Remark

Let A be a ring endowed with an antiautomorphism J, M and N right A-modules, and $\Phi$ a left sesquilinear form for J on $M \times N$ (No. 2, Remark); it therefore satisfies the equality

The mapping $\Phi'$ of $N \times M$ into A defined by $\Phi'(y, x) = \Phi(x, y)^{J'}$ (where $J' = J^{-1}$) is a left sesquilinear form for $J'$, and $\Phi$ and $\Phi'$ identify with bilinear forms on $M^{J'} \times N$ and $N^{J'} \times M$ respectively. The mappings $s_\Phi$ and $s_{\Phi'}$ associated with these bilinear forms are called the left and right associated mappings to the sesquilinear form $\Phi$, and are denoted by $s_\Phi$ and $d_\Phi$. Thus, by definition

and $s_\Phi$ (resp. $d_\Phi$) is a linear mapping of $M^{J'}$ into $N^*$ (resp.

of $N'$ into $M^*$). One would state and prove easily the analogues, for the case considered here, of Definition 7 and of props. 4, 5, 6.