ops sorry, such that the bracket above is an element in the hall basis?

@YCor by the way do you perhaps know how to characterize all the permutations $\sigma\in S_n$ such that $[x_{\sigma(1)},[x_{\sigma(2)},\dots,[x_{\sigma(n-1)},x_{\sigma(n)}]\dots]]$?

Of course these permutations needs to encode information about the positioning of the lie brackets, and this is precisely what I am trying to figuring out. I would like a class of permutations that allows to describe all those 6 generators at once. Hope this helps

Ok maybe $n=3$ is misleading. For $n=4$ you have $6$ such commutators, $[x_3,[x_2,[x_1,x_4]]]$, $[x_4,[x_2,[x_1,x_3]]]$, $[x_4,[x_3,[x_1,x_2]]]$, $[[x_1,x_2],[x_3,x_4]]$, $[[x_1,x_3],[x_2,x_4]]$, and $[[x_1,x_4],[x_2,x_3]]$. Is it possible to characterize all such configurations in terms of a special class of permutations on $4$ elements?

@YCor that would be the first step in my task. But then i would need to take into account different configurations of brackets, e.g., $[[x_1,x_2],[x_3,x_4]]$. In some sense I need a class of permutations which is also able to encode the positioning of brackets. The paper that was suggested is good starting point, even though the authors do not work with a Hall basis. I am trying to figure out if their description in Definition 2 admits a characterization in terms of Lie brackets

Yes sorry: I also forgot to mention that I was speaking of bi-Lipschitz homeomorphisms defined on some open set containing $\Omega$. This should be ok, right?

@Mateusz: I was wondering whether the class of weakly lipschitz domain is at least preserved by lipschitz homeomorphisms. Is this statement true or do you know any counterexample?

@YCor may I kindly ask how you deduce that $\mathfrak{a}$ has real rank zero from the absence of real eigenvalues? I'm trying to understand your argument and this is the only point I'm missing. Many thanks in advance -Guido-

Thank you very much for your answer! From what I understand one considers the Iwasawa decomposition $G=KAN$ of a semisimple group $G$, therefore if we take $M$ the centralizer of $A$ in $K$, then $MAN$ is a minimal parabolic subgroup of $G$. The factor $AN$ now is cocompact solvable and connected, as desired. Am I right? -Guido-

The last part using Lie Kolchin is clear. Many thanks in advance -Guido-

I know that any lie Group contains a cocompact subgroup, but in general we cannot guarantee that it is connected, so this is where I am a bit confused. In the other hand, at the beginning of your last paragraph, it seems implicitly stated that with the previous arguments we should find a cocompact solvable connected real group. Am I wrong? -Guido-

My main issue is the connectedness of the cocompact solvable subgroup that we find. The theorem I found is the following (on the complex field): let H be a closed subgroup of G: then G/H cocompact iff H contains a maximal solvable connected subgroup (Borel-Tits IHS, prop 9.3). The equivalence follows using the fixed point theorem. I understand well that we find what we are looking for on C. How can we guarantee the existence of such a real group?

Dear professor Bader, sorry to bother you again, but I realized that there is still a small point in the proof which puzzles me. Namely, I've understood that we can reduce everything to the case of real algebraic linear groups. Nonetheless to apply correctly the Borel Fixed point theorem we need to complexify to work on an algebraically closed field. The cocompact solvable connected subgroup that we find is thus a complex subgroup. But we would like to have it real. Is it automatic the passage to a real subgroup having all the required properties, or am I missing something? Bests, -Guido-

Not exactly, unfortunately. Indeed the main point of my model is that two switching times differs at least by a time $\tau$. This is why we assume $U_i$ distributed as explained above: notice that $\mathbb{P}(U_i< \tau)=0$. In the paper you suggested I didn't find such restriction, however it is definitely worth a look!

I cannot express enough my gratitude. It was a pleasure to have this discussion with you! My bests -Guido-

Thank you very much. Your explanation clarifies exactly all the points where I was a bit confused. I've got a last (small question). The fact that all these representation are 1 or 2 dimensional does not imply anything on the dimension of the compact orbits. Am I right? I mean, we can have, in general compact orbits of any dimension (I am thinking of irreducible groups where it seems strange to have circles). BTW, I am really grateful for your explanation. Again thank you very much!