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Let $V$ be a vector space over a field of characteristic $0$, and let $L_k(V)$ be the degree $k$ part of the free Lie algebra over $V$. There is an exact sequence $$0\to D_n(V)\to L_1(V)\otimes L_{n+1}(V)\to L_{n+2}V\to 0$$ where the map on the right is the Lie bracket and $D_n(V)$ is defined as the kernel of this map. Now $D_n(V)$ is a $GL(V)$-module, and I'm curious about how it decomposes as a direct sum of irreducibles. By playing around with dimension formulas, I was able to show that $D_1(V)=S^{(1,1,1)}V$, $D_2(V)=S^{(2,2)}V$ and $D_3(V)=S^{(3,1,1)}V$. Here $S^\lambda(V)$ is the irreducible representation of $GL(V)$ corresponding to partition $\lambda$. So in fact, if I haven't made a mistake, $D_k(V)$ is actually irreducible for $k=1,2,3$, with nice symmetric looking partitions.

Is $D_k(V)$ always irreducible, and is there an easy way to tell what the partition is? If not, does anyone know what is known about this?

If it helps, here is an alternate way of characterizing $D_n$. Consider the Lie operad, and look at the part with a total of $n+2$ input/output slots, denoted $Lie((n+2))$. Then $$D_n\cong [V^{\otimes n+2}\otimes Lie((n+2))]_{Sym(n+2)},$$ which denotes the coinvariants under the action of $Sym(n)$ acting simultaneously on $Lie((n+2))$ and $V^{\otimes n}$, in the latter case by permuting the factors. This is isomorphic to a space of planar unitrivalent trees with leaves labeled by vectors from $V$, modulo antisymmetry and Jacobi (IHX) relations.

I don't know very much representation theory, so it's possible my question has an easy answer.

Added 9/20/2011 Morita, Sakasai and Suzuki just posted a preprint proving that $D_{4k+2}$ and $D_{4k+3}$ always have symmetric decompositions, in the sense that there is a symmetry in the corresponding Young diagrams exchanging rows and columns. This is a neat result.

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+1. Here's a comment that maybe you already know, which is that there will be some phase transition near $n = \dim V$, just because there's a phase transition in Schur-Weyl theorem. The $\dim V \to \infty$ limit is going to be a lot like the abstract trees+IHX version of the theory, and might be more combinatorially tractable (or maybe not). Conversely, when $V = k^1$ there's not much to the Lie algebra, so maybe the formulas will remain, but they're somehow more subtle. – Theo Johnson-Freyd May 25 '11 at 1:47
up vote 7 down vote accepted

The representations are not in general irreducible. The second characterisation is the more useful to my mind because one can decompose $Lie((n+2))$ as an $Sym(n+2)$-module, which gives a decomposition of the $GL(V)$-module of interest. Theo already recited the magical words Schur-Weyl.

The dimension of $Lie((n+2))$ as a vector space is $n!$. So the $Sym(6)$-module $Lie((6))$ has dimension 24. However the representation of maximal dimension is given by the partition $(3,2,1)$ which has dimension 16. Playing with the dimensions along with a little knowledge of the Lie operad (that it doesn't have any 1 dimensional modules) gives the decomposition of dimension 24 = 10 + 9 + 5. The 10 and 9 dimensional modules are unique up to transposition. For the 5 dimensional one there are (up to transposition) two possibilities (2,2,2) and (2,1,1,1,1), but we know that it must be the first because restricting (2,1,1,1,1) to $Sym(5)$ gives a one dimensional module in $Lie(5)$.

I think that I've read about the general case somewhere but my preliminary google has drawn a blank. If I can find the reference I'll add it to my answer.


The $Sym(n+2)$-module $Lie((n+2))$ is known as the Whitehouse module, there are some slides on Richard Stanley's website.

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So how do you go from a decomposition of $Lie((n+2))$ to a decomposition of $D_n(V)$? – Jim Conant May 25 '11 at 11:07
BTW, thanks for the link to Stanley's webpage! Those slides are very interesting. – Jim Conant May 25 '11 at 11:14
$V^{\otimes n}$ is a $GL(V)\times Sym(n)$-module, which it's probably nicer to view as a bimodule with a left action of $GL(V)$ and right action of $Sym(n)$, your formula for $D_n(V)$ is tensoring on the right (over $Sym(n+2)$) $V^{\otimes n+2}$ with the $Sym(n+2)$-module $Lie((n+2))$. So a decomposition of $Lie((n+2))$ yields a decomposition of $D_n(V)$ as a $GL(V)$-module. – James Griffin May 25 '11 at 11:39
Oh, and finding that webpage was pure luck. In searching for a reference for this I got distracted and started reading a paper on PreLie algebras, which just happened to mention that $Lie((n+2))$ is known as the Whitehead module. – James Griffin May 25 '11 at 11:42
I hope it's not the WhiteHEAD module. Independency results would be the last I'd want to see in this field. :P – darij grinberg May 25 '11 at 12:14

Hello. I would like to explain an elementary way to decompose $D_n(V)$ as a $GL(V)$-module. (But for large $n$, it is difficult to calculate by hand.)

We will use the following results. From now, we assume $\dim{V} \gg n$.

  1. The multiplicities of an irreducible $GL(V)$-module $S^\lambda{V}$ in $L_n(V)$ is equal to the one of an the irreducible $\mathrm{Cyc}_n$-module $\exp(2\pi{i}/n)$ in the $S_n$-module $\mathrm{Res}_{\mathrm{Cyc}_n}^{S_n}D^\lambda$. Here $D^\lambda$ is the irreducible $S_n$-module corresponding to $\lambda$.

  2. There is a combinatorial way to decompose $\mathrm{Res}_{\mathrm{Cyc}_n}^{S_n}D^\lambda$. We use the notion of the "major index" for a standard tableau of shape $\lambda$. This appears in Stanley's slide which is referd by James Griffin. For more details about 1 and 2, see Garsia's paper "Combinatorics of the free Lie algebra and the symmetric group"(Theorem 8.4) and Reutenauer's book "Free Lie algebras"(Theorem 8.8 and 8.9) etc.

  3. Pieri's rule. We can obtain the irreducible decomposition of the $GL(V)$-module $L_1(V) \otimes S^\lambda{V}$, namely this module is isomorphic to the direct sum of irreducible $GL(V)$-modules $S^\mu{V}$, where $\mu$ runs over the set of partitions by adding a box to $\lambda$. Then, if we know the irreducible decomposition of $L_n(V)$ by 1 (and 2), we can also obtain the irreducible decomposition of $L_1(V) \otimes L_n(V)$.

Example. For simplicity, we denote $\lambda$ by $S^\lambda{V}$.

$L_5(V) \cong (4,1) \oplus (3,2) \oplus (3,1^2) \oplus (2^2,1) \oplus (2,1^3)$.

$L_1(V) \otimes L_5(V)$ is isomorphic to $(5,1) \oplus 2(4,2) \oplus 2(4,1^2) \oplus (3^2) \oplus 3(3,2,1) \oplus 2(3,1^3) \oplus (2^3) \oplus 2(2^2,1^2) \oplus (2,1^4)$.

$L_6(V) \cong (5,1) \oplus (4,2) \oplus 2(4,1^2) \oplus (3^2) \oplus 3(3,2,1) \oplus (3,1^3) \oplus 2(2^2,1^2) \oplus (2,1^4)$. Then we have $D_4(V) \cong (4,2) \oplus (3,1^3) \oplus (2^3)$.

Similarly, I think, we can obtain $D_5(V) \cong (5,1^2) \oplus (4,2,1) \oplus (3^2,1) \oplus (3,2,1^2) \oplus (2^2,1^3)$.

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