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The Cauchy identity states that $$ \prod_{i,j} \frac{1}{1-x_i y_j} = \sum_\lambda s_\lambda(x) s_\lambda(y), $$ where $s_\lambda(x)$ is the Schur function.

Is there a known decomposition of the product $$ \prod_{i,j,k} \frac{1}{1-x_i y_j z_k} $$ as a sum of Schur functions?

Essentially equivalently, let $U$, $V$ and $W$ be finite dimensional complex vector spaces. Is the irreducible decomposition of $Sym(U \otimes V \otimes W)$ known as a $GL(U) \times GL(V) \times GL(W)$-module?

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Yes, up to the hard problem of determining Kronecker coefficients. Let $\Delta^\lambda$ be the Schur functor for the partition $\lambda$ of $r$ and let $S^\lambda$ be the corresponding irreducible representation of $S_r$ with character $\chi^\lambda$. Schur–Weyl duality states that

$$ U^{\otimes r} \cong \bigoplus_{\lambda \in \mathrm{Par}(r)} \Delta^\lambda(U) \otimes S^\lambda $$

as a representation of $\mathrm{GL}(U) \times S_r$.

Take $\dim U, \dim V, \dim W \ge r$. It follows by taking $S_r$ invariants in $U^{\otimes r} \otimes V^{\otimes r} \otimes W^{\otimes r}$ that the multiplicity of the irreducible $\mathrm{GL}(U) \times \mathrm{GL}(V) \times \mathrm{GL}(W)$-module $\Delta^\lambda(U) \otimes \Delta^\mu(V) \otimes \Delta^\nu(W)$ in $\mathrm{Sym}^r (U \otimes V \otimes W)$ is the Kronecker coefficient $g_{\lambda\mu\nu} = \langle \chi^\lambda\chi^\mu\chi^\nu, 1_{S_n} \rangle$. Restated in symmetric polynomials, this says

$$ \prod_{ijk}\frac{1}{1-x_iy_jz_k} = \sum_{\lambda\mu\nu}g_{\lambda\mu\nu} s_\lambda(x)s_\mu(y)s_\nu(z)$$

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    $\begingroup$ For a reference, see Stanley's EC2, exercise 7.78(f). $\endgroup$ Commented May 9, 2017 at 19:57
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    $\begingroup$ These coefficients are also known to be hard to compute, in the sense of #P-complete (up to conventions). The family of Kronecker coefficients contain the Littlewood-Richardson coefficients, which, in turn, contains the (skew) Koskta numbers. $\endgroup$ Commented May 9, 2017 at 20:31
  • $\begingroup$ Blerg, thanks everyone! I had looked at the exercises in EC2, but only the ones surrounding RSK stuffs. The answer is much worse than I hoped! $\endgroup$
    – andrewBee
    Commented May 9, 2017 at 21:11
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    $\begingroup$ On the plus side, if you find a way to prove this identity by an RSK-type correspondence, you'll have solved one of the main open problems in the character theory of the symmetric group. $\endgroup$ Commented May 10, 2017 at 19:04

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