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Let $K$ be the rationalization of a simply-connected finite CW complex. Then the Samelson product gives $\pi_*(\Omega K)$ the structure of a graded Lie algebra, and the Hurewicz map $h: \pi_*(\Omega K) \to H_*(\Omega K)$ carries brackets to Pontrjagin commutators (and if $K$ is a suspension, $h$ is universal enveloping). Thus the image of $h$ is a Lie subalgebra of $H_*(\Omega K)$.

My questions:

  1. Must the Lie algebra $\pi_*(\Omega K)$ be finitely generated as a graded Lie algebra?
  2. Must $h( \pi_*(\Omega K))$ be finitely generated as a graded Lie algebra?
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Not an answer.

$h$ is already universal enveloping provided only that $K$ is the rationalization of a simply connected space; this is, for example, Theorem 21.5 in Felix, Halperin, and Thomas. In particular, $h$ is injective so the answer to the two questions is the same.

As for the actual question, for starters, there is a Lie model of $K$ (which is in particular a dg Lie algebra whose homology Lie algebra is $\pi_{\bullet}(\Omega K)$) whose underlying graded Lie algebra is freely generated by an element in each degree $i - 1$ for each $i$-cell of $K$, with differentials determined by the attaching maps. This is a corollary of Theorem 24.7 in Felix, Halperin, and Thomas. Conversely, given such a dg Lie algebra $L$ we can write down a finite CW complex which has it as a Lie model. So the question reduces to the following purely algebraic question:

Does a connected dg Lie algebra $L$ whose underlying graded Lie algebra is free on finitely many generators have a finitely generated homology Lie algebra?

I imagine someone knows the answer to the analogous question for connected dg algebras; that would be helpful as a guide.

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    $\begingroup$ Jeff knows that it is not finitely generated in the commutative dga case. mathoverflow.net/questions/182437/… $\endgroup$ – Ben Wieland Apr 4 '15 at 15:33
  • $\begingroup$ Kathryn Hess informs me that the answer to this question is also no. I think there is a counterexample in section 37 of Felix, Halperin, and Thomas but I'm not sure how to verify that it is in fact a counterexample. $\endgroup$ – Qiaochu Yuan Apr 8 '15 at 23:59
  • $\begingroup$ I expect that is a counterexample, but if she didn't specifically point you to it, I doubt that it is the simplest counterexample. $\endgroup$ – Ben Wieland Apr 10 '15 at 17:02

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