Differential Operator Invertible Good evening,
I have a little problem with the invertibility of an operator $A : W^{1,2}(S^1,\mathbb{R}^{2n})\longrightarrow L^2(S^1,\mathbb{R^{2n}})$ defined as $$(A\zeta)(t) = \dot{\zeta}(t)-\dot{\Psi}(t)\cdot\Psi^{-1}(t)\cdot\zeta(t),$$
where $\Psi(t)\in \mathbb{R}^{2n\times 2n}$ is a path of symplectic matrices with $\Psi(0)=Id$.
The claim then is that $A$ is invertible if and only if $1$ is not an eigenvalue of $\Psi(t)$ for any $t$.
In the paper, it sounds like this is obvious, so I started to show that $\ker A = \{0\}$: If $A\zeta = 0$ (i.e. $(A\zeta)(t) = 0$ a.e.), we have
$$\dot{\Psi}(t)\cdot\underbrace{\Psi^{-1}(t)\cdot\zeta(t)}_{=:\alpha(t)} =\dot{\zeta}(t)  .$$
So setting $\alpha (t) := \Psi^{-1}(t)\cdot\zeta(t)$, this simplifies to
\begin{align}
\dot{\Psi}(t)\cdot\alpha(t) &=\frac{d}{dt}(\Psi(t)\cdot\alpha(t)) \\ \Leftrightarrow \dot{\Psi}(t)\cdot\alpha(t) &=\dot{\Psi}(t)\cdot\alpha(t)+\Psi(t)\dot{\alpha}(t) \\ \Leftrightarrow \Psi(t)\cdot\dot{\alpha}(t) &= 0
\end{align}
Since $\Psi(t)$ is invertible (as a symplectic matrix), this means $\dot{\alpha}(t) = 0$, so $\alpha(t)$ is a constant vector and so by definition of $\alpha$ we get that
$$ \zeta(t)=\Psi(t)\cdot\alpha $$
is actually an element of the kernel. This seems be the place where the claim with $1$ (not) being an eigenvalue of $\Psi(t)$ should come into play, however I don't really see how. Of course, maybe my steps have an error somewhere, so I would really like to hear other people's opinions on it.
 A: $\Rightarrow$: Let $v \in \mbox{ker}(\Phi(\tilde{t}) - id) \neq \lbrace 0 \rbrace$ for some $\tilde{t} \in [0,1]$ i.e.
\begin{equation}
\Phi(\tilde{t})v = v
\end{equation}
Define
\begin{equation}
v(t) = \Phi(t)v
\end{equation}
assume w.l.o.g. $\tilde{t} = 1$ then
$v(1) = \Phi(1)v = v = v(0)$ hence $v \in W^{1,2}(S^1, \mathbb{R}^{2n})$. Now
\begin{equation}
Av = 0
\end{equation}
means
\begin{equation}
\frac{d}{dt}v = -S v
\end{equation}
where $S = \dot{\Phi}(t)\Phi^{-1}(t)$ or equivalent
\begin{equation}
\frac{d}{dt}v = \left(\frac{d}{dt}\Phi\right) \Phi^{-1}v = -\Phi \frac{d}{dt}(\Phi^{-1})v
\end{equation}
which can be rephrased by saying $\frac{d}{dt}(\Phi^{-1}v) = 0$ or
\begin{equation}
v(t) = \Phi(t)v_0
\end{equation}
for $v \in \mathbb{R}^{2n}$. Since $v$ is a loop we obtain
$v(0) = v(1) = \Phi(1)v(0)$. This shows that $v \in \mbox{ker} A$ hence $\mbox{ker} A \neq \lbrace 0 \rbrace$ and $A$ is not invertible.
$\Leftarrow$: Let $v \in L^2(S^1, \mathbb{R}^{2n})$. We wish to find a unique $u \in W^{1,2}(S^1, \mathbb{R}^{2n})$ such that
\begin{equation}
Au = v.
\end{equation}
This implies
\begin{equation}
\dot{u} =(Su - v)
\end{equation}
We can construct a solution $u_x$ for any $x \in \mathbb{R}^{2n}$ satisfying $Au_x = v$ and $u_x(0) = x$ by defining
\begin{equation}
u_x(t) = \Phi(t)\left(x - \int^{t}_{0}{\Phi^{-1}(s)}v(s)ds \right)
\end{equation}
for any $x \in \mathbb{R}^{2n}$. Now the periodicity of $u_x$ is the case if
\begin{equation}
x = \Phi(1)\left( x- \int^{1}_{0}{\Phi(s)^{-1}v(s) ds} \right).
\end{equation}
Rewriting
\begin{equation}
(\Phi(1) - id)x = \Phi(1)\int^{1}_{0}{\Phi^{-1}(s)v(s)ds}.
\end{equation}
This can be solved by $\Phi(1) - id$ is invertible hence $A$ is invertible.
