Outer induction and restriction make the symmetric-group character ring a graded Hopf algebra

For each $t$, set $G_t^0:=\mathrm{Sym}(\{0,\dots ,t-1\})$ and $G_t^1:=\mathrm{Sym}(\{1,\dots ,t\})$, and let ${\beta }_t(i)=i+1$, with the unique empty bijection for $t=0$. Conjugation $c_t:G_t^0\to G_t^1$, $c_t(\sigma )={\beta }_t\sigma {\beta }_t^{-1}$, is a group isomorphism by [F5, F6] and carries the standard zero-based block embeddings of [F3] to the one-based embeddings used in [F2]. Given a one-based subgroup $H^1\le G_t^1$, set $H^0:=c_t^{-1}(H^1)$ and pull an $H^1$-module $W$ back to $W^0$ by $h{\cdot }_{0}w:=c_t(h){\cdot }_{1}w$. The map from the induced-function model for $(G_t^1,H^1,W)$ to that for $(G_t^0,H^0,W^0)$ [F7], $F^1\mapsto F^0$ with $F^0(g):=F^1(c_t(g))$, is invertible and satisfies $F^0(gh)=c_t(h{)}^{-1}{\cdot }_1{F}^0(g)=h^{-1}{\cdot }_0{F}^0(g)$. It intertwines left translation because $c_t(g_0^{-1}g)=c_t(g_0{)}^{-1}{c}_t(g)$. Pulling class functions back along $c_t$ therefore transports induction characters, and $c_t^{-1}$ maps the one-based outer-product blocks to the zero-based blocks of [F3]. Thus the outer product and restriction coproduct can both be computed in the zero-based realization.

Fix $f\in R(S_n)$ and $a+b+c=n$. Under the iterated external-product isomorphism from [F4], the $(a,b,c)$-component of either $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(f)$ or $(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(f)$ is the restriction of $f$ pulled back along the same embedding of $S_a\times S_b\times S_c$ acting on the three consecutive blocks of sizes $a,b,c$: composing the first two-block inclusions in either parenthesization gives that identical map on each triple $(\sigma ,\tau ,\rho )$. The iterated external-product maps agree on every tensor $\chi \otimes \psi \otimes \eta$ because both evaluate it as $\chi (\sigma )\psi (\tau )\eta (\rho )$, so the triple components are equal. Summing the finitely many triples proves coassociativity. The endpoints in [F3] show $(\epsilon \otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }=\mathrm{i}\mathrm{d}=(\mathrm{i}\mathrm{d}\otimes \epsilon )\mathrm{\Delta }$ for the augmentation in the Statement.

Define ${\mathrm{\Delta }}_{\mathrm{\Lambda }}f=f(X,Y)$ by substituting the disjoint union of finite alphabets into a stable symmetric function; this is well defined by [F16], is an algebra homomorphism, is coassociative by associativity of alphabet concatenation, and has counit evaluation at the empty alphabet, which is the constant term. Let $X=(x_1,\dots ,x_p)$ and $Y=(y_1,\dots ,y_q)$ be disjoint finite alphabets, ordered with every $x_i<y_j$. In a semistandard tableau of straight shape $\lambda$ on $X\bigsqcup Y$, the cells filled from $X$ form an initial segment in each row by [F17, F18]. If a cell in row $i+1$ and column $j$ is filled from $X$, then the cell above it exists by the partition shape [F24]; column strictness forces that upper entry to be smaller, hence it is also in $X$. Thus the row lengths of the $X$-cells are weakly decreasing and form a partition diagram $\mu \subseteq \lambda$. Restriction gives a semistandard tableau of shape $\mu$ on $X$ and one of shape $\lambda \mathrm{/}\mu$ on $Y$. Conversely, such a pair combines to a semistandard tableau of shape $\lambda$: within each piece the inequalities hold, and at each horizontal or vertical boundary every $X$-letter is smaller than every $Y$-letter. This is a weight-preserving bijection, so the diagonal coproduct satisfies ${\mathrm{\Delta }}_{\mathrm{\Lambda }}(s_{\lambda })=s_{\lambda }(X,Y)={\sum }_{\mu \subseteq \lambda }{s}_{\mu }(X)s_{\lambda \mathrm{/}\mu }(Y)$, with stable passage justified by [F16, F17, F19]. This is exactly the skew-Schur expansion in the characteristic coproduct identity [F15], so that identity intertwines the restriction coproduct with ${\mathrm{\Delta }}_{\mathrm{\Lambda }}$ on every Schur function.

Use the common zero-based realization fixed in step 1.1. For honest characters $\chi \in R(S_m)$ and $\psi \in R(S_n)$, let $N=m+n$, let $H=S_m\times S_n$ preserve consecutive source blocks $C_1,C_2$, and let $K=S_a\times S_b$ preserve consecutive target blocks $R_1,R_2$, where $a+b=N$. For $g\in S_N$ define $M(g)=\left(\begin{array}{cc}r& s\\ u& v\end{array}\right)$, where $r=\mathrm{\mid }{C}_1\cap g^{-1}{R}_1\mathrm{\mid }$, $s=\mathrm{\mid }{C}_1\cap g^{-1}{R}_2\mathrm{\mid }$, $u=\mathrm{\mid }{C}_2\cap g^{-1}{R}_1\mathrm{\mid }$, and $v=\mathrm{\mid }{C}_2\cap g^{-1}{R}_2\mathrm{\mid }$. Its row sums are $m,n$ and column sums $a,b$, and left multiplication by $K$ and right multiplication by $H$ preserve these counts. For each matrix with these margins, split $C_1,C_2$ into consecutive pieces of sizes $(r,s)$ and $(u,v)$, and split $R_1,R_2$ into consecutive pieces of sizes $(r,u)$ and $(s,v)$; the increasing bijections between paired pieces define a canonical representative $g_M$. If $g$ has this matrix, then the sets $C_j\cap g_M^{-1}{R}_i$ and $C_j\cap g^{-1}{R}_i$ have equal sizes. On each $C_j$, the unique increasing bijections from $C_j\cap g_M^{-1}{R}_i$ onto $C_j\cap g^{-1}{R}_i$ combine to $h_j\in S_{\mathrm{\mid }{C}_j\mathrm{\mid }}$; these give $h\in H$. Define $k_i$ on the disjoint pieces $gh(C_j\cap g_M^{-1}{R}_i)\subseteq R_i$ by $k_i(gh(x))=g_M(x)$; the pieces partition $R_i$ on both sides, so $k_i$ is a permutation of $R_i$, and together they give $k\in K$ with $kgh=g_M$. Thus the matrices classify $K\mathrm{\backslash }{S}_N\mathrm{/}H$, and the explicitly defined $g_M$ form a complete finite set of Mackey representatives, including zero-size blocks.

For finite alphabets $X,Y$, the coefficient of $t^n$ in $H_X(t)H_Y(t)$ is ${\sum }_{i+j=n}{h}_i(X)h_j(Y)$; by the defining sum for $h_n$ in [F20], this enumerates every degree-$n$ monomial in $X\bigsqcup Y$ once. Thus ${\mathrm{\Delta }}_{\mathrm{\Lambda }}(h_n)={\sum }_{i+j=n}{h}_i\otimes h_j$, and stable specialization is valid by [F16, F21]. Define $T(f)=(-1{)}^d\omega (f)$ for homogeneous $f\in {\mathrm{\Lambda }}^d$. Since $\omega$ is a graded ring automorphism [F23], $T$ is a graded algebra homomorphism and $T(h_i)=(-1{)}^i{e}_i$. The finite-rank identity $E(-t)H(t)=1$ [F22], together with compatibility of the stable generators under specialization [F21], gives coefficientwise in $\mathrm{\Lambda }$ that ${\sum }_{i+j=n}(-1{)}^i{e}_i{h}_j={\delta }_{n0}$; reindexing gives ${\sum }_{i+j=n}{h}_i(-1{)}^j{e}_j={\delta }_{n0}$. Therefore $T\ast \mathrm{i}\mathrm{d}$ and $\mathrm{i}\mathrm{d}\ast T$ agree with the constant-term counit on each generator $h_n$, including $h_0=1$. Both convolutions are algebra homomorphisms: $(T\ast \mathrm{i}\mathrm{d})(fg)=\sum T(f_{(1)})T(g_{(1)})f_{(2)}{g}_{(2)}=(T\ast \mathrm{i}\mathrm{d})(f)(T\ast \mathrm{i}\mathrm{d})(g)$ because ${\mathrm{\Delta }}_{\mathrm{\Lambda }}$ and $T$ are algebra maps and $\mathrm{\Lambda }$ is commutative; the same calculation with $T$ in the second tensor factor proves multiplicativity of $\mathrm{i}\mathrm{d}\ast T$. Since $h_1,h_2,\dots$ freely generate $\mathrm{\Lambda }$ by [F21], both convolutions equal the counit on all of $\mathrm{\Lambda }$. Thus $T$ is a two-sided antipode and $S_{\mathrm{\Lambda }}(f)=(-1{)}^d\omega (f)$ on ${\mathrm{\Lambda }}^d$.

For the representative $g_M$ in step 2.1, the intersection $L=K\cap g_{M}H{g}_M^{-1}$ consists exactly of permutations preserving each of the four intersections $R_1\cap g_M{C}_1$, $R_1\cap g_M{C}_2$, $R_2\cap g_M{C}_1$, and $R_2\cap g_M{C}_2$, hence is $S_r\times S_u\times S_s\times S_v$. Pulling the Mackey conjugate character ${}^{g_M}(\chi ⊠\psi )$ back to the source blocks by $g_M^{-1}$ and using [F10] gives the product of $\alpha ={\mathrm{Res}}_{S_r\times S_s}^{S_m}\chi$ and $\gamma ={\mathrm{Res}}_{S_u\times S_v}^{S_n}\psi$. Expand these honest product-group characters in the external-irreducible bases [F4], say $\alpha ={\sum }_i{c}_i{\xi }_i⊠{\xi }_i^{\mathrm{\prime }}$ and $\gamma ={\sum }_j{d}_j{\eta }_j⊠{\eta }_j^{\mathrm{\prime }}$. Regroup the four factors from source order $(r,s,u,v)$ into target order $(r,u,s,v)$; on $L$ the Mackey input character is then ${\sum }_{i,j}{c}_i{d}_j({\xi }_i⊠{\eta }_j)⊠({\xi }_i^{\mathrm{\prime }}⊠{\eta }_j^{\mathrm{\prime }})$, a character of $(S_r\times S_u)\times (S_s\times S_v)$.

Apply Mackey's formula [F9] to ${\mathrm{Res}}_K^{S_N}{\mathrm{Ind}}_H^{S_N}(\chi ⊠\psi )$. For the $M$-summand, apply the external-factor induction identity [F11] to each summand of the character in step 3.1, inducing from $S_r\times S_u$ to $S_a$ and from $S_s\times S_v$ to $S_b$. By the outer-product definition [F2], this Mackey term becomes under ${\mathrm{\Phi }}_{a,b}$ the image of ${\sum }_{i,j}{c}_i{d}_j({\xi }_i\circ {\eta }_j)\otimes ({\xi }_i^{\mathrm{\prime }}\circ {\eta }_j^{\mathrm{\prime }})$, precisely the product of the $(r,s)$-component of $\mathrm{\Delta }(\chi )$ and the $(u,v)$-component of $\mathrm{\Delta }(\psi )$.

The matrices of step 2.1 are in bijection with all degree allocations $r+s=m$, $u+v=n$, $r+u=a$, $s+v=b$. Hence summing the Mackey terms in step 4.1 gives exactly the $(a,b)$-component of $\mathrm{\Delta }(\chi )\mathrm{\Delta }(\psi )$; injectivity of ${\mathrm{\Phi }}_{a,b}$ in [F3, F4] gives ${\mathrm{\Delta }}_{a,b}(\chi \circ \psi )=(\mathrm{\Delta }(\chi )\mathrm{\Delta }(\psi ){)}_{a,b}$. This holds for every $a+b=m+n$, so $\mathrm{\Delta }(\chi \circ \psi )=\mathrm{\Delta }(\chi )\mathrm{\Delta }(\psi )$. Since the character rings consist of finite integral combinations of honest characters and $\circ ,\mathrm{\Delta }$ are bilinear, the identity extends to all virtual $\chi ,\psi$. The endpoints in [F3] also give $\mathrm{\Delta }(e)=e\otimes e$.

If $f\in R(S_m)$ and $g\in R(S_n)$ are homogeneous, then $\epsilon (f\circ g)=0=\epsilon (f)\epsilon (g)$ whenever $m+n>0$, since the outer product has degree $m+n$ by [F2]. When $m=n=0$, $R(S_0)=\mathbb{Z}e$ and $\circ$ is integer multiplication, so the same equality holds. Bilinearity gives that $\epsilon$ is a unital algebra homomorphism on $R_S$. Together with steps 1.2 and 5.1, this verifies all bialgebra axioms in [F12].

$R_S$ is nonnegatively graded by [F1], and its degree-zero component is exactly $\mathbb{Z}e$, with unit map $\mathbb{Z}\stackrel{\sim }{\to }R(S_0)$; hence the bialgebra of step 6.1 is connected by [F12]. The antipode lemma [F13] gives a unique graded antipode with the reduced-coproduct recursion, proving parts (i) and (ii) of the Statement.

By [F14], $\mathrm{ch}$ is a degree-preserving graded-ring isomorphism sending $e$ to $1$, and [F15] says it intertwines $\mathrm{\Delta }$ with ${\mathrm{\Delta }}_{\mathrm{\Lambda }}$. Since it preserves degree, it also carries the augmentation of step 6.1 to the constant-term counit: degree-zero elements map to constants and positive-degree elements have zero constant term. Thus $\mathrm{ch}$ is a bialgebra isomorphism from the bialgebra of step 6.1 to the diagonal bialgebra on $\mathrm{\Lambda }$.

Transporting the antipode of step 7.1 through the bialgebra isomorphism of step 7.2 gives an antipode on $\mathrm{\Lambda }$; by uniqueness of two-sided convolution inverses it is the map $T$ of step 2.2. Hence $\mathrm{ch}$ intertwines the antipodes and is an isomorphism of graded Hopf algebras, proving part (iii). All double-coset representatives were explicitly determined by a finite matrix, all tableaux sums are finite in each degree, and the recursive antipode uses no selections; no form of the axiom of choice is used.