Lecture 15. The Magnus expansion

Posted on January 3, 2013 by Fabrice Baudoin

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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