TITLE: Breaking up the free Lie algebra into Gl irreps
QUESTION [9 upvotes]: The free Lie algebra $L(V)$ generated by an $r$-dimensional vector space $V$ is, in the
language of https://en.wikipedia.org/wiki/Free_Lie_algebra , 
the free Lie algebra generated by any choice of basis $e_1, \ldots , e_r$ for the vector space $V$. (Work over the field ${\mathbb R}$ or ${\mathbb C}$, whichever you prefer.)
It is a graded Lie algebra
$$L(V) = V \oplus L_2 (V) \oplus L_3 (V) \oplus \ldots .$$
The general linear group $Gl(V)$ of $V$ acts on $L(V)$  by gradation-preserving Lie algebra automorphisms.
Thus  each graded piece  $L_k (V)$ is a finite dimensional
representation space for $GL(V)$. (The  `weight' of $L_k (V)$ is  $k$ in the sense that  $\lambda Id \in Gl(V)$ acts on $L_k (V)$ by scalar multiplication by $\lambda^k$.) 
QUESTION: How does $L_k (V)$ break up into $GL(V)$-irreducibles?
I only really know that $L_2 (V) = \Lambda ^2 (V)$, which  is already irreducible.
To start the game off, perhaps some reader out there already is  friends with  $L_3 (V)$ 
as a $GL(V)$-rep, and can tell me its irreps in terms of the Young diagrams / Schur theory involving 3 symbols?
(My motivation arises from trying to understand some details of the subRiemannian geometry http://en.wikipedia.org/wiki/Sub-Riemannian_manifold of the Carnot group whose Lie algebra is the free k-step Lie algebra, which is $L(V)$-trunated after step $k$. )

REPLY [7 votes]: Fix a primitive $k$-th root of unity $\zeta_k$, and let $\rho\in S_k$ be a $k$-cycle. (I am working over $\mathbb{C}$ here, obviously.) Klyachko [Kl] proved in 1974 that $L_k(V)\simeq \{x\in V^{\otimes k}\,|\,\rho(x) = \zeta_k\cdot x\}$.
From this Frobenius reciprocity gives the following formula for the multiplicity of $\mathbb{S}_\lambda(V)$ in $L_k(V)$ (though I believe this formula is much older):
$\frac{1}{k}\sum_{d|k}\mu(d)\chi_\lambda(\tau^{m/d})$ 
Here $\mu$ denote the Möbius function, $\chi_\lambda\colon S_k\to \mathbb{Z}$ be the character of the corresponding $S_k$-irrep, and $\tau\in S_k$ is a $k$-cycle (so $\tau^{k/d}$ is a product of $k/d$ disjoint $d$-cycles). However this formula is not so helpful, since the irreducible characters of $S_k$ are not easy to compute; in practice, there is a much more useful (positive, bijective) formula as follows.

[As motivation] Since $L_k(V)\subset V^{\otimes k}$, the multiplicity of $\mathbb{S}_\lambda(V)$ in $L_k(V)$ is bounded above by its multiplicity in $V^{\otimes k}$, which is the number of standard tableaux of shape $\lambda$. (A tableau $T$ of shape $\lambda$ is a labeling of the boxes of $\lambda$ by $1,\ldots,k$; it is standard if the labels are increasing within each row and within each column.)
The descent set $D_T\subset \{1,\ldots,k\}$ of a tableau $T$ is the subset   $D_T=\{i\,|$ box $i$ is in a higher row than box $i+1\}$. The major index $\text{maj}(T)$ is the sum $\text{maj}(T):= \sum_{i\in D_T}i$.
The multiplicity of $\mathbb{S}_\lambda(V)$ in $L_k(V)$ is the number of standard tableaux $T$ of shape $\lambda$ with major index $\text{maj}(T)\equiv 1\bmod{k}$.

You should have no trouble working out the computations you want by hand for small $k$ using this result (as F.C.'s computations show, the answer gets large fast as $k$ gets bigger).
[Remark on references: I always thought this formula was due to Klyachko as well; from a brief look at his paper I don't see it there (but it may be implicit there; certainly his "$\text{ind}(\sigma)$" is precisely the major index modulo $k$). It appears explicitly in Kraśkiewicz-Weyman "Algebra of coinvariants and the action of a Coxeter element" which appeared as a preprint in the late 1980s. Credit can be difficult to pin down in this period, especially due to the communication barriers between the Soviet Union and other countries; I make no guarantees about the proper credit for these results.]
[Kl]: Klyachko's article was in Russian; the English translation is
Klyachko, Lie elements in the tensor algebra, Siberian Mathematical Journal 15 (1974) 6, 914-920, online here