## Lecture 15. The Magnus expansion

In the previous lecture, we proved the Chen’s expansion formula which establishes the fact that the signature of a path is the exponential of a Lie series. This expansion is of course formal but analytically makes sense in a number of situations that we now describe. The first case of study are linear equations.

Let us consider matrices $M_1,\cdots,M_d \in \mathbb{R}^{n \times n}$ and let $y_n:[0,T] \to \mathbb{R}^n$ be the solution of the differential equation
$y(t)=y(0)+\sum_{i=1}^d \int_0^t M_i y(s)d x^i(s),$
where $x \in C^{1-var}([0,T],\mathbb{R}^d)$. The solution $y$ admits a representation as an absolutely convergent Volterra series
$y(t)=y(0)+ \sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)} M_{i_1}\cdots M_{i_k} \left( \int_{\Delta^{k}[0,t]}dx^{I} \right) y(0).$
The formal analogy between this expansion and the signature leads to the following result:

Proposition: There exists $\tau > 0$ such that for $0 \le t \le \tau$,
$y(t)= \exp \left( \sum_{k \geq 1} \sum_{I \in \{1,\cdots ,d\}^k}\Lambda_I (x)_{t} M_I \right)y(0),$
where
$M_I = [M_{i_1},[M_{i_2},...,[M_{i_{k-1}}, M_{i_{k}}]...],$
is the iterated Lie bracket and
$\Lambda_I (x)_{t}= \sum_{\sigma \in \mathcal{S}_k} \frac{\left(-1\right) ^{e(\sigma )}}{k^{2}\left( \begin{array}{l} k-1 \\ e(\sigma ) \end{array} \right) } \int_{\Delta^k[0,t]} dx^{\sigma^{-1} \cdot I}.$

Proof: We only give the sketch of the proof. Details can be found in this paper by Strichartz. First, we observe that a combinatorial argument shows that
$\sum_{\sigma \in \mathcal{S}_k} \frac{1}{\left( \begin{array}{l} k-1 \\ e(\sigma ) \end{array} \right) } \le \frac{C}{2^k} k! \sqrt{k}.$
On the other hand, we have the estimate
$\left| \int_{\Delta^k[0,t]} dx^{\sigma^{-1} \cdot I} \right| \le \int_{\Delta^k[0,t]} \| dx^{\sigma^{-1} \cdot I}\| \le \frac{1}{k!} \| x \|^k_{1-var, [0,t]}.$
As a consequence, we obtain
$\left| \Lambda_I (x)_{t} \right| \le \frac{C}{2^k k^{3/2}} \| x \|^k_{1-var, [0,t]}.$
For the matrix norm we have the estimate $\| M_I \| \le C^k,$ so we conclude that for some constant $\tilde{C}$,
$\left\| \sum_{I \in \{1,\cdots ,d\}^k}\Lambda_I (x)_{t} M_I \right\| \le \frac{\tilde{C}^k}{ k^{3/2}} \| x \|^k_{1-var, [0,t]}.$
We deduce that if $\tau$ is such that $\| x \|_{1-var, [0,\tau]} < \frac{1}{\tilde{C}}$, then the series
$\sum_{k \geq 1} \sum_{I \in \{1,\cdots ,d\}^k}\Lambda_I (x)_{t} M_I$
is absolutely convergent on the interval $[0,\tau]$. At this point, we can observe that the Chen's expansion formula is a purely algebraic statement, thus expanding the exponential
$\exp\left( \sum_{k \geq 1} \sum_{I \in \{1,\cdots ,d\}^k}\Lambda_I (x)_{t} M_I \right)y(0)$
and rearranging the terms leads to
$y(0)+ \sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)} M_{i_1}\cdots M_{i_k} \left( \int_{\Delta^{k}[0,t]}dx^{I} \right) y(0)$
which is equal to $y(t)$ $\square$

Another framework, close to this linear case, in which the Chen's expansion makes sense are Lie groups. Let $\mathbb{G}$ be a Lie group acting on $\mathbb{R}^d$. Let us denote by $\mathfrak{g}$ the Lie algebra of $\mathbb{G}$. Elements of $\mathfrak{g}$ can be seen as vector fields on $\mathbb{R}^d$. Indeed, for $X\in \mathfrak{g}$, we can define
$X(x)=\lim_{t \to 0} \frac{ e^{tX}(x)-x}{t},$
where $e^{tX}$ is the exponential mapping on the Lie group $\mathbb{G}$. With this identification, it is easily checked that the Lie bracket in the Lie algebra coincides with the Lie bracket of vector fields and that the exponential map $e^{tX}$ in the group corresponds to the flow generated by the vector field $X$. As above we get then the following result:

Proposition: Let $V_1,\cdots, V_d \in \mathfrak{g}$ and $x \in C^{1-var}([0,T], \mathbb{R}^d)$. Let us consider the differential equation
$y(t)=y(0)+\sum_{i=1}^d \int_0^t V_i(y(s)) dx^i(s).$
There exists $\tau > 0$ such that for $0 \le t \le \tau$,
$y(t)= \exp \left( \sum_{k \geq 1} \sum_{I \in \{1,\cdots ,d\}^k}\Lambda_I (x)_{t} V_I \right)y(0).$

A special case will be of interest for us: The case where the Lie group $\mathbb{G}$ is nilpotent. Let us recall that a Lie group $\mathbb{G}$ is said to be nilpotent of order $N$ if every bracket of length greater or equal to $N+1$ is 0. In that case, the sum in the exponential is finite and the representation is then of course valid on the whole time interval $[0,T]$.

This entry was posted in Rough paths theory. Bookmark the permalink.