Hermitian symmetry of the relative equivariant intersection pairing

On the dual coefficient fibers and their conjugates, the natural evaluation pairing is the ring-valued pairing $(a,\bar{b})\mapsto a\bar{b}$ in a unitary frame. This is transport-invariant: dual transport is $a\mapsto a{u}^{-1}$, conjugate-dual transport on the second factor corresponds to $\bar{b}\mapsto u\bar{b}$, and their product is unchanged. Its transposed conjugate pairing has the same value, since $\overline{b\stackrel{ˉ}{a}}=a\bar{b}$. Thus the reversed-factor cup product followed by involution can be compared with the original product in the same constant abelian-group coefficient system $R$. This does not exchange factors inside a noncommutative product without applying the involution.

Choose relative dual two-cocycles $\alpha ,\beta$ representing $a_p,a_q$. Conjugation turns $\beta$ into a right-coefficient cocycle whose relative cap is $q$. The front-evaluation cap in [F2] retains the back face: $\bar{\beta }$ is evaluated on the front face to give the coefficient of $q$, while $\alpha$ evaluates its retained back face. Consequently the exact formulas, with the transport evaluation of step 1.1, are $\lambda (p,q)=⟨\tau (\bar{\beta }⌣\alpha ),[W,\mathrm{\partial }W]⟩,\overline{\lambda (q,p)}=⟨\alpha ⌣\bar{\beta },[W,\mathrm{\partial }W]⟩\mathrm{.}$ Here $\tau$ means reverse the two coefficient factors and use the pairing of step 1.1, so the coefficient product in the second expression is still ordered as the value of $\alpha$ followed by the conjugate value of $\beta$. Both cocycles vanish on simplices contained in $\mathrm{\partial }W$.

The needed cup interchange is an explicit finite cochain identity. For a singular three-simplex with ordered vertices $0123$, put $H(0123)=\alpha (013)\overline{\beta (123)}-\alpha (023)\overline{\beta (012)}\mathrm{.}$ Every factor is transported to the first vertex before pairing; for the first product this transports the second factor from vertex1 to vertex0. On each singular four-simplex the pulled-back local systems are trivialized by its straight vertex paths, so coefficient transports obey the face identities and the calculation reduces to the same displayed ring products. Expanding the five terms of $\delta H(01234)$ and using $\delta \alpha =\delta \beta =0$ gives $\delta H(01234)=\alpha (012)\overline{\beta (234)}-\alpha (234)\overline{\beta (012)}\mathrm{.}$ For clarity the unreduced expansion is $\alpha (124)\overline{\beta (234)}-\alpha (134)\overline{\beta (123)}-\alpha (024)\overline{\beta (234)}+\alpha (034)\overline{\beta (023)}+\alpha (014)\overline{\beta (134)}-\alpha (034)\overline{\beta (013)}-\alpha (014)\overline{\beta (124)}+\alpha (024)\overline{\beta (012)}+\alpha (013)\overline{\beta (123)}-\alpha (023)\overline{\beta (012)}$; substituting the four-face cocycle identities groups it into the two displayed terms. Coefficient multiplication retains its order throughout. Hence $\delta H=\alpha ⌣\bar{\beta }-\tau (\bar{\beta }⌣\alpha )$. This is the degree-two cup-one homotopy written out locally; no commutative-ring cup-interchange theorem is being applied in a broader scope.

Since $\alpha ,\beta$ are relative, $H$ vanishes on every simplex in $\mathrm{\partial }W$. If $z$ is the finite relative orientation cycle, then $⟨\delta H,z⟩=⟨H,\mathrm{\partial }z⟩=0$. Evaluating the identity of step 3.1 and using step 2.1 proves the asserted Hermitian equality. For the cross terms, apply the equality and $\bar{1}=1,\bar{0}=0$. The cochains were chosen only for the two given classes, each simplex calculation involved finitely many faces, and the evaluation used a single finite orientation chain. All coefficient identifications were relative to the supplied standard form. Thus no choice principle or surface transversality assertion has entered.