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The Littlewood–Richardson coefficients are the multiplicities $$ c(\lambda,\mu,\nu)= \dim_{\mathbb{C}}\operatorname{Hom}_{S_n}(S(\nu),S(\lambda/\mu)) $$ and the Littlewood–Richardson rule says that these coefficients are enumerated by the semistandard Young tableaux satisfying the lattice permutation condition. Moreover, using James' approach one can actually construct the homomorphisms from these tableaux (so we can work on a structural level). There are also plenty of proofs using only combinatorics (of Schur functions).

Is there a proof on a level between these two? The characters of the symmetric groups have "more structure" than simple combinatorics but have less structure than representations. Is there an entirely character-theoretic proof of this rule? I imagine that any such proof would use the Murnaghan–Nakayama rule for (skew representations of) symmetric groups.

P.S. James gave one of the first proofs of the Littlewood–Richardson rule in 1977 (funnily enough, this wasn't proven by Littlewood and Richardson). James' proof calculates these homomorphisms over any field (whereas his contemporaries Schützenberger (1977) and Thomas (1974) only considered the semisimple case using combinatorics) and so his work is far more impressive and benefits from being self-contained. Yet he gets little recognition of this (see Wikipedia for example). His proof can be found here https://doi.org/10.1016/0021-8693(77)90380-5

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  • $\begingroup$ How do you want the characters to be defined then? As symmetric functions in the formal sense, i.e., without reference to an alphabet, but just by Jacobi-Trudi-like identities? $\endgroup$ – darij grinberg May 23 '18 at 14:51
  • $\begingroup$ No, I mean as characters of the symmetric group. As traces of the representing matrices. As in characters in the character table. $\endgroup$ – Chris Bowman May 23 '18 at 16:28
  • $\begingroup$ How do you even define the character corresponding to partition $\lambda$, without defining the Specht module? $\endgroup$ – darij grinberg May 23 '18 at 16:38
  • $\begingroup$ I changed {\rm dim} to \dim and did some other edits. That makes a conspicuous difference in spacing. With \dim and with \operatorname{Hom}, the space to the left and right depends on the context. Remember that MathJax evolved ultimately from software created by Donald Knuth, so one expects it to behave well. $\endgroup$ – Michael Hardy May 23 '18 at 17:50
  • $\begingroup$ Hi Darij, I'm not saying that I hate representations! I don't mind that characters are ultimately defined via representations :) All I was wondering was if there is a character-theoretic proof. I'm a big fan of the Murnaghan--Nakayama rule and I think it would be nice if it can be used to prove this classical result. Hi Michael, thanks for the edits! $\endgroup$ – Chris Bowman May 23 '18 at 17:58
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I might have misunderstood the question, or what it asks for, but doesn't the method of crystals do this? See Prop. 2.62.

The idea is that there is a notion of a crystal structure on a combinatorial set. The cool thing is that it automatically makes your set into an $S_n$-module (so you get a representation).
On the other hand, a crystal structure is very combinatorial, so you have a strong connection to combinatorics.

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  • $\begingroup$ Perhaps I'm misunderstanding what you're saying. But I think crystals enter this picture via the representation theory of ${\rm GL}_n$. I think the relationship with $S_n$ isn't direct, but rather is via Schur--Weyl duality. In particular, I don't think there's any way of interpreting a crystal statement directly in the setting of characters of representations of $S_n$. What I'm really hoping is that the (skew) Murnaghan--Nakayama rule can be used to prove this result. $\endgroup$ – Chris Bowman May 23 '18 at 18:05
  • $\begingroup$ Aha, well, there is a skew MN-rule in Stanleys EC2.. but I suppose you want a less combinatorial one.. $\endgroup$ – Per Alexandersson May 23 '18 at 20:01

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