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Let $\Lambda$ be the ring of symmetric functions in infinitely many variables, $x_1$, $x_2$, .... For $f \in \Lambda$, let $\Delta(f) \in \Lambda \otimes \Lambda$ be $f(x_1 \otimes 1, 1 \otimes x_1, x_2 \otimes 1, 1 \otimes x_2, \ldots)$. It turns out that $\Delta$ of a Schur function is given by Littlewood-Richardson coefficients: $\Delta(s_{\nu}) = \sum_{\lambda, \mu} c_{\lambda \mu}^{\nu} s_{\lambda} \otimes s_{\mu}$. Littlewood-Richardson coefficients also describe multiplication: $s_{\lambda} \cdot s_{\mu} = \sum c_{\lambda \mu}^{\nu} s_{\nu}$. And we have the identity (elementary once you unravel what it means) $\Delta(fg) = \Delta(f) \Delta(g)$.

So, for any partitions $\kappa$, $\lambda$, $\mu$, $\nu$, we must have $$\sum_{\rho} c_{\kappa \lambda}^{\rho} c_{\mu \nu}^{\rho} = \sum_{\sigma_1, \sigma_2, \sigma_3, \sigma_4} c_{\sigma_1 \sigma_2}^{\kappa} c_{\sigma_2 \sigma_3}^{\lambda} c_{\sigma_3 \sigma_4}^{\mu} c_{\sigma_4 \sigma_1}^{\nu}.$$

I'm curious whether anyone knows a bijective proof of this? No motivation right now, but I find it pretty useful to know which algebraic facts have good combinatorial explanations.

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  • $\begingroup$ Is at least the associativity/coassociativity obvious/clear combinatorially? $\endgroup$ Feb 2, 2012 at 17:53
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    $\begingroup$ Depends which combinatorial rule you use. If you are using jdt then yes, the proof that jdt is well defined also shows that it is associative. If you are using hives/honeycombs/puzzles, see arxiv.org/abs/math/0306274 $\endgroup$ Feb 2, 2012 at 18:40
  • $\begingroup$ Forgive me for a stupid question, but what about coassociativity? Is it follow from that, proved similarly, or? $\endgroup$ Feb 2, 2012 at 19:23
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    $\begingroup$ If you use the same combinatorial rule to describe LR coefficients in both cases, then any combinatorial proof of associativity is also a proof of co-associativity. $\endgroup$ Feb 2, 2012 at 19:29

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