It is well-known in geometric analysis that one can use curve-shortening flow to prove the isoperimetric inequality (where the general result requires curve-shortening flow for non-convex curves).

I was wondering if it might be possible to also prove such an inequality using Ricci flow (given hypotheses on convexity at the boundary and total curvature constraints if necessary).


1 Answer 1


Anthony Manning proved that the volume entropy decreases under volume normalized Ricci flow on surfaces of negative curvature. Question 4 at the end of his paper asks whether the Cheeger isoperimetric constant is a strictly increasing function of Ricci flow. So it looks like this is an open question.

There is a curious sort of topological isoperimetric inequality in three dimensions. Starting with a hyperbolic metric on an acylindrical 3-manifold with minimal boundary, one may run Ricci flow on the doubled manifold normalized by the minimum of the scalar curvature. Then this normalized volume is minimized by the hyperbolic metric with maximal area (totally geodesic) of the boundary. So the ratio of area to volume with sectional curvature at least $-1$ and minimal boundary is minimized in the totally geodesic case.

This is proved by monotonicity formulas of Hamilton and Perelman. Let $\lambda(g)$ be the minimal eigenvalue of the operator $-4\Delta_g + R$, where $R$ is the scalar curvature and $\Delta_g$ is the Laplacian. Define the quantity $$V_{\lambda}(g) = Vol(M,g)(-\frac16 \min\{\lambda(g),0\})^{\frac32}$$ (there is a similar formula holding in any dimension). Then $V_{\lambda}(g)$ is monotonically decreasing in dimension 3 for Ricci flow with surgery (the analogous quantity in dimension 2 will also be monotonic). One can actually see that all of the normalized eigenvalues are monotonic. In particular, if $R(g_0)$ is constant, then the first eigenvalue $\lambda_1(g_0)$ will be determined by $R(g_0)$ and the second eigenvalue of $-4\Delta_g+R$. For a surface of negative curvature, one also has monotonicity of $$V_R(g) = V(M,g) (-\frac16 \min\{R_{min}(g),0\})^{\frac32}.$$ As $t\to \infty$, $g_t$ will approach a constant curvature metric, so I think this should give a relation between the eigenvalue of the Laplacians for the initial and final metrics. In turn, the first eigenvalue of the Laplacian is related to the Cheeger constant by the Cheeger and Buser inequalities. But this is probably not the sort of relation you're looking for.

  • $\begingroup$ Hi Ian, that's interesting, thanks for this. For example, in this paper homepages.warwick.ac.uk/~maseq/topping_isoineq_ManuMath.pdf one proves an isoperimetric inequality on curved surfaces, I was wondering if it would be possible to use Ricci flow on surfaces to get the same result (probably with hypotheses if necessary). $\endgroup$ May 9, 2020 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.