Tensor Product Multiplicities and Littlewood Richardson — Examples

Fix integers $a,b\ge 0$ and let $c\ge 0$ be an integer; the coefficient to compute is $c_{a\omega ,b\omega }^{c\omega }$. By [F1] its value is the alternating sum of the multiplicities $m_b(\phi )=1$ over the regular weights $\phi$ of $L(b)$ with $\nu (\phi )=c\omega$; recall $\lambda =a\omega$, $\mu =b\omega$, $\rho =\omega$.

Regularity and the value of $u$. For a weight $\phi ={\phi }_0\omega$ of $L(b)$, the shifted weight $\phi +a\omega +\rho =({\phi }_0+a+1)\omega$ is fixed by $s$ exactly when it is zero, i.e. ${\phi }_0=-(a+1)$. Such a weight exists in $L(b)$ exactly when $a+1\le b$ and $b+a+1$ is even, by [F3]; it is then the unique irregular weight. For every other weight, ${\phi }_0+a+1\mathrm{\ne }0$, so $u(\phi )=1$ if ${\phi }_0+a+1>0$ and $u(\phi )=s$ if ${\phi }_0+a+1<0$ [F2].

Contributions of the regular weights. Write $\phi =(b-2j)\omega$, $j=0,\dots ,b$, and put $t={\phi }_0+a+1=b-2j+a+1$. If $t>0$, then $\nu (\phi )=(b-2j+a)\omega$ and the sign is $(-1{)}^0=+1$; this contributes $+1$ to $c_{ab}^{\nu }$ with $\nu =b+a-2j$, i.e. to $c=a+b-2j$ with $0\le j<(a+b+1)\mathrm{/}2$. If $t<0$, then $u(\phi )=s$, $\nu (\phi )=s(t\omega )-\omega =-t\omega -\omega =(-t-1)\omega =(2j-a-b-2)\omega$ and the sign is $(-1{)}^1=-1$; this contributes $-1$ to the coefficient of $c=2j-a-b-2$. The inequality $t<0$ means $j>(a+b+1)\mathrm{/}2$, so $j$ ranges over the integers from $\lfloor (a+b+1)\mathrm{/}2\rfloor +1$ to $b$ (if any), and the corresponding $c=2j-a-b-2$ are exactly the integers congruent to $a+b$ modulo $2$ lying in the interval $0\le c\le b-a-2$ when $a+b$ is even and $1\le c\le b-a-2$ when $a+b$ is odd, the empty interval when the bound $b-a-2$ is negative.

The $+1$ family has $0\le j\le \min (b,\lfloor (a+b)\mathrm{/}2\rfloor )$, so its output weights are exactly the integers $c\equiv a+b(\mathrm{m}\mathrm{o}\mathrm{d}2)$ with $\max (0,a-b)\le c\le a+b$. If $b\le a$, there is no $-1$ family, and the least output is $a-b$. If $b>a$, the $-1$ family of step 2.1 cancels exactly the same-parity $+1$ outputs below $b-a$, namely $0\le c\le b-a-2$. In either case the survivors are precisely $\mathrm{\mid }a-b\mathrm{\mid }\le c\le a+b$ with the stated parity, each with coefficient one; all other coefficients are zero.

Reading the multiplicities: the values $c$ with $c_{ab}^c=1$ are $a+b,a+b-2,\dots ,\mathrm{\mid }a-b\mathrm{\mid }$, exactly $\min (a,b)+1$ values, with extremes $a+b$ and $\mathrm{\mid }a-b\mathrm{\mid }$; hence $L(a)\otimes L(b)\cong {⨁}_{j=0}^{\min (a,b)}L(a+b-2j)$, and the dimension check is ${\sum }_{j=0}^{\min (a,b)}(a+b-2j+1)=(a+1)(b+1)=\dim L(a)\dim L(b)$.

By [F1] the tensor product $L({\omega }_1)\otimes L({\omega }_1)$ is the direct sum of the $L({\omega }_1+\gamma )$ over the three elements $\gamma \in W{\omega }_1=\{{\omega }_1,{\omega }_2-{\omega }_1,-{\omega }_2\}$, with terms labeled by non-dominant weights dropped. The three translates are ${\omega }_1+{\omega }_1=2{\omega }_1$, ${\omega }_1+({\omega }_2-{\omega }_1)={\omega }_2$ and ${\omega }_1-{\omega }_2$; by [F2] the first two are dominant integral and the third is not. Hence $c_{{\omega }_1{\omega }_1}^{2{\omega }_1}=c_{{\omega }_1{\omega }_1}^{{\omega }_2}=1$ and all other tensor-product multiplicities vanish.

Identification with symmetric and exterior squares: by [F3] the submodule ${\mathrm{Sym}}^{2}V\subseteq V\otimes V$ is nonzero of dimension $6$ and has highest weight $2{\omega }_1$, so it contains $L(2{\omega }_1)$; similarly ${\mathrm{\Lambda }}^{2}V$ is nonzero of dimension $3$ with highest weight ${\omega }_2$ and contains $L({\omega }_2)$. The decomposition of step 1.1 has exactly the two summands $L(2{\omega }_1)$ and $L({\omega }_2)$, so $9=\dim (V\otimes V)=\dim L(2{\omega }_1)+\dim L({\omega }_2)$ with $\dim L(2{\omega }_1)\le 6$ and $\dim L({\omega }_2)\le 3$; hence $\dim L(2{\omega }_1)=6=\dim {\mathrm{Sym}}^{2}V$ and $\dim L({\omega }_2)=3=\dim {\mathrm{\Lambda }}^{2}V$, and the inclusions are equalities: ${\mathrm{Sym}}^{2}V=L(2{\omega }_1)$ and ${\mathrm{\Lambda }}^{2}V=L({\omega }_2)$.

The trivial module is not a summand: it would have to be one of the $L({\omega }_1+\gamma )$ with ${\omega }_1+\gamma =0$, i.e. $\gamma =-{\omega }_1\in W{\omega }_1$; but the three elements of $W{\omega }_1$ listed in [F1] are distinct from $-{\omega }_1$ (equality would force ${\omega }_2=0$, ${\omega }_1={\omega }_2$ or ${\omega }_1=0$). Hence $L(0)$ does not occur, consistent with the dimension count $6+3=9$.

Apply [F1] with $\lambda =(2,1)$ and $d=1$, using $S_{(1)}(V)=V$ and $h_1=s_{(1)}$: the summands are the partitions $\nu$ of $4$ with $[(2,1)]\subseteq [\nu ]$ whose complement is a horizontal strip of one box.

By [F2] those partitions are exactly $(3,1)$, $(2,2)$ and $(2,1,1)$, and the LR coefficient for each is one. The character identity of the Statement includes all three terms, with $s_{(2,1,1)}=0$ at rank $2$; the module decomposition has only the nonzero Schur summands, so the final term is omitted there by [F3].

Dimension check at $r=3$: using the values of [F3], $8\cdot 3=24=15+6+3$; dimension check at $r=2$: $2\cdot 2=4=3+1$. Both identities match the displayed decomposition.

The boxes of $(3,2,1)\mathrm{/}(2,1)$ 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 $2$.

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 $c_{(2,1),(2,1)}^{(3,2,1)}=2$.

Rank and the full expansion. For $r\ge 3$, [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 $3$.

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.

The tableau $T$ is a semistandard skew tableau of shape $\nu \mathrm{/}\lambda =(2,1)$ and content $\mu =(1,1,1)$: its first row is weakly increasing ($1\le 3$), its first column is strictly increasing ($1<2$) and its second column is a single cell, and its entries are exactly one $1$, one $2$ and one $3$.

The reading word of $T$ is obtained by reading each row from right to left starting with the top row: the first row contributes $3,1$ and the second row contributes $2$, so $w(T)=312$ (Littlewood--Richardson tableaux and coefficients).

The first prefix of $w(T)$ is the single letter $3$; it contains zero copies of the letter $2$ and one copy of the letter $3$, so the prefix condition of a lattice word fails for $i=2$. Hence $T$ is a semistandard skew tableau of shape $(2,1)$ and content $(1,1,1)$ whose reading word is not a lattice word, and by definition $T$ is not a Littlewood--Richardson tableau; this refutes the statement.

The instance is sharp: the semistandard tableaux of shape $(2,1)$ and content $(1,1,1)$ are exactly $T$ and the tableau with first row $12$ and second row $3$. Indeed the top-left entry must be $1$: were it $2$, the cell below it in the first column would carry the only remaining letter larger than $2$, namely $3$, leaving $1$ in the top-right cell in violation of weak row increase; were it $3$, no letter would remain below it in strict column increase. With $1$ in the top-left cell, the remaining letters $2$ and $3$ may be placed in the other two cells in either order, since $2,3>1$ and each row and column condition involves at most those two cells, giving exactly the two tableaux. The second of these has reading word $213$, which also fails the lattice condition at its first letter $2$; consistently, $c_{\mathrm{\varnothing },(1,1,1)}^{(2,1)}=0$ by the defining count of LR tableaux, while $s_{\mathrm{\varnothing }}=1$ and $s_{(2,1)}(x_1,x_2,x_3)=x_1^2{x}_2+x_1^2{x}_3+x_1{x}_2^2+2x_1{x}_2{x}_3+x_1{x}_3^2+x_2^2{x}_3+x_2{x}_3^2$ is not the squarefree polynomial $s_{(1,1,1)}=e_3$ (Littlewood--Richardson tableaux and coefficients, Skew Jacobi–Trudi and tableau expansion, Stable Schur functions from bialternants).