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. "


Sorry about the left-right confusion (I think I am left-right-illiterate...).

If we have a left action $(g,x)\mapsto g.x$, but write it as a map $b:M\times G\to M$, $b(x,g)=g.x$, is that a left or a right action?

The reason for our choice of notion in the co- or bialgebraic part is time-ordering, we want the time to run from the left to the right, i.e. increments from the future are multiplied from the right. Mixing this with the usual preference for left action in the "classical" discussion in the introduction probably wasn't a good idea.

Your action of $G$ on $V$ is a map $\rho:G\times V\to V$. You can extend it linearly to the group algebra, $\rho:\mathbb{C}G\otimes V\to V$.

When you want to pass to a coaction, you take the dual $\rho^*:V^*\to (\mathbb{C} G\otimes V)^*\cong F(G)\otimes V^*$, since the dual of the group algebra $\mathbb{C}G$ is the algebra of function on $G$. On linear functional $\phi\in V^*$, you have $\rho^*(\phi)=\phi\circ \rho$. To get a coaction on $V$, you have to identify $V$ with its dual space. For $V=\mathbb{C}^M$ we can do this by considering $V$ as the Hilbert space $L^2(M)$ with the counting measure.

So we have $F(G)\otimes V^*\cong \mathbb{C}^G\otimes\mathbb{C}^M=\mathbb{C}^{G\times M}$ and then $\rho^*(\mathbf{1}_x)(g,y)=\langle \mathbf{1}_x, \alpha_g(\mathbf{1}_y)\rangle=\langle \mathbf{1}_x, \mathbf{1}_{g^{-1}.y})\rangle = \delta_{g.x,y}=\left(\sum_{h\in G}\mathbf{1}_h\otimes\mathbf{1}_{h.x}\right)(g,y)=\left(\sum_{h\in G}\mathbf{1}_h\otimes \alpha_{h^{-1}}(\mathbf{1}_{x})\right)(g,y)$. Which is your formula (except that I switched left and right back).

This coaction satisfies $({\rm id}\otimes \rho^*)\circ\rho^*(v) = \sum_{g,h\in G} \mathbf{1}_g\otimes \mathbf{1}_h \otimes \alpha_{(gh)^{-1}}(v)= (\Delta\otimes {\rm id})\circ\rho^*(v)$.

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  • $\begingroup$ Thank you this is great. $b:M\times G\rightarrow M$, $b(x,g)=g\cdot x$ would be a left action. What cleared a lot of this up for me was the calculation, where the $j_n$ are coming from the regular corepresentation, and $f=\sum_ia_i \mathbf{1}_{\{ i \}}$ that $\Psi(j_n(f))=\sum_{g\in G}a_g\varepsilon(\mathbf{1}_{\{gg_1^{-1}\cdots g_n^{-1}\}})\prod_{i=1}^n\phi(\mathbf{1}_{\{g_i\}})$. $\endgroup$ – JP McCarthy Mar 19 '14 at 18:38
  • $\begingroup$ ...so to get corepresentations on $A$ you don't look at representations on $A$ but rather representations on $A'$! $\endgroup$ – JP McCarthy May 2 '14 at 18:49

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