Let be a field and let be the free associative -algebra generated by indeterminates . Let be the Lie subalgebra of with the same generating set. Both and are canonically graded by : let denote the graded components consisting of homogeneous elements of total degree .

A useful result credited variously to Dynkin, Specht and Wever, gives an element such that and . In particular, it follows that if the characteristic of does not divide , then is a direct summand of . The purpose of this post is to record a short proof of the DSW Lemma, closely following this paper of Lyndon.

It suffices to define on monomials: we do this by

It is clear that , and it is not hard to prove, using the Jacobi identity, that is spanned by left-normed commutators, of the type in the right-hand side above. So it only remains to prove that . A direct proof of this seems hard. For example, when we have , corresponding to the identity

in which there is already quite a lot of non-obvious cancellation. However, as Lyndon shows, there is a short algebraic proof of the DSW Lemma.

We need some notation. Given , let denote the -linear map defined by . We write elements of as or simply , bearing in mind that the variables do not commute. (In this post is never a single variable.) We denote by the linear endomorphism of obtained by substituting the linear map in place of in the polynomial . We say that is a *Lie polynomial* if . For example, is *not* a Lie polynomial, because

has two summands, whereas

has four. Any Lie polynomial can be obtained from the by repeated bracketing operations.

First of all one shows that and is a Lie polynomial then

by a routine induction on the degree of . This identity is closely related to the fact that is a Lie algebra homomorphism (which in turn is equivalent to the Jacobi identity). Indeed, if we take then the identity above becomes .

We now prove the DSW Lemma by induction on the degree. For the inductive step, we observe that if and are *any* polynomials such that has degree and has degree then

.

(Proof: reduce to the case where and are monomials, where it’s obvious.) Hence if and are Lie polynomials, then, by the first step and induction, we have

Similarly we have . Hence

which gives the inductive step.

Finally we note that if we let the symmetric group act on by place permutation, then is in the image of the action map . In fact

.

It seems to be a definite (and clearly unnecessary) challenge to prove that working straight from this definition, within the group ring . A naive inductive proof would require, for its inductive step, the identity

which seems hard to prove directly.