# Commutator relations in associated algebra.

Consider the free associated algebra over some field $$k$$ $$k\langle x_1,\cdots,x_n\rangle.$$ Order the generators $$x_1<\cdots.

I can define a complete bracket of an ordered monomial $$x_{i_1}\cdots x_{i_k}$$ with $$i_1\leq\cdots\leq i_k$$ by $$[x_{i_1}\cdots x_{i_k}]:=[\cdots[x_{i_1},x_{i_2}],x_{i_3}],\cdots],x_{i_k}]$$ where $$[-,-]$$ is the binary Lie bracket. If there is only one $$x_i$$, then we define $$[x_i]:=x_i$$.

Here are some observations

• $$2x_1x_2=[x_1x_2]+\big(x_2[x_1]+x_1[x_2]\big)$$
• $$3x_1x_2x_3=[x_1x_2x_3]+\big(x_1[x_2x_3]+x_2[x_1x_3]+x_3[x_1x_2]\big)+\big(x_1x_2[x_3]+x_1x_3[x_2]+x_2x_3[x_1]\big)$$

The obvious guess would be $$n\;x_1\cdots x_n=\sum_{m=1}^n\sum_{i_1<\cdots

This looks very symmetric and I tend to believe it, and I also want to believe there is a ''symmetric'' proof.

One motivation for this definition comes from the notion of a universal enveloping algebra $$U(\mathfrak g)$$ of a Lie algebra $$\mathfrak g$$. By the Poincaré-Birkhoff-Witt theorem, if we choose an ordered basis $$\xi_1<\cdots<\xi_n$$ of $$\mathfrak g$$, then $$\xi_1^{k_1}\cdots\xi_n^{k_n}$$ with $$(k_1,\cdots,k_n)\in\mathbb N^n$$ form a basis of $$U(\mathfrak g)$$. The complete bracket of such a base vector is always in $$\mathfrak g$$.

• Is there a name for what I call complete bracket? I think people should have considered it.
• Of course a proof of $$(*)$$, or a counterexample.

It turns out the proof of the formula is not difficult by induction on $$n$$.

I would like to post and prove a more flexible formula $$(w_1+\cdots+w_n)x_1\cdots x_n=\sum_{m=1}^n\sum_{i_1<\cdots and we recover the equation in the question by letting $$w_1=\cdots=w_n=1$$.

The formula $$(**)$$ for $$n=2$$ is the easy-to-check identity $$(w_1+w_2)x_1x_2=w_1x_2[x_1]+w_2x_1[x_2]+w_1[x_1x_2].$$

To make equations look more compact, I introduce some notations:

• $$[n]:=\{1,\cdots,n\}$$;
• for a nonempty subset $$S\subset[n]$$, it is $$S=\{i_1<\cdots, and denote $$x^S:=x_{i_1}\cdots x_{i_s}$$, and $$[x^S]:=[x_{i_1}\cdots x_{i_s}]$$.

With above notations, the formula we will prove is $$$$(w_1+\cdots+w_n)x^{[n]}=\sum_{\emptyset\ne S\subset[n]}w_{\min(S)}x^{[n]\setminus S}[x^S].$$$$

Assume we have shown the equation for $$n$$. For $$n+1$$, we use variables $$y_1,\cdots,y_{n+1}$$, and let $$x$$ be the truncation $$y_{\leq n}$$, i.e. $$(x_1,\cdots,x_n)=(y_1,\cdots,y_n)$$. We have $$$$\begin{split} &\sum_{\emptyset\ne S\subset[n+1]}w_{\min(S)}y^{[n+1]\setminus S}[y^S]\\ =&\sum_{\substack{S\subset[n]\\S\ne\emptyset}}+\sum_{S=\{n+1\}}+\sum_{\substack{S=\{n+1\}\sqcup S'\\S'\subset [n]\\S'\ne\emptyset}}\\ =&\sum_{\substack{S\subset[n]\\S\ne\emptyset}}w_{\min(S)}x^{[n]\setminus S}y_{n+1}[x^S]+w_{n+1}y^{[n+1]}+\sum_{\substack{S'\subset[n]\\S'\ne\emptyset}}w_{\min(S')}x^{[n]\setminus S'}\big[[x^{S'}],y_{n+1}\big]\\ =&\sum_{\substack{S\subset[n]\\S\ne\emptyset}}w_{\min(S)}x^{[n]\setminus S}[x^S]y_{n+1}+w_{n+1}y^{[n+1]}\\ =&(w_1+\cdots+w_n)x^{[n]}y_{n+1}+w_{n+1}y^{[n+1]}\\ =&(w_1+\cdots+w_{n+1})y^{[n+1]} \end{split}$$$$