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.