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For which triples $\{A,B,C\}$ of positive integers does there exist a (finite or compact) group $G$ with unitary irreducible representations of dimensions $A$,$B$, and $ C$ whose tensor product contains the trivial representation?

Is this known in general? If not, what useful necessary and/or sufficient conditions are known?

For example, it is necessary that $A \le BC$ , $B \le AC$, and $C \le AB$. This is clear, since the tensor product of any two of the representations must contain the dual of the third (whose dimension therefore can't be larger than the product of dimensions of the other two).

Given allowed triples $(A_1,B_1,C_1)$ and $(A_2,B_2,C_2)$ corresponding to groups $G_1$ and $G_2$ and representations $(R_1,R_2,R_3)$ and $(R'_1,R'_2,R'_3)$, we have that $(A_1 A_2, B_1 B_2, C_1 C_2)$ is also allowed, since we can take the product group and the combined representations $((R_1,R'_1),(R_2,R'_2),(R_3,R'_3))$. So we could equivalently try to characterize the `elementary' triples that cannot be obtained via this product construction.

Update: It is possible to prove that a necessary condition is that

$ABC - A^2 - B^2 - C^2 + gcd(A,B) + gcd(A,C) + gcd(B,C) - gcd(A,B,C) \ge 0$

For $(A,B,C) = (2,B,C)$, this can also be shown to be sufficient. In this case, the allowed cases are $(2,N,N)$ and $(2,NK,(N+1)K)$. See https://arxiv.org/abs/1801.03508. It would be interesting to understand whether or not this condition is sufficient in general.

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    $\begingroup$ I think the question is too general to answer explicitly. For example, in the finite case, it is equivalent to asking what are the degrees of the irreducible constituents of $\chi \mu$ when $\chi,\mu$ are irreducible characters of respective degrees $A$ and $B$? $\endgroup$
    – Geoff Robinson
    Commented Nov 12, 2016 at 19:25
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    $\begingroup$ For one (or two) particular groups, the condition is straightforward. If $G$ is SO$(3)$, the conditions are essentially the triangle inequality. The dimensions of the irreducibles are precisely the odd integers, and there is only one irreducible for each odd dimension. Suppose $C \leq B \leq A$ (all being odd). Then the trivial character appears in the product $\chi_A \chi_B \chi_C$ iff $A \leq B +C$. The argument is an immediate consequence of the Clebsch-Gordan thingy. A similar thing is doable with SU$(2)$, but now parity conditions will also apply. $\endgroup$
    – David Handelman
    Commented Nov 12, 2016 at 21:22
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    $\begingroup$ I see that I reversed the ordering on $A,B,C$ from that of the OP. Sorry. $\endgroup$
    – David Handelman
    Commented Nov 12, 2016 at 21:28
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    $\begingroup$ Just a remark: For semisimple compact connected Lie groups, the question boils down to the complexification (and arbitrary finite-dim reps). In particular for simply connected compact Lie groups, it reduces to a questions about finite-dimensional reps of the corresponding semisimple complex Lie algebra. $\endgroup$
    – YCor
    Commented Nov 13, 2016 at 4:13
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    $\begingroup$ A second remark is that Geoff's comment holds in general (no need to restrict to the finite case). Also, as regards the question, the answer doesn't change if compactness is dropped: if $(A,B,C)$ is achieved by a representation of an arbitrary group, replacing it by the Zariski closure of the image of the rep yields a complex reductive linear algebraic group, and replacing it by a maximal compact subgroup (which is Zariski-dense then) yields a compact group action, which we can make unitary. $\endgroup$
    – YCor
    Commented Nov 13, 2016 at 9:23

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