The irreducible decomposition of the tensor product of two irreducible representations of GL(n) is described by the Littlewood-Richardson rule. This same rule also governs the decomposition of the product of two Schur polynomials into a linear combination of Schur polynomials. In both cases, we label the components of the product and the decomposition with Young diagrams, or integer partitions.
We noticed that the Young diagrams in the decomposition always seem to form a lattice w.r.t. dominance order. That is, for each pair of diagrams in the decomposition, the least upper bound ("join") and the greatest lower bound ("meet") of the pair is also part of the decomposition.
For example, looking at the following decomposition (where the Young diagrams are denoted by the corresponding integer partitions):
$$(2,1) \otimes (2,1)=(4,2) \oplus (4,1,1) \oplus (3,3) \oplus 2 (3,2,1) \oplus (3,1,1,1) \oplus (2,2,2) \oplus (2,2,1,1)$$
The two unordered pairs in this decomposition are:
(2,2,2), (3,1,1,1); and (3,3), (4,1,1).
The join and meet of the first pair are (3,1,1,1) and (2,2,1,1) respectively.
For the second pair, these are (4,2) and (3,2,1). In both cases, the join and meet diagrams are parts of the decomposition. We checked it for many larger examples with extensive symbolic calculations, and the product had this lattice property in all cases.
Thus my question is: Applying the Littlewood-Richardson rule to a pair of Young diagrams, does the set of Young diagrams appearing in the result (with nonzero multiplicity) always form a lattice wrt dominance order?
Best Answer
Please excuse that I answer with a link, I only have a phone right now. http://www.findstat.org/MapsDatabase/Mp00192/