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\title{Quasiconformal classification of 3D contact Lie groups}
\author{Pierre Pansu}
\date{August 15th, 2023}

\usepackage[T1]{fontenc}

\mode<presentation>{\usetheme{Warsaw}}

\begin{document}

\begin{frame}
\titlepage
\end{frame}

\section{The problem}


\begin{frame}

\begin{multicols}{2}

We take the metric definition of a \textbf{quasiconformal homeomorphism}: at a.e. point $x$, 
$$
\limsup_{r\to 0}\frac{L(f,x,r)}{\ell(f,x,r)}\le H.
$$

We focus on contact subRiemannian spaces, i.e. smooth contact manifolds with subRiemannian metrics.

\begin{center}

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\end{center}

\end{multicols}

\pause

\begin{prob}
How can one prove that two contact subRiemannian manifolds are not quasiconformally equivalent?
\end{prob}\pause

We call a \emph{contact Lie group} the data of a connected Lie group $G$ and a left-invariant contact structure $\xi$ on $G$. \pause

\medskip

Any two choices of left-invariant subFinsler metrics on $\xi$ are biLipschitz equivalent, in particular quasiconformally equivalent. Therefore one can discuss wether two contact Lie groups are quasiconformally equivalent or not.

\end{frame}

\subsection{$1$-dimensional tools}

\begin{frame}

We are concerned first with the quasiconformal classification of $3$-dimensional contact Lie groups.

\begin{thm}
\begin{enumerate}
  \item The left invariant contact structures on $U(1)\times Dil(\R)$ and $PSl(2,\R)$ are quasiconformally equivalent, as are their coverings.
  \item In all other cases, two contact 3-dimensional Lie groups are quasiconformally equivalent if and only if they are isomorphic as Lie groups.
\end{enumerate}
\end{thm}\pause

\medskip

All contact structures encountered (on $\R^3$, $\R^2\times S^1$) are smoothly contactomorphic. So no interference with contact topology. Nevertheless, a trick is borrowed from contact topology.

\end{frame}

\subsection{Higher dimensional tools}

\begin{frame}

The previous classification relies on the classical tools of quasiconformal geometry: the modulus of a curve family and the capacity of a condenser. They are $1$-dimensional invariants, expressible in terms of lengths of curves or $L^p$ norms of closed $1$-forms. \pause

\medskip

We shall define higher dimensional avatars of such invariants, based on differential forms on higher degree, and provide examples where these avatars seem to be necessary. \pause


\begin{thm}[Work in progress]
There exist (high-dimensional) pairs of contact subRiemannian manifolds $(M,N)$ such that
\begin{enumerate}
  \item Both $M$ and $N$ admit cocompact isometry groups.
  \item $M$ and $N$ are smoothly contactomorphic.
  \item $M$ and $N$ cannot be distinguished using classical $1$-dimensional tools.
  \item $M$ and $N$ are distinguished by a $2$-dimensional invariant.
\end{enumerate}
\end{thm}

\pause

The use of higher degree differential forms in the quasiworld is not so frequent:

\hskip1em Donaldson-Sullivan, \emph{Quasiconformal 4-manifolds}. Acta Math. (1989).

\hskip1em Iwaniec-Martin, \emph{Quasiregular mappings in even dimensions}. Acta Math. (1993).


\end{frame}

\section{$1$-dimensional techniques}

\subsection{Parabolicity}

\begin{frame}

Say a $2n+1$-dimensional contact subRiemannian manifold is \emph{conformally parabolic} if there exist compactly supported functions $u$ which are $\ge 1$ on a fixed ball but with arbitrarily small $\int|\nabla_{\HH} u|^{2n}$.\pause

\begin{lem}[Parabolicity of 3D contact Lie groups]
Here is the list of conformally parabolic $3$-dimensional contact groups, grouped in topological classes:
\begin{itemize}
  \item $SU(2)$.
  \item $SO(3)$.
  \item Proper quotients of $\widetilde{Mot(\R^2)}$ and $Heis$.
  \item $\widetilde{Mot(\R^2)}$ and $Heis$ themselves.
\end{itemize} 
\end{lem}

Since qc homeos lift to qc homeos on coverings, it suffices to show that $\widetilde{Mot(\R^2)}$ and $Heis$ are not quasiconformally equivalent. 

\end{frame}

\subsection{Teichm\"uller condensers}

\begin{frame}

\begin{thm}[F\"assler-Koskela-Le Donne (2015)]
$\widetilde{Mot(\R^2)}$ and $Heis$ are not quasiconformally equivalent. 
\end{thm}\pause

\medskip

Their argument uses capacities of condensers of Teichm\"uller's type: formed by two unbounded curves. 

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\end{center}

\end{frame}


\begin{frame}

\begin{defi}[Ferrand (1972)]
Say that a noncompact subRiemannian manifold \textbf{has a Teichm\"uller distance} if 
\begin{align*}
d_T(A,B)=(\inf \int|du|^{2n+2}\,;\, u=0~ (\text{resp. }1)&\text{ on an unbounded curve containing }A\\
&(\text{resp. }B)\})^{-1/(2n+1)}
\end{align*}
is positive. If so, it defines a distance.
\end{defi}\pause

\medskip

Because volume growth in $\widetilde{Mot(\R^2)}$ is slow (cubic), there exist functions whose horizontal gradient has finite $L^4$ norm, but which take values 1 (resp. 0) along two geodesic rays. Therefore $\widetilde{Mot(\R^2)}$ has a Teichm\"uller distance.

\medskip

However, by self-similarity, $Heis$ cannot have a Teichm\"uller distance.

\end{frame}

\subsection{Gr\"otzsch condensers}

\begin{frame}

\begin{prop}[qc $\implies$ qi principle]
Let $M$ and $N$ be bounded geometry $2n+1$-dimensional contact subRiemannian manifolds. If both have isoperimetric dimensions $>2n+2$, then every quasiconformal mapping $M\to N$ is a quasiisometry.
\end{prop}\pause

The prototype of such distorsion estimates is due to Gr\"otzsch (1928). It is based on capacities of condensers formed by $\infty$ and a curve.


\begin{center}

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\end{center}

Parabolicity means that the capacities of all such condensers vanish.

\end{frame}

\begin{frame}

\begin{defi}[Gal (1960), Vuorinen (1988)]
On a nonparabolic contact subRiemannian manifold, the \textbf{Gr\"otzsch distance} is
\begin{align*}
d_G(A,B)=\inf \int|du|^{2n+2}\,;\, u\text{ compactly supported, }u=1\text{ on a curve joining }A\text{ to }B\}.
\end{align*}
\end{defi}

\medskip

Isoperimetry (in fact, Sobolev inequality) implies that $d_G$ grows at least like some function of the distance. Bounded geometry implies that $d_G$ grows at most like some function of the distance. Whence the qi estimate on qc mappings.

\pause

\medskip

\textbf{Remark}.
For $3$-dimensional contact Lie groups,
$$
\text{conformal parabolicity}\iff IsopDim\leq 4,
$$
so the qc $\implies$ qi principle applies to all nonparabolic $3$-dimensional contact Lie groups.

\end{frame}

\begin{frame}

\textbf{Completion of the classification}

\medskip

1. One produces a list of 3D contact Lie groups.\pause

\medskip

2. The qc classification splits into parabolic and nonparabolic.\pause

\medskip

3. The qc classification of parabolic 3D contact Lie groups follows from topology and Fässler-Koskela-Le Donne 2015.\pause

\medskip

4. The qc classification of nonparabolic 3D contact Lie groups follows from the qc $\implies$ qi principle combined with the qi classification of 3D Lie groups (which can be found in Fässler-Le Donne 2021).\pause

\medskip

5. One shows that all quasiisometric pairs are indeed quasiconformally equivalent.

\medskip

For instance, $PSL_2(\R)$ carries two $PSL_2(\R)$-equivariantly nonequivalent left invariant contact structures, they turn out to be $Dil(\R)$-equivariantly isomorphic (mirror + isotopy).




%Fässler, Katrin; Le Donne, Enrico Geom. Dedicata 210 (2021), 27?42.

\end{frame}


\section{Conformal cohomology}

\begin{frame}

\textbf{Cohomological interpretation of parabolicity}

\smallskip

Let $X$ be a Riemannian $n$-manifold. Fix a ball $B$. This creates a relative $1$-cohomology class in $H_c^1(X,B;\R)$, represented by differentials of smooth compactly supported functions that take value $1$ on $B$. $X$ is conformally parabolic $\iff$ the $L^n$ norm of this cohomology class vanishes.\pause

\medskip

\textbf{Question}. Can one use higher degree differential forms instead?\pause

\medskip

\textbf{Answer}. In Riemannian geometry, Goldshtein-Troyanov define \emph{conformal cohomology} as the cohomology of the de Rham complex with decay conditions: 
\begin{center}
$k$-forms are assumed to belong in $L^{n/k}$.
\end{center}
\pause

\medskip

Vanishing of Gr\"otzsch distance (i.e. conformal parabolicity) is close to (but not exactly the same as) $L^{\infty,n}H^1$ not being Hausdorff. It implies that for every $q<\infty$, $L^{q,n}H^1$ is not Hausdorff.\pause

\medskip

Vanishing of Teichm\"uller distance is close to (but not exactly the same as) $L^{\infty,n}H^1\not=0$. For every $q<\infty$, it follows from $L^{q,n}H^1=0$.

\end{frame}

\begin{frame}

We want to use a subRiemannian avatar, in dimension $2n+1$, with $\frac{n}{k}$ replaced with $\frac{2n+2}{\alpha(k)}$, where $\alpha(k)$ is the minimal Hausdorff dimension of $k$-dimensional submanifolds,
$$
\alpha(k)=\begin{cases}
  k    & \text{if }k\le n, \\
 k+1     & \text{otherwise},
\end{cases}
$$
and de Rham's complex replaced with \emph{Rumin}'s complex. Let us call it \emph{Rumin conformal cohomology}.

\end{frame}

\subsection{Quasiconformal invariance}

\begin{frame}

\begin{thm}[Kleiner-M\"uller-Xie]
Rumin conformal cohomology is a quasiconformal invariant of contact subRiemannian manifolds.
\end{thm}\pause

\medskip

Margulis-Mostow 1995: if a homeomorphism of subRiemannian manifolds is quasiconformal, so is its inverse.\pause

\medskip

Heinonen-Koskela 1998: a quasiconformal homeomorphism of subRiemannian Heisenberg group belongs to $W^{1,p}$ for some $p>2n+2$.\pause

\medskip

Kleiner-M\"uller-Xie 2021: $W^{1,p}$ mappings, $p>2n+2$, induce chain maps on Rumin's complex.\pause

\medskip

Therefore quasiconformal mappings induce functorial chain maps between Rumin conformal complexes, hence isomorphisms between Rumin conformal cohomologies.

\end{frame}

\subsection{$L^{q,p}$ cohomology calculations}

\begin{frame}

\begin{prop}[Work in progress]
There exist (high-dimensional) pairs of contact subRiemannian manifolds $(M,N)$ such that
\begin{enumerate}
  \item Both $M$ and $N$ admit cocompact isometry groups.
  \item $M$ and $N$ are smoothly contactomorphic.
  \item $M$ and $N$ are conformally parabolic. 
  \item $M$ and $N$ have vanishing Teichm\"uller distances.
  \item $M$ and $N$ differ in $2$-dimensional Rumin conformal cohomology $L^{2n+2,n+1}H^2$.
\end{enumerate}
\end{prop}\pause

The construction relies on facts about the $L^{q,p}$ cohomology of Carnot groups.

\begin{fac}[Pansu-Rumin 2018]
Let $G$ be a Carnot group with homogeneous dimension $Q$. 
\begin{enumerate}
  \item $L^{q,p}H^1(G)=0\iff \frac{1}{p}-\frac{1}{q}\ge\frac{1}{Q}$.
  \item Let $[w_{min},w_{max}]$ denote the range of weights occurring in the Lie algebra cohomology $H^{2}(\mathfrak{g})$. Then
  \begin{enumerate}
  \item $\frac{1}{p}-\frac{1}{q}\ge\frac{w_{max}-1}{Q}\implies L^{q,p}H^2(G)=0$.
  \item $\frac{1}{p}-\frac{1}{q}<\frac{w_{min}-1}{Q}\implies L^{q,p}H^2(G)\not=0$.
\end{enumerate}
\end{enumerate}
\end{fac}

\end{frame}

\subsection{The construction}

\begin{frame}

The basic bricks are pairs of step-$3$ Carnot Lie algebras $\mathfrak{f}$ and $\mathfrak{g}$ such that
\begin{enumerate}
  \item $\mathrm{dim}(\mathfrak{f})=\mathrm{dim}(\mathfrak{g})=m$.
  \item $Q(\mathfrak{f})=Q(\mathfrak{g})=2n+2$.
  \item $w_{min}(\mathfrak{f})\ge 3$, $w_{max}(\mathfrak{g})=2$.
\end{enumerate}

Then one forms products with spheres $V_1=F\times S^{n+1-m}$, $V_2=G\times S^{n+1-m}$, and one takes projectivized cotangent bundles $M=P(T^*V_1)$, $N=P(T^*V_2)$, in order to get contact manifolds with cocompact $F$- (resp. $G$-) actions.

\medskip

By quasiisometry invariance of $L^{q,p}$ Rumin cohomology (Baldi-Franchi-Pansu-Tripaldi (2021)), the cohomological properties, including parabolicity and vanishing of Teichm\"uller distance, pass from $F,G$ to $M,N$.\pause

\medskip

\begin{lem}
A Carnot Lie algebra $\mathfrak{f}$ has $w_{min}(\mathfrak{f})\ge 3$ if and only if its $2$-step quotient $\mathfrak{f}/\mathfrak{f}^{(3)}$ is free.
\end{lem}

\end{frame}


\begin{frame}

\begin{lem}
A $3$-step Carnot Lie algebra $\mathfrak{g}$ is determined by the choices of
\begin{enumerate}
  \item A vectorspace $E_1$.
  \item A subvectorspace $E_2$ of $\Lambda^2 E_1$.
  \item A subvectorspace $E_3$ of the kernel of the tautological map
  $$
  E_1\otimes E_2 \to E_1\otimes \Lambda^2 E_1 \to \Lambda^3 E_1.
  $$
\end{enumerate}
Then $w_{max}(\mathfrak{g})=2$ if and only if 
\begin{enumerate}
  \item $E_3=\mathrm{Ker}(E_1\otimes E_2 \to \Lambda^3 E_1)$.
  \item $(E_1\otimes E_3)\cap(S^2 E_1\otimes E_2)=\{0\}$.
\end{enumerate}
\end{lem}\pause
So $E_3$ is uniquely determined by $E_2$, and it is subject to an open condition, which holds for generic choices of $E_2$ in $\Lambda^2 E_1$ provided dimensions match:
\begin{align}
\label{1}
   \frac{(n_1-1)(n_1-2)}{6}\le n_2 \le \frac{2}{n_1+1}{n_1 \choose 3}. 
\end{align}


\end{frame}

\begin{frame}

For every integers $p$, and $r$ such that
\begin{align}
\label{2}
   0\le r\le \frac{p^3-p}{6},  
   \end{align}
there exists step-$3$ Lie algebras $\mathfrak{f}$ such that $\mathfrak{f}/\mathfrak{f}^{(3)}$ is free on $p$ generators and $\mathrm{dim}(\mathfrak{f}^{(3)})=r$.

\medskip

The equations for $\mathrm{dim}(\mathfrak{g})=\mathrm{dim}(\mathfrak{f})$ and $Q(\mathfrak{g})=Q(\mathfrak{f})$ boil down to

\begin{align*}
\begin{cases}
n_2 &=~\frac{1}{2}p^2+\frac{3}{2}p-2n_1    \\
r &=~ (1+n_1)(\frac{1}{2}p^2+\frac{3}{2}p-2n_1)-{n_1 \choose 3}+n_1-\frac{1}{2}p^2-\frac{1}{2}p.  
\end{cases}
\end{align*}

If one picks $p$ and $n_1$ such that $\frac{n_1}{p}$ belongs to the interval $(x_0,\sqrt{3})$, where $x_0=1.53...$ is the largest root of equation $x^3-3x+1=0$, conditions (1) and (2) hold. This yields the required Lie algebras.

\end{frame}

\end{document}

\begin{frame}

\begin{lem}
A $3$-step Carnot Lie algebra $\mathfrak{g}$ is determined by the choices of
\begin{enumerate}
  \item A vectorspace $E_1$.
  \item A subvectorspace $E_2$ of $\Lambda^2 E_1$.
  \item A subvectorspace $E_3$ of the kernel of the tautological map
  $$
  E_1\otimes E_2 \to E_1\otimes \Lambda^2 E_1 \to \Lambda^3 E_1.
  $$
\end{enumerate}
Then $w_{min}(\mathfrak{g})=2$ if and only if 
\begin{enumerate}
  \item $E_3=\mathrm{Ker}(E_1\otimes E_2 \to \Lambda^3 E_1)$.
  \item $(E_1\otimes E_3)\cap(S^2 E_1\otimes E_2)=\{0\}$.
\end{enumerate}
\end{lem}\pause
In other words, $E_3$ is uniquely determined by $E_2$, and its codimension (in $E_1\otimes E_2$) is typically equal to that of $\mathrm{Ker}(E_1\otimes \Lambda^2 E_1 \to \Lambda^3 E_1$, i.e. $\mathrm{dim}(\Lambda^3 E_1)$. So typically, $n_3:=\mathrm{dim}(E_3)=n_1 n_2 - {n_1 \choose 3}$. Assuming the map $E_2\to E_1\otimes E_3$ between Grassmannians is transverse to the Schubert cell defined by intersecting $S^2 E_1\otimes E_2$, the intersection is generically trivial as soon as $n_1 n_3 +{n_1+1 \choose 2}n_2 \le n_1^2 n_2$.  
Whence the numerical conditions
\begin{align}
\label{1}
   \frac{(n_1-1)(n_1-2)}{6}\le n_2 \le \frac{2}{n_1+1}{n_1 \choose 3}. 
\end{align}


\end{frame}

\begin{frame}

For every integers $p$, and $r$ such that
\begin{align}
\label{2}
   0\le r\le \frac{p^3-p}{6},  
   \end{align}
there exists step-$3$ Lie algebras $\mathfrak{f}$ such that $\mathfrak{f}/\mathfrak{f}^{(3)}$ is free on $p$ generators and $\mathrm{dim}(\mathfrak{f}^{(3)})=r$.

The equations for $\mathrm{dim}(\mathfrak{g})=\mathrm{dim}(\mathfrak{f})$ and $Q(\mathfrak{g})=Q(\mathfrak{f})$ boil down to

\begin{align*}
\begin{cases}
n_2 &=~\frac{1}{2}p^2+\frac{3}{2}p-2n_1    \\
r &=~ (1+n_1)(\frac{1}{2}p^2+\frac{3}{2}p-2n_1)-{n_1 \choose 3}+n_1-\frac{1}{2}p^2-\frac{1}{2}p.  
\end{cases}
\end{align*}

Let us pick $p$ and $n_1$ such that $\frac{n_1}{p}$ tends to some real number $x$. Then $n_2\sim\frac{1}{2x^2}n_1^2$ and $r\sim (\frac{x}{2}-\frac{x^3}{6})p^3$. Therefore conditions (\ref{1}) and (\ref{2}) boil down to
$$
\frac{1}{6}\le \frac{1}{2x^2}\le\frac{1}{3},\quad 0\le\frac{x}{2}-\frac{x^3}{6}\le \frac{1}{6}.
$$
equivalent to
$$
1.53...=x_0 \le x\le \sqrt{3}=1.73...
$$
where $x_0$ is the largest root of equation $x^3-3x+1=0$.
\end{frame}

\end{document}


\begin{thm}[Baldi-Franchi-Pansu-Tripaldi 2021]
The conformal cohomology of bounded geometry contact subRiemannian manifolds coincides with the quasiisometry invariant $\ell^{2n+2;(2n+2)/\alpha(k)}$ cohomology, provided $1<k<2n+1$.
\end{thm}


Hence $\widetilde{Mot(\R^2)}$ is not quasiconformally equivalent to $Heis$.\pause

\medskip

\textbf{Remark}. Not so many uses of higher degree differential forms in quasiconformal theory:\\
Donaldson-Sullivan, \emph{Quasiconformal 4-manifolds}. Acta Math. 1989.\\
Iwaniec-Martin, \emph{Quasiregular mappings in even dimensions}. Acta Math. 1993.

\end{frame}

\begin{frame}

Thanks to Xiangdong Xie, Stefan M\"uller and
\begin{center}
\includegraphics[width=2in]{Kleiner_Bruce.jpg}
\end{center}
for providing us with tools to analyse subRiemannian quasiconformal mappings.

\bigskip

Happy birthday, Bruce!

\end{frame}

\begin{frame}

Finally, we examine pairs of quasiisometric conformally hyperbolic contact groups.

We have seen that $(\widetilde{SL(2,\R)},\xi_2)$ is quasiconformal to $\R\times Dil(\R)$ and so do their quotients. 

There remains to compare $(\widetilde{SL(2,\R)},\xi_1)$ with $(\widetilde{SL(2,\R)},\xi_2)$. In the smooth category, $(\widetilde{SL(2,\R)},\xi_2)$ is contactomorphic to the standard contact structure of $Heis$ (by continuously deforming $Dil(\R)$ to $\R^2$). Similarly, by continuously deforming hyperbolic plane to Euclidean plane, $(SL(2,\R),\xi_1)$ is contactomorphic to $Mot(\R^2)$, which in turn is contactomorphic to $Heis/K$, where $K$ is a horizontal cyclic subgroup, see
{\color{blue} \href{https://www.imo.universite-paris-saclay.fr/~patrick.massot/exposition/gallerie_contact/liens/}{there}}. In formulae, the diffeo
$$
(p,q,\theta)\mapsto (x=\theta,y=p\cos\theta-q\sin\theta,z=p\sin\theta+q\cos\theta)
$$
pulls back $dz-y\,dx$ to $\sin\theta\,dp+\cos\theta\,dq$. So no obstruction from contact geometry.

\medskip


\textbf{Questions}. 
\begin{enumerate}
  \item Are the two invariant contact structures on $\widetilde{SL(2,\R)}$ quasiconformally equivalent?
  \item If so, what about their quotients?
\end{enumerate}

\textbf{Remark}. $Heis/K$ is not a group, hence not covered by our problem, but the problem can be broadened to include 3D contact \emph{homogeneous spaces}. In 1994, R. Miner showed that $Heis/K$ is not quasiconformal to $Heis/Z$ where $Z$ is a cyclic subgroup of the center.

\end{frame}

\begin{frame}

\textbf{Classification of $3$-dimensional real Lie algebras}

\medskip

Based on the dimension of the derived Lie algebra $\mathfrak{g}^2:=[\mathfrak{g},\mathfrak{g}]$.

0. If $\mathrm{dim}(\mathfrak{g}^2)=0$, $\mathfrak{g}$ is abelian.

\smallskip

1. If $\mathrm{dim}(\mathfrak{g}^2)=1$, either $\mathrm{dim}(\mathrm{Ker}(ad_{\mathfrak{g}^2}))=3$ and $\mathfrak{g}=\mathfrak{Heis}$ or $\mathrm{dim}(\mathrm{Ker}(ad_{\mathfrak{g}^2}))=2$ and $\mathfrak{g}=\R\oplus\mathfrak{Dil}(\R)$.

\smallskip

2. If $\mathrm{dim}(\mathfrak{g}^2)=2$, $\mathfrak{g}$ is the semidirect product of $\R^2$ with $\R$ acting via a matrix $M$ of rank $2$, unique up to similarity and scaling, leading to a family $\mathfrak{g}_\mu$, $\mu\in[-1,0)\cup(0,1]$, when $M$ is a diagonal matrix with eigenvalues $1$ and $\mu$, and exceptions $\mathfrak{g}_u$ when $M$ is unipotent and $\mathfrak{g}_e$ when $M$ is a skewsymmetric matrix.

\smallskip

3. If $\mathrm{dim}(\mathfrak{g}^2)=3$, $\mathfrak{g}$ is semisimple, hence $\mathfrak{g}=\mathfrak{so}(3)$ or $\mathfrak{sl}(2,\R)$.

\end{frame}



\begin{frame}

\begin{thm}
Let $Z<Heis$ be a cyclic subgroup of the center. Let $K<Heis$ be the cyclic subgroup generated by a horizontal translation. Then the 3D contact homogeneous spaces $Heis/Z$ and $Heis/K$ are not quasiconformal.
\end{thm}

\textbf{Proof}. Assume $f:Heis\to Heis$ intertwines generators $k\in K$ and $z\in Z$ and fixes the origin. Then $f$ is quasiconformal, hence globally quasisymmetric (Gr\"otsch 1928): for every integer $r$, it maps an $r$-ball to a quasi-$r'$-ball. Since $k^r$ moves the origin to a distance $\sim r$ and $z^r$ moves the origin to a distance $\sim r^2$, $r'\sim \sqrt{r}$.

Consider the curve families 
$$
\Gamma_r=\{k^r\text{-loops contained in }B(r)\},\quad \Gamma'_{r}=\{z^{r}\text{-loops contained in }B(C\,\sqrt{r})\}.
$$
Then $f$ maps $\Gamma_r$ to $\Gamma'_{r}$, so the conformal moduli satisfy
$$
M(\Gamma_r)\le H(f)\,M(\Gamma'_{r}).
$$
However, the horizontal projection of $B(r)$ has area $\sim r^3$, so the subfamily of horizontal segments in $\Gamma_r$, of length $r$, has modulus $\ge r^3/r=r^2$. A fortiori, $M(\Gamma_r)\ge r^2$. Whereas $M(\Gamma'_{r}\le C\,r$.

\end{frame}


\end{document}

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\title{Quasiconformal classification of 3D contact Lie groups}
\author{Pierre Pansu}
\date{June 21st, 2023}

\usepackage[T1]{fontenc}

\mode<presentation>{\usetheme{Warsaw}}

\begin{document}

\begin{frame}
\titlepage
\end{frame}

\section{Results}

\begin{frame}

\begin{lem}[List of 3D Lie groups admitting left-invariant contact structures]
Every $3$-dimensional Lie group admits a left-invariant contact structure, except for the abelian $\R^3$ and its quotients, and the group $Dil(\R^2)$ of translations and dilations of the Euclidean plane.
\end{lem}

\begin{lem}[List of 3D contact Lie groups]
Left-invariant contact structures on a $3$-dimensional Lie group fall into a finite number of classes mod automorphisms. This number is $2$ for $\widetilde{Sl(2,\R)}$ and its quotients, and $1$ otherwise.
\end{lem}
According to (the left-invariant version of) Gray's theorem, any two structures belonging to the same connected component of the set of left-invariant contact structures on $G$ differ by an inner automorphism. This set is the complement of a conic in the real projective plane $P(\mathfrak{g}^*)$, so it has at most 2 connected components. These components are permuted by automorphisms, except in the case of $\mathfrak{sl}(2,\R)$. 

\end{frame}

\begin{frame}

Here are two obvious left invariant contact structures on $PSL(2,\R)$:
\begin{enumerate}
  \item The tautological contact structure on the projectivized cotangent bundle $\xi_1$.
  \item The homogeneous connection $\xi_2$ on this bundle.
\end{enumerate}
$\xi_1$ and $\xi_2$ define opposite orientations and lift to distinct components of left invariant contact structures on $\widetilde{SL(2,\R)}$.

\begin{lem}[$U(1)\times Dil(\R)$]
The left invariant contact structures on $U(1)\times Dil(\R)$ are quasiconformally equivalent to $(PSl(2,\R),\xi_2)$.
\end{lem}

\textbf{Proof}. $\xi_2$ is invariant under $U(1)\times PSl(2,\R)$ acting on $PSl(2,\R)$. The subgroup $U(1)\times Dil(\R)$ acts simply transitively, whence a pulled-back contact structure on $U(1)\times Dil(\R)$ which is left invariant. So suitable left invariant subRiemannian metrics make both groups isometric.

\end{frame}


\begin{frame}

\begin{lem}[qc $\implies$ qi]
Let $M$ and $N$ be bounded geometry $3$-dimensional subRiemannian contact manifolds. If both have isoperimetric dimensions $>4$, then every quasiconformal mapping $M\to N$ is a quasiisometry.
\end{lem}

\begin{lem}[IsopDim of 3D contact Lie groups]
Let $G$ be a $3$-dimensional connected Lie group. Then
\begin{itemize}
  \item If $G=SU(2)$ and quotients, IsopDim $=0$.
  \item If $G=Mot(\R^2)$ or $G=Heis/\Z$, $\Z\subset Z(Heis)$, IsopDim $=2$.
  \item If $G=\widetilde{Mot(\R^2)}$, IsopDim $=3$.
  \item If $G=Heis$, IsopDim $=4$.
  \item Otherwise, IsopDim $>4$.
\end{itemize}

\end{lem}

\textbf{Proof}. Isoperimetric dimension $\le d$ $\implies$ volume growth $\le d$.

\end{frame}

\begin{frame}

\begin{lem}[QI classification of 3D Lie groups]
Two noncompact $3$-dimensional connected Lie groups are quasiisometric if and only if they are isomorphic, with the following exceptions: 
\begin{itemize}
  \item $\widetilde{Mot(\R^2)}$ is quasiisometric to $\R^3$.
  \item All proper quotients of $\widetilde{Mot(\R^2)}$ and all proper quotients of $Heis$ are quasiisometric.
  \item $\widetilde{SL(2,\R)}$ is quasiisometric to $\R\times Dil(\R)$.
  \item All proper quotients of $\widetilde{SL(2,\R)}$ and all proper quotients of $\R\times Dil(\R)$ are quasiisometric.
\end{itemize}
\end{lem}

\textbf{Proof}. One can use:
\begin{itemize}
  \item Isoperimetry: polynomial (4 classes), solvable unimodular, solvable nonunimodular or semisimple.
  \item Dimension of asymptotic cone: $2$ for $\widetilde{SL(2,\R)}$ and $\R\times Dil(\R)$, and $<2$ for others.
  \item Hyperbolicity, dimension of ideal boundary, rigidity of Heintze groups.
  \item Eskin-Fisher-Whyte for nonhyperbolic solvable nonunimodular groups.
\end{itemize}

\end{frame}

\begin{frame}

\begin{lem}[Parabolicity of 3D contact Lie groups]
If $G$ is a $3$-dimensional contact Lie group, $G$ is conformally parabolic $\iff$ IsopDim$(G)\le 4$.
\end{lem}

\textbf{Proof}. For general 3D contact subRiemannian manifolds, Isoperimetric dimension $>4$ $\implies$ conformal hyperbolicity. Converse handled case by case.

\medskip

Next we sort out among the conformally parabolic contact groups. Topologically, they are divided into 4 classes:
\begin{itemize}
  \item $SU(2)$.
  \item $SO(3)$.
  \item Proper quotients of $\widetilde{Mot(\R^2)}$ and $Heis$.
  \item $\widetilde{Mot(\R^2)}$ and $Heis$ themselves.
\end{itemize} 
$\widetilde{Mot(\R^2)}$ and $Heis$ are distinguished by $L^{4,2}H^2$, which is a quasiconformal invariant. It follows that quotients of these groups cannot be quasiconformal either.

\end{frame}

\begin{frame}

Finally, we examine pairs of quasiisometric conformally hyperbolic contact groups.

We have seen that $(\widetilde{SL(2,\R)},\xi_2)$ is quasiconformal to $\R\times Dil(\R)$ and so do their quotients. 

There remains to compare $(\widetilde{SL(2,\R)},\xi_1)$ with $(\widetilde{SL(2,\R)},\xi_2)$. In the smooth category, $(\widetilde{SL(2,\R)},\xi_2)$ is contactomorphic to the standard contact structure of $Heis$ (by continuously deforming $Dil(\R)$ to $\R^2$). Similarly, by continuously deforming hyperbolic plane to Euclidean plane, $(SL(2,\R),\xi_1)$ is contactomorphic to $Mot(\R^2)$, which in turn is contactomorphic to $Heis/K$, where $K$ is a horizontal cyclic subgroup, see
{\color{blue} \href{https://www.imo.universite-paris-saclay.fr/~patrick.massot/exposition/gallerie_contact/liens/}{there}}. In formulae, the diffeo
$$
(p,q,\theta)\mapsto (x=\theta,y=p\cos\theta-q\sin\theta,z=p\sin\theta+q\cos\theta)
$$
pulls back $dz-y\,dx$ to $\sin\theta\,dp+\cos\theta\,dq$. So no obstruction from contact geometry.

\medskip


\textbf{Questions}. 
\begin{enumerate}
  \item Are the two invariant contact structures on $\widetilde{SL(2,\R)}$ quasiconformally equivalent?
  \item If so, what about their quotients?
\end{enumerate}

\textbf{Remark}. $Heis/K$ is not a group, hence not covered by our problem, but the problem can be broadened to include 3D contact \emph{homogeneous spaces}. In 1994, R. Miner showed that $Heis/K$ is not quasiconformal to $Heis/Z$ where $Z$ is a cyclic subgroup of the center.

\end{frame}

\begin{frame}

\textbf{Classification of $3$-dimensional real Lie algebras}

\medskip

Based on the dimension of the derived Lie algebra $\mathfrak{g}^2:=[\mathfrak{g},\mathfrak{g}]$.

0. If $\mathrm{dim}(\mathfrak{g}^2)=0$, $\mathfrak{g}$ is abelian.

\smallskip

1. If $\mathrm{dim}(\mathfrak{g}^2)=1$, either $\mathrm{dim}(\mathrm{Ker}(ad_{\mathfrak{g}^2}))=3$ and $\mathfrak{g}=\mathfrak{Heis}$ or $\mathrm{dim}(\mathrm{Ker}(ad_{\mathfrak{g}^2}))=2$ and $\mathfrak{g}=\R\oplus\mathfrak{Dil}(\R)$.

\smallskip

2. If $\mathrm{dim}(\mathfrak{g}^2)=2$, $\mathfrak{g}$ is the semidirect product of $\R^2$ with $\R$ acting via a matrix $M$ of rank $2$, unique up to similarity and scaling, leading to a family $\mathfrak{g}_\mu$, $\mu\in[-1,0)\cup(0,1]$, when $M$ is a diagonal matrix with eigenvalues $1$ and $\mu$, and exceptions $\mathfrak{g}_u$ when $M$ is unipotent and $\mathfrak{g}_e$ when $M$ is a skewsymmetric matrix.

\smallskip

3. If $\mathrm{dim}(\mathfrak{g}^2)=3$, $\mathfrak{g}$ is semisimple, hence $\mathfrak{g}=\mathfrak{so}(3)$ or $\mathfrak{sl}(2,\R)$.

\end{frame}
\end{document}

\subsection{List of contact Lie groups}

\begin{frame}

\begin{lem}[List of 3D Lie groups admitting left-invariant contact structures]
Every $3$-dimensional Lie group admits a left-invariant contact structure, except for the abelian $\R^3$ and its quotients, and the group $Dil(\R^2)$ of translations and dilations of the Euclidean plane.
\end{lem}\pause

\begin{lem}[List of 3D contact Lie groups]
Left-invariant contact structures on a $3$-dimensional Lie group fall into a finite number of classes mod automorphisms. This number is $2$ for $\widetilde{Sl(2,\R)}$ and its quotients, and $1$ otherwise.
\end{lem}
According to (the left-invariant version of) Gray's theorem, any two structures belonging to the same connected component of the set of left-invariant contact structures on $G$ differ by an inner automorphism. 

\medskip

This set is the complement of a conic in the real projective plane $P(\mathfrak{g}^*)$, so it has at most 2 connected components. 

\medskip

These components are permuted by automorphisms, except in the case of $\mathfrak{sl}(2,\R)$. 

\end{frame}

\begin{frame}

Here are two obvious left invariant contact structures on $PSl(2,\R)$:
\begin{enumerate}
  \item The tautological contact structure on the projectivized cotangent bundle $\xi_1$.
  \item The homogeneous connection $\xi_2$ on this bundle.
\end{enumerate}
$\xi_1$ and $\xi_2$ define opposite orientations and lift to distinct components of left invariant contact structures on $\widetilde{SL(2,\R)}$, even mod automorphisms.\pause

\begin{lem}[$U(1)\times Dil(\R)$]
The left invariant contact structures on $U(1)\times Dil(\R)$ are quasiconformally equivalent to $(PSl(2,\R),\xi_2)$.
\end{lem}

\textbf{Proof}. $\xi_2$ is invariant under $U(1)\times PSl(2,\R)$ acting on $PSl(2,\R)$. The subgroup $U(1)\times Dil(\R)$ acts simply transitively, whence a pulled-back contact structure on $U(1)\times Dil(\R)$ which is left invariant. So suitable left invariant subRiemannian metrics make both groups isometric.

\end{frame}

