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I haven't seen this question in standard textbooks, so I decide to give it a try here. It might relate to deeper structures of certain TQFTs, but I'm not sure.

Let $G$ be a finite group. Its finite-dimensional complex representations are well understood, the regular representation $\mathbb{C}[G]$ being the most important one. Note that it's not only a vector space acted by $G$: it also has a natural algebra structure and naturally compatible with $G$-action (EDIT: .. in a less natural fashion --- $g(ab)= (ga)b$, instead of $g(ab) = gagb$.)

Question

Is there a classification of group representations that have compatible algebra structures?

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  • $\begingroup$ "algebra" = "associative unital algebra"? $\endgroup$
    – YCor
    Commented Nov 21, 2019 at 10:29
  • $\begingroup$ Yes! sorry to not have mentioned that explicitly $\endgroup$
    – Student
    Commented Nov 21, 2019 at 14:34
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    $\begingroup$ Your condition implies $$ ga=(g1)a $$ and it is also clear that $(g1)(h1)=(gh)1$, so the representations you are interested in are the same as associative unital algebras $A$ together with a group homomorphism from $G$ to the group of invertible elements of $A$. A classification of such would subsume classification of all associative algebras (with a trivial homomorphism of $G$ to invertibles). $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 21, 2019 at 18:24

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A classification is too much to hope for, but the representation theory tells you whether such an algebra structure can exist: if $V$ is your $G$-representation, then an algebra product corresponds to a non-zero element in $\operatorname{Hom}(V \otimes V, V)$.

The best way to figure out whether such an element exists, is by decomposing $V$ into irreducibles $V_1,\dots,V_n$ and then verifying whether one of the representations $V_i \otimes V_j$ contains some $V_k$ as a constituent. This will very often (but not always) be the case, so many group representations do indeed admit compatible algebra structures.

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    $\begingroup$ Hmm.. I did not mention explicitly, but as my example $\mathbb{C}[G]$ suggests, the algebra is compatible in the way that $g(ab) = (ga)b$. In your answer, I suppose you mean $Hom_G(V\otimes V,V)$. However, $G$ acts on $V\otimes V$ by $g(v\otimes w) = gv\otimes gw$. $\endgroup$
    – Student
    Commented Nov 21, 2019 at 14:33
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    $\begingroup$ Oh, I see, I should have looked at your example more carefully. (I also see now that you want associative unital algebras, which is also not what I had in mind.) I think that your definition of "compatible" is less natural (because the elements of $G$ do not induce automorphisms of this algebra structure), but of course your question makes perfect sense. $\endgroup$
    – Tom De Medts
    Commented Nov 21, 2019 at 16:33
  • $\begingroup$ Hmm yes it is not natural. The instance in my mind comes from Witten-Dijkgraaf 2d-TQFT, where representations are assigned to points. It seems to me that the algebra structure of $\mathbb{C}[G]$ plays an important role.. so I'd like to know more examples.. or even a classification if any. $\endgroup$
    – Student
    Commented Nov 21, 2019 at 17:38
  • $\begingroup$ @Student Could you please add to the question what exactly do you mean by compatibility? For example, $\mathbb C[G]$ also has a right $G$-action and satisfies $(ab)g=a(bg)$ and $(ag)b=a(gb)$, do you want these too? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 21, 2019 at 17:41
  • $\begingroup$ Added! Thanks for the comment. At least point I'm not sure if I want those too. But since the picture comes from TQFT, I'd suppose to! My original intention was to get some pointers to useful references, but it hasn't been done yet, I should dig more into my question and make it more precise. $\endgroup$
    – Student
    Commented Nov 21, 2019 at 17:45

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