Definition 8

Let $\Phi$ be a bilinear form on $E \times F$ whose associated mappings $s$ and $d$ are bijective. The bilinear form $\hat{\Phi}$ on $F^* \times E^*$ defined by (27) is called the inverse form of $\Phi$.

Let now $\hat{s}$ and $\hat{d}$ be the linear mappings from $F^*$ into $E^{**}$ and from $E^*$ into $F^{**}$ associated on the left and on the right with $\hat{\Phi}$. Since, for $x' \in E^*$ and $y' \in F^*$, one has by definition

one sees, by comparing with (27), that the linear form $\hat{d}(x')$ on $F^*$ is equal to that defined by the element $d^{-1}(x')$ of $F$. It follows that the composed mapping $\hat{d} \circ d$ is the canonical mapping of $F$ into its bidual $F^{**}$, and that this mapping is bijective since $d$ and $\hat{d}$ (the latter by transport of structure) are bijective; therefore, if one identifies canonically $F$ with $F^{**}$, one has $\hat{d} = d^{-1}$. Analogously $E$ is identified canonically with $E^{**}$, the canonical mapping of $E$ into $E^{**}$ is $\hat{s} \circ s$, and one has $\hat{s} = s^{-1}$. It follows from this that the inverse form of $\hat{\Phi}$ is $\Phi$.

Consider now a ring $A$ endowed with an antiautomorphism $J$, two left $A$-modules $E$ and $F$, and a right sesquilinear form $\Phi$ for $J$ on $E \times F$ such that the mappings associated with $\Phi$, which will be denoted by $s$ and $d$, are bijective. Define a mapping $\widehat{\Phi}$ from $F^* \times E^*$ into $A$ by the first equation (27). This mapping satisfies, by (24) (No. 6), the relation

The mapping $\widehat{\Phi}$ is obviously $\mathbf{Z}$-bilinear; moreover, one has, for $a,\ b$ in $A$, $x' \in E^*$ and $y' \in F^*$, and by virtue of the definitions of $s$ and $d$,

hence $\widehat{\Phi}$ is a left sesquilinear form for $J$ on $F^* \times E^*$ (No. 2).