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By "Cauchy decomposition" I mean the following identity, both sides in which are representations of $GL_n(\mathbb C)\times GL_m(\mathbb C)$: $$\mathrm{Sym}^p(V\otimes W)=\bigoplus_{\lambda\vdash p} V^\lambda\otimes W^\lambda.$$ In the above $V=\mathbb C^n$ with $GL_n(\mathbb C)$ acting on in the natural way, similarly $W=\mathbb C^m$. The sum is over all Young diagrams with $p$ squares and of height no more than $\mathrm{min}(n,m)$. Finally, $V^\lambda$ and $W^\lambda$ denote the irreducible representation of the corresponding $GL$ with its highest weight given by $\lambda$. (Is there a more appropriate name for this fact?)

Well, an analogous identity also holds: $$\mathrm{Sym}^p(V^*\otimes W)=\bigoplus_{\lambda\vdash p} (V^\lambda)^*\otimes W^\lambda.$$ Here the asterisk simply denotes taking the dual space equipped with the structure of a representation in the canonical way.

What would be the best (or any) paper to cite in this case? Of course, I'm hoping to see precisely this dual version written down somewhere.

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    $\begingroup$ I might be totally wrong, but can't the analogous identity be derived from the first one? Define an automorphism $w$ of $\mathrm{GL}_n\left(\mathbb C\right)$ by $w\left(A\right)=A^{T-1}$ for all matrices $A\in \mathrm{GL}_n\left(\mathbb C\right)$. Then, it is easy to see that $V^{\ast}\cong V^w$ as representations of $\mathrm{GL}_n\left(\mathbb C\right)$. Hence, $V^{\ast}\otimes W \cong V^w \otimes W \cong \left(V\otimes W\right)^{w\times \mathrm{id}}$. So much for the left hand side. For the right hand side, ... $\endgroup$
    – darij grinberg
    Commented May 1, 2012 at 17:45
  • $\begingroup$ ... it is enough to show that every partition $\lambda$ of $p$ satisfies $\left(V^{\lambda}\right)^{\ast} \cong \left(V^{\lambda}\right)^w$. But this follows from $V^{\ast} \cong V^w$ using $\left(V^{\lambda}\right)^{\ast} \cong \left(V^{\ast}\right)^{\lambda}$ (this should be pretty easy to check) and $\left(V^{\lambda}\right)^w \cong \left(V^{w}\right)^{\lambda}$ (this follows from functoriality of Schur functors). $\endgroup$
    – darij grinberg
    Commented May 1, 2012 at 17:47
  • $\begingroup$ Remark: To prove $\left(V^{\lambda}\right)^{\ast} \cong \left(V^{\ast}\right)^{\lambda}$, it is enough to recall that representations of $S_n$ are self-dual (since all irreducible representations are real) and use the definition of Schur functors as Hom's from irreducible representations of $S_n$. $\endgroup$
    – darij grinberg
    Commented May 1, 2012 at 18:07
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    $\begingroup$ I admit that the above argument was not very canonical, though (in the sense that my above automorphism $w$ was defined using the standard basis of $V$, so I do not know how to define something like this naturally for $\mathrm{GL}\left(V\right)$ instead of $\mathrm{GL}_n$), so I do not think it is optimal. $\endgroup$
    – darij grinberg
    Commented May 1, 2012 at 18:09
  • $\begingroup$ Yes, the second identity can certainly be derived from the first in one way or another. It can also be proven in almost any fashion the first one can. I was just hoping to avoid any kind of proof by giving a reference. Well, no such luck. $\endgroup$
    – Igor Makhlin
    Commented May 2, 2012 at 9:38

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