Dual of a $\mathbb{C}G$-module, can we get $g^{-1}$ outside $\theta(\cdot)$?

Question

I am a bit lost with dual modules.

Let $G$ be a group, $\mathbb{C}$ be the complex numbers and let $V$ be a $\mathbb{C}G$-module. Then as I understand it, the dual is $V^*=\operatorname{Hom}_{\mathbb{C}G}(C,\mathbb{C}G)$. So if we have $\theta\in V^*$ the action is given by $(g\cdot \theta)(v)=\theta(g^{-1}\cdot v)=g^{-1}\theta(v)$ where after the last equality we get the product inside $\mathbb{C}G$. Is this correct?

Answer 1

There is a difference between the dual of a $kG$-module, and the dual of an $R$-module, where $R=kG$. If $R$ is a ring and $M$ is a right $R$-module, then the dual of $M$, $\mathrm{Hom}_R(M,R)$, is a left $R$-module.

This is not the same as the dual of a right (or left) $kG$-module, which is the dual of $M$ as a vector space $\mathrm{Hom}_k(M,k)$, with the group action as you have described, notably on the same side as the original action.

There are two different concepts, one used in representation theory of groups and one used in module theory for arbitrary rings.

Answer 2

The dual should be $V^\ast = \hom_{\mathbb C}(V, \mathbb C)$ with $g\theta$ defined by $(g\theta)(v) = \theta(g^{-1}v)$. Note the maps are just $\mathbb C$-linear, not $\mathbb CG$-linear so you can't pull out $g^{-1}$.