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\title{Conformal Hölder exponents}
\author{P. Pansu}
\date{\today}

\newtheorem{thm}{Theorem}
\newtheorem{lem}{Lemma}
\newtheorem{prop}[lem]{Proposition}
\newtheorem{cor}[lem]{Corollary}
\newtheorem{defi}[lem]{Definition}
\newtheorem{rem}[lem]{Remark}
\newtheorem{nota}[lem]{Notation}
\newtheorem{ques}[lem]{Question}
\newtheorem{prob}[lem]{Problem}
\newtheorem{sol}[lem]{Solution}
\newtheorem{conc}[lem]{Conclusion}
\newtheorem{ex}[lem]{Example}
\newtheorem{exs}[lem]{Examples}
\newtheorem{res}[lem]{Results}
\newtheorem{plan}[lem]{Plan of lecture}
\newtheorem{scheme}[lem]{Scheme of proof}

\def\proof{\par\medskip\noindent {\bf Proof}}
\def\qed{~q.e.d.}

\def\R{\mathbb{R}}
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\def\Q{\mathbb{Q}}
\def\n#1{{|\hskip-1pt | #1 |\hskip-1pt |}}

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\begin{document}


\begin{frame}
\titlepage
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\section{Gromov's H\"older homeomorphism problem}

\begin{frame}

\begin{ques}
\textcolor{red}{\em (Gromov 1993)}. Let $M$ be an $n$-dimensional subRiemannian manifold. For which $\alpha\in(0,1)$ does there exist locally a homeomorphism $\R^n \to M$ which is $C^{\alpha}$-H\"older continuous ?
\end{ques}\pause

\begin{defi}
Let $\alpha(M)=\sup\{\alpha\in(0,1)\,|\,\exists$ locally a homeomorphism $\R^n \to M\}$.
\end{defi}\pause

\begin{ex}
If $G$ is a $r$-step Carnot group, the exponential map $\mathfrak{g}=Lie(G)\to G$ is locally $C^{1/r}$-H\"older continuous. Thus $\alpha(M)\geq 1/r$.
\end{ex}\pause

\begin{prop}
Let $M$ have Hausdorff dimension $Q$. Then $\alpha(M)\leq\frac{n}{Q}$.
\end{prop}

\end{frame}

\subsection{Isoperimetric inequality}

\begin{frame}

\begin{prop}
Let $M$ be equiregular, of dimension $n$ and Hausdorff dimension $Q$. Then $\alpha(M)\leq\frac{n-1}{Q-1}$.
\end{prop}\pause

\textbf{Proof}. Use the isoperimetric inequality for piecewise smooth domains $D\subset M$,
\begin{eqnarray*}
vol(D)^{Q-1/Q}\leq const. \,\mathcal{H}^{Q-1}(\partial D).
\end{eqnarray*}\pause

It follows that the boundary of any non smooth domain $\Omega$ has Hausdorff dimension at least $Q-1$. Indeed, cover $\partial \Omega$ with balls $B_j$ and apply (*) to $\Omega\cup\bigcup B_j$. This gives a lower bound on $\mathcal{H}^{Q-1}(\partial(\bigcup B_j))\leq\sum\mathcal{H}^{Q-1}(\partial B_j)\leq const.\,\sum\mathrm{diameter}(B_j)^{Q-1}$.

\begin{center}
\includegraphics[width=2in]{isop.pdf}
\end{center}

\end{frame}

\subsection{Horizontal submanifolds}

\begin{frame}

\begin{prop}
\textcolor{red}{\em (Gromov 1993)}. Let $\mathbb{H}^{m}$ denote $2m+1$-dimensional Heisenberg group. Let $V\subset\mathbb{H}^{m}$ be a subset of topological dimension $m+1$. Then the Hausdorff dimension of $V$ is at least $m+2$. It follows that $\alpha(\mathbb{H}^{m})\leq\frac{m+1}{m+2}$.
\end{prop}\pause

\textbf{Proof}. According to topological dimension theory (Alexandrov), there exists an $m$-dimensional polyhedron $P$ and a continuous map $f:P\to\mathbb{H}^{m}$ such that every map sufficiently $C^0$-close to $f$ hits $V$. 

\medskip

Gromov approximates $f$ with piecewise {\em horizontal} maps which sweep an open set $U$. This gives rise to a local projection $p:U\to\R^{m+1}$ such that for every ball $B$, the tube $p^{-1}(p(B))$ has volume $\leq$ const. $\mathrm{diameter}(B)^{m+2}$. 

\medskip

Cover $V$ with balls $B_j$. The corresponding tubes $T_j = p^{-1}(p(B_j))$ cover $U$. Then the volume of $U$ is less than $\sum\mathrm{diameter}(B_j)^{m+2}$, which shows that $dim_{Hau}(V)\geq m+2$. \pause

\begin{thm}
\textcolor{red}{\em (Gromov 1993)}. Let $M$ be a generic subRiemannian manifold of dimension $n$, Hausdorff dimension $Q$, with an $h$-dimensional distribution. Let $k\leq h$ be such that $h-k\geq (n-h)k$. Then $\alpha(M)\leq \frac{n-k}{Q-k}$.\end{thm}

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\section{Curvature pinching}

\begin{frame}

\frametitle{Curvature pinching}

\begin{defi}
Let $M$ be a Riemannian manifold. Let $-1\leq\delta<0$. Say $M$ is {\em $\delta$-pinched} if sectional curvature ranges between $-1$ and $\delta$. 
Define the {\em optimal pinching} $\delta(M)$ of $M$ as the least $\delta\geq-1$ such that $M$ is biLipschitz to a $\delta$-pinched simply connected Riemannian manifold. \pause 
\end{defi}

\begin{ex}
Rank one symmetric spaces of noncompact type are hyperbolic spaces over the reals $H^{n}_{\R}$, the complex numbers $H^{m}_{\C}$, the quaternions $H^{m}_{\H}$, and the octonions $H^{2}_{\O}$.

Real hyperbolic space has sectional curvature $-1$. Other rank one symmetric spaces are $-\frac{1}{4}$-pinched. \pause
\end{ex}

\begin{ques}
What is the optimal pinching of $H^{m}_{\C}$ ?
\end{ques}

\end{frame}

\subsection{Visual metrics}

\begin{frame}

\begin{defi}
Say two geodesic rays in a Riemannian manifold are {\em asymptotic} if their Hausdorff distance is finite. The {\em visual boundary} of a negatively curved manifold is the set of asymptoticity classes of geodesic rays. 
\end{defi}\pause

\textbf{Facts}. \begin{itemize}
  \item The visual boundary, seen from a point $o$, is a sphere (use polar coordinates).
  \item It carries a {\em visual metric} $d_{o}$.
  \item Different visual metrics $d_{o}$ and $d_{o'}$ are equivalent.
  \item BiLipschitz maps between negatively curved Riemannian manifolds induce {\em quasisymmetric} maps between ideal boundaries.
\end{itemize}

\begin{center}
\includegraphics[width=2in]{visual.pdf}\pause\hskip1cm\includegraphics[width=1.5in]{qs.pdf}
\end{center}

\end{frame}


\subsection{Curvature versus H\"older homeomorphisms}

\begin{frame}

\begin{ex}
If $M$ is a rank one symmetric space, the visual metrics on its ideal boundary are equivalent to subRiemannian metrics.
\end{ex}\pause

\begin{prop}
Let $M$ be a simply connected $\delta$-pinched Riemannian manifold. Equip the ideal boundary $X$ of $M$ with a visual metric. The natural homeomorphism $S^{n-1}\to X$ is $C^{\alpha}$ with $\alpha=\sqrt{-\delta}$, and its inverse is Lipschitz.
\end{prop}
Indeed, geodesics from a unit ball to a point come together exponentially fast, with exponents ranging from $\sqrt{-\delta}$ to $1$ (Rauch comparison theorem, 1950's).\pause

\begin{ques}
Let $X$ be a nonRiemannian subRiemannian manifold. Let $\alpha>1/2$. Does there exist quasisymmetricly equivalent metrics on $X$ which locally admit $C^{\alpha}$ homeomorphisms from Euclidean space ? With Lipschitz inverses ?
\end{ques}

\end{frame}

\section{Conformal dimension}

\begin{frame}

\begin{defi}
Let $X$ be a compact metric space. The {\em conformal dimension} of $X$ is the infimum of Hausdorff dimensions of metric spaces quasisymmetricly equivalent to $X$.
\end{defi}\pause

\begin{prop}
Assume $(X,d)$ has topological dimension $n$ and conformal dimension $Q>n$. Pick a quasisymmetricly equivalent metric $d'$ on $X$. If $\alpha>\frac{n}{Q}$, there are no local $C^{\alpha}$ homeomorphisms $\R^n \to (X,d')$.
\end{prop}

\proof.
Indeed, Hausdorff dimension of $d'$ is $\geq Q$.\pause

\begin{ex}
Carnot groups have conformal dimension equal to their Hausdorff dimension. Heisenberg group $\mathbb{H}^m$ has conformal dimension $2m+2$. Even up to a quasisymmetric change of metric, no $C^{\alpha}$ homeomorphisms from $\R^{2m+1}$ if $\alpha>\frac{2m+1}{2m+2}$.
\end{ex}
Note: no restriction on inverse mapping.

\end{frame}

\subsection{Conformal dimension of hypersurfaces}

\begin{frame}

\textbf{Aim}: For subRiemannian manifolds, improve $\frac{n}{Q}$ into $\frac{n-1}{Q-1}$.\pause

\medskip

\textbf{Strategy}: Show that hypersurfaces must have Hausdorff dimension $\geq Q-1$ in quasisymmetric metrics.\pause

\begin{rem}
Vertical dilation invariant subgroups often have conformal dimension $Q-1$.
\end{rem}

\begin{ex}
In $\mathbb{H}^1$, subgroup $\{y=0\}$ is a {\em Rickman rug}, i.e. a product $(\R,d)\times(\R,d^{1/2})$, conformal dimension is $1+2=3$. 
\end{ex}

\begin{ex}
In $\mathbb{H}^m$, $m\geq 2$, subgroup $\{y_m =0\}$ is a Carnot group isometric to $\mathbb{H}^{m-1}\times\R$.
\end{ex}\pause

This is not sufficient: a H\"older map might pull back all of these to fractal subsets of $\R^n$.

\end{frame}

\section{Unsuccessful attempt}

\begin{frame}

\textbf{Leading intuition}: Given a function $u:X\to\R$, $\int|\nabla u|^{Q}$ is a conformal invariant. It may be written
\begin{eqnarray*}
\int|\nabla u|^{Q}=\int_{\R}\left(\int_{u^{-1}(t)}|\nabla u|^{Q-1}\right)\,dt.
\end{eqnarray*}\pause

\textbf{Technique}: Replace integrals with Hausdorff measures.\pause

\begin{nota}
Given a set function $\phi$, a subset $Y\subset X$, let 
\begin{eqnarray*}
\Phi^{p;\epsilon}(Y)=\inf\{\sum_{i}\phi(B_{i})^p \,;\,B_{i}\,\mathrm{countable~
covering~of }\,Y \,\mathrm{by}\,\mathrm{balls~of~diameters}\,\leq\epsilon\}.
\end{eqnarray*}
Let 
\begin{eqnarray*}
\Phi^{p}(Y)=\lim_{\epsilon\to 0}\Phi^{p;\epsilon}(Y).
\end{eqnarray*}

Denote by
\begin{eqnarray*}
osc_u (B)=\max_{B}u-\min_{B}u.
\end{eqnarray*}
\end{nota}

\end{frame}

\subsection{Capacities}

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\begin{lem}
On a Carnot group, given two open sets $A$ and $B$ with disjoint closures, $cap_{Q}(A,B):=\inf\{OSC_{u}^{Q}(X)\,;\,u~\textrm{continuous},\,u(A)=0,\,u(B)=1\}>0$.
\end{lem}\pause

\textbf{Observation}: Let $B_i$ be a countable covering of $X$.
\begin{eqnarray*}
\sum_{i}osc_u (B_i)^{Q}=\int_{\R}\left(\sum_{\{t\,;\,B\cap u^{-1}(t)\not=\emptyset\}}osc_u (B_i)^{Q-1}\right)\,dt.
\end{eqnarray*}\pause

\begin{lem}
\label{level}
Let $A\subset X$, let $u:A\to[0,1]$ be continuous. Then
\begin{eqnarray*}
OSC_{u}^{Q}(X)\geq\int_{\R}OSC_{u}^{Q-1}(u^{-1}(t))\,dt.
\end{eqnarray*}
\end{lem}

\end{frame}

\subsection{Conformal H\"older exponent}

\begin{frame}

Given a homeomorphism $f:\R^n \to X$, let $u=v\circ f^{-1}$, $v$ linear. Want to relate $OSC^{Q-1}_{u}(u^{-1}(t))$ to $\mathcal{H}^{\alpha(Q-1)}(v^{-1}(t))$. \pause
This suggests

\begin{defi}
Let $f:X\to Y$ be a homeomorphism. The \emph{conformal H\"older exponent}
$CH(f)$ of $f$ is the supremum of $\alpha$'s such that for all $\ell>0$, there exists
$L>0$ such that for all $x$, $x'$, $x''$ in $X$, 
\begin{eqnarray*}
d(f(x),f(x''))\leq \ell \,d(f(x),f(x'))\Rightarrow d(x,x'')\leq L\,d(x,x')^{\alpha}.
\end{eqnarray*}\pause
Say that $f$ is a \emph{conformally H\"older homeomorphism} if $CH(f)>0$. Let $CH(X,Y)$ denote the supremum of $\alpha$'s such
that there locally exist homeomorphisms $X\to Y$ with conformal H\"older exponents $\geq\alpha$.\pause
\end{defi}

\begin{ex}
If $f:X\to Y$ is $C^{\alpha}$ and $f^{-1}$ is $C^{\beta}$, then $CH(f)\geq\alpha\beta$.
\end{ex}\pause

\begin{lem}
Let $f:X\to Y$ and $g:Y\to Z$ be homeomorphisms. Assume that $g$ is
quasisymmetric. Then $CH(g\circ f)=CH(f)$.
\end{lem}

\end{frame}

\begin{frame}

\begin{lem}
Let $f:X\to Y$ be a homeomorphism, $v:X\to\R$ a Lipschitz function. Let $\alpha<CH(f)$. Then
\begin{eqnarray*}
OSC^{Q-1}_{u}(u^{-1}(t))\leq const.\,\mathcal{H}^{\alpha(Q-1)}(v^{-1}(t)).
\end{eqnarray*}
\end{lem}\pause

Compare to previous inequalities:

\begin{eqnarray*}
\inf\{OSC_{u}^{Q}(X)\,;\,u~\textrm{continuous},\,u(A)=0,\,u(B)=1\}>0.
\end{eqnarray*}
\begin{eqnarray*}
OSC_{u}^{Q}(X)\geq\int_{\R}OSC_{u}^{Q-1}(u^{-1}(t))\,dt.
\end{eqnarray*}

\end{frame}

\subsection{Advertisement}

\begin{frame}

\textbf{Metric geometry, algorithms and groups}
\medskip

Paris, january 10th - april 8th, 2011                

\medskip

Institut Henri Poincaré

\medskip

Organizers: Guy Kindler (Jerusalem), James Lee (U. Washington), Claire Mathieu (Brown), Ryan O'Donnell (Carnegie Mellon), Pierre Pansu (Paris-Sud/ENS), Nicolas Schabanel (LIAFA-CNRS), Lior Silberman (Vancouver) 

\medskip

Workshops: january 17-21, march 21-25

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search the web: \texttt{pansu metric}
\end{frame}


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