The restriction coproduct of the character ${\chi }^{(3,1)}$ of $S_4$

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.