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.

  • 2
    $\begingroup$ This question is very well-asked. I wish I could upvote it more than once. ... also, if you don't get any answers soon, it's probably specialized enough that you can take it to MO. $\endgroup$ – Neal Mar 10 '14 at 11:03
  • $\begingroup$ Could you define what you mean by the quantum group $F(G)$? $\endgroup$ – Tobias Kildetoft Mar 10 '14 at 11:57
  • $\begingroup$ The algebra of functions on $G$ with the comultiplication given by the coproduct and the counit given by $\varepsilon(f)=f(e)$. $\endgroup$ – JP McCarthy Mar 10 '14 at 13:02
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    $\begingroup$ What is called a "(left) action" in your quote is called a "right action" everywhere else... Also, you should give a full citation and quote also what "it satisfies", for the off chance that it does introduce a left action but, for some reasons unknown, writes it on the right. $\endgroup$ – darij grinberg Mar 11 '14 at 1:52
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    $\begingroup$ @darijgrinberg The quote is from link.springer.com/chapter/10.1007/11376637_1 and the corepresentation definition is from ems-ph.org/books/book.php?proj_nr=72 $\endgroup$ – JP McCarthy Mar 11 '14 at 9:24

This question has been answered over on Mathoverflow... here.

My confusion came from confusing left and right...


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