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2 Added update section, no other changes

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

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

1

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.