Murphy cells are the same-shape contragredient Specht duals

The correctly scaled universal trace formula. Work first over $A=\mathbb{Z}[v^{\pm 1}]$, using $q=v$ as the formal parameter. For $d,e\in D_{\lambda }$, length additivity gives $b_d={\sum }_{u\in S_{\lambda }}{T}_{ud}$, and [F1] gives $\tau (b_d^{\ast }{b}_e)=0$ when $d\mathrm{\ne }e$, while $\tau (b_d^{\ast }{b}_d)={\sum }_{u\in S_{\lambda }}{q}^{\mathrm{\ell }(ud)}=q^{\mathrm{\ell }(d)}{P}_{\lambda }(q)$, where $P_{\lambda }(q):={\sum }_{u\in S_{\lambda }}{q}^{\mathrm{\ell }(u)}$. By bilinearity, $\tau (a^{\ast }b)=P_{\lambda }(q)\beta (a,b)(a,b\in M_A(\lambda ))\mathrm{.}$ The factor $P_{\lambda }$ is essential; it is not generally a unit and can vanish after specialization. It is nonzero in the domain $A$, since its constant coefficient is $1$.

Orthogonality before specialization. Put $K_A=M_A(\lambda )\cap I_A^{⊳\lambda }$ and $p=z_{\lambda }h$. For a higher Murphy element $m_{uv}^{\nu }=T_{d_{\nu }(u)}^{\ast }{x}_{\nu }{T}_{d_{\nu }(v)}$ with $\nu ⊳\lambda$, its product with $p$ after star is $(m_{uv}^{\nu }{)}^{\ast }p=T_{d_{\nu }(v)}^{\ast }{x}_{\nu }(T_{d_{\nu }(u)}{x}_{\lambda }{T}_{w_{\lambda }})y_{{\lambda }^{\mathrm{\prime }}}h=0,$ because $x_{\nu }H{y}_{{\lambda }^{\mathrm{\prime }}}=0$ by [F4]. The outer factor multiplies a zero vector; no factor is silently discarded. Expand $k\in K_A$ in the higher Murphy basis using [F3]. Then $k^{\ast }p=0$ and step 1.1 gives $P_{\lambda }(q)\beta (k,p)=0$. Cancel this nonzero polynomial only in the universal domain $A$, proving $\beta (K_A,P_A)=0$. The quotient and its kernel in [F3] are base-change compatible: the quotient is free on the supplied standard basis, its kernel is the split kernel of that quotient, and the independent Murphy construction identifies its specialization with $M_R(\lambda )\cap I_R^{⊳\lambda }$. Thus tensoring the zero pairing identity gives $\beta (K_R,P_R)=0$ for every $R$ and unit $q$. No cancellation is performed in $R$, including when $P_{\lambda }(q)=0$.

The descended pairing is perfect. Define $B_{\lambda }(a+K_R,p)=\beta (a,p)$; step 2.1 proves this is well defined. Contravariance gives $B_{\lambda }(ch,p)=B_{\lambda }(c,p{h}^{\ast })$, so its adjoint $C_{\mathrm{M}}(\lambda )\to P(\lambda {)}^{\circ }$ is an $H$-homomorphism by [F5]. The map $M(\lambda )\to M(\lambda {)}^{\circ }$, $a\mapsto \beta (a,-)$, is an isomorphism with unit diagonal $q^{\mathrm{\ell }(d)}$. Compose this map itself with restriction $M(\lambda {)}^{\circ }\to P(\lambda {)}^{\circ }$, which is onto because $P(\lambda )$ is split in $M(\lambda )$. The resulting surjection kills $K_R$ and therefore factors onto $C_{\mathrm{M}}(\lambda )\to P(\lambda {)}^{\circ }$. Both are free of rank $f^{\lambda }$. Choosing preimages of their finitely many basis vectors gives a right inverse of its square matrix; the determinant is a unit, so the adjoint is an isomorphism. This proves clause 1 without any nondegeneracy of the DJ self-form.

Kernel, finite formulas and every base change. The kernel of $M(\lambda )\to P(\lambda {)}^{\circ }$ is $P(\lambda {)}^{\perp }$ by its definition. Since the factored adjoint is an isomorphism, this is exactly $K_R=M(\lambda )\cap I^{⊳\lambda }$. Write $e_t={\sum }_d{a}_{d,t}{b}_d$ with the finite Laurent-polynomial coefficients of the integral polytabloid basis. Then $B_{\lambda }(m_s,e_t)=q^{\mathrm{\ell }(d_{\lambda }(s))}{a}_{d_{\lambda }(s),t}$. Compute these coefficients by the finite signed-symmetrizer expansion and standard-basis multiplication; the inverse pairing matrix is $\det (B_{\lambda }{)}^{-1}\mathrm{adj}(B_{\lambda })$. Over $A$ its determinant is a Laurent unit by step 3.1, so both the pairing and its inverse specialize to every $R$, retaining perfectness and the kernel identity. Empty shapes have the rank-one pairing $1$. No generic semisimplicity, nondegeneracy of the DJ self-form, division by a specialized Poincaré polynomial or choice principle is used.