$ be a metric product of two manifolds $(\mathbb{M}\_1^{k},g\_1,A\_1)$ and $(\mathbb{M}\_2^{n-k},g,A\_2)$ both with nonnegative curvature and cone-like. Does the cone angle of the product manifold only depends on $k$, $n$, $A\_1$ and $A\_2$. It sounds like the new cone angle will be some average of $A\_1$ and $A\_2$, but I failed to get a clean formula even assuming product of two rotationally symmetric cones in $R^3$ (like surface of evolution). Is there something simple I miss? Thanks!



cone angle at infinity for product of cones
I claim that $A=A\_1A\_2$.


Indeed, the distance in the product is given by the standard Pythagorean formula, therefore
$$
 B(o,r) = \bigcup\_{x\in B\_1(o\_1,r)} B\_2(o\_2,\sqrt{r^2-|o\_1x|^2})
$$
where $B$, $B\_1$ and $B\_2$ denote metric balls in $M$, $M\_1$ and $M\_2$, $o=(o\_1,o\_2)$ and $|o\_1x|$ denotes the Riemannian distance from $o\_1$ to $x$. Hence by Fubini
$$
 Vol(B(o,r)) = \int\_{B\_1(o\_1,r)} Vol\_2(B\_2(o\_2,\sqrt{r^2-|o\_1x|^2})) \ dVol\_1(x)
$$
where $Vol\_1$ and $Vol\_2$ are the volume forms in $M\_1$ and $M\_2$.
Using the co-area formula, this can be rewritten as
$$
 Vol(B(o,r)) = \int\_0^r dt \int\_{S(t)} Vol\_2(B\_2(o\_2,\sqrt{r^2-t^2})) \ dVol^{k-1}\_1(x) .
$$
where $S(t)$ denotes the sphere of radius $t$ centered at $o\_1$ in $M\_1$ and $Vol\_1^{n-1}$ its $(k-1)$-dimensional volume form. For most values of $t$, the $Vol\_2(...)$ under the integral equals $A\_2$ times the similar Euclidean volume, with a small relative error. Hence
$$
 Vol(B(o,r)) \sim A\_2 \int\_0^r dt \int\_{S(t)} \omega\_{n-k}(r^2-t^2)^{(n-k)/2} \ dVol^{k-1}\_1(x) 
$$
(you need Bishop-Gromov inequality here, to estimate the volumes that are not among those "most"). The right-hand side equals to
$$
A\_2 \int\_0^r \omega\_{n-k}(r^2-t^2)^{(n-k)/2} \ Vol^{k-1}\_1(S(t)) \ dt .
$$
The growth of volumes of spheres in $M\_1$ is the same as that of balls, i.e.
$$
 Vol^{k-1}\_1(S(t)) \sim k\omega\_k t^{k-1} .
$$
Therefore
$$
Vol(B(o,r)) \sim A\_1A\_2 \int\_0^r \omega\_{n-k}(r^2-t^2)^{(n-k)/2} k\omega\_k t^{k-1} \ dt .
$$
By the same argument applied to $\mathbb R^n$, the latter integral equals the volume of the Euclidean $r$-ball, hence the result.


