# Conjugate linear maps between $*$-algebra modules

Let $A$ be a $*$-algebra, $E,$ and $F$ two $A$-modules, and a map $f:E \to F$ such that $$f(ae) = a^*f(e), ~~~~~~~ a \in A.$$ This seems to me to be the natural generalisation of a conjugate linear map. However, it seems to be too restrictive. Since for, $b$ also in $A$, we must have $$f(abe)=a^*f(be) = a^*b^*f(e),$$ and $$f(abe) = (ab)^*f(e) = b^*a^*f(e).$$ Assuming that we have no zero multiplication, this implies that $a^*$ and $b^*$ commute, which means, since $*$ is invertible, that every element of $A$ commutes. Is there something wrong with my reasoning here, or is my definition just too restrictive?

-

The basic idea is this: given a complex vector space $V$, define the complex conjugate vector space $\overline{V}$ to be $$\overline{V} = \{ \overline{v} \mid v \in V \}$$ with addition and scalar multiplication given by $$\overline{v} + \overline{w} = \overline{v+w}, \quad \lambda \cdot \overline{v} = \overline{\overline{\lambda} v}.$$ So $\overline{V}$ is the same as $V$ as an abelian group (even as a real vector space) but the scalar multiplication is conjugated. Then the point is that conjugate-linear maps $V \to W$ correspond to linear maps $\overline{V} \to W$.
Now if $A$ is a $\ast$-algebra and $V$ is a left $A$-module, the question is: how do we turn $\overline{V}$ into an $A$-module? We can't just define $a \cdot \overline{v} = \overline{av}$, because that's not linear in $A$, rather it is conjugate-linear in $A$. We have the $\ast$-structure, which can fix that problem: we define $$a \cdot \overline{v} = \overline{a^\ast v}.$$ What you wrote in your question amounts to the fact that this is not multiplicative in $a$, i.e. we find that $$(a b) \cdot \overline{v} = \overline{b^\ast a^\ast v} = b \cdot (a \cdot \overline{v}).$$ What this means is that the complex conjugate of a left $A$-module is really a right $A$-module, so we really should write the action as $$\overline{v} \cdot a = \overline{a^\ast v}.$$ Similarly, the complex conjugate of a right module is a left module.
So if you want to talk about conjugate-linear maps between modules, really you should have one of them be a left module and one of them a right module. Say $f : V \to W$ is a conjugate-linear map, where $V$ is a left module and $W$ is a right module. Then the corresponding linear map is $$\hat{f} : \overline{V} \to W, \quad \hat{f}(\overline{v}) = f(v).$$ Now we ask what is the correct compatibility criterion for $f$ in order that the linear map $\hat{f}$ is a map of right $A$-modules. We want $$\hat{f}(\overline{v} \cdot a) = \hat{f}(\overline{v}) a,$$ which amounts to $$f(a^\ast v) = f(v) a, \quad \text{or} \quad f(av) = f(v)a^\ast.$$