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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!

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up vote 5 down vote accepted

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.

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Thanks for your patience! It is very helpful! –  Bo_Y Jul 29 '12 at 3:30
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