For the almost Mathieu operator, which can be described by the following SL$(2,R)$ matrix, dubbed the transfer matrix, $$ T(E, n;\omega) = \begin{pmatrix} E - 2\lambda \cos(2\pi\omega n) & -1 \\ 1 & 0 \end{pmatrix} $$ the spectrum is a Cantor set for any irrational number $\omega$.
The way I understand this statement is the following. We first consider a sequence of rational numbers $\{p_n/q_n\}$ whose limit is the irrational number we want to discuss. E.g. $p_n/q_n$ can be chosen as the continued fraction approximation to that number. Accordingly, we can consider a sequence of periodic dynamical system $\omega_n=p_n/q_n$. For each periodic system, we can study the trace of the transfer matrix for one period $$ T_{\text{one period}}(E;\omega_n)=\prod_{k=1}^{q_n} T(E, k,;\omega_n)\,. $$
Those energy with Tr $T_{\text{one period}}(E;\omega_n)>2$ determines which part should be removed from the spectrum. The limit yields the spectrum for the irrational case we are interested in.
I'm curious whether the conclusion for the spectrum is sensitive to the detailed form of the transfer matrix.
More concretely, I have another transfer matrix $T_{my}(E,n,\omega)$ of the category SU(1,1), with $T_{my}(E=0)=T_{my}(E=1)$ so that $E$ lives on a circle. For $\omega = \omega_n = p_n/q_n$, there are $q_n-1$ disconnected parts to be removed from the spectrum, which are centering at $r/q_n,r=1,2,...q_n-1$. Namely, the spectrum consists of $q_n-1$ 'bands'.
For example, let us choose the sequence to approach the inverse golden ratio. When $\omega_1 = 1/2$, we remove a small interval $(1/2-\epsilon, 1/2+\epsilon)$; when $\omega = 2/3$, we retake the whole interval $[0,1]$ and remove two intervals $(1/3-\epsilon',1/3+\epsilon')$ and $(2/3-\epsilon',2/3+\epsilon')$. Thus this is not a monotonic process as the standard Cantor set. I also add prime to imply that those $\epsilon$ are not necessarily equal.
The length of each removed part is different and has a $n$ dependence in such a way that as $n\rightarrow\infty$, the measure of the spectrum goes to zero.
Then I'm wondering as $\omega$ approaches an irrational number along the sequence of continued fractions, is the final spectrum still a Cantor-type set? I will appreciate even if someone could give some intuitive explanation for the standard almost Mathieu operator.