The Magnus expansion $\Omega(t, t\_0) = \sum^\infty\_k \Omega\_k(t,t\_0)$ is so that the solution


$$
Y(t) = e^{\Omega(t,t\_0)}\,Y\_0,
$$


solves an ODE


$$
Y'(t) = A(t)\,Y(t), \qquad Y(t\_0) = Y\_0.
$$


Consequently, we know that $d e^{\Omega(t,t\_0)}/dt = A(t)\, e^{\Omega(t,t\_0)}$.


I am asking if the following properties hold in general and how to prove if true:


1. $d \Omega(t,t\_0)/dt = A(t)$? (I think no)
2. $e^{\Omega(t,t\_0)} = e^{\Omega(t, s)}\,e^{\Omega(s, t\_0)}$?
3. $d e^{\Omega(t,t\_0)}/dt\_0 = -e^{\Omega(t,t\_0)}\,A(t\_0)$?




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References of Magnus expansion: [Wiki](https://en.wikipedia.org/wiki/Magnus_expansion)



Properties of Magnus expansion
To answer these questions it it helpful to write
$$e^{\Omega(t,t\_0)}={\cal T}\exp\left(\int\_{t\_0}^t A(t')dt'\right),\;\;t\geq t\_0,$$
in terms of the time ordering operator ${\cal T}$, such that a product of noncommuting operators $A(t\_1)A(t\_2)\ldots$ is ordered in the way that $t\_1\geq t\_2\geq\cdots$. By construction, the time ordering operator commutes with $d/dt$ or $d/dt\_0$, and we can differentiate the exponent of the integral without worrying about the operator ordering.


1. We thus have
$$\frac{d}{dt} e^{\Omega(t,t\_0)}=A(t)e^{\Omega(t,t\_0)},$$
but *not* $d\Omega/dt=A(t)$.
2. If $t\geq s\geq t\_0$, we have indeed
$$e^{\Omega(t,t\_0)} = e^{\Omega(t, s)}\,e^{\Omega(s, t\_0)}.$$
3. We also have
$$\frac{d}{dt\_0} e^{\Omega(t,t\_0)}=-e^{\Omega(t,t\_0)}A(t\_0).$$


