## 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.