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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? |
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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 want to read more then you need to learn about operads, and in particular the Lie operad. 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. |
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