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In $1969$, Margulis proved, for suitable constant $h>0$and $r$ is a positive constant that :

$a(p)=\lim_{r\to \infty} \frac{VolS(p,r)}{e^{h r}}$ with ($(S(p,r)$ is geodesic spheres), exists at each point $p$ in manifolds of negative curvature which it is the main result implies purely exponential growth of volume of geodesic spheres.Really I'am curiouse to know if this limit does exist for Manifolds with positive curvature ,we may look to an example for application this in unit tangent bundle of $S^4$ and probably cohomology $CP^3$ which they admit Riemannian metrics with positive sectional curvature almost everywhere ? and what about continuity of $a(p)$ in this case ?

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  • $\begingroup$ What is $S(p,r)$? What is $\operatorname{exp}(hr)$? $\endgroup$
    – Ben McKay
    Commented Apr 12, 2020 at 14:43
  • $\begingroup$ I meant Volume of spheres $\endgroup$
    – zeraoulia rafik
    Commented Apr 12, 2020 at 14:45
  • $\begingroup$ I think volume of balls is governed by Ricci, so should be no problem. Try Gromov, Sign and geometric meaning of curvature. $\endgroup$
    – Ben McKay
    Commented Apr 12, 2020 at 16:40
  • $\begingroup$ @BenMcKay Surely you mean the Bishop-Gromov comparison theorem? The denominator usually is the analogous volume in a comparison space. So I believe that the denominator $e^{hr}$ will produce less significant results ($0$ whenever $\operatorname{Ric}\ge 0$ if I am not mistaken). And am I right that $r$ is not meant to be constant? $\endgroup$
    – Sebastian Goette
    Commented Apr 12, 2020 at 16:45

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When $M$ has positive curvature, the limit should always be 0. Indeed if $M$ has non-negative Ricci curvature, Bishop-Gromov tells you that $\frac{Vol S(p, r)}{n\omega_nr^{n-1}}\leq 1$, so the volumes grow at most polynomially.

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Bishop's inequality: in any complete Riemannian manifold of Ricci nonnegative curvature, if balls around a point have radii $r$ and $r'=\lambda r$, then their volumes have $\lambda^n V \ge V'$. This is for balls, not for spheres, but I think it is what you are looking for.

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