Can the "generalized commutator bracket" $[(a,b,c)]$ be written using only the usual commutator bracket and $\mathbb{Q}$-linear combinations?

Question

Let $\mathbb{K}$ denote a field.

Background conventions.

1. All my $\mathbb{K}$-algebras are associative.

2. Given a finite set $X$, write $E(X)$ for the set of even permutations of $X$, and $O(X)$ for the set of odd permutations.

3. This question is easier to state if we identify each natural number $n$ with the set of strictly smaller naturals. For instance, $3 = \{0,1,2\}$. Furthermore, we think of $A^n$ as the set of all functions $n \rightarrow A.$ Furthermore, an $n$-tuple of elements of $A$ is thought of as a function $n \rightarrow A$.

4. If $A$ is a monoid and $a \in A^n,$ write $\mathrm{prod}(a)$ for the product of the elements of $a$ in the order they appear. For example, $\mathrm{prod}(3,4) = 3 \times 4 = 12$ in the monoid $(\mathbb{N},\times,1).$

Main definition. (Generalized Commutator Bracket) Suppose $A$ is a $\mathbb{K}$-algebra. Then given a natural number $n$ and a function $a \in A^n$, define:

$$[a] = \left(\sum_{\pi:\mathrm{E}(n)}\mathrm{prod}(a \circ \pi)\right) - \left(\sum_{\pi:\mathrm{O}(n)}\mathrm{prod}(a \circ \pi)\right)$$

For example, letting $\tau$ denote the unique non-trivial permutation of $2$, we may write:

$$[(a,b)] = \left(\sum_{\pi:\mathrm{E}(2)}\mathrm{prod}((a,b) \circ \pi)\right) - \left(\sum_{\pi:\mathrm{O}(2)}\mathrm{prod}((a,b) \circ \pi)\right)$$

$$= \mathrm{prod}((a,b) \circ \mathrm{id}_2) - \mathrm{prod}((a,b) \circ \tau) = \mathrm{prod}(a,b)-\mathrm{prod}(b,a) = ab-ba$$

So $[(a,b)] = ab-ba$. Hence $[-]$ can be viewed as a generalization of the usual commutator bracket.

Questions.

• Can the expression $[(a,b,c)]$ be written using only the usual (2-place) commutator bracket and $\mathbb{Q}$-linear combinations? For example $[(a,b,c)] = [a,[b,c]]+3[b,[a,a]]$ is a valid candidate solution. (But false, of course.)

• How about the general case where $n$ is an arbitrary positive integer?

• What is this generalized commutator bracket thingo usually called? If the generalized commutator bracket doesn't have a standard name, then where can I go to learn more about it?

Answer 1

Suppose $n>2$ and $[(a_1,\dots,a_n)]$ can be expressed (for general $a_1,\dots,a_n$) as $\mathbb Q$-linear combinations of brackets. This is equivalent to say $[(a_1,\dots,a_n)]$ is in the Lie algebra generated by $a_i$'s in $R:=\mathbb{Q}\langle a_1,\dots,a_n\rangle$ (the free $\mathbb Q$-algebra generated by $a_1,\dots,a_n$).

But there is a standard Hopf algebra structure on $R$, which is induced by defining co-multiplication and co-unit as $\Delta(a_i)=1\otimes a_i+a_i\otimes 1, \epsilon(a_i)=0$. One can easily show that every element $x$ in the Lie algebra generated by $a_i$'s is primitive, i.e. $\Delta(x) = x\otimes 1 +1\otimes x$. (In fact this also a sufficient condition, but I don't need it.)

An easy computation using the multiplicativity of $\Delta$ shows that for every permutation $\sigma\in S_n$ : $$\Delta(a_{\sigma(1)}\cdots a_\sigma(n)) = \sum_{k\leq n, i\in \Sigma(k,n-k)} a_{\sigma(i_1)}\cdots a_{\sigma(i_k)}\otimes a_{\sigma(i_{k+1})}\cdots a_{\sigma(i_n)},$$ where $\Sigma(k,l):= \{i\in S(k+l): i_1<\cdots<i_k, i_{k+1}<\cdots<i_{k+l}\}$.

Now we compute the coefficient of $a_1a_2\otimes a_3\cdots a_n$ in the expansion of $$\Delta\left([(a_1,\dots,a_n)]\right)= \Delta\left(\sum_{\sigma\in S(n)} \mathrm{sgn}(\sigma)\; a_{\sigma(1)}\cdots a_{\sigma(n)}\right).$$

it is equal to the sum of signs of all permutations $\sigma$ s.t. $12$ and $34\dots n$ appear in this order in $\sigma(1)\dots \sigma(n)$ or equivalently $\sigma^{-1}\in \Sigma(2,n-2)$. Sign is invariant under inversion, so this coefficient is equal to $$\sum_{\sigma\in \Sigma(2,n-2)} \mathbb{sgn}(\sigma) = \sum_{1\leq i<j\leq n} \mathbb{sgn}(\sigma_{ij}),$$ where $\sigma_{ij}$ is the only permutation in $\Sigma(2,n-2)$ with $\sigma(1) = i, \sigma(2)=j$. We have $\mathrm{sgn}(\sigma_{ij}) = (-1)^{i+j+1},$ so

$$\sum_{1\leq i<j\leq n} \mathbb{sgn}(\sigma_{ij}) = \sum_{1\leq i<j\leq n} (-1)^{i+j+1} = \sum_{1<j\leq n} \frac{(-1)^j+1}{2} = \lfloor\frac{n}{2}\rfloor >0.$$

This computation shows that $[(a_1,\dots,a_n)]$ is not primitive (there are terms like $a_1a_2\otimes a_3\cdots a_n$ in $\Delta([(a_1,\dots,a_n)])$ ) and so it can not be written in terms of usual brackets.

Answer 2

Note that

$$[a,b,c] = abc + bca + bca - bac - acb - cba = a [b,c] - b [a,c] + c [a,b]$$

And recursively, I think we have

$$[a_1, a_2, \ldots, a_n] = \sum_{i = 1}^n (-1)^{i + 1} a_i [a_1, \ldots, a_{i-1}, a_{i+1} \ldots, a_n]$$

You can also check that if $A$ is $n \times n$ a matrix with entries $a_i$ on the whole line $i$, then your quantity is $\det A$ (but you'd have to be careful about what result about the determinant still hold in a non commutative ring).

Now there might exist some cases where for some specific values of $a$, $b$ and $c$, the three-fold bracket $[a,b,c]$ lies in the linear span of the two-fold brackets (for example if they commute, everything is zero). But the coefficients, when the exist, depend on $a$, $b$ and $c$ (there cannot exist coefficients independent of $a$, $b$ and $c$ because of homogeneity, since $[a,b,c]$ is homogeneous of degree $3$, while the 2-fold ones are homogeneous of degree $2$).