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\newtheorem{rem}[lem]{Remark}
\newtheorem{ntn}[lem]{Notation}
\newtheorem{que}[lem]{Question}
\newtheorem{pbm}[lem]{Problem}
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\def\R{\mathbb{R}}
\def\C{\mathbb{C}}
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\title{Large scale geometry of nilpotent Lie groups}
\author{Pierre Pansu}
\date{June 19-20th, 2023}

\usepackage[T1]{fontenc}

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

\begin{document}

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

\begin{frame}

\textbf{Goal}. The large scale behaviour of nilpotent Lie groups is governed by self-similarity:
\begin{itemize}
  \item There is a subclass of nilpotent Lie groups, called Carnot groups, which admit (subFinsler) metrics, which are exactly self-similar. 
  \item Every nilpotent Lie group is asymptotic to such a Carnot group.
\end{itemize}

This behaviour is specific to nilpotent groups: self-similarity and isometric homogeneity characterize subFinsler Carnot groups.  \pause

\begin{plan}
\begin{enumerate}
  \item Large scale geometry of the Heisenberg group
  \item Generalization to nilpotent groups
  \item GH-asymptotic cones
  \item Characterization of self-similar isometrically homogeneous spaces
\end{enumerate}
\end{plan}\pause

Lectures based on Enrico Le Donne's {\color{blue} \href{https://cvgmt.sns.it/media/doc/paper/5339/sub-Riem_notes\%202023-5\%20submitted_for_proposal.pdf}{book project}}.

Ask him for updates.

\end{frame}

\section{Large scale geometry of the Heisenberg group}

\begin{frame}

We are interested in large scale invariants like
\begin{itemize}
  \item Volume growth: $v(R)=$ number of unit balls needed to cover a ball of radius $R$.
  \item Isoperimetry: $I(v)=$ boundary volume needed to enclose volume at least $v$.
  \item More general filling inequalities: $Fill(L)=$ area needed to fill all loops of length at most $L$.
\end{itemize}\pause

\medskip

For abelian Lie groups, i.e. Euclidean spaces $(\R^n,g)$, scale invariance implies that each function is a power function, and reduces to settling the cases where $R$, $v$ or $L$ is equal to $1$. \pause Indeed, abelian Lie groups possess rescaling automorphisms $\delta_\eps : x\mapsto\eps x$ such that $\delta_\eps^*g=\eps^2 g$, \pause whence
\begin{align*}
dist(\delta_\eps (x),\delta_\eps(x'))&=\eps \,dist(x,x'),\\
vol(\delta_\eps(A))&=\eps^n \,vol(A),\\
vol(\delta_\eps(\partial A))&=\eps^{n-1}\,vol(\partial A),\\
length(\delta_\eps(\ell))&=\eps \,length(\ell),\\
area(\delta_\eps(S))&=\eps^2 \,area(S)
\end{align*} 
for surfaces.\pause

\medskip

\textbf{Question}. Are there other groups for which such scale invariance applies?

\end{frame}

\subsection{Riemannian Heisenberg group}

\begin{frame}

The $3$-dimensional Heisenberg group is
$$
\HH=\{\begin{pmatrix}
1 & x & z   \\
0 &  1 & y \\
0 & 0 & 1
\end{pmatrix}\,;\,x,y,z\in\R   \}.
$$
Setting $x\to\eps x$ and $y\to \eps y$ can be completed into a $1$-parameter group of automorphisms $\delta_\eps$ by setting $z\to \eps^2 z$.\pause

\medskip

If $M=\begin{pmatrix}
1 & x & z   \\
0 &  1 & y \\
0 & 0 & 1
\end{pmatrix}$, then $M^{-1}=\begin{pmatrix}
1 & -x & -z+xy   \\
0 &  1 & -y \\
0 & 0 & 1
\end{pmatrix}$ and
$$
M^{-1}dM=\begin{pmatrix}
1 & dx & dz-xdy   \\
0 &  1 & dy \\
0 & 0 & 1
\end{pmatrix}
$$
gives us a basis of left invariant $1$-forms. \pause So a typical left-invariant metric is 
$$
g_1=dx^2+dy^2+(dz-xdy)^2.
$$
Then 
$$
\delta_\eps^*g_1=\eps^2(dx^2+dy^2+\eps^2(dz-xdy)^2).
$$

\end{frame}

\subsection{SubRiemannian Heisenberg group}

\begin{frame}

\textbf{Question}. How does $g_\eps=dx^2+dy^2+\eps^2(dz-xdy)^2$ behave as $\eps\to\infty$ ?

\medskip


\begin{minipage}{5cm}

The \emph{horizontal plane field} $\mathrm{Ker}(dz-xdy)$ is generated by left-invariant vector fields $\xi=\frac{\partial}{\partial x}$ and $\eta=\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}$. %La {\em distance de Carnot} $d(p,p')$ est l'$\inf$ des longueurs des courbes horizontales joignant $x$ à $x'$. La finitude de la distance de Carnot résulte de la figure ci-contre.

\end{minipage}\hskip.5cm\begin{minipage}{5cm}

\begin{center}
\includegraphics[width=1.7in]{contact_std1.png}
\end{center}

\end{minipage}

\vskip3.35cm

\end{frame}

\begin{frame}

\textbf{Question}. How does $g_\eps=dx^2+dy^2+\eps^2(dz-xdy)^2$ behave as $\eps\to\infty$ ?

\medskip


\begin{minipage}{5cm}

The \emph{horizontal plane field} $\mathrm{Ker}(dz-xdy)$ is generated by left-invariant vector fields $\xi=\frac{\partial}{\partial x}$ and $\eta=\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}$. The \emph{subRiemannian} or \emph{Carnot-Carath\'eodory distance} $d_\infty(p,p')$ is the $\inf$ of lengths of horizontal curves joining $p$ to $p'$. Its finiteness follows from 
$$
\zeta=[\xi,\eta]=\frac{\partial}{\partial z},
$$
see the picture.

\end{minipage}\begin{minipage}{5cm}

\begin{center}

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\end{center}

\end{minipage}\pause

The family of distances $d_\eps$ defined by $g_\eps$ is increasing and bounded above by $d_\infty$. Therefore it converges.

\begin{pro}
$$
\lim_{\eps\to\infty}d_\eps = d_\infty.
$$
\end{pro}

Proof given soon.

\end{frame}

\subsection{Quasiisometry}

\begin{frame}

\begin{lem}[$\pi$ is a submetry]
The projection $\pi:\HH\to\R^2$, $(x,y,z)\mapsto (x,y)$ is a submetry, i.e. it maps $d_\eps$ balls to Euclidean balls of equal radii.
\end{lem}\pause

Indeed, $g_\eps\ge dx^2+dy^2$, so $\pi$ is $1$-Lipschitz. Conversely, along horizontal curves, $g_\eps = dx^2+dy^2$, so $\pi$ preserves the length of horizontal curves. Since all plane curves have horizontal lifts (setting $z(t)=\int_{0}^{t}x(x)y'(s)ds$), the image of a ball is exactly a ball.\pause

\begin{lem}[Upper bound on $d_\infty$]
The Carnot-Carath\'eodory distance satisfies
$$
d_\infty((x,y,z),(0,0,0))\le \sqrt{x^2+y^2}+4\sqrt{|z|}.
$$
\end{lem}\pause

We already featured a horizontal path of length $4\sqrt{|z|}$ joining $(0,0,0)$ to $(0,0,z)$. Then the horizontal lift $t\mapsto (tx,ty,z+\frac{t^3}{3}xy)$ of the line segment from $(0,0)$ to $(x,y)$ joins $(0,0,z)$ to $(x,y,z)$ and has length $\sqrt{x^2+y^2}$.\pause

\medskip

\textbf{Remark}. By compactness, there exists a constant $c>0$ such that $$c(\sqrt{x^2+y^2}+4\sqrt{|z|})\le d_\infty((x,y,z),(0,0,0)).$$

\end{frame}

\begin{frame}

\begin{cor}[$d_\infty$ and $d_1$ are locally $\frac{1}{2}$-H\"older equivalent]
There exists a constant $C$ such that, on the $d_\infty$ unit ball,
$$
d_\infty\le C\,d_1^{1/2}.
$$
\end{cor}\pause

Indeed, on a bounded set, $d_1 \sim 2\sqrt{x^2+y^2}+32|z|\ge d_\infty^2$.\pause

\begin{lem}[$d_\infty$ and $d_1$ are quasiisometric]
For the same constant $C$,
$$
d_1\le d_{\infty}\le C(d_1+\sqrt{d_1}).
$$
\end{lem}\pause

Indeed, if two points sit at $d_1$ distance $L$, cutting a $d_1$ geodesic in pieces of length $1$, one sees that their $d_\infty$-distance is at most $\lfloor L \rfloor C+C\sqrt{\{L\}}$.

\begin{center}

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\end{center}

\end{frame}

\begin{frame}

\begin{thm}[Large scale geometry of $d_\infty$]
\begin{itemize}
  \item Volume growth: $v(R)=O(R^4)$.
  \item Isoperimetry: $I(v)=O(v^{3/4})$.
  \item Horizontal loop filling: $Fill(L)=O(L^3)$.
\end{itemize}
\end{thm}

\begin{itemize}
  \item Volume growth is easy: $\delta_R$ maps the unit ball to the $R$-ball, and multiplies volumes by exactly $R^4$. \pause

  \item Surfaces have Hausdorff dimension at least $3$, this explains the exponents. Area needs be taken as $3$-dimensional Hausdorff measure. Dilates of a horizontal loop of length $L$ based at some point of the loop sweep a surface whose area is $O(L^3)$. \pause

  \item Isoperimetry is most easily proven using harmonic analysis. Ask me.\pause
\end{itemize}

\begin{cor}[Large scale geometry of $d_1$]
\begin{itemize}
  \item Volume growth: $v(R)=O(R^4)$.
  \item Isoperimetry: $I(v)=O(v^{3/4})$.
  \item Loop filling: $Fill(L)=O(L^3)$.
\end{itemize}
\end{cor}

Indeed, $d_1$ is quasiisometric to $d_\infty$.

\end{frame}

\subsection{Sublinear equivalence}

\begin{frame}

\begin{thm}[Sharper  - still unsharp - comparison between $d_1$ and $d_\infty$]
$$
d_\infty=d_1+O(d_1^{1/2}).
$$
In particular, $d_\eps\to d_\infty$ as $\eps\to\infty$.
\end{thm}\pause

It is indeed sharper, since 
$$
\frac{d_\eps(e,p)}{d_\infty(e,p)}=\frac{d_1(e,\delta_\eps(p))}{d_\infty(e,\delta_\eps(p))}
$$
tends to $1$ as $\eps\to\infty$.\pause

\medskip

\textbf{Proof of Theorem}

\smallskip

Let $p\in\HH$ at distance $L=d_1(e,p)$. Let $\gamma$ be a minimizing geodesic from $e$ to $p$, with constant speed $L$. Let $\gamma_\infty$ be the horizontal lift from $e$ of $\pi\circ\gamma$, let $\phi=\delta_{1/L}\circ\gamma$ and $\phi_\infty=\delta_{1/L}\circ\gamma_\infty$. Both have length $1$. Then
\begin{align*}
\frac{1}{L}d_\infty(e,p)&= d_\infty(e,\phi(1))\\
&\le d_\infty(e,\phi_{\infty}(1))+d_\infty(\phi_{\infty}(1),\phi(1))\\
&\le 1+C\,d_1(\phi_{\infty}(1),\phi(1))^{1/2}.
\end{align*}

\end{frame}

\begin{frame}

One views $\phi$ and $\phi_\infty$ as trajectories of left invariant time dependent vector fields $X$ and $Y$: $X(q,t)$ is the left invariant vector field which equals $\phi'(t)$ at $\phi(t)$, idem for $Y$.

\begin{lem}[Gr\"onwall-type Lemma]
Let $X$ and $Y$ be time dependent vector fields on the unit ball of Euclidean space. Assume that for all points $q,q'$ and all positive times $t$,
$$
\|X(q,t)-Y(q',t)\| \le C'\,(\eps+\|q-q'\|).
$$
Then their trajectories $\phi$ and $\psi$ from the origin satisfy
$$
\|\phi(t)-\psi(t)\|\leq (e^{C't}-1)\eps,
$$
whenever both are defined.
\end{lem}\pause

\medskip

Here, for all $t\in[0,1]$, $\gamma(t)$ and $\gamma_\infty(t)$ both lie in the $d_1$-ball of radius $L$, hence $\phi(t)$ and $\phi_\infty(t)$ both lie in a compact part of $\HH=\R^3$ where $g_1$ upper bounded by a multiple of the Euclidean norm and where left invariant vector fields are Lipschitz. 

\end{frame}

\begin{frame}

\begin{align*}
\frac{1}{L}d_\infty(e,p)
\le 1+C\,d_1(\phi_{\infty}(1),\phi(1))^{1/2}.
\end{align*}

\begin{lem}[Gr\"onwall-type Lemma]
$$
\|X(q,t)-Y(q',t)\| \le C'\,(\eps+\|q-q'\|)\implies \|\phi(t)-\psi(t)\|\leq (e^{C't}-1)\eps,
$$
\end{lem}\pause

Here, for all $q$, $Y(q,t)$ is the horizontal component of $X(q,t)= \phi'(t)=\delta_{1/L}(\gamma'(t))$ left-translated to $q$. Since $Z=\delta_L(Y(q,t)-X(q,t))$ is vertical,
$$
\|Y(q,t)-X(q,t)\|_1=\|\delta_{1/L}(Z)\|_1=\frac{1}{L^2}\|Z\|_1\le \frac{1}{L},
$$ 
so a similar inequality holds for the Euclidean norm. Changing one of the $q$ to $q'$ adds a Lipschitz term $O(\|q-q'\|)$. The Lemma yields
$$
d_1(\phi_{\infty}(1),\phi(1))\le (e^{C'}-1)\frac{\text{const.}}{L},
$$
and $\frac{1}{L}d_\infty(e,p)\le 1+O(L^{-1/2})$, where $L=d_1(e,p)$.

\end{frame}

\begin{frame}

The sharper comparison yields

\begin{cor}
Let $\omega$ denote the volume of the unit $d_\infty$ ball. Then the volumes of $d_1$ balls satisfy
$$
vol(B_1(R))=\omega\,R^{4}+O(R^{3.5}).
$$
\end{cor}\pause
Indeed, the volume of the $d_\infty$ shell $B_\infty(R)\setminus B_\infty(R-C\,\sqrt{R})$ is $\omega(R^4 - (R-C\,\sqrt{R})^4)\sim R^{3.5}$.\pause

\medskip

\textbf{Remark}. By the same method,  similar asymptotic expansions hold for periodic metrics on $\HH$, for word metrics on lattices of $\HH$.\pause

\begin{que}
Can one refine this asymptotic expansion?
\end{que}\pause
Certainly yes. There is an exact analytic expression for $d_\eps$...\pause

\medskip

\textbf{Remark}. The sharper comparison does not directly yield asymptotics for the isoperimetric profile nor the filling function. Nevertheless, the strategy of approximating Riemannian by subRiemannian quantities should apply there as well.

\end{frame}

\section{Generalization to nilpotent groups}

\subsection{SubFinsler metrics}

\begin{frame}

SubFinsler metrics generalize the subRiemannian metric encountered on Heisenberg group.\pause

\bigskip

On a smooth connected manifold $M$, a \emph{distribution} is a smooth subbundle $\Delta$ of the tangent bundle. \pause

\medskip

We always assume that it satisfies \emph{H\"ormander's condition}: iterated brackets of sections of $\Delta$ eventually span the whole tangent space at each point. This garantees (Chow's 193? theorem, with ancestors going back to C. Carath\'eodory's 1911 interpretation of S. Carnot's second principle of thermodynamics) that curves tangent to $\Delta$ join every pair of points in $M$.\pause

\medskip

A subFinsler metric on $(M,\Delta)$ is the data of a continuous family on norms on $\Delta$. Minimizing the lengths of horizontal curves yields a distance, the subFinsler distance.

\end{frame}

\subsection{Selfsimilar subFinsler metrics on Carnot groups}

\begin{frame}

Heisenberg group is a prototype in a broader family.

\begin{dfi}[Carnot group]
A \textbf{Carnot grading} on a Lie algebra $\mathfrak{g}$ is a gradation
$$
\mathfrak{g}=\bigoplus_{i=1}^{s}\mathfrak{g}_i,
$$
such that $[\mathfrak{g}_i,\mathfrak{g}_j] \subset \mathfrak{g}_{i+j}$, and $\mathfrak{g}_1$ generates $\mathfrak{g}$. $s$ is called the \textbf{step} of $\mathfrak{g}$.

A \textbf{Carnot group} is the data of a simply connected Lie group with a Carnot grading on its Lie algebra.
\end{dfi}\pause

The maps such that $\delta_\eps=\eps^i$ on $\mathfrak{g}_i$ are Lie algebra (and thus group via $\exp$) automorphisms. A norm on $\mathfrak{g}_1$ defines a left invariant subFinsler metric on the group, which is selfsimilar: $\delta_\eps$ multiplies it exactly by $\eps$.


\end{frame}

\begin{frame}

\begin{thm}[Sublinear equivalence, Breuillard-Le Donne 2013, Giannella 2017]
Let $G$ be a Carnot group of step $s$, let $d_1$ be a left invariant subFinsler metric on $G$. There exists a left invariant \textbf{selfsimilar} subFinsler metric $d_\infty$ on $G$ such that
$$
d_\infty =d_1+O(d_1^{1-(1/s)}).
$$
\end{thm}\pause

\bigskip

Take the norm on $\mathfrak{g}_1\simeq G/[G,G]$ that makes the projection $G\to G/[G,G]$ a submetry. Then same proof as for Heisenberg group. 

\end{frame}

\subsection{NonCarnot nilpotent Lie groups}

\begin{frame}

\begin{dfi}
Let $\mathfrak{g}$ be a Lie algebra. Define recursively $\mathfrak{g}^1=\mathfrak{g}$ and $\mathfrak{g}^{i+1}=[\mathfrak{g}^{i},\mathfrak{g}]$. A Lie algebra is \textbf{nilpotent} if there exists $s$ such that $\mathfrak{g}^{s+1}=\{0\}$. The smallest such $s$ is the \textbf{step} of $\mathfrak{g}$.
\end{dfi}\pause

\begin{ntn}
Let $G$ be a simply connected nilpotent Lie group with Lie algebra $\mathfrak{g}$. The number
$$
Q=\sum_{i=1}^{s}i\,\mathrm{dim}(\mathfrak{g}^{i}/\mathfrak{g}^{i+1})=\sum_{i=1}^{s}\mathrm{dim}(\mathfrak{g}^{i})
$$
is called the \textbf{homogeneous dimension} of $G$.
\end{ntn}
We shall see below that $Q$ is the degree of polynomial volume growth of $G$.\pause

\bigskip

Every Carnot Lie algebra is nilpotent. Not every nilpotent Lie algebra admits a Carnot grading.

\begin{exa}
The Lie algebra $\mathfrak{n}$ with basis $X_1,\ldots,X_5$ and nonzero brackets $[X_1,X_2]=X_3$, $[X_1,X_3]=X_4$, $[X_1,X_4]=X_5=[X_2,X_3]$ does not admit a Carnot grading.
\end{exa}

\end{frame}

\begin{frame}

To see this, we first define

\begin{dfi}[The associated Carnot graded Lie algebra $Car(\mathfrak{g})$]
Let $\mathfrak{g}$ be a nilpotent Lie algebra. Let
$$
Car(\mathfrak{g})=\bigoplus_{i=1}^{s}\mathfrak{g}^{i}/\mathfrak{g}^{i+1},
$$
with the induced brackets $\mathfrak{g}^{i}/\mathfrak{g}^{i+1} \times \mathfrak{g}^{j}/\mathfrak{g}^{j+1}\to \mathfrak{g}^{i+j}/\mathfrak{g}^{i+j+1}$. It is a Lie algebra, which comes with a tautological Carnot grading.
\end{dfi}\pause
If a Lie algebra $\mathfrak{g}$ admits a Carnot grading, then $\mathfrak{g}$ and $Car(\mathfrak{g})$ are isomorphic as Carnot Lie algebras. Here, $Car(\mathfrak{n})$ has one less nonzero bracket, $[X_2,X_3]=0$, so it is not isomorphic to $\mathfrak{n}$.\pause

\begin{dfi}[Associated Carnot group]
Let $G$ be a simply connected nilpotent Lie group. The \textbf{associated Carnot group} $Car(G)$ is the simply connected Lie group with Lie algebra $Car(\mathfrak{g})$.
\end{dfi}

\end{frame}

\begin{frame}

\begin{thm}[Breuillard-Le Donne 2013, Gianella 2017]
Let $G$ be a simply connected nilpotent Lie group of step $s$, let $d_1$ be a left invariant subFinsler metric on $G$. There exists a left invariant selfsimilar subFinsler metric $d_\infty$ on $Car(G)$ and a diffeomorphism $f:G\to Car(G)$ such that
$$
f^*d_\infty =d_1+O(d_1^{1-(1/s)}).
$$
\end{thm}\pause

The same method applies, with a change: the vector fields $X$ and $Y$ are now left invariant with respect to distinct group structures on the vector space $\mathfrak{g}$, the $G$ and $Car(G)$ multiplications, transported to $\mathfrak{g}$ and $Car(\mathfrak{g})$ by exponentials, and a linear isomorphism $\mathfrak{g}\to Car(\mathfrak{g})$ that preserves filtrations by commutators (but not gradings).\pause

\medskip

We now show that this change is insignificant.

Everything is carried to the normed vector space $\mathfrak{g}$. Recall that $p\in \mathfrak{g}$ lies at $G$-distance $L=d_1(e,p)$. Let $\gamma$ be a $d_1$-minimizing geodesic from $0$ to $p$, with constant speed $L$. Let $\gamma_\infty$ be the $Car(G)$-horizontal lift from $0$ of $\pi\circ\gamma$, let $\phi=\delta_{1/L}\circ\gamma$ and $\phi_\infty=\delta_{1/L}\circ\gamma_\infty$.


\end{frame}

\begin{frame}

$X(q,t)$ is the $G$-left invariant vector field which equals $\phi'(t)$ at $\phi(t)$, and $Y(q,t)$ is the $Car(G)$-left invariant vector field which equals $\phi'_{\infty}(t)$ at $\phi_{\infty}(t)$.\pause

\medskip

$X$ and $Y$ still have the same projection on $\mathfrak{g}/\mathfrak{g}^2$, so $X-Y$ takes values in $\mathfrak{g}^2$. 
$$
Z=\delta_L(Y(q,t)-X(q,t))=\gamma'_{\infty}(t)-\gamma'(t) = O(L).
$$\pause
Also, $\phi(t)$ and $\phi_{\infty}(t)$ stay bounded, so the restriction of the norm to $\mathfrak{g}^2$, left-translated to these points, is still multiplied by $\frac{1}{L^2}$ by $\delta_{1/L}$, up to a bounded multiplicative error. \pause Therefore
$$
\|Y(q,t)-X(q,t)\|=\|\delta_{1/L}(Z)\|\le \text{const. }\frac{1}{L^2}\|Z\|\le \frac{\text{const.}}{L},
$$ \pause
so Gr\"onwall's Lemma applies and shows that 
$$
d_1(\phi_{\infty}(1),\phi(1))\le (e^{C'}-1)\frac{\text{const.}}{L},
\quad \text{and hence}\quad
\frac{1}{L}d_\infty(0,p)
\le 1+C\,d_1(0,p)^{-1/s}.
$$\pause
Since the inequality $d_1\le d_{\infty}$ need not hold, one must exchange $G$ and $Car(G)$ and repeat the argument to get the opposite inequality, for $L'=d_{\infty}(0,p)$,
\begin{align*}
\frac{1}{L'}d_1(0,p)
\le 1+C\,d_\infty(0,p)^{-1/s}.
\end{align*}

\end{frame}

\section{GH-asymptotic cones}


\subsection{Gromov-Hausdorff convergence}

\begin{frame}

\begin{dfi}[Gromov-Hausdorff convergence]
Given two metric spaces $X$ and $Y$, an \textbf{$\eps$-approximation} is a map $\phi:X\to Y$ whose image is $\eps$-dense and which preserves distances up to an additive error of $\eps$:
$$
|d(\phi(x),\phi(x'))-d(x,x')|<\eps.
$$\pause
The \textbf{Gromov-Hausdorff distance} $dist_{GH}(X,Y)$ is the infimal $\eps$ such that there exists $\eps$-approximations $X\to Y$.\pause

The \textbf{Gromov-Hausdorff topology} on based metric spaces $(X,x_0)$ requires Gromov-Hausdorff convergence of all balls centered at the basepoint.
\end{dfi}\pause

\begin{exa}
Let $G$ be a Carnot group with selfsimilar left invariant subFinsler metric $d$. Assume $G$ admits a discrete cocompact subgroup $\Gamma$. Then the family of subgroups $(\delta_\eps(\Gamma),d_{|\delta_\eps(\Gamma)})$, with the induced distances, converges to $(G,d)$ as $\eps\to 0$.
\end{exa}\pause

\begin{exa}
The map which to a triple $([\cdot,\cdot],\Delta,\|\cdot\|)$ of a Lie algebra structure $[\cdot,\cdot]$ on $\R^n$, a generating linear subspace $\Delta\subset\R^n$ and a norm $\|\cdot\|$ on $\Delta$ associates the induced subFinsler metric on the corresponding simply connected Lie group is continuous.
\end{exa}

\end{frame}

\begin{frame}

Start from a metric space $X$ and look at it from farther and farther away. Does the picture stabilize?

\begin{dfi}[GH asymptotic cone]
Let $(X,d,x_0)$ and $(K,d_K,k_0)$ be based metric spaces. Say that $K$ is a Gromov-Hausdorff asymptotic cone if there exists a sequence $\eps_j\to 0$ such that the sequence of based metric spaces $(X,\eps_j d,x_0)$ converges to $K$.\pause

Say that $X$ admits a unique Gromov-Hausdorff asymptotic cone if the whole family of based metric spaces $(X,\eps d,x_0)$ converges in Gromov-Hausdorff topology.
\end{dfi}

It is easy to draw pictures of spaces $X$ with nonunique GH-asymptotic cones. \pause Also, if $X$ admit a unique GH-asymptotic cone $K$, then $K$ is selfsimilar.\pause

\begin{cor}
Let $(G,d_1)$ be a simply connected nilpotent Lie group with a left invariant subFinsler metric. Then $G$ admits a unique Gromov-Hausdorff asymptotic cone, which is isometric to $Car(G)$ equipped with a selfsimilar left invariant subFinsler metric.
\end{cor}\pause

GH-asymptotic cones are a special case of the general notion of asymptotic cone of a based metric space. Asymptotic cones always exist, but they are GH-asymptotic cones and locally compact only for space of polynomial volume growth.


\end{frame}

\subsection{Geometric consequences}

\begin{frame}

\begin{cor}[Volume growth]
Let $G$ be a simply connected nilpotent Lie group of homogeneous dimension $Q$ and step $s$, let $d_1$ be a left invariant subFinsler metric on $G$. There exists a positive number $\omega$ such that the volume of balls satisfies
$$
v(R) =\omega R^Q+ O(R^{Q-\frac{1}{s}}).
$$
\end{cor}\pause

However filling functions of nilpotent groups are not fully determined by their asymptotic cones. The following general inequality holds (Gromov, Papasoglu): define the filling degree of a Lie group $G$ as the infimal $\nu$ such that $Fill^{G}(L)=O(L^\nu)$ as $L\to+\infty$. Then 
$$
FillDeg(G)\le FillDeg(\text{asymptotic cone}(G)).
$$\pause

However, it is not always an inequality.

\begin{exa}[Llosa-Isenrich, Pallier, Tessera 2023]
There exist simply connected nilpotent Lie groups whose filling degree differs from that of their asymptotic cone.
\end{exa}
%Ask Gabriel Pallier.



\end{frame}

\subsection{Quantitative aspects}

\begin{frame}

Given a based metric space $(X,d)$, the speed of convergence to the asymptotic cone $K$ is measured by the function
$$
R\mapsto \sigma(R)=dist_{GH}((B^X(R),\frac{1}{R}d),(B^K(1),d_K)).
$$\pause

The following is a consequence of the asymptotic expansion of distances on nilpotent Lie groups.

\begin{cor}
Let $G$ be a simply connected nilpotent Lie group of step $s$, equipped with a left invariant subFinsler metric on $G$. Then
$$
\sigma(R)=O(R^{-\frac{1}{s}}).
$$
\end{cor}
\pause

\textbf{Question}. How sharp is this estimate?

\begin{exa}[Breuillard-Le Donne 2013]
Let $G=\R\times\HH$. This is a step $2$ nilpotent group, which satisfies $\sigma(R)\sim R^{-\frac{1}{2}}$.
\end{exa}

The authors relate this slow convergence phenomenon to the presence of abnormal curves ($\R$ factors) for the limiting subFinsler metric.

\end{frame}

\begin{frame}

Abnormality of curves is defined in general for smooth distributions $\Delta$. In the Carnot case, horizontal lifting defines a map from curves starting from $0$ in $G/[G,G]$ to curves starting from $e$ in $G$, and in particular an endpoint map $H^1([0,1],G/[G,G])\to G$. A curve is \emph{abnormal} if this map is not a submersion. \pause

\medskip

\begin{dfi}
Say a $2$-step nilpotent Lie group is \textbf{nonsingular} if in its Lie algebra $\mathfrak{g}$, for every $X\notin\mathfrak{g}^2$, $ad_X:\mathfrak{g}\to\mathfrak{g}^2$ is onto.
\end{dfi}\pause

\begin{lem}[Nicolussi Golo]
A $2$-step Carnot group is nonsingular if and only if its selfsimilar left-invariant subFinsler metrics admit no abnormal geodesics.
\end{lem} 

The geodesics in question are merely horizontal $1$-parameter groups, whence the algebraic characterization.\pause

\begin{thm}[Breuillard-Le Donne 2013, Tashiro 2021]
If $G$ is a nonsingular $2$-step Carnot group, for any left invariant subFinsler metric $d_1$ on $G$ there exists a selfsimilar left invariant Finsler metric $d_\infty$ on $G$ such that $d_1-d_\infty$ is bounded.
\end{thm}\pause

\end{frame}


\begin{frame}

\textbf{Goal}. The large scale behaviour of nilpotent Lie groups is governed by self-similarity:
\begin{itemize}
  \item There is a subclass of nilpotent Lie groups, called Carnot groups, which admit (subFinsler) metrics, which are exactly self-similar. 
  \item Every nilpotent Lie group is asymptotic to such a Carnot group.
\end{itemize}

This behaviour is specific to nilpotent groups: self-similarity and isometric homogeneity characterize subFinsler Carnot groups. 

\begin{plan}
\begin{enumerate}
  \item Large scale geometry of the Heisenberg group
  \item Generalization to nilpotent groups
  \item GH-asymptotic cones
  \item Characterization of self-similar isometrically homogeneous spaces
  \item {\color{red} Quasiisometries of nilpotent groups}
\end{enumerate}
\end{plan}

Lectures based on Enrico Le Donne's {\color{blue} \href{https://cvgmt.sns.it/media/doc/paper/5339/sub-Riem_notes\%202023-5\%20submitted_for_proposal.pdf}{book project}}.

Ask him for updates.

\end{frame}

\section{Characterization of self-similar isometrically homogeneous spaces}

\begin{frame}

\begin{thm}[Berestovskii 2004, Le Donne 2015]
If a metric space is
\begin{enumerate}
  \item locally compact,
  \item geodesic,
  \item isometrically homogeneous,
  \item self-similar, i.e. it admits a homeomorphism that multiplies distance by a number $>1$,
\end{enumerate}
then it is isometric to a subFinsler Carnot group.
\end{thm}\pause

The proof relies on
\begin{enumerate}
  \item Gleason-Montgomery-Zippin's theorem providing sufficient topological and metric conditions for the isometry group of a general metric space to be a Lie group.\pause
  \item Berestovskii's complement in the geodesic case.\pause
  \item Mitchell's description of metric tangents to equiregular subFinsler manifolds: they are Carnot groups equipped with selfsimilar subFinsler metrics. 

\end{enumerate}

\end{frame}

\subsection{Solution of Hilbert's  5th problem}

\begin{frame}

\begin{thm}[Gleason, Montgomery-Zippin 1952]
Let $X$ be a metric space. Assume that 
\begin{itemize}
  \item  $X$ is connected and locally connected,       
  \item $X$ is locally compact, 
  \item $X$ has finite topological dimension, 
  \item the isometry group $Isom(X)$ acts transitively. 
\end{itemize}
Then $Isom(X)$ has the structure of a Lie group with finitely many connected components, and $X$ has the structure of an analytic manifold.
\end{thm}\pause

\bigskip

In our setting, finite dimensionality follows from self-similarity, via \emph{doubling}.

\begin{dfi}
A metric space $X$ is \textbf{doubling} if there exists a $N$ such that every $r$-ball can be covered with $N$ $r/2$-balls.
\end{dfi}\pause

\medskip

self-similar $+$ locally compact $\implies$ doubling $\implies$ finite Hausdorff dimension $\implies$ finite topological dimension.

\end{frame}

\subsection{Geodesic and isometry homogeneous metrics}

\begin{frame}

Geodesic and isometry homogeneous metrics have been shown to be subFinsler.

\begin{thm}[Berestovskii 1988]
Let $d$ be a geodesic invariant distance on some homogeneous space of a Lie group. Then $d$ is subFinsler.
\end{thm}\pause

\bigskip

Indeed, use local coordinates near some point $x_0$ to define self-homeomorphisms $\delta_\eps$ of space $X$. Consider rescaled small balls $\beta_r=\delta_{1/r}(B(x_0,r))$ and take the intersection $\beta=\bigcap_{r\to 0}\beta_r$. 

\medskip

One shows that $\beta$ is a compact convex and centrally symmetric set. Thus it is a convex body in a linear subspace $L\subset T_{x_0}X$. It is invariant under the stabilizer $G_{x_0}$. Therefore, by translation, it defines a subFinsler metric which coincides with $d$.

\end{frame}

\subsection{Metric tangents in subFinsler geometry}

\begin{frame}

The last step exploits self-similarity, using the fact that smooth subFinsler manifolds admit metric tangents.\pause 
Indeed, a selfsimilar space is isometric to its metric tangents.

\begin{dfi}[Metric tangents]
Let $(X,d)$ be a metric space. A metric tangent at $x_0$ is a Gromov-Hausdorff limit of rescaled based spaces $(X,\eps d,x_0)$ as $\eps\to\infty$.
\end{dfi}\pause

\begin{dfi}[Equiregular distribution]
A smooth distribution $\Delta$ is \textbf{equiregular} if the subspaces of the tangent bundle spanned by 1st, 2nd... order brackets have constant rank.
\end{dfi}

\begin{thm}[Mitchell 1985]
Let $X$ be a smooth equiregular subFinsler manifold. Then $X$ admits at every point a unique metric tangent: it is a Carnot group equipped with a selfsimilar subFinsler metric. 
\end{thm}\pause

\bigskip

We see that tangents happen to be groups. Why?

\end{frame}

\begin{frame}

Here is a heuristic suggested by Bella\"iche and implemented by O. Mohsen.

\bigskip

Let $X$ be a topological space. Then $X\times X$ is a pseudogroup, i.e. there exists a partially defined multiplication rule, here
$$
(x_1,x_2),\,(x_2,x_3) \mapsto (x_1,x_3).
$$\pause
Assuming that $X$ admits a metric tangent $T_x$ at each point, then there exist approximate isometries $\phi_\eps^x:(X,\eps d)\to T_x$. One can introduce a topology on the disjoint union
$$
X\times X\times(0,1] \cup TX,\quad \text{where }TX:=\bigcup_{x\in X} T_x.
$$
where $(x,x',\eps)$ converges to $(z,v)$ when $\eps\to 0$, $x\to z$ and $\phi_\eps^x(x')\to v$. \pause

\medskip

Then the pseudogroup structure on $X\times X$ converges to a pseudogroup structure on $TX$, where elements $(x,v)$ and $(x',v')$ can be multiplied if and only if $x=x'$. Therefore each metric tangent $T_x$ carries a multiplication. Inverses exist in $X\times X$, so they exist in $T_x$, therefore $T_x$ is a group.

\end{frame}

\subsection{Quasiisometries of nilpotent groups}

\begin{frame}

\begin{thm}
If two nilpotent Lie groups $G$ and $G'$ are quasiisometric, then their associated Carnot groups $Car(G)$ and $Car(G')$ are isomorphic (as groups with gradings).
\end{thm}\pause

\medskip

\textbf{Step 1}. Every quasiisometry $f:G\to G'$ induces a (nonunique) biLipschitz map $Car(G)\to Car(G')$ with respect to subFinsler metrics.\pause

\medskip

Indeed, setting $d_\eps=\eps d_1$, $d'_\eps=\eps d'_1$, 
$$
-C+\frac{1}{L}\,d_1(x,x')\le d'_1(f(x),f(x')) \le Ld_1(x,x')+C
$$ 
$$
\implies -\eps C+\frac{1}{L}\,d_\eps(x,x')\le d'_\eps(f(x),f(x')) \le Ld_\eps(x,x')+\eps C.
$$ \pause 

Composing with $\eps$-approximations yields maps $Car(G)\to Car(G')$ which are biLipschitz up to small additive error. Arzela-Ascoli $\implies$ a (nonunique) limiting biLipschitz map.


\end{frame}

\begin{frame}



\textbf{Step 2}. From now on, $G,G'$ are Carnot, $f:G\to G'$ is biLipschitz with respect to selfsimilar subFinsler metrics. We would like that $\delta_{\eps}\circ f\circ \delta_{1/\eps}$ converges as $\eps\to 0$. If so, the limit should be a selfsimilar, i.e. graded, group homomorphism. This suggests the

\begin{dfi}[Differentiabilty]
Say a map $f$ between neighborhoods of $e$ in Carnot groups is \textbf{differentiable} at $e$ if $\delta_{1/\eps}\circ f\circ \delta_{\eps}$ converges uniformly on compact sets as $\eps\to 0$ to a graded group homomorphism. Using left translations, one defines differentiability at other points.
\end{dfi}\pause


\begin{thm}[Rademacher-Stepanov]
Lipschitz and quasiconformal maps between open sets of Carnot groups are differentiable almost everywhere. In the quasiconformal case, differentials are graded group isomorphisms.
\end{thm}\pause

This proves the above quasiisometric rigidity theorem.


\end{frame}

\begin{frame}

\textbf{Case where $f:\HH\to\R$ is Lipschitz}. 

A Lipschitz map admits horizontal partial derivatives almost everywhere. These derivatives are approximately continuous almost everywhere.

\smallskip

\begin{minipage}{5cm}
Write $e=p_0$, $p=p_6$, $\partial_i=$ successively $\xi,\eta,-\xi,-\eta,\xi,\eta$,
\begin{align*}
&f(p)-f(e)=\sum_{i=1}^6 f(p_i)-f(p_{i-1})\\
&=\sqrt{|z|}\sum_{i=0}^3 (\partial_i f)(p_i)+x(\xi f)(p_4)+y(\eta f)(p_5)+\eps\\
&=\sqrt{|z|}\sum_{i=0}^3 (\partial_i f)(e)+x(\xi f)(e)+y(\eta f)(e)+\eps\\
&=x(\xi f)(e)+y(\eta f)(e)+\eps
\end{align*}


\end{minipage}\begin{minipage}{5cm}

\begin{center}

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\end{center}

\end{minipage}

where $\eps=o(|x|+|y|+\sqrt{|z|})$, provided $e$ is a point of existence and approximate continuity of all partial derivatives, the Lipschitz bound allowing the needed flexibility in the choice of break points. 

\end{frame}

\begin{frame}

\textbf{Case where $f:\R\to\HH$ is Lipschitz}. 

\smallskip

Assume that $f(0)=e$, $f(t)=(x(t),y(t),z(t))$. One shows that $f$ is horizontal, hence
$$
z(t)-\frac12 x(t)y(t)=\int_{0}^{t}\frac12(x(s)y'(s)-x'(s)y(s))\,ds,
$$
is the area enclosed by the projection $s\mapsto(x(s),y(s))$ and the line segment from $(x(t),y(t))$ to $(0,0)$.

\begin{center}

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\draw (-2.0761828727643485,-2.6792128625140723) node[anchor=north west] {$(0,0)$};
\draw (7.988558612930328,-2.712104174558826) node[anchor=north west] {$(x(t),y(t))$};
\begin{scriptsize}
\draw [fill=qqqqff] (-0.398725958481902,-3.7317348479461954) circle (1.5pt);
\draw [fill=qqqqff] (7.626754180438038,-3.6988435359014415) circle (1.5pt);
\end{scriptsize}
\end{tikzpicture}

\end{center}\pause

So, if $x$ and $y$ have derivatives at $0$, $z(t)-\frac12 x(t)y(t)=o(t^2)$.
\begin{align*}
\delta_{1/\eps}(x(\eps),y(\eps),z(\eps))
&=(\frac{1}{\eps}x(\eps),\frac{1}{\eps}y(\eps),\frac{1}{\eps^2}z(\eps))
\end{align*}
tends to $(x'(0),y'(0),\frac12 x'(0)y'(0))$.

\end{frame}

\begin{frame}

\textbf{General case} is a mixture.\pause

\medskip

\textbf{Absolute continuity}. In order to show that differentials are bijective, one needs show that for a.e. $p\in G$, $f^{-1}$ is differentiable at $f(p)$. I.e. that $f^{-1}$ of the differentiability locus of $f^{-1}$ has full measure. This is obvious for biLipschitz maps, but not for quasiconformal maps.

\begin{lem}[Mostow 1970]
Quasiconformal homeomorphisms between open sets in Carnot groups are absolutely continuous on horizontal lines, and hence map null sets to null sets.
\end{lem}\pause

\medskip

The Rademacher-Stepanov type theorem and the absolute continuity theorem have been extended to smooth equiregular distributions by Margulis and Mostow: differentials become graded homomorphisms between metric cones. The theorems hold under weaker differentiability assumptions on distributions, see work by Vodopjanov and his school.

Recently, differentiability loci of Lipschitz mapping on Carnot groups have been investigated by De Philippis, Marchese, Merlo, Pinamonti and Rindler: the Rademacher-Stepanov theorem fails for all measure classes but Haar's.

\end{frame}


\end{document}


















Pat Ballard's Mister Sandman, sung by The Chordettes
https://www.youtube.com/watch?v=CX45pYvxDiA
https://www.youtube.com/watch?v=KCGJWvHJTCI

Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk
Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk

Mister Mario, I had a dream (Bonk, Bonk, Bonk, Bonk)
I dreamed that bright minds would join our team (Bonk, Bonk, Bonk, Bonk)
You're so talented as an analyst (Bonk, Bonk, Bonk, Bonk)
You've done so much that I cannot make a list
You're dynamic and sometimes complex (Bonk, Bonk, Bonk, Bonk)
For you I lay a red Sierpinski carpet (Bonk, Bonk, Bonk, Bonk)
Thank you for joining the team
Mario, you did fulfill my dream

Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk
Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk

Mister Kleiner, I had a dream
I dreamed that bright minds would join our team
You were so great when you were Riemannian
I am so grateful you turned subRiemannian
You're so singular, you keep flowing away
Without embedding your feet in the clay
Thank you for joining the team
Bruce, I am glad you fulfilled my dream

Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk
Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk, Bonk

Bruce and Mario (yes), you've brought us a dream
You've brought our subject to the mainstream
You have connected it to topology
Dynamical systems and dendrology
For analysts, space is metric
Maps are no better than quasisymmetric,  
Here is the goal of today's venue
Making the quasiworld, please, please, please
Making the quasi... world... come... true






Mr. Sandman, bring me a dream (bung, bung, bung, bung)
Make him the cutest that I've ever seen (bung, bung, bung, bung)
Give him two lips like roses and clover (bung, bung, bung, bung)
Then tell him that his lonesome nights are over
Sandman, I'm so alone (bung, bung, bung, bung)
Don't have nobody to call my own (bung, bung, bung, bung)
Please turn on your magic beam
Mr. Sandman, bring me a dream

Mr. Sandman, bring me a dream
Make him the cutest that I've ever seen
Give him the word that I'm not a rover
Then tell him that his lonesome nights are over
Sandman, I'm so alone
Don't have nobody to call my own
Please turn on your magic beam
Mr. Sandman, bring me a dream

Mr. Sandman (yes) bring us a dream
Give him a pair of eyes with a "come-hither" gleam
Give him a lonely heart like Pagliacci
And lots of wavy hair like Liberace
Mr. Sandman, someone to hold (someone to hold)
Would be so peachy before we're too old
So please turn on your magic beam
Mr. Sandman, bring us, please, please, please
Mr. Sandman, bring us a dream





Il Ristoro della Pe'
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Téléphone : 050 554873
