MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Is at least the associativity/coassociativity obvious/clear combinatorially? – Vladimir Dotsenko Feb 2 '12 at 17:53
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 – David Speyer Feb 2 '12 at 18:40
Forgive me for a stupid question, but what about coassociativity? Is it follow from that, proved similarly, or? – Vladimir Dotsenko Feb 2 '12 at 19:23
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. – David Speyer Feb 2 '12 at 19:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.