# Breaking up the free Lie algebra into Gl irreps

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 $k$. )

The Whitehouse module referred to in one of the other answers is not necessary, since it is related to the cyclic operad Lie, that is to the representation of $S_{n+1}$ in $Lie(n)$.

The decomposition in terms of Young diagrams is, as far as I understand, first done in a paper of Kraskiewicz and Weyman (preprint W. Kraskiewicz, J. Weyman. Algebra of Invariants and the Action of a Coxeter Element. Math. Inst. Copernicus Univ. Chopina, Torun (1987), published W. Kraskiewicz, J. Weyman. Algebra of invariants and the action of a Coxeter element. Bayreuth. Math. Schr., 63 (2001)). Neither the preprint nor the published version are easily available, but lots of sources online discuss this topic. One set of slides that I found on Google right now which seems to contain all the statements you might possibly need is http://personalpages.manchester.ac.uk/staff/Marianne.Johnson/Antalyaslides.pdf

• Thank you V. Dotsenko! Those slides by M Johnson almost completely answer my question. I will have to learn /review how to build a rep from a Young Tableaux but I can see it is all there. – Richard Montgomery Nov 19 '14 at 20:18

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.

1. [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.)

2. 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 $\text{maj}(T):= \sum_{i\in D_T}i$.

3. 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

• Wow! that Klyachago theorem is cool. -thanks for stating it and the info on it. – Richard Montgomery Nov 20 '14 at 7:06

Here is the result, computed using sage:

sage: def lie(n):
....:     p = SymmetricFunctions(QQ).p()
....:     return p.sum_of_terms((Partition([d for j in range(ZZ(n / d))]),
....:                            moebius(d) / n) for d in divisors(n))
sage: s = SymmetricFunctions(QQ).schur()
sage: [s(lie(i)) for i  in range(1,8)]
[s,
s[1, 1],
s[2, 1],
s[2, 1, 1] + s[3, 1],
s[2, 1, 1, 1] + s[2, 2, 1] + s[3, 1, 1] + s[3, 2] + s[4, 1],
s[2, 1, 1, 1, 1] + 2*s[2, 2, 1, 1] + s[3, 1, 1, 1] + 3*s[3, 2, 1] + s[3, 3] + 2*s[4, 1, 1] + s[4, 2] + s[5, 1],
s[2, 1, 1, 1, 1, 1] + 2*s[2, 2, 1, 1, 1] + 2*s[2, 2, 2, 1] + 2*s[3, 1, 1, 1, 1] + 5*s[3, 2, 1, 1] + 3*s[3, 2, 2] + 3*s[3, 3, 1] + 3*s[4, 1, 1, 1] + 5*s[4, 2, 1] + 2*s[4, 3] + 2*s[5, 1, 1] + 2*s[5, 2] + s[6, 1]]
`

This uses a classical formula for the character of the symmetric group acting on Lie(n).

• s.lie(n) can also be accessed by calling s.gessel_reutenauer(n) once ticket #17125 is merged into Sage (which is slated to happen in the next beta). – darij grinberg Nov 19 '14 at 21:11

What you are describing is the algebraic operad Lie. More details can be found here. The Whitehouse modules are exactly what you get when you take Lie onto the other side of Schur Weyl duality. The sequence of partitions John describes in the link are the Whitehouse modules tensored with the alternating representation.

• Well, the Whitehouse module is given by the action of $S_{n+1}$ on Lie(n), so it's something else. – F. C. Nov 19 '14 at 19:31