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In a couple articles I've read lately, I've seen it mentioned that the cigar soliton has linear volume growth. What does this mean? I thought maybe, if you compute the volume of geodesic balls and increase their radius, their volumes grow linearly, but using the metric for the cigar in polar coordinates ($g = \dfrac{dr^2 + r^2 d\theta^2}{1+r^2}$), one sees that the volume of a ball of radius $r$ is given by: $$\int_{0}^{2\pi} \int_{0}^{R} \frac{r}{1+r^2} dr \ d\theta = \pi \log(R^2 + 1)$$

where I used that the matrix of $g$ is given by $$ \left(\begin{array}{cc}{\frac{1}{1+r^2}} & {0} \\ {0} & {\frac{r^2}{1+r^2}}\end{array}\right) $$

so it would appear that this is not what's meant. What's going on then? What does it mean for the cigar soliton to have linear volume growth?

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    $\begingroup$ Volume grows logarthimically in the coordinate $r$, but $r$ is not the distance to the tip. $\endgroup$
    – RBega2
    Commented Oct 1, 2019 at 2:33
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    $\begingroup$ If you change variables to $ds^2=dr^2/(1+r^2)$, then the volume of $R$-ball can be computed by integrating $s$ over $[0,R]$ in the metric of the form $ds^2+f(s)^2d\theta^2$ where you need to compute $f$. In general, the volume of an $R$-ball in a complete open $n$-manifold of nonnegative Ricci curvature is between $cR$ and $CR^n$ for some positive constants $c, C$. The cigar is asymptotic to a cylinder so its volume grows linearly. $\endgroup$
    – Igor Belegradek
    Commented Oct 1, 2019 at 2:56
  • $\begingroup$ @RBega2 what's the distance to the tip then? $\endgroup$
    – Matheus Andrade
    Commented Oct 1, 2019 at 17:15
  • $\begingroup$ @MatheusAndrade It's a pretty straightforward integral. In fact, it's $d(p)=log(r(p)+\sqrt{1+r(p)^2})$ so is asymptotically a logarithm in $r$. Since volume and distance both grow like $log(r)$ that means the volume growth is linear. This is really more appropriate for math.stackexchange. $\endgroup$
    – RBega2
    Commented Oct 1, 2019 at 17:37
  • $\begingroup$ @RBega2 that's true! Thanks for the patience, my doubts are all solved now. If you want to post a short answer I'll accept it. $\endgroup$
    – Matheus Andrade
    Commented Oct 1, 2019 at 17:50

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