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May 25, 2011 at 14:28 vote accept Jim Conant
May 25, 2011 at 12:55 comment added James Griffin Whoops, yes that would be unfortunate.
May 25, 2011 at 12:14 comment added darij grinberg I hope it's not the WhiteHEAD module. Independency results would be the last I'd want to see in this field. :P
May 25, 2011 at 11:42 comment added James Griffin 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.
May 25, 2011 at 11:39 comment added James Griffin $V^{\otimes n}$ 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.
May 25, 2011 at 11:14 comment added Jim Conant BTW, thanks for the link to Stanley's webpage! Those slides are very interesting.
May 25, 2011 at 11:07 comment added Jim Conant So how do you go from a decomposition of $Lie((n+2))$ to a decomposition of $D_n(V)$?
May 25, 2011 at 10:50 history edited James Griffin CC BY-SA 3.0
Added update section, no other changes
May 25, 2011 at 10:29 history answered James Griffin CC BY-SA 3.0