Let us consider a complete Riemannian manifold $M$ of dimension $n$ with $Ric \geq 0$. Then the Bishop-Gromov volume comparison theorem says that for any $p \in M$, the function $$ \frac{\text{Vol}(B(p, r))}{r^n}$$ is monotonically decreasing on $r \in (0, \infty)$. My question is, how global is this result?

More concretely, let's say that we perturb $M$ so that $M \setminus K$ now has $Ric \geq 0$, where $K$ is compact. Does the same conclusion hold here? To be more specific, consider the case where $p \in K$. Then do we still have that $$ \frac{\text{Vol}(B(p, r))}{r^n}$$ is monotonically decreasing for large enough $r$? Thanks in advance for any insight!

  • $\begingroup$ What does "here" refer to exactly? $\endgroup$ Oct 18, 2017 at 8:53
  • $\begingroup$ @MikhailKatz By "here", I wanted to mean complete $M$ with compactly supported negative Ricci curvature. Edited the question. $\endgroup$
    – user116108
    Oct 18, 2017 at 8:54
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    $\begingroup$ You should first ask your question for sectional curvature, where I think it is pretty clear that the answer is negative: when a ball encounters a region where the curvature bound does not hold, the volume behavior can change drastically. That's my impression though I am not 100% sure. $\endgroup$ Oct 18, 2017 at 8:58
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    $\begingroup$ I think this holds but haven't worked out the details. Bishop-Gromov is proved using the Sturm comparison theorem, where the volume form along a geodesic is compared to that of a flat metric. If you simply replace the flat volume form by the solution to $u'' + ku = 0$, where $k$ is a function of distance from the center of a geodesic ball, and becomes $0$ outside some distance $R$, then you get the desired conclusion. $\endgroup$
    – Deane Yang
    Oct 18, 2017 at 13:41

1 Answer 1


If $K\subset B(x,R)$ then $$ \frac{\text{Vol}(B(p, r))}{(r-R)^n}$$ is monotonic for $r>R$.

The proof is the same and it is about maximum one could expect.


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