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The Pieri formula gives a decomposition

$$\textrm{Sym}^d V \otimes \textrm{Sym}^d V = \bigoplus \mathbb{S}_{(d+a, d-a)}V,$$

the sum over $0\leq a \leq d$. The left-hand side decomposes into a direct sum of $\textrm{Sym}^2(\textrm{Sym}^d V)$ and $\bigwedge^2(\textrm{Sym}^d V)$. Show that, in fact,

$\textrm{Sym}^2(\textrm{Sym}^d V)=\mathbb{S}_{(2d,0)}V\bigoplus \mathbb{S}_{(2d-2,2)}V\bigoplus \mathbb{S}_{(2d-4,4)}V \bigoplus...$


$\bigwedge^2(Sym^d V)=\mathbb{S}_{(2d-1,1)}V\bigoplus \mathbb{S}_{(2d-3,3)}V\bigoplus \mathbb{S}_{(2d-5,5)}V \bigoplus...$

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Are you referring to this: Pieri formula? –  draks ... Oct 10 '12 at 19:01
yes, exactly. I feel like the idea is to show that the sum of characters on the right hand side in either case is the same as the character on the left, I'm just not sure exactly what the character on the left should be. –  Chet Oct 11 '12 at 1:00
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