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Do we have topological counterparts of solvability and nilpotency, which are central concepts for (finite-dimensional) Lie algebras, for infinite dimensional Banach algebras with the commutator bracket?

Let $A$ be a Banach algebra. The usual bracket $[x,y]=xy-yx$ denotes the commutator of $x,y\in A$. $[V,W]$ denotes the closed linear span of $\{[v,w] : v\in V, w\in W\}$ for any two subsets $V,W\subseteq A$.

Let $A^{(1)} = [A,A]$ and $A^{(k+1)} = [A^{(k)},A^{(k)}]$ for $k\geq 1$. Let's say that $A$ is $\omega$-solvable if $$\bigcap_{k=1}^{\infty} A^{(k)} = \{0\}.$$ Similarly, let $N^{(1)} = [A,A]$ and $N^{(k+1)} = [N^{(k)},A]$ for $k\geq 1$. Let's say that $A$ is $\omega$-nilpotent if $$\bigcap_{k=1}^{\infty} N^{(k)} = \{0\}.$$ It is natural to consider the two properties above for Banach algebras, and perhaps there's a well-developed theory in a different context. I'll be grateful if you write a brief explanation and/or direct me to the relevant literature.


Footnote: The commutators of several full operator algebras $B(X)$ on a Banach space $X$, group algebras, and several $C^{*}$-algebras (like finite/infinite von Neumann-algebras, several group $C^{*}$-algebras) as well as tracial states on those are well-known. There is a characterization of the commutator subspace $[A,A]$ for zero product determined algebras (property $\mathbb{B}$ is the keyword). However, I am not aware of a general theory of the commutator subspaces for arbitrary Banach algebras, which explains the in-between relationship of the commutator subspaces considered as a chain. My humble search of the literature about infinite-dimensional Lie algebras didn't take me far, either. I suspect that there exist hidden gems somewhere deep in the oceans of the Banach algebra co/homology or inf-dim Lie algebra theory that are known to the experts but are mystery to the eyes of an amateur outsider.

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  • $\begingroup$ Just to clarify: you are asking about Banach Lie algebras, right? Banach algebras are things with an associative product; Lie algebras are things with a non-associative product. You seem to be asking about the following: any associative algebra gives rise to a Lie algebra using the commutator as the Lie bracket; what happens when we apply this to a Banach algebra? $\endgroup$
    – Yemon Choi
    Commented Apr 22, 2022 at 19:07
  • $\begingroup$ @YemonChoi Sir, please kindly see the edited post. Inasmuch as I admire the way you paraphrased my inquiry in such a way that no layperson would be left clueless outside, I believe the stars of their respective fields like yourself have a better command of the respective literature. Kind regards and thanks in advance for any input you'd like to share. $\endgroup$
    – Onur Oktay
    Commented Apr 23, 2022 at 6:17

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