# A positive formula for the dimensions of homogeneous components of free Lie algebras

The homogeneous component of degree $k$ in the free Lie algebra $\mathfrak{Lie}(x_1,\dots,x_n)$ in $n$ letters is of dimension $$g_n(k)=\frac{1}{k}\sum_{d|k}\mu(d)n^{k/d}.$$ This is also the number of Lyndon words of length $k$ in $n$ letters, and of a few other things...

Question: Is there a positive formula for this number?

As an aside,

Question: Is there a corresponding formula for the dimensions of the homogeneous components of the free Lie triple system on $n$ letters?

• Perhaps you meant the mu function instead of the phi function and n instead of m? In any case, I don't see a reason to expect a positive formula here; look at the case when n is prime. – Qiaochu Yuan May 6 '10 at 20:44
• Er, whoops; I meant the case where k is prime. You get (n^k - n)/k and there's really no sensible way around that minus sign that I can see. – Qiaochu Yuan May 6 '10 at 20:52
• Andy- positive formulae are like pornography, I know them when I see them. In general it's a sum where all the terms are positive and hopefully in bijection with known sets. – Ben Webster May 6 '10 at 22:45
• @Ben: your comment really confuses me... since for a basis of the free Lie algebra we can take Lyndon words (a very combinatorial object), the above formula can be thought of as a sum of one term which has a clear combinatorial meaning! – Vladimir Dotsenko May 7 '10 at 5:27
• I second Vladimir's comment. I'm also confused: as Mariano mentions, Lyndon words (or any Hall set in the free magma, for that matter, as in chapter II of Bourbaki's Lie groups and Lie algebras) are a basis for the free Lie algebra. Why isn't this a "positive formula"? I guess the point must really be the second question about triple systems? – GS May 11 '10 at 10:59

This doesn't answer the question, but might still be of interest to you. Let $V$ be the $n$-dimensional vector space spanned by your $n$ letters.

The vector space $V^{\otimes k}$ has a natural $S_k$ action. There exists an $S_k$ module, which I will denote $\text{Lie}(k)$, such that the $k$th homogenous component of the free Lie algebra on $V$ is isomorphic to

$V^{\otimes k} \otimes_{S_k} \text{Lie}(k)$.

And this module has dimension $(k-1)!$. This wont help you with the dimensions you want, but I think that it's interesting.

If you just want to know the $S_k$-module structure on $\text{Lie}(k)$ then it can be given as follows: Let $C_k$ be a subgroup of $S_k$ generated by a $k$-cycle. Let $W$ be a 'primitive' representation of $C_k$. (this requires a primitive $k$th root of unity in your field). Then the module we are looking for is $W$ induced up to $S_k$.

This last bit is a bit mysterious to me.

• This might not be an answer to OP's question but it answers a question I had about the dimension of that module, so +1 – Javi Feb 3 '20 at 17:58

Classical result 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)) proves that the multiplicity of the Specht module $$S^{\lambda}$$ in $$\mathrm{Lie}(n)$$ is given by an elegant combinatorial formula: it is equal to the dimension of standard Young tableaux of shape $$\lambda$$ and major index congruent to $$1$$ modulo $$n$$. Recall that the major index is equal to the sum of all $$i$$ such that $$i+1$$ is strictly below $$i$$ in the tableau. By Schur-Weyl duality, this means that the irreducible representation $$S^\lambda(V)$$ of $$GL(V)$$ appears in the free Lie algebra generated by $$V$$ with that same multiplicity, the major index. Since dimensions of irreducible representations of $$GL(V)$$ admit well-known positive formulas, you get a positive formula for the dimension of the homogeneous component of the free Lie algebra.