## Notes of Gilles Courtois’ lecture

Poincaré inequalities and Ricci curvature

Joint work with G. Besson and S. Hersonsky.

1. Poincaré inequalities

1.1. Definition

Definition 1 Say a metric measure space ${(X,d,\mu)}$ satisfies a Poincaré inequality if there exist constants ${C}$, ${C'}$, ${\sigma\geq 1}$ such that for all ${R>0}$, all functions ${u:X\rightarrow R}$ and all balls,

$\displaystyle \begin{array}{rcl} \int_{B(R)}|u-u_{B(R)}|^{\sigma}\leq C\,R^{\sigma}\int_{B(C'R)}|\nabla u|^{\sigma}. \end{array}$

Here, ${u_B}$ is an average. Tthe length of the gradient is defined as an upper gradient, i.e. the least function ${\rho}$ such that for all ${x}$, ${y\in X}$ and any rectifiable path ${\gamma}$ from ${x}$ to ${y}$,

$\displaystyle \begin{array}{rcl} |u(x)-u(y)|\leq\int_{\gamma}\rho . \end{array}$

Example 1 ${{\mathbb R}^n}$, ${{\mathbb Z}^n}$; Riemannian manifolds with nonnegative Ricci curvature, nilpotent Lie groups satisfy Poincaré inequalities. Hyperbolic space, trees do not.

1.2. What are they good for, I

Colding and Minicozzi’s proof of a conjecture of Yau stating that when ${Ricci\geq 0}$, the space of harmonic functions of polynomial growth is finite dimensional goes as follows.

Pick ${R}$ very large, ${\epsilon\in(0,1)}$. Cover optimally ${B(R)}$ with ${N}$ balls of radius ${\epsilon R}$. Then ${N\leq (\frac{1+\epsilon}{\epsilon})^n}$ by Bishop-Gromov. Let ${V}$ be a finite dimensional vector space of harmonic functions of growth ${\leq R^d}$. Estimate the dimension of ${V}$ as follows. Consider the map ${\Phi_R :V\rightarrow{\mathbb R}^N}$, mapping ${f}$ to its averages on ${\epsilon R}$ balls. One can choose ${\epsilon}$ such that for any ${V}$, there exists ${R}$ such that ${\Phi_R}$ is injective. Indeed, let ${f}$ be in the kernel. Then

$\displaystyle \begin{array}{rcl} \int_{B(R)}f^2 \leq \sum\int_{B_j}f^2 \leq C\,(\epsilon R)^2 \sum\int_{C'B_j}|\nabla f|^2 \leq \mathrm{const.}(\epsilon R)^2\int_{B(C'R)}|\nabla f|^2. \end{array}$

Since ${f}$ is harmonic, the reverse Poincaré inequality holds,

$\displaystyle \begin{array}{rcl} \int_{B(C'R)}|\nabla f|^2\leq \frac{\mathrm{const.}}{R^2}\int_{B(C''R)}f^2. \end{array}$

(this does not require any curvature assumption). Combining these inequalities shows that ${\int_{B(R)}f^2}$ grows fast, which contradicts the assumption that ${f}$ has polynomial growth, unless ${f=0}$.

Question. Is the above property (finitely many harmonic functions of polynomial growth) a quasi-isometry invariant ? What about rough isometries ?

There are counter examples for bounded harmonic functions (Lyons-Sullivan).

1.3. What are they good for, II

Theorem 2 (Bonk-Kleiner) Let ${X}$ be a compact, ${Q}$-Ahlfors regular metric space which is homeomorphic to the 2-sphere. Assume ${X}$ satisfies a ${(1,Q)}$-Poincaré inequality. Then ${Q=2}$ and ${X}$ is quasi-symmetric to the round 2-sphere.

As a special case, let ${X}$ be the ideal boundary of a hyperbolic group ${\Gamma}$. In general, Poincaré inequality is not satisfied. Otherwise, one could conclude that ${\Gamma}$ is virtually a lattice in hyperbolic 3-space, thus solving Cannon’s conjecture.

2. Attempt : a weaker inequality

Question. Does polynomial growth plus a lower bound on Ricci curvature imply a Poincaré inequality ?

Example 2 Let ${M}$ be a compact negatively curved manifold. The horospheres in the universal cover have bounded sectional curvature and polynomial volume growth.

The following result can be extracted from works by Th. Coulhon and L. Saloff-Coste.

Theorem 3 Let ${X}$ be a Riemannian manifold with Ricci curvature bounded below and polynomial growth ${V(R)\leq \mathrm{const.}R^{\alpha}}$. Then there exist ${C}$, ${C'}$, ${R_0 \geq 1}$, such that for all ${\sigma\geq 1}$, for all ${R>R_0}$, all functions ${u:X\rightarrow R}$ and all balls,

$\displaystyle \begin{array}{rcl} \int_{B(R)}|u-u_{B(R)}|^{\sigma}\leq C\,R^{\sigma+\alpha-1}\int_{B(C'R)}|\nabla u|^{\sigma}. \end{array}$

The weakness is the ${\alpha-1}$ in the exponent. Nevertheless, it is sharp. Indeed, consider the stupid comb-shaped tree (of quadratic growth), and a function ${u}$ with ${du}$ concentrated on one edge. Note that horosphere should not look like that. Indeed, they are kind of quasi-periodic.

The proof is done first for graphs, in which case it is trivial (sometimes called Poincaré’s duplication principle). Then manifolds are approximated by graphs.

Question. Let ${M}$ be a compact hyperbolic manifold. Change the metric. Compare respective horospheres : are they quasi-symmetrically equivalent ?

Question. Let ${M}$ be a compact negatively curved manifold. Do horospheres satisfy two-sided volume bounds

$\displaystyle \begin{array}{rcl} c\,R^{\alpha}\leq V(R)\leq C\,R^{\alpha} ? \end{array}$

Advertisements

## About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri PoincarĂ©, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
This entry was posted in Workshop lecture and tagged . Bookmark the permalink.