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.

  • $\begingroup$ Is at least the associativity/coassociativity obvious/clear combinatorially? $\endgroup$ – Vladimir Dotsenko Feb 2 '12 at 17:53
  • 1
    $\begingroup$ 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 $\endgroup$ – David E Speyer Feb 2 '12 at 18:40
  • $\begingroup$ Forgive me for a stupid question, but what about coassociativity? Is it follow from that, proved similarly, or? $\endgroup$ – Vladimir Dotsenko Feb 2 '12 at 19:23
  • 2
    $\begingroup$ 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. $\endgroup$ – David E Speyer Feb 2 '12 at 19:29

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.