@ThorbenK: done! Thanks for the suggestion.

As you suspect, in dimension 3 the answer to your first question should be "no". Indeed, if one deletes any collection of circles from $M$ then the complement is a 3-manifold with toroidal boundary. Since the fundamental group is free, no boundary component can $\pi_1$-embed. But then, by Dehn's lemma, every boundary component is compressible. There's something to check here, but I think it follows that the complement is a connect sum of solid tori and $S^1\times S^2$'s, whence $M$ has to be a connect sum of lens spaces and $S^1\times S^2$'s.

I learned a lot from this answer!

@MoisheKohan: well I think that’s the answer to the first part of the question!

It's perfectly natural to wonder whether the work of Goldman, Magee etc can be generalised from surfaces to arbitrary Fuchsian groups. You could try combing through the 355 (!) papers on MathSciNet that cite Goldman's article. You could also try contacting Magee directly.

It may be that, in order to think about this group, one needs to use the techniques referenced in the answers to this MSE question: math.stackexchange.com/questions/667035/… .

@CorentinB: It's natural to suspect that it's a graph, but I don't think it is. The product topology is different from the weak topology. For instance, let $e_n$ be the $n$th unit vector. The sequence $(e_n)$ has no convergent subsequence in the weak topology. But $[0,1]^{\mathbb{N}}$ is compact by Tychonoff's theorem, so the sequence does have a convergent subsequence in the product topology. Now, it may still be an increasing union of graphs, and that may be enough to get the conclusion, but it doesn't follow immediately, I think.

You say you're interested in the co-compact case. Perhaps it's worth pointing out that, in this case, the torsion-free case that you ask about at the end is precisely the case of a surface group $\Sigma_g$.

It's great that you're thinking about these things! 'he title of this post is a big problem, though. You have not simplified the proof, because you do not have a complete proof. You have to show us that you have simplified the proof, not the other way around. By the way, there are loads of variations on this construction. I believe the original idea for doing it like this is due to Cameron Gordon. Your idea may be related to his proof.

I believe the question of whether or not every finitely presented Poincaré Duality group is the fundamental group of a closed aspherical manifold is open in all dimensions >2. So the "obstructions that can arise that stop a CW complex being a manifold" are not fully understood (though, as you say, Poincaré duality is certainly a necessary condition).

@Fougeron: if that works for you, then fine. It sounds like you are only interested in membership of finite subgroups: this is an important extra qualification; it would have been nice if you had mentioned it.

To address the specific issue of undecidability raised by @Carl-FredrikNybergBrodda (assuming your matrices are somehow given to you so you can compute exactly).... The membership problem for subgroups of $O(6)$ (say) is certainly undecidable, since it contains a a product of two non-abelian free groups. Finiteness is probably decidable, using the fact that every finite subgroup of a linear group has an abelian subgroup of bounded index, but requires more thought.

Your question reminded me of one more reference (apologies if it turns out to be irrelevant). In this recent arXiv preprint -- arxiv.org/abs/2404.12943 -- the authors develop statistical techinques to spot symmetries in data sets. They may have to deal with some similar issues to the ones you need to worry about.

I think many people on this forum will be very unhappy with your statement "it is still very easy to detect that an 8th turn rotation generates a finite group of order 8 .... up to say $10^{−15}$ error." To whatever accuracy you like, you can't know if such a specified subgroup is finite or infinite!