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If $R$ is an irreducible representation of a simple Lie-groups $G$ I assume there is always a lowest integer $n$ such that the tensor product representation $R \otimes R \otimes \ldots \otimes R$ (n times) contains the trivial (or singlet) representation. I can more or less obtain $n$ on a case-by-case basis (for example with $G=SU(N)$ and $R=Fund$ it appears $n=N$) but first of all is there an established name for $n(G,R)$? And is there a generic method to find it for any $G$ and $R$?

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    $\begingroup$ There’s an easy obstruction: Rep(G) is graded by central character, and the trivial character has trivial grading. So for your example of SU(n) with its standard rep this tells you only powers which are multiples of n can possibly contain the trivial. $\endgroup$
    – Noah Snyder
    Commented Nov 26, 2020 at 16:41
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    $\begingroup$ @SamHopkins: Not quite, because when $\lambda$ is in the root lattice the answer isn't 1! $\endgroup$
    – Noah Snyder
    Commented Nov 26, 2020 at 16:53
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    $\begingroup$ The other key point is that if you don't contain the trivial and you do contain a self-dual irrep, then the answer is 2. This actually covers most cases. All that's left is $SU(n)$, reps of $SO(4n+2)$ whose highest weights are multiples of the spin weight, and some reps of $E_6$. I don't see a particularly easy way to handle $SU(n)$ other than just using central character, and somehow checking you don't get very very unlucky and miss the trivial in the smallest non-trivial multiple landing in the root lattice. $\endgroup$
    – Noah Snyder
    Commented Nov 26, 2020 at 17:10
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    $\begingroup$ A really good exercise is to see that taking duals of irreps of a Lie algebra is always implemented by an order 2 Dynkin diagram automorphism. So if your Dynkin diagram doesn't have an involution then every rep is self-dual. Even when it does have an involution it still might be that every rep is self-dual (this happens the even type D's) or it might be that most of the vertices are fixed (happens for the odd type D's). The cases I listed are the only actual non-self-dual irreps. $\endgroup$
    – Noah Snyder
    Commented Nov 26, 2020 at 20:37
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    $\begingroup$ Moishe’s right that my list was too restrictive in the type D odd case. That case is also going to be trickier like the SU(n) case, except that now I think you’re only looking at like 4th powers (or maybe just 2nd? what’s the center?) so it should be easier. $\endgroup$
    – Noah Snyder
    Commented Nov 26, 2020 at 23:20

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