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].

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