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Equivalently, is the free Lie algebra on finitely many generators over a fixed field $k$ (say of characteristic not equal to $2$) residually finite-dimensional in the sense that any nonzero element remains nonzero in some finite-dimensional quotient? Some quick Googling on my part was not successful here.

Motivation: If this is false, an identity holding in all finite-dimensional Lie algebras which doesn't hold in all Lie algebras isolates a potentially interesting class of infinite-dimensional Lie algebras (namely the ones satisfying the identity).

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Why would a finitely generated Lie algebra not be finite dimensional? I assume the similarity type includes the ground field F and that the dimension is measured "over F". Gerhard "Quite Ignorant Of Lie Algebras" Paseman, 2012.08.11 – Gerhard Paseman Aug 12 '12 at 6:41
To respond to my comment above, it is conceivable that the bracket could be used to create many independent elements. Even so, I have a hard time imagining of an identity which would not hold in an infinte dimensional algebra yet would hold in any finite dimensional subalgebra. Gerhard "Still Ignorant Of Lie Algebras" Paseman, 2012.08.11 – Gerhard Paseman Aug 12 '12 at 6:49

2 Answers 2

up vote 9 down vote accepted

Maybe I am missing something but let me try. The free Lie algebra is a subalgebra of the free associative algebra made Lie via the bracket. The free associative algebra is residually finite dimensional by truncating polynomials. Hence its Lie algebra is residually finite dimensional. Here I assume finitely generated but that is enough.

clarification. I wrote this answer before going to bed so let me say it better. There is a functor from associative algebras to Lie algebras which sends A to the vector space A with the commutator bracket $[a,b]=ab-ba]$. It is then clear that this functor preserves residual finite dimensionality.

It is well known that the free Lie algebra is obtained by applying this Functor to the free associative algebra and generating a sub-Lie algebra by the letters. Since the free associative algebra is residually finite by truncating polynomials we are done.

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I don't see how the first claim follows. As far as I can tell, the universal properties involve only give the universal enveloping algebra of a free Lie algebra as a quotient, not a subalgebra, of the free associative algebra. If you're not using a universal property, what are you using? – Qiaochu Yuan Aug 12 '12 at 4:29
@Qiaochu: I believe Benjamin means that the free Lie algebra is a Lie subalgebra of the underlying (i.e., commutator) Lie algebra of the free associative algebra, no universal enveloping algebra involved. – Evan Jenkins Aug 12 '12 at 5:05
@Evan: I understand the claim that's being made, I just don't see why it's true. – Qiaochu Yuan Aug 12 '12 at 5:12
@Qiaochu: Oh, I misunderstood your comment. Page 31 of Shlomo Sternberg's Lie algebra notes ( has a nice proof of this. The PBW theorem is precisely the answer to your question of what this uses aside from universal properties. – Evan Jenkins Aug 12 '12 at 5:32
@Evan: oh, of course. The free associative algebra and the universal enveloping algebra of the free Lie algebra must be isomorphic because they represent the same functor... thanks for the reference! – Qiaochu Yuan Aug 12 '12 at 14:12

I believe this is entirely equivalent to Benjamin Steinberg's answer, but one way to see this is that the quotient of a free Lie algebra on finitely many generators by the $n$th term $\mathfrak{F}_n$ of the lower central series (i.e., the ideal generated by $n$-fold bracketings of generators) is a finite dimensional Lie algebra, namely, the free Lie algebra of nilpotence class $n$ on the same generating set. Each element of the free Lie algebra lies outside of $\mathfrak{F}_{d + 1}$, where $d$ is the maximal bracket-nesting depth of the element.

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