Levy-Gromov Isoperimetric Inequality

Question

In his paper "Paul Levy's Isoperimetric Inequality", Gromov gives the following isoperimetric inequality:

Let $V$ be a closed $(n+1)$-dimensional Riemannian Manifold with $\mathrm{Ric}(V) \geq n \space (= \mathrm{Ric}(S^{n+1}))$. Let $V_0 \subset V$ be a domain with smooth boundary and let $B$ be a round ball in $S^{n+1}$ such that $$ \frac{ Vol(V_0)}{Vol(V)}= \frac{Vol(B)}{Vol(S^{n+1})}.$$ Then it follows that $$ \frac{Vol( \partial V_0)}{Vol(V)} \geq \frac{Vol(\partial B)}{Vol(S^{n+1})}. $$

Now my question: in a (slightly earlier) article 'Isoperimetric Inequalities In Riemannian Manifolds', Gromov states that the above inequality will still be true even if $V$ only admits a negative lower bound on its Ricci curvature. Does anyone have a reference for a proof of this, or is the statement obvious? It just seems to me that the hypothesis compares the curvature of $V$ to that of $S^{n+1}$, so allowing $\mathrm{Ric}(V)$ to be negative will obscure this.

Answer 1

I guess Gromov wanted to say that there is a lower bound for $\mathop{\rm vol}\partial V_0$ in terms of $\mathop{\rm vol} V_0/\mathop{\rm Vol} V$, $\mathop{\rm diam}V$ and lower bound for Ricci curvature. The same proof as in "Paul Levy's Isoperimetric Inequality", gives such a bound, but it is not longer sharp.

BTW, there is an analog of Levy--Gromov for open manifolds with $\mathop{\rm Ricc}\ge 0$. It is sharp and gives a lower bound for $\mathop{\rm vol}\partial V_0$ in terms of $\mathop{\rm vol} V_0$ and the volume growth of $V$, BUT as far as I know it is not written. (Please correct me if I am wrong.)

Answer 2

There's a paper of Berard Besson Gallot who generalize the Levy--Gromov result to have a diameter dependence as well as allowing for negative lower curvature bounds:

"Sur une inégalité isopérimétrique qui généralise celle de Paul Lévy-Gromov." Invent. Math 1985

http://www.ams.org/mathscinet-getitem?mr=788412

In particular, given a (negative/zero) bound on Ric, there is some C so that if $diam(M) \leq C$, then your desired inequality is true.

An alternative way to state this is that there is some $R = R(\min Ric, diam(M))$ so that your inequality is true if you use a sphere of radius $R$, instead of $1$.

The proof proceeds via the Heintze--Karcher inequality like Gromov, but then they make a more precise analysis of the lower bound.

Answer 3

I'm not sure about who did it first and how sharp the results are, but you can find isoperimetric inequalities for a negative lower bound on Ricci in a paper of Croke. My recollection is that Gallot also proved a similar isoperimetric inequality, but I can't seem to find the paper.

Answer 4

See Cavaletti-Mondino's paper for the most general version that also works for metric measure spaces satisfying Riemannian weal lower Ricci bounds!