I am wondering if there exists a complete Riemannian manifold (noncompact) such that $$\lim\limits_{r\rightarrow \infty}\frac{vol(B(x,r))}{e^{r^3}}=\infty$$

where the denominator denotes the volume of a geodesic ball around $x$.

Does anyone know an example?

Best wishes


1 Answer 1


You can construct metrics with essentially any volume growth you like. Consider

$$ g = dr^2 + f^2(r) d\theta^2 $$ in polar coordinates - so long as $f$ is of the form $f(r) = r + o(r)$ as $r \to 0$, this defines an honest metric on $\mathbb R^2$. Such a metric has volume form $ f(r) dr \wedge d \theta,$ and balls $B(0,R) = \{ r< R \}$,

and thus $$ {\rm Vol}(B(0,R)) = 2 \pi \int_0^R f.$$

Thus (away from $r=0$) any smooth increasing function is ${\rm Vol\ }B(R)$ for some metric. These metrics are all complete, since any curve escaping to infinity has length at least $\int_{r_0}^\infty dr = \infty.$

  • $\begingroup$ Nice answer. But the integral should be from $0$ to $R$, not $0$ to $\infty$. $\endgroup$
    – Jack Lee
    Dec 21, 2016 at 16:06
  • $\begingroup$ Thanks for the correction Jack. $\endgroup$ Dec 22, 2016 at 6:35
  • $\begingroup$ Is it correct that any curve escaping to infinity as length at least $\int_{r_{0}}^{\infty}{1\times dr}$ because the metric is given by $g=1\times dr^{2}+...$? $\endgroup$
    – Shaq155
    Nov 11, 2021 at 15:56
  • 1
    $\begingroup$ @Shaq155: if you parametrize the curve as $r(t), \theta(t)$, then its length over any interval $(t_0, t_1)$ is $\int_{t_0}^{t_1} \sqrt{ r'(t)^2 + f(r(t))^2 \theta'(t)^2 } dt \ge \int |r'(t)| dt \ge \int r'(t) dt = r(t_1) - r(t_0).$ If the curve escapes to infinity then we have $r(t_1) \to \infty$ and thus the length is infinite. $\endgroup$ Nov 12, 2021 at 0:41
  • $\begingroup$ @Anthony Carpetis: Thank you! $\endgroup$
    – Shaq155
    Nov 12, 2021 at 9:37

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