Induction formula for Lie operad

Statement
The Lie operad in a field $$K$$ of characteristic zero is a permutative (i.e., symmetric) operad. Thus, the arity $$r$$ component of this operad is equipped with a natural action of the symmetric group $$S_r$$. We can thus consider it as a $$KS_r$$-module, or equivalently, as a representation of $$S_r$$.

The goal is to describe this representation explicitly. The significance of this would be that if we had such an explicit description, the corresponding Schur functor applied to a $$n$$-dimensional vector space would give the degree $$r$$ graded component of the free $$K$$-Lie algebra on $$n$$ variables.

The answer is as follows: let $$C_r$$ be a cyclic subgroup of $$S_r$$ generated by a $$r$$-cycle and let $$\varphi$$ be any irreducible representation that sends the generator of $$C_r$$ to a primitive $$r^{th}$$ root of unity. Then, the representation of $$S_r$$ we desire is the induced representation from $$C_r$$ to $$S_r$$ of $$\varphi$$, i.e., it is the representation $$\operatorname{Ind}_{C_r}^{S_r} \varphi$$. Note that the representation has degree $$(r - 1)!$$.

Note that, thanks to Dynkin's lemma, the Lie operad is explicitly given by the Dynkin operator, which is a more precise description because it describes the left ideal in $$\mathbb{Q}[S_n]$$ whereas the induction formula only describes the corresponding two-sided ideal.

Related facts

 * Formula for dimension of graded component of free Lie algebra
 * Basic Lie products form a freely generating set of graded component of free Lie algebra