(Topological) Perfection of an infinite-dimensional Lie group implies the same for its Lie algebra?

Question

In the answer https://math.stackexchange.com/a/2888901/ it is claimed that since the connected component of the identity of the infinite-dimensional Lie group $\mathrm{Diff}(M)$ (with $M$ compact) is known to be perfect, then its Lie algebra $\mathrm{Vect}(M)$ (vector fields on $M$) is also perfect. I worry about the technical details of this. I'm happy just to know that $\mathrm{Vect}(M)$ is only topologically perfect (the derived algebra is merely dense in the whole algebra).

In what generality can you say this claim is true? Let us restrict to Milnor-regular Lie groups, to avoid any of the nastier examples of infinite-dimensional Lie groups and algebras.

Answer 1

I'm still not sure about about the implication asked about, it's probably not a formal statement ($\mathrm{Diff}(M)$ is not a locally exponential group, so I believe one needs to do something other than rely on perfectness of the group to get perfectness of the algebra).

But, the result claimed is true, by Theorem 1.4.3 in

• Augustin Banyaga, The Structure of Classical Diffeomorphism Groups, Mathematics and Its Applications 400, Springer 1997, https://doi.org/10.1007/978-1-4757-6800-8

and in fact more: for non-compact $M$, the Lie algebra of compactly-supported vector fields on $M$ is perfect.