Remark

Let A be an involutive algebra and B a maximal commutative involutive subalgebra of A. Then B is a maximal commutative subalgebra. In particular, if A is unital, then the algebra B is full.

In fact, let $x\in A$ be an element permutable with B. Then $x^*$ is permutable with B. Write $x=x_1+ix_2$ with $x_1$ and $x_2$ hermitian; the elements $x_1$ and $x_2$ are permutable with B (lemma 2). The subalgebra of A generated by B and $x_1$ is therefore commutative and involutive. Consequently, it is equal to B, so that $x_1\in B$. Analogously, one has $x_2\in B$, and therefore finally $x\in B$.

Let A be an involutive algebra. If $f$ is a linear form on A, the mapping $x\mapsto \overline{f(x^*)}$ on A is a linear form on A, denoted by $f^*$. The mapping $f\mapsto f^*$ is a semilinear involution on $A'$. One says that $f$ is hermitian if $f=f^*$. By lemma 1 of I, p. 95, every linear form $f$ on A has a unique representation $f=f_1+if_2$ where $f_1$ and $f_2$ are hermitian, namely $f_1=\frac{1}{2}(f+f^*)$ and $f_2=\frac{1}{2i}(f-f^*)$.

For a linear form $f$ to be hermitian, it is necessary and sufficient that the restriction of $f$ to $A_h$ be real-valued (proposition 1 of I, p. 95). The mapping $f\mapsto f|A_h$ is an isomorphism of the real vector space of hermitian forms onto the dual vector space of the real vector space $A_h$.

In particular, $\mathsf{X}'(A)_h$ (resp. $\mathsf{X}(A)_h$) will denote the set of hermitian characters of A (resp. the set of nonzero hermitian characters of A). A character is therefore hermitian if its restriction to $A_h$ is real-valued.

If A is commutative and if $\chi$ is a character of A, then $\chi^*$ is a character of A, and the mapping $\chi \mapsto \chi^*$ is a homeomorphism of $\mathsf{X}'(A)$ onto $\mathsf{X}'(A)$.