Dynkin–Specht–Wever Lemma

Let K be a field and let A be the free associative K-algebra generated by indeterminates x_1, \ldots, x_n. Let L be the Lie subalgebra of A with the same generating set. Both A and L are canonically graded by \mathbf{N}_0: let L_n \subseteq A_n denote the graded components consisting of homogeneous elements of total degree n.

A useful result credited variously to Dynkin, Specht and Wever, gives an element \Omega_n \in \mathrm{End}_K(A_n) such that \Omega^2 = n\Omega and A_n \Omega = L_n. In particular, it follows that if the characteristic of K does not divide n, then L_n is a direct summand of A_n. 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 \Omega on monomials: we do this by

x_{i_1} x_{i_2} \ldots x_{i_n} \Omega_n = [[x_{i_1},x_{i_2}], \ldots, x_{i_n}].

It is clear that A_n \Omega \subseteq L_n, and it is not hard to prove, using the Jacobi identity, that L_n is spanned by left-normed commutators, of the type in the right-hand side above. So it only remains to prove that \Omega^2 = n \Omega. A direct proof of this seems hard. For example, when n=3 we have [[x,y],z] \Omega_3 = 3[[x,y],z], corresponding to the identity

[[x,y],z] - [[y,x],z] - [[z,x],y] + [[z,y],x] = 3[[x,y],z]

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 u \in A, let \mathrm{ad}\; u: A \rightarrow A denote the K-linear map defined by y (\mathrm{ad}\; u) = [y,u]. We write elements of A_n as F(x_1,\ldots,x_n) or simply F(x), bearing in mind that the variables x_1, \ldots, x_n do not commute. (In this post x is never a single variable.) We denote by F(\mathrm{ad}\; x) the linear endomorphism of A_n obtained by substituting the linear map \mathrm{ad}\; x_i : A_n \rightarrow A_n in place of x_i in the polynomial F(x) \in A_n. We say that F(x) \in A_n is a Lie polynomial if \mathrm{ad}\; F(x) = F(\mathrm{ad}\;x). For example, x_1x_2 is not a Lie polynomial, because

y\; \mathrm{ad}\; F(x) = y (\mathrm{ad}\; x_1x_2) = [y,x_1x_2]

has two summands, whereas

y\; F(\mathrm{ad}\; x) = y\; \mathrm{ad}\; x_1 \; \mathrm{ad}\; x_2 = [[y,x_1],x_2]]

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

First of all one shows that u \in A_n and F(x) is a Lie polynomial then

u(\mathrm{ad}\; F(x)) = u F(\mathrm{ad}\; x)

by a routine induction on the degree of F. This identity is closely related to the fact that \mathrm{ad} : A_n \rightarrow \mathrm{End}(A_n) is a Lie algebra homomorphism (which in turn is equivalent to the Jacobi identity). Indeed, if we take F(x_1,x_2) = [x_1,x_2] \in L_2 then the identity above becomes [u, [x_1,x_2]] = u [\mathrm{ad}\; x_1, \mathrm{ad}\; x_2].

We now prove the DSW Lemma by induction on the degree. For the inductive step, we observe that if G(x) and H(x) are any polynomials such that G(x) has degree p and H(x) has degree q then

G(x)H(x) \Omega_{p+q} = G(x) \Omega_p H(\mathrm{ad}\; x).

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

G(x)H(x) \Omega_{p+q} = pG(x) (\mathrm{ad}\; H(x)) = p[G(x), H(x)].

Similarly we have H(x)G(x) \Omega_{p+q} = q[H(x), G(x)]. Hence

[G(x),H(x)] \Omega_{p+q} = G(x)H(x) \Omega_{p+q} - H(x)G(x)\Omega_{p+q} \\ = p[G(x),H(x)] - q[H(x),G(x)] = (p+q)[G(x),H(x)]

which gives the inductive step.

Finally we note that if we let the symmetric group S_n act on A_n by place permutation, then \Omega_n is in the image of the action map S_n \rightarrow \mathrm{End}_K(A_n). In fact

\Omega_n = (1-(1,2))(1-(1,2,3)) \ldots (1-(1,2,\ldots,n)).

It seems to be a definite (and clearly unnecessary) challenge to prove that \Omega_n^2 = n\Omega_n working straight from this definition, within the group ring KS_n. A naive inductive proof would require, for its inductive step, the identity

x_n [[x_1, x_2]..., x_{n-1}] \Omega_n (1-(1,2,\ldots,n)) = -[[x_1,x_2], \ldots, x_n]

which seems hard to prove directly.

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