## Lecture 16. Free Carnot groups

We introduce here the notion of Carnot group, which is the correct structure to understand the algebra of the iterated integrals of a path up to a given order. It is worth mentioning that these groups play a fundamental role in sub-Riemannian geometry as they appear as the tangent cones to sub-Riemannian manifolds.

Definition: A Carnot group of step (or depth) $N$ is a simply connected Lie group $\mathbb{G}$ whose Lie algebra can be written
$\mathcal{V}_{1}\oplus...\oplus \mathcal{V}_{N},$
with
$\lbrack \mathcal{V}_{i},\mathcal{V}_{j}]=\mathcal{V}_{i+j}$
and $\mathcal{V}_{s}=0,\text{ for }s > N.$

There are some basic examples of Carnot groups.

Example 1: The group $\left( \mathbb{R}^d ,+ \right)$ is the only commutative Carnot group.

Example 2: (Heisenberg group) Consider the set $\mathbb{H}_n =\mathbb{R}^{2n} \times \mathbb{R}$ endowed with the group law
$(x,\alpha) \star (y, \beta)=\left( x+y, \alpha + \beta +\frac{1}{2} \omega (x,y) \right),$
where $\omega$ is the standard symplectic form on $\mathbb{R}^{2n}$, that is
$\omega(x,y)= x^t \left( \begin{array}{ll} 0 & -\mathbf{I}_{n} \\ \mathbf{I}_{n} & ~~~0 \end{array} \right) y.$
On $\mathfrak{h}_n$ the Lie bracket is given by
$[ (x,\alpha) , (y, \beta) ]=\left( 0, \omega (x,y) \right),$
and it is easily seen that $\mathfrak{h}_n=\mathcal{V}_1 \oplus \mathcal{V}_2,$ where $\mathcal{V}_1 =\mathbb{R}^{2n} \times \{ 0 \}$ and $\mathcal{V}_2= \{ 0 \} \times \mathbb{R}$. Therefore $\mathbb{H}_n$ is a Carnot group of depth 2.

The Carnot group $\mathbb{G}$ is said to be free if $\mathfrak{g}$ is isomorphic to the nilpotent free Lie algebra with $d$ generators. In that case, $\dim \mathcal{V}_{j}$ is the number of Hall words of length $j$ in the free algebra with $d$ generators. A combinatorial argument shows then that:
$\dim \mathcal{V}_{j}= \frac{1}{j} \sum_{i \mid j} \mu (i) d^{\frac{j}{i}}, \text{ } j \leq N,$
where $\mu$ is the Möbius function. A consequence from this is that when $N \rightarrow +\infty$,
$\dim \mathfrak{g} \sim \frac{d^N}{N}.$
The free Carnot groups are the ones that will be the most relevant for us, so from now on, we will restrict our attention to them.

Let $\mathbb{G}$ be a free Carnot group of step $N$. Notice that the vector space $\mathcal{V}_{1}$, which is called the basis of $\mathbb{G}$, Lie generates $\mathfrak{g}$, where $\mathfrak{g}$ denotes the Lie algebra of $\mathbb{G}$. Since $\mathbb{G}$ is step $N$ nilpotent and simply connected, the exponential map is a diffeomorphism and the Baker-Campbell-Hausdorff formula therefore completely characterizes the group law of $\mathbb{G}$ because for $U,V \in \mathfrak{g}$,
$\exp U \exp V = \exp \left( P (U,V) \right)$
for some universal Lie polynomial $P$ whose first terms are given by
$P (U,V) = U+V+\frac{1}{2} [U,V] +\frac{1}{12} [[U,V],V]-\frac{1}{12}[[U,V],U]$
$-\frac{1}{48} [V,[U,[U,V]]]-\frac{1}{48} [U,[V,[U,V]]]+\cdots.$
On $\mathfrak{g}$ we can consider the family of linear operators $\delta_{t}:\mathfrak{g} \rightarrow \mathfrak{g}$, $t \geq 0$ which act by scalar multiplication $t^{i}$ on $\mathcal{V}_{i}$. These operators are Lie algebra automorphisms due to the grading. The maps $\delta_t$ induce Lie group automorphisms $\Delta_t :\mathbb{G} \rightarrow \mathbb{G}$ which are called the canonical dilations of $\mathbb{G}$.

It is an interesting fact that every free Carnot group of step $N$ is isomorphic to some $\mathbb{R}^m$ endowed with a polynomial group law. Indeed, let $X_1,\cdots,X_d$ be a basis of $\mathcal{V}_{1}$. From the Hall-Witt theorem we can construct a basis of $\mathfrak{g}$ which is adapted to the grading
$\mathfrak{g}=\mathcal{V}_{1}\oplus \cdots \oplus \mathcal{V}_{N},$
and such that every element of this basis is an iterated bracket of the $X_i$‘s. Such basis, which can be made quite explicit, will be referred to as a Hall basis over $X_1,\cdots,X_d$. Let $\mathcal{B}$ be such a basis. For $X \in \mathfrak{g}$, let $[X]_\mathcal{B}$ be the coordinate vector of $X$ in the basis $\mathcal{B}$. If we denote by $m$ the dimension of $\mathfrak{g}$, we see that we can define a group law $\star$ on $\mathbb{R}^m$ by the requirement that for $X,Y \in \mathfrak{g}$,
$[X]_\mathcal{B} \star [Y]_\mathcal{B} =[ P_N(X,Y) ]_\mathcal{B}=[ \ln (e^X e^Y) ]_\mathcal{B}.$
It is then clear that $(\mathbb{R}^m, \star)$ is a Carnot group of step $N$ whose Lie bracket is given by:
$[ [X]_\mathcal{B} , [Y]_\mathcal{B}] =[ [X,Y] ]_\mathcal{B}.$
Therefore, every free Carnot group of step $N$ such that $\dim \mathcal{V}_{1}=d$ is isomorphic to $(\mathbb{R}^m, \star)$. Another representation of the free Carnot group of step $N$ which is particularly adapted to rough paths theory is given in the framework of formal series. As before, let us denote by $\mathbb{R}[[X_1, \cdots, X_d ]]$ the set formal series. Let us denote by $\mathbb{R}_N[X_1,\cdots,X_d]$ the set of truncated series at order $N$, that is $\mathbb{R}[[X_1, \cdots, X_d ]]$ quotiented by $X_{i_1}\cdots X_{i_k}=0$ if $k \ge N+1$. In this context, the free nilpotent Lie algebra of order $N$ can be identified with the Lie algebra generated by $X_1,\cdots,X_d$ inside $\mathbb{R}_N[X_1,\cdots,X_d]$, where the bracket is of course given by the anticommutator. This representation of the free nilpotent Lie algebra of depth $N$ shall be denoted by $\mathfrak{g}_N(\mathbb{R}^d)$ in the sequel of the course. The free nilpotent group of step can then be represented as $\mathbb{G}_N(\mathbb{R}^d)=\exp ( \mathfrak{g}_N(\mathbb{R}^d))$ where the exponential map is the usual exponential of formal series.

We are now ready for the definition of the lift of a path in $\mathbb{G}_N(\mathbb{R}^d)$.

Definition: Let $x \in C^{1-var}([0,T],\mathbb{R}^d)$. The $\mathbb{G}_N(\mathbb{R}^d)$ valued path
$\sum_{k=0}^{N} \int_{\Delta^k [0,t]} dx^{\otimes k}, \quad 0 \le t \le T,$
is called the lift of $x$ in $\mathbb{G}_N(\mathbb{R}^d)$ and will be denoted by $S_N(x)$.

It is worth noticing that $S_N(x)$ is indeed valued in $\mathbb{G}_N(\mathbb{R}^d)$ because from the Chen’s expansion formula:
$S_N(x)(t)=\exp \left( \sum_{k = 1}^N \sum_{I \in \{1,\cdots ,d\}^k}\Lambda_I (x)_{t} X_I \right),$
where the notations have been introduced before. The multiplicativity property of the signature also immediately implies that for $s \le t$,
$S_N(x)(t)=S_N(x)(s)\exp\left( \sum_{k = 1}^N \sum_{I \in \{1,\cdots ,d\}^k}\Lambda_I (x)_{s,t} X_I\right).$

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