If you don't have a lot of dimensions, even integral eigenvalues give examples. For example, if your eigenvalues are 2, and 3, you satisfy the no-resonance condition.

From context it seems that by $3A$ you mean $A+A+A=\{a+a'+a''\mid a, a', a'' \in A\}$, and not $\{3a\mid a\in A\}$?

Reading a proof and verifying all the details is the mathematical equivalent of replication. I would like to believe that most mathematicians do this for most of the results they cite. There are certainly results that are long and complicated and out of the reach for most people in a field to check in detail, but I would also like to believe that people avoid using them until experts in their field say they have checked the proof. As long as this replication is cheap and easy compared to physical experiments, our replication issues will be smaller.

Since $Z$ has $0$ differential, the differential of $Z\otimes C$ is $(-1)^a_Z\otimes \Delta_C$, and since everything is flat, you should get $H(Z\otimes C)\cong Z\otimes H(C)$. I cannot figure out what else might be going on.

Note that $(Xa)^T(X^Tb)\neq a^Tb$.You don't end up with $X^TX$, you get $(X^T)^2$.

It's been too long since I've thought about this stuff, but I want to recommend the book "Free Lie Algebras" by Christophe Reutenauer. It was an absolute pleasure to read, and I'm certain that the book has a good discussion of why there is a basis in correspondence with Lyndon words.

@GA316 In this context, I believe it means a short exact sequence of the form $0\to M\to M'\to N\to 0$ With a split short exact sequence, $M'\cong M\oplus N$, but if you do not have complete reducibility, then there will exist non-split short exact sequences.

@JimHumphreys Yes, that makes sense. The original post wasn’t making it clear that this was a special case that was being reduced to, but rather a generic thing being started with.

I don’t have a copy of Humpreys on hand right now, so I can’t check (though could you say where in the book this is?), but it seems odd without context to start with talk of codimension 1 submodules, as there is a priori no reason why any codimension 1 subspace should be a sub module.

There is a filtration on the Weyl algebra such that the associated graded ring is just a polynomial algebra in $2n$ variables, and so I suspect that the non-commutative analogue of Groebner bases works out very nicely. Have you had any luck doing computations when $|\mathcal D|=1$?

Physics is probably why people first looked at symplectic manifolds (namely cotangent bundles), but why we continue to study them is deeper. Somewhat, it is just to see the consequences of a simple condition (e.g., there are actual topological consequences to having a symplectic form on a manifold), somewhat it is because that condition appears in surprising places (e.g., co-adjoint orbits carry a symplectic structure), somewhat it is because there are continuing applications to physics (e.g., understanding quantization of classical systems), but there is more out there.