2
$\begingroup$

In the paper

The volume of a small geodesic ball in a Riemannian manifold, Michigan Mathematical Journal 20 (1973), doi: 10.1307/mmj/1029001150

A. Gray proved the following result:

Let $M$ be a Riemannian manifold of dimension $n$ with positive scalar Ricci curvature at $x \in M$. If $R$ is a sufficiently small real number, calling $B(x, \, R)$ a geodesic ball of radius $R$ centered at $x$, we have $$\mathrm{vol} \,(B(x, \, R)) \leq C R^n,$$ where $C$ is a constant depending only on $n$.

My question involves balls of sufficiently big radius, instead of sufficiently small one. In other words, let me as what follows:

Let $M \subset \mathbb{R}^N$ be an (unbounded) embedded submanifold of dimension $n$. Under which conditions on $M$ and $x \in M$ it is possible to find constants $C$ and $\gamma$ such that $$\mathrm{vol} \,(B(x, \, R)) \leq C R^{\gamma}$$ for all $R >>0$?

I am not an expert on differential geometry, so I apologize in advance if my formulation of the problem is not optimal, or if the answer turns out to be somehow trivial.

Any reference to the relevant literature will be greatly appreciated.

Improve this question
$\endgroup$
4
  • 6
    $\begingroup$ Since volume is an intrinsic geometric invariant, the fact that $M$ is a submanifold is irrelevant. If you assume that Ricci curvature is nonnegative, then the inequality follows from the Bishop-Gromov inequality (in fact, just the Bishop inequality). See en.wikipedia.org/wiki/Bishop%E2%80%93Gromov_inequality $\endgroup$
    – Deane Yang
    Commented Oct 25, 2017 at 18:27
  • $\begingroup$ A more general condition is given here: mathoverflow.net/questions/283748/… . Even this condition can probably be weakened more, but there has to be an assumption of "not too much negative Ricci curvature". $\endgroup$
    – Deane Yang
    Commented Oct 26, 2017 at 14:19
  • 1
    $\begingroup$ @Deane Yang: Thank you for the precious information. May I suggest that you expand your comments into an answer, so that the question will not look unanswered anymore? $\endgroup$
    – Francesco Polizzi
    Commented Oct 26, 2017 at 14:24
  • 1
    $\begingroup$ OK. Answer below. $\endgroup$
    – Deane Yang
    Commented Oct 26, 2017 at 17:02

1 Answer 1

Reset to default
3
$\begingroup$

Since volume is an intrinsic geometric invariant, the fact that $M$ is a submanifold is irrelevant. If you assume that Ricci curvature is nonnegative, then the inequality follows from the Bishop-Gromov inequality (in fact, just the Bishop inequality).

Also, a more general condition is given in another MathOverflow question. Even this condition can probably be weakened more, but there has to be some kind of assumption saying "not too much negative Ricci curvature".

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .