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A Littlewood--Richardson coefficient greater than one

Example

Assume the Axiom of Choice. For partitions $λ, μ, ν$ , let $c_{λ μ}^{ν}$ be the Littlewood--Richardson coefficient of Littlewood--Richardson tableaux and coefficients, the number of LR tableaux of shape $ν / λ$ and content $μ$ ; by The Littlewood--Richardson tensor-product rule it is the multiplicity of $S_{ν} (V)$ in $S_{λ} (V) \otimes S_{μ} (V)$ for $V = C^{r}$ when $ℓ (λ), ℓ (μ), ℓ (ν) \leq r$ ; if $ℓ (ν) > r$ , the coefficient remains the same LR tableau count but $S_{ν} (V) = 0$ (Schur modules and their characters). Then $c_{(2, 1), (2, 1)}^{(3, 2, 1)} = 2.$ Explicitly, the skew diagram $(3, 2, 1) / (2, 1)$ consists of one box in each of the three rows, at $(1, 3)$ , $(2, 2)$ and $(3, 1)$ in English row-column coordinates (Partitions, English diagrams, and conjugation), so the semistandard skew tableaux of content $(2, 1)$ are simply the three words of content $(2, 1)$ ; their reading words, taken right to left in each row starting with the top row, are $1, 1, 2$ ; $1, 2, 1$ ; and $2, 1, 1$ , and the first two are lattice words while $2, 1, 1$ fails at its first letter. The corresponding expansion in Schur functions is $s_{(2, 1)}^{2} = s_{(4, 2)} + s_{(4, 1, 1)} + s_{(3, 3)} + 2 s_{(3, 2, 1)} + s_{(3, 1, 1, 1)} + s_{(2, 2, 2)} + s_{(2, 2, 1, 1)},$ which at rank $3$ gives $8^{2} = 27 + 10 + 10 + 2 \cdot 8 + 1$ , the two shapes with four rows contributing $0$ (Stable Schur functions from bialternants, Semistandard tableaux expand Schur characters).

Facts & Assumptions

Given: AC, the partitions $λ = μ = (2, 1)$ and $ν = (3, 2, 1)$ , and the skew diagram $ν / λ$ .

[F1]

A semistandard skew tableau of shape $ν / λ$ weakly increases along rows and strictly increases down columns, and has content $μ$ when each letter $i$ occurs $μ_{i}$ times; its reading word reads the rows from right to left starting with the top row, and it is an LR tableau exactly when that word is a lattice word (Skew diagrams and semistandard skew tableaux, Semistandard tableaux and Kostka numbers, Littlewood--Richardson tableaux and coefficients).

[F2]

The LR coefficient is the tableau count of Littlewood--Richardson tableaux and coefficients; it is the multiplicity of $S_{ν} (V)$ in $S_{λ} (V) \otimes S_{μ} (V)$ when $ℓ (ν) \leq r$ , while $S_{ν} (V) = 0$ when $ℓ (ν) > r$ . At rank $r$ , the character of $S_{η} (V)$ is $s_{η} (x_{1}, \dots, x_{r})$ (The Littlewood--Richardson tensor-product rule, Schur modules and their characters, Semistandard tableaux expand Schur characters).

[F3]

For a three-row partition $(a, b, c)$ , padded by zeros, deleting all entries $3$ from a semistandard tableau leaves a two-row shape $(p, q)$ with $a \geq p \geq b \geq q \geq c$ : equal entries $3$ cannot share a column. Conversely these inequalities make the removed boxes a horizontal strip, so any tableau of shape $(p, q)$ on ${1, 2}$ extends uniquely by filling the removed boxes with $3$ . Its $q$ columns of height two are forced to be $1$ above $2$ , and the remaining $p - q$ first-row boxes contain a weakly increasing string of $1$ 's followed by $2$ 's, with $p - q + 1$ choices. Thus $s_{(a, b, c)} (1, 1, 1) = \sum_{p = b}^{a} \sum_{q = c}^{b} (p - q + 1) = (a - b + 1) (b - c + 1) (a - c + 2) / 2$ . At rank two the same column argument gives $s_{(a, b)} (1, 1) = a - b + 1$ . This gives the rank-three values $27, 10, 10, 8, 1, 8$ for $(4, 2), (4, 1, 1), (3, 3), (3, 2, 1), (2, 2, 2), (2, 1)$ respectively; the two four-row shapes give zero. The unique tableau of shape $(2, 2, 2)$ has two columns, both $1, 2, 3$ (Semistandard tableaux expand Schur characters, Stable Schur functions from bialternants).

Verification

1.1F1givenalgebra

The boxes of $(3, 2, 1) / (2, 1)$ are $(1, 3)$ in the first row, $(2, 2)$ in the second and $(3, 1)$ in the third: each row of the diagram contains exactly one box, and no two boxes share a column. Hence a filling of these three boxes is semistandard if and only if it is a word of content $(2, 1)$ , with no further condition, so there are exactly three semistandard tableaux, obtained by choosing the box that carries $2$ .

2.1F1step 1.1algebra

The reading word of a filling with the box $(i, j)$ read in the order $(1, 3), (2, 2), (3, 1)$ is the displayed triple of letters; the three possibilities are $1, 1, 2$ , $1, 2, 1$ and $2, 1, 1$ . A word is a lattice word when each prefix contains at least as many $1$ 's as $2$ 's; this holds for $1, 1, 2$ and $1, 2, 1$ but fails for $2, 1, 1$ , whose first prefix has one $2$ and no $1$ . Hence exactly two of the three semistandard tableaux are LR tableaux, and $c_{(2, 1), (2, 1)}^{(3, 2, 1)} = 2$ .

3.1F1F2step 1.1step 2.1algebra

Rank and the full expansion. For $r \geq 3$ , [F2] and step 2.1 identify the coefficient $2$ with the multiplicity of $S_{(3, 2, 1)} (V)$ ; at rank $2$ this Schur module is zero, although the LR coefficient remains $2$ . To check the displayed stable expansion, apply the Littlewood--Richardson rule at rank $6$ , so every partition of $6$ is within the rank bound. The only partitions of $6$ containing $(2, 1)$ are $(5, 1), (4, 2), (4, 1, 1), (3, 3), (3, 2, 1), (3, 1, 1, 1), (2, 2, 2), (2, 2, 1, 1), (2, 1, 1, 1, 1)$ . Their LR reading-word counts for content $(2, 1)$ are respectively $0, 1, 1, 1, 2, 1, 1, 1, 0$ : the nonzero words are $112$ for $(4, 2), (4, 1, 1), (3, 1, 1, 1), (2, 2, 1, 1)$ , $121$ for $(3, 3), (2, 2, 2)$ , and both $112, 121$ for $(3, 2, 1)$ . For $(5, 1)$ the top row forces reading word $211$ , which is not lattice; for $(2, 1, 1, 1, 1)$ the first column has three boxes but the content supplies only two distinct letters, so no semistandard filling exists. The remaining partitions $(6)$ and $(1, 1, 1, 1, 1, 1)$ do not contain $(2, 1)$ , so their coefficients vanish by definition. These counts give the stated expansion, with the two four-row terms vanishing at rank $3$ .

4.1F2F3step 3.1algebra∎

Rank- $3$ consistency. Evaluating the expanded identity at $x_{1} = x_{2} = x_{3} = 1$ and using [F3] gives $s_{(2, 1)} (1, 1, 1)^{2} = 8^{2} = 64 = 27 + 10 + 10 + 2 \cdot 8 + 1,$ the terms of the two four-row shapes $(3, 1, 1, 1)$ and $(2, 2, 1, 1)$ vanishing because no semistandard tableau with entries in ${1, 2, 3}$ can have four rows. This checks the expansion numerically.

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Sources

J. R. Stembridge, A Concise Proof of the Littlewood--Richardson Rule, Electronic Journal of Combinatorics 9 (2002), #N5, 4 pp. (standard reference, not scraped)
R. Goodman and N. R. Wallach, Symmetry, Representations, and Invariants, Graduate Texts in Mathematics 255, Springer 2009 (standard reference, not scraped)