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](http://arxiv.org/abs/1109.3963) 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.



GL(V)-representation theory for a Lie bracket kernel
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.


Update
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The $Sym(n+2)$-module $Lie((n+2))$ is known as the Whitehouse module, there are some slides on Richard Stanley's [website](http://www-math.mit.edu/~rstan/trans.html).


