The restriction coproduct on the graded symmetric-group character ring

Let $R_S={⨁}_{n\ge 0}R(S_n)$ be the graded abelian group of ordinary symmetric-group character rings (The graded ordinary representation ring of the symmetric groups). For $n\ge 0$ and $a,b\ge 0$ with $a+b=n$, regard $S_n$ as the permutations of $\{0,1,\dots ,n-1\}$ (The finite symmetric group $S_n$, one-line notation, and cycle notation) and set

$A_{a,b}:=\{0,1,\dots ,a-1\},B_{a,b}:=\{a,a+1,\dots ,n-1\},$

with an empty block when its size is zero. Define

${\iota }_{a,b}:S_a\times S_b⟶S_n,{\iota }_{a,b}(\sigma ,\tau )(i):=\{\begin{array}{ll}\sigma (i),& i\in A_{a,b},\\ a+\tau (i-a),& i\in B_{a,b}\mathrm{.}\end{array}$

The map is injective and respects composition: it acts as $\sigma$ on the first block and as the translated permutation $\tau$ on the second. Its image

$H_{a,b}:=\{g\in S_n:g(A_{a,b})=A_{a,b},g(B_{a,b})=B_{a,b}\}$

is a subgroup (Subgroup) identified with $S_a\times S_b$ by ${\iota }_{a,b}$; indeed the identity, products and inverses in the image are respectively ${\iota }_{a,b}(1,1)$, ${\iota }_{a,b}(\sigma {\sigma }^{\mathrm{\prime }},\tau {\tau }^{\mathrm{\prime }})$, and ${\iota }_{a,b}({\sigma }^{-1},{\tau }^{-1})$ (The external direct product $G\times H$ with componentwise multiplication, $G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections, Monoid homomorphism and group homomorphism).

For $f\in R(S_n)$, restrict its virtual character to $H_{a,b}$ and pull it back along ${\iota }_{a,b}$. This is in $R(S_a\times S_b)$: write $f={\sum }_j{m}_j{\chi }_{V_j}$ with $m_j\in \mathbb{Z}$ and $V_j$ finite-dimensional complex representations of $S_n$ (Virtual characters and the character ring $R(G)$ of a finite group, A finite-dimensional representation $\rho :G\to \mathrm{GL}(V)$ over a field, and its degree, The character ${\chi }_V(g)=\mathrm{tr}({\rho }_V(g))$ of a finite-dimensional complex representation); each $V_j$ restricts to a representation of $H_{a,b}$ and, after pullback, of $S_a\times S_b$ (The sign representation of $S_n$ and the restriction ${\mathrm{Res}}_H^G(V)$ of a representation to a subgroup). By Maschke's theorem it is a finite direct sum of irreducible representations, and additivity of characters then puts its character in the integral span of irreducible characters, namely $R(S_a\times S_b)$ (Maschke's theorem for finite groups over fields whose characteristic does not divide $\mathrm{\mid }G\mathrm{\mid }$, Characters add on direct sums, multiply on tensor products, and conjugate on duals). Denote the resulting character by

$f\circ {\iota }_{a,b}:S_a\times S_b\to \mathbb{C},(f\circ {\iota }_{a,b})(\sigma ,\tau )=f({\iota }_{a,b}(\sigma ,\tau ))\mathrm{.}$

The external-product map

${\mathrm{\Phi }}_{a,b}:R(S_a){\otimes }_{\mathbb{Z}}R(S_b)⟶R(S_a\times S_b),\chi \otimes \psi ⟼\chi ⊠\psi ,$

is an isomorphism (The character ring of a direct product is the tensor product of the factor character rings). Define ${\mathrm{\Delta }}_{a,b}(f)$ to be the unique element satisfying ${\mathrm{\Phi }}_{a,b}({\mathrm{\Delta }}_{a,b}(f))=f\circ {\iota }_{a,b}$. The restriction coproduct is the $\mathbb{Z}$-linear map

$\mathrm{\Delta }:R_S⟶R_S{\otimes }_{\mathbb{Z}}{R}_S,\mathrm{\Delta }(f):={\sum }_{a=0}^n{\mathrm{\Delta }}_{a,n-a}(f)(f\in R(S_n)),$

where the restriction and pullback maps and ${\mathrm{\Phi }}_{a,b}^{-1}$ are $\mathbb{Z}$-linear, and each summand is included in the $(a,n-a)$ graded component. Extend this formula linearly to an arbitrary $f={\sum }_n{f}_n\in R_S$; the sum is finite because $R_S$ is a direct sum. Thus $\mathrm{\Delta }(R(S_n))\subseteq {⨁}_{a+b=n}R(S_a){\otimes }_{\mathbb{Z}}R(S_b)$, so $\mathrm{\Delta }$ has degree zero and is a finite sum in each degree, with no completion (The graded ordinary representation ring of the symmetric groups, The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums). At the endpoints, ${\mathrm{\Delta }}_{0,n}(f)=\mathbf{1}\otimes f$ and ${\mathrm{\Delta }}_{n,0}(f)=f\otimes \mathbf{1}$; in particular $\mathrm{\Delta }(c\mathbf{1})=c(\mathbf{1}\otimes \mathbf{1})$ in degree zero.

The coproduct is cocommutative. Let $\tau (x\otimes y)=y\otimes x$. The ordered block subgroups need not be equal: in $S_3$, the transposition $(01)$ belongs to $H_{2,1}$ but not to $H_{1,2}$. They are conjugate by the block-swap permutation ${\rho }_{a,b}\in S_n$ given by ${\rho }_{a,b}(i)=b+i$ for $0\le i<a$ and ${\rho }_{a,b}(a+j)=j$ for $0\le j<b$. Directly from the block actions,

${\rho }_{a,b}{\iota }_{a,b}(\sigma ,\tau ){\rho }_{a,b}^{-1}={\iota }_{b,a}(\tau ,\sigma )\mathrm{.}$

Since every virtual character of $S_n$ is a class function, $f(g)=f({\rho }_{a,b}g{\rho }_{a,b}^{-1})$; consequently the two restrictions, after exchanging factors, agree. Applying the injective maps ${\mathrm{\Phi }}_{a,b}$ and ${\mathrm{\Phi }}_{b,a}$ gives ${\mathrm{\Delta }}_{b,a}(f)=\tau ({\mathrm{\Delta }}_{a,b}(f))$. Summing over all $a+b=n$ proves $\tau \mathrm{\Delta }(f)=\mathrm{\Delta }(f)$. No form of the axiom of choice is used.