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Consider the category $\mathrm{Rep}_{\mathbb{C}}(GL_n)$ of representations of GL_n over complex numbers. Then a theorem of Rajan(See https://doi.org/10.4007/annals.2004.160.683) says that if $V_1, V_2, \dots, V_n$ and $W_1, W_2, \dots, W_m$ are irreducible representations of GL_n of nonzero highest weights such that $V_1 \otimes \dots \otimes V_n \cong W_1 \otimes \dots \otimes W_m$, then $m=n$ and $V_i \simeq W_{\sigma(i)} \otimes \det^{\alpha_i}$, where $\alpha_i$ are integers and $\sigma$ is a permutation of $n$.

Recall that Deligne had constructed the semisimple rigid abelian tensor categories(See https://publications.ias.edu/deligne/paper/438). Can we say something similar in the context of $\mathrm{Rep}(GL_t)$ where $t \in \mathbb{C} \setminus \mathbb{N}$? More precisely I have the following question

$\it{Question}$ : Let $X_1, X_2, \dots, X_n$ and $Y_1, Y_2, \dots, Y_m$ be irreducible objects of the category $\mathrm{Rep}(GL_t)$(for $t \in \mathbb{C} \setminus \mathbb{Z}$) over some algebraically closed field of characteristic zero. Suppose $X_1 \otimes \dots \otimes X_n \cong Y_1 \otimes \dots \otimes Y_m$. Is $m=n$? Are there any relation between individual objects $X_i$ and $Y_j$ as in the above mentioned result?

More generally, tet $(\mathcal{C}, \otimes)$ be a rigid abelian category. You may even assume semisimple. I wonder if we impose the condition of "unique factorization" as above then would it be possible to say that there cannot be too many such categories. Perhaps something in terms of growth of lengths of tensor power of objects?

I have found following related query but with no answers : Has the notion of a unique factorization category been defined and studied?.

Apologies if the question is too naive. Any comments, references and suggestions are welcome. Thank you!

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    $\begingroup$ What is $\mathrm{GL}_t$ for $t\not\in\mathbb N$? $\endgroup$ Commented Apr 25, 2022 at 15:56
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    $\begingroup$ @VladimirSotirov There does not exists such a group, but there are categories $\mathrm{Rep}(GL_t)$ constructed by Deligne which is loosely speaking constructed by interpolation of categories $\mathrm{Rep}(GL_n)$. Please see the article publications.ias.edu/deligne/paper/438. $\endgroup$
    – random123
    Commented Apr 25, 2022 at 16:28
  • $\begingroup$ Are you sure there isn't a condition missing in your statement of Rajan's theorem in the first paragraph? The way it is written now I feel I can use it to prove that $1 = 2$ $\endgroup$
    – Vincent
    Commented Apr 25, 2022 at 18:39
  • $\begingroup$ @Vincent $V_i$ and $W_j$ are irreducible. Does that help? Maybe you can let me know as to what condition I am missing and I will edit the question accordingly. Thanks! $\endgroup$
    – random123
    Commented Apr 26, 2022 at 2:55
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    $\begingroup$ @Vincent Thank you for pointing this out. I will make this edit. I forgot this very important condition. I agree that the result is quite interesting. It seems that such conditions should force some strong constraint on any tensor category. Wondered is this has been explored and hence the question. $\endgroup$
    – random123
    Commented Apr 26, 2022 at 12:15

1 Answer 1

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Sure, this follows from the case for $GL_n$ basically for free. Tensor product multiplicities in $Rep(GL_t)$ are the same as for $GL_n$ for all $n \gg 0$. So any counterexample to the above theorem for $t \in \mathbb{C} - \mathbb{N}$ would produce a counterexample for $GL_n$ for all large $n$.

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  • $\begingroup$ Thank you very much. That is a nice argument. Would you also like to comment on the vague question in the next paragraph to the Question? $\endgroup$
    – random123
    Commented Apr 26, 2022 at 2:59
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    $\begingroup$ For the vague question: This property can fail for finite groups -- for example it fails for $A_5$, there we have $V_4 \otimes V_5 \otimes V_3 \cong V_4 \otimes V_5 \otimes V'_3$ where the subscript denotes the dimension (there are two non-isomorphic representations of dimension 3). Characterizing exactly when it holds is probably hard. $\endgroup$
    – Nate
    Commented Apr 26, 2022 at 14:36
  • $\begingroup$ Thanks again for your comment. This is very helpful. I will think a little more and maybe in future will have a better question. Thanks again for your nice answer and helpful comments. $\endgroup$
    – random123
    Commented Apr 27, 2022 at 3:54
  • $\begingroup$ @Nate Thanks, this also provides a counterexample to my question here: math.stackexchange.com/a/4936189/491450 . $\endgroup$
    – Smiley1000
    Commented Jun 22 at 14:41

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