Consider matrix ODE:
$latex \dot{\phi}(t)=A(t)\phi(t)$
Where $latex A(t)$ is a given periodic matrix with period $latex T$, i.e. $latex A(x)=A(x+T), \forall x\in R$.
Then the solution $\phi(t)$ satisfied identity:
$latex \phi(t+T)=\phi(t)\phi^{-1}(0)\phi(T)$.
This could be explained as $latex \phi^{-1}\phi(T)=\int_{0}^T\phi(t)$.
Now we consider to solve the equation: $latex e^{TB}=\phi^{-1}(0)\phi(T)$. At least formally it could be solved:
$latex B=\frac{1}{T}log(\frac{\phi(T)}{\phi(0)})$.
(Unfortunately $latex log$ is a multi-value function so $latex B=B_0+2\pi ik I$, where $latex I$ is the identity matrix and $latex B$ is a solution of $latex e^{TB}=\phi^{-1}(0)\phi(T)$.) This argument is false.
In fact matrix is not like numbers, the $latex log$ function is much more complicated. we have,
$latex log(A)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{A^n}{n} …(*)$
So to solve $latex e^{TB}=\frac{1}{T}(\frac{\phi(T)}{\phi(I)})$, it is equivalent to :
$latex B=\frac{1}{T}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(\frac{\phi(T)}{\phi(I)})^n}{n}$
But this type of identity only meaningful when $latex ||\frac{\phi(T)}{\phi(I)}||<1$, so is it true that for $latex ||\frac{\phi(T)}{\phi(I)}||<1$ the equation is solved by $latex (*)$, and for $latex ||\frac{\phi(T)}{\phi(I)}||\geq 1$ it do not have solution?
The naive inspirit is wrong, the situation is similar to the $latex \mathbb Q_p$ case while $latex log_p$ could extend to $latex D(p^{\frac{-1}{p-1}-})$ and the identity
$latex exp_p(log_p(1+x))=1+x$
always holds for $latex x\in D(p^{\frac{-1}{p-1}-})$. The key observation is $latex log[(1+Y)(1+Y)]=log(1+X)+log(1+Y)$ always holds when $latex ||X||,||Y||<1$, this will lead to a reasonable value of
$latex log[(1+X)(1+Y)]=log(1+X+Y+XY)$
even when $latex ||X+Y+XY||\geq 1$ and this process could be continue to the whole matrix space and the identity enjoy the accosted principle so $latex log(X)$ is well-defined for all $latex X\in M_{2\times 2}$.
Now it is time to consider the rotation number, which is defined by $latex \lim_{n\to \infty}\frac{f^{n}(x)-x}{n}$ for $latex f:R\to R$ is a continuous increasing function.
And I do not know how to associated a dynamic system for the matrix $latex B$ given here, but in any case it seems iff it is given by a hemoermorphifm then the rotation number is zero due to the following reason:
Consider $latex \mathbb S^1$ as the quotient $latex \mathbb R/\mathbb Z$. Your homeomorphism $latex f$ lifts to a homeomorphism
$latex \phi : \mathbb R \to \mathbb R$ such that $latex \phi(x+1)=\phi(x)+1$.
Form the map $latex h:=\frac{1}{q} \sum _{n=1} ^q (\phi^{\circ n}-pn)$, where $latex \phi ^{\circ n}$ is the composition $latex n$ times of $latex \phi$ with itself. By construction $latex h\circ \phi = h+\frac{p}{q} $ and $latex h(x+1)=1+h(x)$, so that $latex h$ factors as a homeomorphism of the circle conjugating $latex f$ to the rotation.
By the way this approach wors in $latex \mathbb R^n$ too.
Maslov index of a holomorphic disk
A natural way to understand the rotation number here is according the way of maslov index, we have the following formula:
$latex f(A)=\int_{\Gamma}\frac{1}{2\pi i}\frac{f(\lambda)}{\lambda I-A}f(\lambda)d\lambda$
$latex TB=log(\frac{\phi(T)}{\phi(0)})=log(\int_0^T \phi'(\lambda)d\lambda)=\int_0^T log(\phi'(\lambda))d\lambda=\int_0^T log(A+F(t))d\lambda$
Proof sketch:
1.$latex B=\frac{1}{T}log(e^{\int_0^T A+f(t)dt})= \frac{1}{T}(\int_{0}^T A+f(t)dt)$.
2. The dynamic system is defined by : $latex W: R^2-\{0\} \to R^2-\{0\}$, $latex W( x)=B x$.
3. this dynamic system $latex (R^2-\{0\},W)$ is conjugate to the dynamic system $latex T:S_1\to S_1$, Not difficult to proof it is a homomorphism on $latex S_1$ and it is zero entropy by Pesin’s formula
If $latex T: S_1\to S_1$ could lifting to $\hat T:R \to R$ the rotation number is defined as :
$latex \lim_{n\to \infty}\frac{\hat T^n(x)}{n}$
This problem is not a good problem due to the philosophy, i.e. use rotation number to describe the information of a hamiltonian flow is not satisfied, in fact it is difficult to establish a suitable definition of “rotation number”! But this is the first crucial thing to establish a theorem!
Hamiltonian flow
In mathematics and physics, a Hamiltonian vector field on a symplectic manifold is a vector field, defined for any energy function or Hamiltonian. A Hamiltonian vector field is a geometric manifestation of Hamilton’s equations in classical mechanics. The integral curves of a Hamiltonian vector field represent solutions to the equations of motion in the Hamiltonian form. The diffeomorphisms of a symplectic manifold arising from the flow of a Hamiltonian vector field are known as canonical transformations in physics and (Hamiltonian) symplectomorphisms in mathematics.[1]
Hamiltonian vector fields can be defined more generally on an arbitrary Poisson manifold. The Lie bracket of two Hamiltonian vector fields corresponding to functions f and g on the manifold is itself a Hamiltonian vector field, with the Hamiltonian given by the Poisson bracket of f and g.