Power sums are orthogonal for the Hall form

Fix $d\ge 0$, whose partition set is finite, and let $V={\mathrm{\Lambda }}_{\mathbb{Q}}^d$; the Hall form extends to $V$ by scalar extension. Let $(u_i)$ and $(v_i)$ be any two bases and write $u_i={\sum }_{\alpha }{a}_{i\alpha }{h}_{\alpha }$ and $v_j={\sum }_{\beta }{b}_{j\beta }{m}_{\beta }$ using the dual bases from [F1]. With $A=(a_{i\alpha })$ and $B=(b_{j\beta })$, [F1] gives $⟨u_i,v_j{⟩}_H=(A{B}^{\mathsf{T}}{)}_{ij}$, while the coefficient of $h_{\alpha }\otimes m_{\beta }$ in ${\sum }_i{u}_i(x)v_i(y)$ is $(A^{\mathsf{T}}B{)}_{\alpha \beta }$. If this tensor equals the complete–monomial kernel ${\sum }_{\alpha ⊢d}{h}_{\alpha }(x)m_{\alpha }(y)$ from [F2], then $A^{\mathsf{T}}B=I$; invertibility yields $B=(A^{\mathsf{T}}{)}^{-1}$ and $A{B}^{\mathsf{T}}=I$, hence $⟨u_i,v_j{⟩}_H={\delta }_{ij}$. This finite dual-kernel criterion makes no symmetry assumption on the Hall form.

By [F3], $u_{\nu }=p_{\nu }$ is a basis of $V$. A partition has finitely many parts, so only finitely many $m_r(\nu )$ are nonzero; thus $z_{\nu }$ is a positive integer and $v_{\nu }=p_{\nu }\mathrm{/}{z}_{\nu }$ is also a basis. The power-sum expansion in [F2] is ${\mathrm{\Omega }}_d={\sum }_{\nu ⊢d}{u}_{\nu }(x)v_{\nu }(y)$, so step 1.1 gives $⟨p_{\lambda },p_{\mu }\mathrm{/}{z}_{\mu }{⟩}_H={\delta }_{\lambda \mu }$. Bilinearity yields $⟨p_{\lambda },p_{\mu }{⟩}_H=z_{\mu }{\delta }_{\lambda \mu }$, which equals $z_{\lambda }$ on the diagonal and zero off it. For $d=0$, $p_{\mathrm{\varnothing }}=1$, $z_{\mathrm{\varnothing }}=1$, and $⟨1,1{⟩}_H=1$; for the one-part partition $(r)$, $z_{(r)}=r$ and $⟨p_r,p_r{⟩}_H=r$.

If $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\ne }\mathrm{\mid }\mu \mathrm{\mid }$, the graded definition in [F1] gives $⟨p_{\lambda },p_{\mu }{⟩}_H=0$, and bilinearity makes a zero input pair to zero. Each fixed degree has finitely many partitions, and the proof uses finite basis changes and sums; no arbitrary choices are made and the axiom of choice is not used.