# Linear independence in (graded) Lie algebras

I asked a mixed-up version of this question earlier.

The Lie algebras I have in mind are the homotopy Lie algebras of wedges of finitely many spheres (in dimensions greater than $1$). Thus each element has a degree and the bracket is the Whitehead product, which satisfies a graded version of the Jacobi identity, etc. The homotopy Lie algebra of a wedge of spheres is free on generators $x_1, \ldots , x_n$ in bijective correspondence with the spheres in the wedge.

Inside of any (graded) Lie algebra $L$ we may form the span of the iterated brackets of the $x_i$ with no repeated factors -- call any element in this span "square-free". Note that Jacobi rewrites of square-free brackets are also square-free.

A sum of $i$-fold brackets is said to have weight $i$.

So let $L = L(x_1, \ldots, x_n, x_{n+1})$ be the free (graded) Lie algebra on the given generators, and let $L_i \subset L$ be the subalgebra $L(x_1, \ldots,\widehat{x_i}, \ldots x_n, x_{n+1})$. For each $i = 1, \ldots , n$, choose a square-free weight $n$ element $w_i \in L_i$, not all of which are zero. Is it true that
$$\mathcal{Z} = \{ [w_i, x_i]\ | \ i = 1, \ldots, n\}$$ is a linearly independent set?

I have verified this with MAPLE for $n\leq 4$, and I'd be happy even for an answer for the special case of ordinary Lie algebras. An answer to the general question would be helpful in resolving some questions about the Lusternik-Schnirelmann category of rational spaces.

• Your claim about wedges of spheres can't be true integrally; do you mean it rationally? – Qiaochu Yuan Jul 1 '14 at 1:09
• Yes, the whole question was intended to be rational. – Jeff Strom Jul 1 '14 at 2:44

Second, the dimension of that space (multilinear elements in the free Lie algebra with $n$ generators, or alternatively, the $n$th component of the operad Lie) is $(n-1)!$, as one can read in almost any source on free Lie algebras or operads.
Everything is now ready to prove your claim. If we let $w_i$ range through spanning sets of the multilinear spaces of respective $L_i$'s, we see that the cardinality of your set is $(n-1)!\cdot n=n!$, that is precisely the dimension of the space of multilinear elements in $L$. Hence it is enough to prove that it is a spanning set. It suffices to prove that each multilinear element is a span of "left-normed Lie monomials" $$[\dots[[x_{i_1},x_{i_2}],x_{i_3}],\dots, x_{i_{n+1}}],$$ where $i_{n+1}\ne n+1$. For that, take some Lie monomial $m=[m_1,m_2]$, where both $m_1, m_2$ are Lie monomials. One of them does not contain $x_{n+1}$, WLOG it is $m_1$. It is of the form $[m_{11},m_{12}]$. Using Jacobi identity, write $$m=[[m_{11},m_{12}],m_2]=[[m_{11},m_2],m_{12}]-[[m_{12},m_2],m_{11}].$$ This procedure represents $m$ as a combination of Lie monomials where $x_{n+1}$-free part is shorter, and hence by induction we shall eventually arrive at a representation where the $x_{n+1}$-free part is of weight 1.