Schur functions on sums and products of alphabets

Genericity. Fix $d\ge 0$ and $N\ge d$. The substitution ${\sigma }_d$ from $\mathbb{Q}[A_1,\dots ,A_d;B_1,\dots ,B_d]$ to $\mathbb{Q}[x_1,\dots ,x_N,y_1,\dots ,y_N]$ with $A_k\mapsto p_k(x)$, $B_k\mapsto p_k(y)$ is injective. Indeed, by [F4] over $\mathbb{Q}$ the identities $k{e}_k=(-1{)}^{k-1}{p}_k+{\sum }_{i=1}^{k-1}(-1{)}^{i-1}{e}_{k-i}{p}_i$ express each $e_k$ recursively in $p_1,\dots ,p_k$ and each $p_k$ recursively in $e_1,\dots ,e_k$ for $k\le N$, so $\mathbb{Q}[e_1,\dots ,e_N]=\mathbb{Q}[p_1,\dots ,p_N]$ as subrings of $\mathbb{Q}[x_1,\dots ,x_N]$; by [F5] that ring is a free polynomial ring on $N$ generators, so $p_1,\dots ,p_N$, and hence $p_1,\dots ,p_d$, are algebraically independent. The same holds in the $y$-variables, and the tensor product over $\mathbb{Q}$ of two injective substitutions is injective.

Multiplicativity. For all specializations $U,V$ and all $n\ge 0$ one has $h_n[U+V]={\sum }_{i+j=n}{h}_i[U]h_j[V]$; consequently $\sigma [U+V]=\sigma [U]\sigma [V]$ coefficientwise. By [F3], $h_n={\sum }_{\lambda ⊢n}{p}_{\lambda }\mathrm{/}{z}_{\lambda }$ involves only $p_1,\dots ,p_n$, so both sides of the claimed identity are the value at $(p_1[U],\dots ,p_n[U],p_1[V],\dots ,p_n[V])$ of the universal polynomial $D_n(A;B):={\sum }_{\lambda ⊢n}{\prod }_i(A_{{\lambda }_i}+B_{{\lambda }_i})\mathrm{/}{z}_{\lambda }-{\sum }_{i+j=n}({\sum }_{\mu ⊢i}{p}_{\mu }(A)\mathrm{/}{z}_{\mu })({\sum }_{\nu ⊢j}{p}_{\nu }(B)\mathrm{/}{z}_{\nu })$, where $p_{\mu }(A)={\prod }_i{A}_{{\mu }_i}$. At honest alphabets $x_1,\dots ,x_n$ and $y_1,\dots ,y_n$, the generating function ${\sum }_k{h}_k(\mathrm{\ell })t^k={\prod }_{\mathrm{\ell }}(1-\mathrm{\ell }t{)}^{-1}$ of [F3] gives ${\sum }_k{h}_k(x\bigsqcup y)t^k={\prod }_{\mathrm{\ell }}(1-\mathrm{\ell }t{)}^{-1}=({\sum }_i{h}_i(x)t^i\left)\right({\sum }_j{h}_j(y)t^j)$, so $D_n(p(x);p(y))=0$. By step 1.1 with $d=N=n$ the substitution is injective, so $D_n=0$, and the identity holds universally; since $\sigma ={\sum }_n{h}_n$, this is multiplicativity of $\sigma$, coefficient by coefficient finite.

Cauchy identities for arbitrary specializations. For all specializations $X,Y$ and all $d\ge 0$, $h_d[XY]={\sum }_{\lambda ⊢d}{s}_{\lambda }[X]s_{\lambda }[Y]={\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }[X]p_{\rho }[Y]$, equivalently $\sigma [XY]={\sum }_{\lambda }{s}_{\lambda }[X]s_{\lambda }[Y]={\sum }_{\lambda }{z}_{\lambda }^{-1}{p}_{\lambda }[X]p_{\lambda }[Y]$ coefficientwise. The second equality is [F3] applied to the alphabet $XY$: $h_d[XY]={\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }[XY]={\sum }_{\rho }{z}_{\rho }^{-1}{p}_{\rho }[X]p_{\rho }[Y]$ by [F1]. For the first, fix $d$; both ${\sum }_{\lambda ⊢d}{s}_{\lambda }[X]s_{\lambda }[Y]$ and ${\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }[X]p_{\rho }[Y]$ are values of universal polynomials in $(p_k[X],p_k[Y]{)}_{k\le d}$, by [F2]. At honest alphabets $x_1,\dots ,x_N$, $y_1,\dots ,y_N$ with $N\ge d$, apply the projection ${\mathrm{\Lambda }}^d(X)\otimes {\mathrm{\Lambda }}^d(Y)\to A_N^d(x)\otimes A_N^d(y)$ of [F10] to the degree-$(d,d)$ component of the Cauchy identity [F6]: it maps the two sides to ${\sum }_{\lambda ⊢d}{s}_{\lambda }(x)s_{\lambda }(y)$ and ${\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }(x)p_{\rho }(y)$, which are therefore equal. By step 1.1 with $d=N$ the universal difference vanishes identically, proving the first equality for all $X,Y$.

Coproduct (a). Apply step 2.2 to the pair of specializations $(U+V,Z)$, where $Z$ denotes the tautological specialization into a fresh copy of $\mathrm{\Lambda }$; the resulting chain of identities is ${\sum }_{\lambda }{s}_{\lambda }[U+V]s_{\lambda }[Z]=\sigma [(U+V)Z]=\sigma [UZ+VZ]=\sigma [UZ]\sigma [VZ]={\sum }_{\mu ,\nu }{s}_{\mu }[U]s_{\nu }[V]s_{\mu }[Z]s_{\nu }[Z]$, where the second equality uses $p_k[(U+V)Z]=p_k[UZ]+p_k[VZ]$ from [F1], the third is step 2.1, and the last is step 2.2 applied twice. By [F7], $s_{\mu }[Z]s_{\nu }[Z]={\sum }_{\lambda }{c}_{\mu \nu }^{\lambda }{s}_{\lambda }[Z]$, so the right side equals ${\sum }_{\lambda }({\sum }_{\mu ,\nu }{c}_{\mu \nu }^{\lambda }{s}_{\mu }[U]s_{\nu }[V])s_{\lambda }[Z]$, each coefficient being a finite sum. The $s_{\lambda }[Z]$ form a $\mathbb{Z}$-basis of ${\mathrm{\Lambda }}^d(Z)$ by [F8] and [F10], so comparing coefficients of $s_{\lambda }[Z]$ gives $s_{\lambda }[U+V]={\sum }_{\mu ,\nu }{c}_{\mu \nu }^{\lambda }{s}_{\mu }[U]s_{\nu }[V]$ for every $\lambda$.

Kernel identity. For $f,g\in {\mathrm{\Lambda }}_{\mathbb{Q}}^d$ and independent formal alphabet variables $U,V,W$, with the Hall form of The Hall inner product on symmetric functions extended $\mathbb{Q}$-bilinearly to $\mathrm{\Lambda }(U)\widehat{\otimes }\mathrm{\Lambda }(V)$, one has $⟨f[U]g[V]\mathrm{\mid }\sigma [UVW]{⟩}_{U,V}=(f\ast g)[W]$. By step 2.2 applied to the pair $(UV,W)$, $\sigma [UVW]={\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }[U]p_{\rho }[V]p_{\rho }[W]$ in the relevant degree, and the pairing is a finite sum because only $\rho ⊢d$ matches the bidegree $(d,d)$ of $f[U]g[V]$. Writing $f={\sum }_{\rho }{a}_{\rho }{p}_{\rho }$ and $g={\sum }_{\rho }{b}_{\rho }{p}_{\rho }$, so that $⟨f,p_{\rho }{⟩}_H=a_{\rho }{z}_{\rho }$ and $⟨g,p_{\rho }{⟩}_H=b_{\rho }{z}_{\rho }$ by [F8], the left side is ${\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }[W]a_{\rho }{b}_{\rho }{z}_{\rho }^2={\sum }_{\rho }{a}_{\rho }{b}_{\rho }{z}_{\rho }{p}_{\rho }[W]$, which is $(f\ast g)[W]$ by the power-sum diagonalization [F9].

Plethysm (b). Keep $U,V,W$ independent formal alphabet variables. Apply step 2.2 to the pair $(UV,W)$: $\sigma [UVW]={\sum }_{\lambda }{s}_{\lambda }[UV]s_{\lambda }[W]$. Pairing with $f[U]g[V]$ and using step 3.2 gives ${\sum }_{\lambda }⟨f[U]g[V],s_{\lambda }[UV]{⟩}_{U,V}{s}_{\lambda }[W]=(f\ast g)[W]={\sum }_{\lambda }⟨f\ast g,s_{\lambda }{⟩}_H{s}_{\lambda }[W]$, the last equality being the orthonormal expansion [F8]. Comparing coefficients of $s_{\lambda }[W]$, legitimate exactly as in step 3.1, yields $⟨f[U]g[V],s_{\lambda }[UV]{⟩}_{U,V}=⟨f\ast g,s_{\lambda }{⟩}_H$. Taking $f=s_{\mu }$, $g=s_{\nu }$ of degree $d$ gives $⟨s_{\mu }[U]s_{\nu }[V],s_{\lambda }[UV]{⟩}_{U,V}=g_{\mu \nu }^{\lambda }$ by [F9] and [F8]. The products $s_{\mu }[U]s_{\nu }[V]$ with $\mu ,\nu ⊢d$ are an orthonormal basis of the bidegree-$(d,d)$ piece of $\mathrm{\Lambda }(U)\widehat{\otimes }\mathrm{\Lambda }(V)$ for the tensor Hall form, because tensor products of orthonormal families are orthonormal and each bidegree piece is finite dimensional; expanding $s_{\lambda }[UV]$ in that basis gives $s_{\lambda }[UV]={\sum }_{\mu ,\nu }{g}_{\mu \nu }^{\lambda }{s}_{\mu }[U]s_{\nu }[V]$. This is a finite identity in the independent power sums of $U,V$, so applying the two specialization homomorphisms proves (b) for arbitrary $A,B$.

Finiteness, empty case and choice. In (a), $c_{\mu \nu }^{\lambda }=0$ unless $\mu \subseteq \lambda$ and $\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\nu \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }$, so only finitely many pairs contribute, and the identity is verified coefficient of $s_{\lambda }[Z]$ by coefficient of $s_{\lambda }[Z]$. In (b), $g_{\mu \nu }^{\lambda }=0$ unless $\mathrm{\mid }\mu \mathrm{\mid }=\mathrm{\mid }\nu \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }$ by [F9], so again only finitely many pairs contribute. Both identities hold for arbitrary specializations into arbitrary commutative $\mathbb{Q}$-algebras, and the proofs use only finite sums, finite bases and the substitution of step 1.1; no choice principle is used. The empty case is included: $s_{\mathrm{\varnothing }}=1$ and $c_{\mathrm{\varnothing },\mathrm{\varnothing }}^{\mathrm{\varnothing }}=g_{\mathrm{\varnothing },\mathrm{\varnothing }}^{\mathrm{\varnothing }}=1$, so both displays hold for $\lambda =\mathrm{\varnothing }$.