Definition 9

Let $\Phi$ be a right sesquilinear form for $J$ on $E \times F$, whose associated mappings $s$ and $d$ are bijective. The left sesquilinear form $\widehat{\Phi}$ for $J$ on $F^* \times E^*$ is called the inverse form of $\Phi$.

We leave it to the reader to define and study the inverse form of a left sesquilinear form. This inverse form is a right sesquilinear form.

Let $\widehat{s}$ and $\widehat{d}$ be the mappings associated with $\widehat{\Phi}$; according to (26) (No. 6) we have

Since $s$ is bijective, and from the equality $\langle s^{-1}(y'),\ x' \rangle = \langle y',\ \widehat{d}(x') \rangle^J$ which follows from (28) and (29), we deduce that $\widehat{d}$ is bijective; hence $\widehat{d} \circ d$ is bijective. Now the equality $\langle d^{-1}(x'),\ y' \rangle = \langle y',\ \widehat{d}(x') \rangle$, which follows from (28) and (29), shows that $\widehat{d} \circ d$ is the canonical mapping from $F$ into its bidual $F^{**}$. One sees analogously that $\widehat{s}$ is bijective and that $\widehat{s} \circ s$ is the canonical mapping from $E$ into $E^{**}$. Hence, if one identifies $E^{**}$ with $E$ and $F^{**}$ with $F$ by means of these canonical mappings, one has $\widehat{s} = s^{-1},\ \widehat{d} = d^{-1}$, and $\Phi$ is the inverse form of $\widehat{\Phi}$.

With the same notation and hypotheses let $a$ be an invertible element of the center of $A$. Then the associated mappings of the form $a\Phi$ are, according to (23) (resp. (24)) equal to $a.d$ and $a.s$ (resp. $a^{J'}.s$), and are therefore bijective. It follows thus from (27) that the inverse form of $a\Phi$ is $a^{-1}\widehat{\Phi}$ (resp. $(a^{J'})^{-1}\widehat{\Phi}$).