Outer Products, Skew Specht Modules, and Littlewood–Richardson Coefficients — Examples

For each $t$, put $G_t^0=\mathrm{Sym}(\{0,\dots ,t-1\})$ and $G_t^1=\mathrm{Sym}(\{1,\dots ,t\})$, and let ${\beta }_t(i)=i+1$ and $c_t(\sigma )={\beta }_t\sigma {\beta }_t^{-1}$, with the unique empty bijection for $t=0$. The map $c_t$ is a group isomorphism because it is bijective and $c_t(\sigma \tau )=c_t(\sigma )c_t(\tau )$ by [F6]. For $a+b=t$, $c_t^{-1}$ carries the one-based block subgroup in [F2] to the zero-based block subgroup in [F5]. Relabeling every entry of a tableau by ${\beta }_t$ sends its tabloid to the corresponding tabloid and conjugates its row and column stabilizers; inversion pairs, hence permutation signs, are preserved by the shift. Thus it sends each polytabloid to the relabeled polytabloid by [F8] and intertwines the Specht actions. If $H^1\le G_t^1$ and $H^0=c_t^{-1}(H^1)$, pull an $H^1$-module $W$ back by $h{\cdot }_{0}w=c_t(h){\cdot }_{1}w$. The induced-function map $F^1\mapsto F^0$, $F^0(g)=F^1(c_t(g))$, preserves covariance since $F^0(gh)=c_t(h{)}^{-1}{\cdot }_1{F}^0(g)=h^{-1}{\cdot }_0{F}^0(g)$, and it intertwines left translation because $c_t(g_0^{-1}g)=c_t(g_0{)}^{-1}{c}_t(g)$. Hence the one-based outer-product and Specht-module calculation transports to the global zero-based convention.

Let $f^{\lambda }$ count standard tableaux of shape $\lambda$. In a nonempty standard tableau the largest entry lies in a removable corner: a node with a node to its right or below would have to carry a larger entry by [F12]. Deleting that largest entry gives a bijection with the disjoint union of standard tableaux on the shapes obtained by deleting one removable corner; conversely, appending the largest entry at any removable corner reverses the deletion. Hence $f^{\mathrm{\varnothing }}=1$ and $f^{\lambda }={\sum }_{x\in \mathrm{Rem}(\lambda )}{f}^{\lambda -x}$. The one-row and one-column shapes each have one standard tableau, and the recurrence gives $f^{(2,1)}=f^{(2)}+f^{(1,1)}=1+1=2$, $f^{(3,1)}=f^{(2,1)}+f^{(3)}=2+1=3$, $f^{(4,1)}=f^{(3,1)}+f^{(4)}=3+1=4$, $f^{(5,1)}=f^{(4,1)}+f^{(5)}=4+1=5$, $f^{(2,2)}=f^{(2,1)}=2$, $f^{(3,2)}=f^{(2,2)}+f^{(3,1)}=2+3=5$, $f^{(4,2)}=f^{(3,2)}+f^{(4,1)}=5+4=9$, $f^{(5,2)}=f^{(4,2)}+f^{(5,1)}=9+5=14$, $f^{(3,3)}=f^{(3,2)}=5$, $f^{(4,3)}=f^{(3,3)}+f^{(4,2)}=5+9=14$, $f^{(1,1,1)}=1$, $f^{(2,1,1)}=f^{(1,1,1)}+f^{(2,1)}=1+2=3$, $f^{(3,1,1)}=f^{(2,1,1)}+f^{(3,1)}=3+3=6$, $f^{(4,1,1)}=f^{(3,1,1)}+f^{(4,1)}=6+4=10$, $f^{(2,2,1)}=f^{(2,1,1)}+f^{(2,2)}=3+2=5$, $f^{(3,2,1)}=f^{(2,2,1)}+f^{(3,1,1)}+f^{(3,2)}=5+6+5=16$, $f^{(2,2,2)}=f^{(2,2,1)}=5$, $f^{(4,2,1)}=f^{(3,2,1)}+f^{(4,1,1)}+f^{(4,2)}=16+10+9=35$, $f^{(3,3,1)}=f^{(3,2,1)}+f^{(3,3)}=16+5=21$, and $f^{(3,2,2)}=f^{(3,2,1)}+f^{(2,2,2)}=16+5=21$. By [F9], the input dimensions are $5$ and $1$, and the five output dimensions are $14,14,35,21,21$.

The second factor $S^{(2)}$ is the trivial Specht module, so the outer Pieri rule [F1], with the block induction convention [F2] and the label bridge in step 1.1, expresses the induced module as the multiplicity-one sum over all partitions $\lambda ⊢7$ containing $(3,2)$ for which $\lambda \mathrm{/}(3,2)$ is a horizontal $2$-strip.

In the one-based block model of [F2], $H=S_5\times S_2$ stabilizes $B=\{6,7\}$. The map from left cosets $gH$ to two-element subsets $g(B)$ is a bijection: $H$ is exactly the setwise stabilizer of $B$, and permutations act transitively on two-element subsets. Hence $[S_7:H]=\left(\genfrac{}{}{0px}{}{7}{2}\right)=21$. By [F10] and [F11], the dimension of the induced module is $21\cdot 5\cdot 1=105$ using the input dimensions from step 1.2; the output dimensions in that step sum to $14+14+35+21+21=105$.

If $\lambda ⊢7$ contains $(3,2)$, then ${\lambda }_1\ge {\lambda }_2\ge 2$ and ${\lambda }_1\ge 3$. The case ${\lambda }_2\ge 4$ is impossible because the first two rows would contain at least eight nodes. If ${\lambda }_2=3$, the only possibilities are $(4,3)$ and $(3,3,1)$. If ${\lambda }_2=2$, the remaining nodes give $(5,2)$, $(4,2,1)$, $(3,2,2)$, or $(3,2,1,1)$. The added nodes in these six cases are respectively in columns $\{4,3\}$, $\{3,1\}$, $\{4,5\}$, $\{4,1\}$, $\{1,2\}$, and $\{1,1\}$. By [F4], exactly the last shape fails the horizontal-strip condition. Thus the five valid shapes are precisely $(5,2),(4,3),(4,2,1),(3,3,1),(3,2,2)$, with the column pairs stated.

Step 2.1 gives the induction decomposition over horizontal strips, and step 3.1 lists exactly the five such strips, each with multiplicity one. This proves the displayed module isomorphism; step 2.2 verifies its dimension independently and the result agrees with James's classical computation.

Listing the partitions of $7$ and retaining those with ${\lambda }_1\ge 2$ and ${\lambda }_2\ge 1$ from [F1] gives $(6,1),(5,2),(5,1,1),(4,3),(4,2,1),(4,1,1,1),(3,3,1),(3,2,2),(3,2,1,1),(3,1,1,1,1),(2,2,2,1),(2,2,1,1,1),(2,1,1,1,1,1)$. The two remaining partitions of $7$, $(7)$ and $(1^7)$, do not contain $\mu$ and have coefficient zero by [F4].

Let $f^{\lambda }$ count standard tableaux. In every nonempty standard tableau the largest entry is at a removable corner, that is, a node with no box to its right or below, by [F11]; deleting it leaves a partition and gives a bijection with the standard tableaux of the predecessor shapes, while appending the largest entry at any removable corner reverses the deletion. Thus $f^{\mathrm{\varnothing }}=1$ and $f^{\lambda }={\sum }_{x\in \mathrm{Rem}(\lambda )}{f}^{\lambda -x}$. The one-row and one-column shapes each have count $1$. Repeated application gives $f^{(2,1)}=1+1=2$, $f^{(3,1)}=2+1=3$, $f^{(4,1)}=3+1=4$, $f^{(5,1)}=4+1=5$, $f^{(2,2)}=2$, $f^{(3,2)}=2+3=5$, $f^{(4,2)}=5+4=9$, $f^{(5,2)}=9+5=14$, $f^{(3,3)}=5$, $f^{(4,3)}=5+9=14$, $f^{(1,1,1)}=1$, $f^{(2,1,1)}=1+2=3$, $f^{(3,1,1)}=3+3=6$, $f^{(4,1,1)}=6+4=10$, $f^{(2,2,1)}=3+2=5$, $f^{(3,2,1)}=5+6+5=16$, $f^{(2,2,2)}=5$, $f^{(2,1,1,1)}=1+3=4$, $f^{(2,2,1,1)}=4+5=9$, $f^{(3,1,1,1)}=4+6=10$, $f^{(4,1,1,1)}=10+10=20$, $f^{(5,1,1)}=10+5=15$, $f^{(4,2,1)}=16+10+9=35$, and $f^{(3,2,1,1)}=9+10+16=35$. By [F7], the input dimensions are $2,3$ and the eight summand dimensions in statement order are $14,15,14,35,20,21,21,35$, where $f^{(3,3,1)}=16+5=21$ and $f^{(3,2,2)}=16+5=21$.

Each retained skew diagram has four boxes, so by [F2] a filling is determined by the position of its unique $2$. Checking the row and column inequalities [F3], the semistandard positions for that $2$, in the same order as step 1.1, are $\{(1,6)\};\{(1,5),(2,2)\};\{(1,5),(3,1)\};\{(2,3)\};\{(1,4),(2,2),(3,1)\};\{(4,1)\};\{(2,3)\};\{(3,2)\};\{(4,1)\}$; the last four shapes admit no semistandard filling. In $(4,1,1,1)\mathrm{/}(2,1)$, $(3,2,2)\mathrm{/}(2,1)$ and $(3,2,1,1)\mathrm{/}(2,1)$, the column with two skew boxes forces the $2$ into its lower box, respectively $(4,1)$, $(3,2)$ and $(4,1)$. For $(3,1,1,1,1)$, $(2,2,1,1,1)$ and $(2,1^5)$ a column has at least three boxes, which cannot be strictly filled with only $1$'s and $2$'s; for $(2,2,2,1)$ each of columns $1$ and $2$ forces a separate $2$.

For a word with one $2$ and three $1$'s, the lattice condition in [F4] holds exactly when at least one $1$ is read before the $2$: after that first $1$, every prefix has at least as many $1$'s as $2$'s, and no letters exceed $2$. Removing the first-read top-right position from the semistandard-position lists in step 2.1 when it occurs, the LR-valid $2$-positions are $(5,2):(2,2)$, $(5,1,1):(3,1)$, $(4,3):(2,3)$, $(4,2,1):(2,2),(3,1)$, $(4,1,1,1):(4,1)$, $(3,3,1):(2,3)$, $(3,2,2):(3,2)$, and $(3,2,1,1):(4,1)$. Thus the coefficients in the order of step 1.1 are $0,1,1,1,2,1,1,1,1,0,0,0,0$, with zero also for $(7)$ and $(1^7)$.

Applying the outer Littlewood–Richardson theorem [F6] to the coefficient list in step 3.1 gives exactly the direct sum in the Statement; all other partitions of $7$ have coefficient zero. In particular, the two LR-valid positions $(2,2)$ and $(3,1)$ for $(4,2,1)$ yield $c_{(2,1),(3,1)}^{(4,2,1)}=2$.

By [F10] the global $S_7$ acts on $\{0,1,\dots ,6\}$, while [F5] gives the one-based block realization. The shift $\beta (i)=i+1$ is a bijection from $\{0,1,\dots ,6\}$ to $\{1,2,\dots ,7\}$; $\sigma \mapsto \beta \sigma {\beta }^{-1}$ preserves products and carries the initial three-letter block to $A=\{1,2,3\}$, so it preserves the subgroup index. In the one-based realization the block subgroup $H$ stabilizes $A$, and the map from its left cosets $gH$ to the images $g(A)$ is a bijection with the three-element subsets: every such subset is an image of $A$ under a permutation (extend bijections on $A$ and its four-element complement), and if $g(A)=g^{\mathrm{\prime }}(A)$ then $g^{-1}{g}^{\mathrm{\prime }}$ preserves $A$ and lies in $H$, so $gH=g^{\mathrm{\prime }}H$. Hence $[S_7:S_3\times S_4]=\left(\genfrac{}{}{0px}{}{7}{3}\right)=\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}=35$. By [F8] and [F9], the induced module has dimension $35\cdot (2\cdot 3)=210$. Step 4.1 and the dimensions in step 1.2 give the right-hand side dimension $14+15+14+2\cdot 35+20+21+21+35=210$. The LR enumeration and coset bijection are finite; any left transversal in [F8] is obtained by finitely many selections, which does not require the axiom of choice.

The diagrams give $[\lambda \mathrm{/}\mu ]=\{(1,3),(1,4),(2,2),(3,1)\}$, and the content condition [F3] requires three $1$'s and one $2$.

A filling is determined by the position of its unique $2$. Placing it at $(1,3)$ is impossible: weak increase in the first row would force the entry at $(1,4)$ to be at least $2$, requiring a second $2$. The other three assignments, listed as values in the box order $(1,3),(1,4),(2,2),(3,1)$, are $(1,2,1,1)$, $(1,1,2,1)$, and $(1,1,1,2)$. Each is semistandard: the first row is weakly increasing and no two boxes lie in the same column. Thus there are exactly three semistandard fillings.

Their reading words, in that order, are $2,1,1,1$, $1,1,2,1$, and $1,1,1,2$. The first fails the lattice condition at its first prefix; in each of the other two, every prefix has at least as many $1$'s as $2$'s. Since no entries exceed $2$, the other lattice inequalities are automatic. Hence exactly two of the three semistandard fillings are LR tableaux by [F4].

Therefore $c_{(2,1),(3,1)}^{(4,2,1)}=2$ by the LR-coefficient definition [F4], and the outer-induction multiplicity is also $2$ by [F5], while the semistandard skew-tableau count is $3$ by step 2.1. These unequal counts refute the stated claim.

The subpartitions of $(3,1)$ are $\mathrm{\varnothing },(1),(2),(1,1),(3),(2,1),(3,1)$: if the second row is empty, the first has length $0,1,2,$ or $3$, and if it has length $1$, the first has length $1,2,$ or $3$. Their sizes give the first tensor-factor degrees $0,1,2,2,3,3,4$; the corresponding skew diagrams are finite by [F5] and [F6]. The group convention is $S_4=\mathrm{Sym}(\{0,1,2,3\})$ by [F13].

For $\mu =(1)$, write the skew cells as $A=(1,2)$, $B=(1,3)$, and $C=(2,1)$. Semistandardness requires $A\le B$ and the reading word is $B,A,C$. Content $(3)$ gives the unique filling $A=B=C=1$, whose word $111$ is lattice. Content $(2,1)$ gives exactly one semistandard lattice filling, $A=B=1,C=2$, with word $112$; the other possible locations of the $2$ either violate $A\le B$ or make the word start with $2$. For content $(1,1,1)$, the distinct entries in the row must be one of $(A,B)=(1,2),(1,3),(2,3)$, so in every semistandard filling the reading word starts with $B\ge 2$ and fails the first-prefix lattice inequality. Thus $s_{(3,1)\mathrm{/}(1)}=s_{(3)}+s_{(2,1)}$.

For $\mu =(2)$, the two skew cells $(1,3)$ and $(2,1)$ have no row or column comparison, and their reading order is $(1,3),(2,1)$. Content $(2)$ has the unique filling $1,1$, and content $(1,1)$ has the unique lattice filling $1,2$; the reversed filling has a word beginning with $2$. Hence $s_{(3,1)\mathrm{/}(2)}=s_{(2)}+s_{(1,1)}$. For $\mu =(1,1)$, the remaining cells $(1,2),(1,3)$ satisfy the row inequality $(1,2)\le (1,3)$ and are read in the reverse order. Content $(2)$ gives one lattice filling, while the only semistandard filling of content $(1,1)$ has word $2,1$ and fails the lattice condition. Hence $s_{(3,1)\mathrm{/}(1,1)}=s_{(2)}$.

For $\mu =(3)$ and $\mu =(2,1)$ the skew diagram consists of one box, so its sole filling has content $(1)$ and its word is lattice, giving $s_{(3,1)\mathrm{/}(3)}=s_{(1)}=s_{(3,1)\mathrm{/}(2,1)}$ by [F3]–[F5] and [F14]. For the empty inner shape, [F17] and [F15] give $s_{(3,1)\mathrm{/}\mathrm{\varnothing }}={\sum }_{\nu ⊢4}⟨s_{(3,1)},s_{\nu }{⟩}_H{s}_{\nu }=s_{(3,1)}$ by orthonormality [F16]. For $\mu =(3,1)$, [F6] leaves only $\nu =\mathrm{\varnothing }$ in [F17], and [F15]–[F16] give $s_{(3,1)\mathrm{/}(3,1)}=⟨s_{(3,1)},s_{(3,1)}{⟩}_H{s}_{\mathrm{\varnothing }}=1$. These are all the remaining subshapes from step 1.1.

Applying the skewing formula [F2], expanding by [F3], using the tableau counts from [F4] and [F14] in steps 1.1–2.1, and translating $s_{\lambda }$ back to ${\chi }^{\lambda }$ by [F8] gives the terms grouped by first-factor degree: ${\mathrm{\Delta }}_{0,4}=1\otimes {\chi }^{(3,1)}$, ${\mathrm{\Delta }}_{1,3}={\chi }^{(1)}\otimes ({\chi }^{(3)}+{\chi }^{(2,1)})$, ${\mathrm{\Delta }}_{2,2}={\chi }^{(2)}\otimes ({\chi }^{(2)}+{\chi }^{(1,1)})+{\chi }^{(1,1)}\otimes {\chi }^{(2)}$, ${\mathrm{\Delta }}_{3,1}=({\chi }^{(3)}+{\chi }^{(2,1)})\otimes {\chi }^{(1)}$, and ${\mathrm{\Delta }}_{4,0}={\chi }^{(3,1)}\otimes 1$, as stated; the character labels are those of the Specht modules in [F7], and the endpoint factors use [F1].

Let $f^{\lambda }$ count standard tableaux. The one-row and one-column shapes each have one standard tableau by [F18], so $f^{(1)}=f^{(2)}=f^{(1,1)}=f^{(3)}=1$. Removing the largest entry gives $f^{(2,1)}=f^{(2)}+f^{(1,1)}=2$ and $f^{(3,1)}=f^{(2,1)}+f^{(3)}=3$: the largest entry is at a removable corner by [F18], and deletion and addition there are inverse operations. Thus [F9] gives dimensions $3,1,2,1,1,1$ for $(3,1),(3),(2,1),(2),(1,1),(1)$. By [F10], [F11], and [F12], the five component values at the identity are $3$, $1(1+2)=3$, $1(1+1)+1\cdot 1=3$, $(1+2)1=3$, and $3$, respectively; this also verifies the empty-factor endpoints. The enumeration is finite and uses no choice.