Remark 1

If $\Phi$ is an $\varepsilon$-hermitian form on E, one has $\Phi(x, y) = \varepsilon \overline{\Phi(x, y)} \overline{\varepsilon}$ whatever $x, y$ in E may be. Hence, if $\Phi$ takes invertible values, one has $\varepsilon \overline{\varepsilon} = 1$. 2) If there exists an invertible element $i$ of the center of A such that $\overline{i} = i \varepsilon$, then, for $\Phi$ to be $\varepsilon$-hermitian, it is necessary and sufficient that $i \Phi$ be hermitian.

The mapping $(y, x) \to \overline{\Phi(x, y)}$ being sesquilinear for J, in order that $\Phi$ be $\varepsilon$-hermitian, it is necessary and sufficient that one have $\Phi(y, x) = \varepsilon \overline{\Phi(x, y)}$ when $x$ and $y$ run through a system of generators of E. In particular, if E admits a finite basis $(e_i)_{1 \leq i \leq n}$, in order that a sesquilinear form $\Phi$ on E be $\varepsilon$-hermitian, it is necessary and sufficient that its matrix $R = (\rho_{ij}) = (\Phi(e_i, e_j))$ satisfy the relations $\rho_{ji} = \varepsilon \overline{\rho_{ij}}$ whatever $i, j$ may be, that is to say $'R = \varepsilon \overline{R}$ ; a matrix $R$ possessing this property is said to be $\varepsilon$-hermitian. When $\varepsilon = 1$ (resp. $-1$) one says that $R$ is hermitian (resp. antihermitian) relative to the antiautomorphism J. When J is the identity (hence A commutative), a hermitian (resp. antihermitian) matrix $R$ is such that $'R = R$ (resp. $'R = -R$) ; it is then said that $R$ is a symmetric (resp. antisymmetric) matrix. In order that $\Phi$ be an alternating form, it is necessary and sufficient that its matrix be antisymmetric and, in addition, that the diagonal terms of $R$ all be zero; a matrix possessing these properties is said to be alternating.

Let $\Phi$ be a sesquilinear form on $E$, and let $s_\Phi$ and $d_\Phi$ be the mappings of $E$ into $E^*$ associated with $\Phi$ on the left and on the right ($§ 1$, n° 6). In order that $\Phi$ be $\varepsilon$-hermitian, it is necessary and sufficient that $\langle x, s_\Phi(y) \rangle = \bar{\varepsilon} \langle x, d_\Phi(y) \rangle$ for all elements $x, y$ of $E$, hence that $s_\Phi = \bar{\varepsilon} d_\Phi$, or again that $\langle x, d_\Phi(y) \rangle = \varepsilon \langle x, s_\Phi(y) \rangle$, hence that $d_\Phi = \varepsilon s_\Phi$.

Let $\Phi$ be an $\varepsilon$-hermitian form such that the mapping $d_\Phi$ of $E$ into $E^*$ associated on the right with $\Phi$ is bijective. For every endomorphism $u$ of $E$ one then has

Indeed, for all elements $x$ and $y$ of $E$, one has hence $u^{**(x)} = \varepsilon \bar{\varepsilon} u(x)$ since $\Phi$ is non-degenerate.

Si $\Phi$ est une forme $\varepsilon$-hermitienne telle que les applications $s_\Phi$ et $d_\Phi$ soient bijectives, alors la forme inverse $\widehat{\Phi}$ de $\Phi$ ($§ 1$, No. 7) est une forme $\bar{\varepsilon}$-hermitienne. En effet, en posant $s = s_\Phi$, $d = d_\Phi$ pour abréger, on déduit de $d = \varepsilon s$ que $s^{-1} = \bar{\varepsilon} d^{-1}$, $s$ étant semilinear. Par suite, quels que soient $u, v$ dans $E$, on a

d’où puisque, $\varepsilon$ est dans le center de $A$.

Enfin, lorsque le ring $A$ est commutative, les prolongements canoniques d’une forme $\varepsilon$-hermitienne $\Phi$ aux puissances tensorielle et extérieure $\bigotimes^p E$ et $\wedge^p E$ de $E$ sont des formes $\varepsilon^p$-hermitiennes, comme il résulte aussitôt des formules (35) et (37) du $§ 1$, No. 9.