I asked this question over on Math.Stack --- where it has a bounty --- but I didn't really get a helpful response so I am asking the question here. One commenter suggests that I am confusing left- and right-actions but perhaps some of ye might be able to help.

" Hi folks I am trying to prove what I think should be a straightforward enough result but I am having to make a somewhat unnatural definition to do it. This unnatural definition is hinted at in a paper by Franz & Gohm:

If $G$ is a group, then $b:M\times G\rightarrow M$ is called a (left) action of $G$ on $M$, if it satisfies...

...as before we have the unital *-homomorphisms $\alpha_g:F(G)\rightarrow F(G)$ defined by $\alpha_g(f)(x):=f(b(x,g))$. Actually, in order to get a representation of $G$ on $F(G)$, i.e. $\alpha_g\alpha_h=\alpha_{gh}$, we must modify the definition and use $\alpha_g(f)(x):=f(b(x,g^{-1}))$. Otherwise we get an

anti-representation.

Let $G$ be a finite group and let $\rho:G\rightarrow GL(V)$ be a representation of $G$. I had wanted to prove that if we define a map by

$$\chi(v)=\sum_{g\in G}u\otimes\mathbf{1}_{\{g\,:\,\rho(g)u=v\}}=\sum_{g\in G}\rho(g^{-1})v\otimes\mathbf{1}_{\{g\}},$$

that $\chi$ would be a corepresentation of the quantum group $F(G)$ on $V$. Something that will ultimately fix my problem, in line with Franz & Gohm's comments above, is if I define

$$\chi_0(v)=\sum_{g\in G}u\otimes\mathbf{1}_{\{g\,:\,\rho(g^{-1})u=v\}}=\sum_{g\in G}\rho(g)v\otimes\mathbf{1}_{\{g\}}.$$

The reason I am uneasy about this is because it destroys a lot of the understanding I thought I had... briefly, if we consider the representation to be an action of $G$ on $V$ such that $u\overset{g}{\longrightarrow}v$ I wanted $\chi$ to encode all of this by saying look all of the things that bring you to $v$: something that looks or feels like $\coprod_i(u_i\overset{g_i}{\longrightarrow} v)$.

To be a corepresentation we need $(I_V\otimes \varepsilon)\circ\chi=I_V$ where $\varepsilon$ is the counit. There is no problem showing this with either definition.

The other property we need is

$$(I_V\otimes \Delta)\circ \chi=(\chi\otimes I_A)\circ\chi,$$
which works fine for $\chi_0$ but for what I want the best I can do is
$$\sum_{g,t\in G}\rho(gt)^{-1}\otimes \mathbf{1}_{\{gtg^{-1}\}}\otimes\mathbf{1}_{\{g\}}=\sum_{g,t\in G}\rho(gt)^{-1}\otimes \mathbf{1}_{\{t\}}\otimes\mathbf{1}_{\{g\}},$$
which only works if $G$ is abelian.

The map $\Delta$ is the coproduct given by $\Delta: F(G)\rightarrow F(G)\otimes F(G)$ $$\Delta(\mathbf{1}_{\{g\}})=\sum_{t\in G}\mathbf{1}_{\{gt^{-1}\}}\otimes \mathbf{1}_{\{t\}}.$$

I suppose I am a little uneasy about letting go of the very little intuition that I have in the realm of quantum groups and I am wondering is there a better reason for using $\chi_0$ over $\chi$ apart from "it works".

Thank you for your time. "