The series $\sigma$: product formula, Cauchy identity, adjunction and reproducing kernel

Transfer principle. Let $d\ge 0$. If an identity between polynomials in the values $p_k[X_1],\dots ,p_k[X_r]$, $k\le d$, holds for all honest alphabets $X_1,\dots ,X_r$ of at least $d$ letters, then it holds for all specializations. Indeed, the substitution $A_k^{(i)}\mapsto p_k(x_1^{(i)},\dots ,x_d^{(i)})$ from $\mathbb{Q}[A_k^{(i)}]$ to $\mathbb{Q}[x^{(1)},\dots ,x^{(r)}]$ is injective: by [F3] over $\mathbb{Q}$ one has $\mathbb{Q}[e_1,\dots ,e_d]=\mathbb{Q}[p_1,\dots ,p_d]$ inside the symmetric polynomials in $d$ variables, and the elementary symmetric polynomials are algebraically independent by the fundamental theorem; tensor products of injective substitutions over the field $\mathbb{Q}$ are injective.

Product formula and negation (1). For an honest alphabet $X$, ${\sum }_n{h}_n[X]t^n={\prod }_{x\in X}(1-xt{)}^{-1}$ by [F2], so the series $\sigma [X]$ equals ${\prod }_{x\in X}(1-x{)}^{-1}$ coefficientwise at $t=1$. Next, $h_n[-U]=(-1{)}^n{e}_n[U]$ for every specialization $U$: both sides are polynomials in the values $p_k[U]$, $k\le n$, by [F4], and at honest alphabets the generating function of [F2] gives ${\sum }_n{h}_n[-x]t^n=\exp (-{\sum }_k{p}_k(x)t^k\mathrm{/}k)={\prod }_x(1-xt)={\sum }_n(-1{)}^n{e}_n(x)t^n$, so the identity follows from step 1.1 with $d=n$. Hence $\sigma [-U]={\sum }_n(-1{)}^n{e}_n[U]$, which at an honest alphabet is ${\prod }_{x\in X}(1-x)$; at the one-letter alphabet $t$, $h_n[-t]=0$ for $n\ge 2$ and $h_1[-t]=-t$, so $\sigma [-t]=1-t$, while $h_n[t]=t^n$ gives $\sigma [t]={\sum }_n{t}^n=(1-t{)}^{-1}$ as a formal series.

Reduction to honest alphabets. Every value $f[U]$ of $f\in {\mathrm{\Lambda }}_{\mathbb{Q}}^n$ is the value at $(p_1[U],\dots ,p_n[U])$ of a universal polynomial with rational coefficients, by the power-sum basis [F4]; the same holds for $h_n[U+V]$, $h_n[UV]$ and $s_{\lambda }[U+V]$. Hence by step 1.1 such identities may be verified on honest alphabets with at least $n$ letters, and that is how the remaining steps proceed.

(2) Multiplicativity and the one-letter scaling. For honest alphabets $X,Y$, ${\sum }_n{h}_n[X\bigsqcup Y]t^n={\prod }_{\mathrm{\ell }\in X\bigsqcup Y}(1-\mathrm{\ell }t{)}^{-1}=({\sum }_i{h}_i[X]t^i\left)\right({\sum }_j{h}_j[Y]t^j)$ by [F2], so $h_n[A+B]={\sum }_{i+j=n}{h}_i[A]h_j[B]$ holds on honest alphabets and hence, by step 2.2 with $d=n$, for all specializations; summing over $n$ gives $\sigma [A+B]=\sigma [A]\sigma [B]$. Since $p_k[tA]=t^k{p}_k[A]$ and $h_n$ is homogeneous of weight $n$ in the power sums, $h_n[tA]=t^n{h}_n[A]$, so $\sigma [tA]={\sum }_n{t}^n{h}_n[A]$.

(3) Cauchy identity. For honest alphabets $X,Y$ and $d\ge 0$, apply the degree-$d$ rank-$N$ projection of [F5] (with $N\ge d$) to the two Cauchy expansions of $\mathrm{\Omega }$: it produces $h_d[XY]={\sum }_{\lambda ⊢d}{s}_{\lambda }[X]s_{\lambda }[Y]={\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }[X]p_{\rho }[Y]$, the outer equality by [F2]. By step 2.2 with $d$, the identity $h_d[XY]={\sum }_{\lambda ⊢d}{s}_{\lambda }[X]s_{\lambda }[Y]$ holds for all specializations, and summing over $d$ gives $\sigma [XY]={\sum }_{\lambda }{s}_{\lambda }[X]s_{\lambda }[Y]$ by degree.

(4) Adjointness. By step 3.2, $\sigma [AX]={\sum }_{\lambda }{s}_{\lambda }[A]s_{\lambda }[X]$, so the adjoint of multiplication by $\sigma [AX]$ is ${\sum }_{\lambda }{s}_{\lambda }[A]s_{\lambda }^{\perp }$ degree by degree. On the other hand $p_{\rho }[X^{\perp }]=p_{\rho }^{\perp }={\prod }_i(p_{{\rho }_i}{)}^{\perp }=(p_{\rho }{)}^{\perp }$ by [F1] and [F7], so $h_n[A{X}^{\perp }]={\sum }_{\rho ⊢n}{z}_{\rho }^{-1}{p}_{\rho }[A]p_{\rho }^{\perp }=h_n[AX{]}^{\perp }$ and $\sigma [A{X}^{\perp }]=\sigma [AX{]}^{\perp }$. To evaluate this adjoint, compute on the Schur basis: by [F6], $s_{\lambda }[X+A]={\sum }_{\rho ,\nu }{c}_{\rho \nu }^{\lambda }{s}_{\rho }[X]s_{\nu }[A]$, so $⟨s_{\lambda }[X+A],s_{\mu }[X]{⟩}_X={\sum }_{\nu }{c}_{\mu \nu }^{\lambda }{s}_{\nu }[A]$; and $⟨\sigma [AX{]}^{\perp }{s}_{\lambda },s_{\mu }{⟩}_X=⟨s_{\lambda },\sigma [AX]s_{\mu }{⟩}_X={\sum }_{\rho }{s}_{\rho }[A]⟨s_{\lambda },s_{\rho }{s}_{\mu }{⟩}_X={\sum }_{\rho }{c}_{\rho \mu }^{\lambda }{s}_{\rho }[A]$ by [F6] and [F4]. The two results agree for every $\mu$, so by nondegeneracy of the Hall form $\sigma [A{X}^{\perp }]s_{\lambda }=s_{\lambda }[X+A]$ for every $\lambda$, and by linearity for all $f$.

(5) Reproducing kernel. For every partition $\lambda$, $⟨\sigma [AX],s_{\lambda }[X]{⟩}_X={\sum }_{\mu }{s}_{\mu }[A]⟨s_{\mu }[X],s_{\lambda }[X]{⟩}_X=s_{\lambda }[A]$ by step 3.2 and orthonormality [F4]; by linearity in $f$ this gives $⟨\sigma [AX]\mathrm{\mid }f[X]{⟩}_X=f[A]$ for all $f\in \mathrm{\Lambda }$.