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Is there a criterion (or, more generally, an algorithm) to decide whether given non-negative integers $n_1, n_2$ and $m_1,...,m_k$ there is a group $G$ with irreducible reps $V_1, V_2$ over $\mathbb C$ of dimensions $n_1$ and $n_2$ whose tensor product $V_1\otimes V_2$ decomposes into $k$ irreducible components of dimensions $m_1,....,m_k$? (I am also interested in versions of this problem when $G$ is required to be finite or Lie.)

I imagine someone studied that in the 100+ years of history of representation theory.

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    $\begingroup$ There is an obvious necessary condition coming from the equality of the dimensions. Perhaps, the theory of symmetric monoidal categories can be used to show that it is also sufficient in the case of finite groups. For compact connected Lie groups (or equivalently, reductive Lie algebras), I am doubtful that anything significantly better than exhaustive search using Weyl character formula is available in general. If you impose further restrictions, such as multiplicity-freeness (e.g., all $m_i$s are distinct), the situation changes dramatically: full classification is available. $\endgroup$ – Victor Protsak Dec 9 '16 at 23:29
  • $\begingroup$ In the case of simple Lie groups (or rather their Lie algebras), given a sequence of dimensions you'd probably need to rely on the Weyl dimension formula coupled with the known methods of decomposing tensor products. Presumably there is an algorithm lurking here, if one starts with the well-known classifications of simple Lie algebras and their irreducible finite dimensional representations over $\mathbb{C}$. Like Victor I wouldn't expect anything easy if the $n_i$ and $m_j$ can be arbitrary. $\endgroup$ – Jim Humphreys Dec 10 '16 at 16:03
  • $\begingroup$ Related: the more specific mathoverflow.net/questions/254537/… $\endgroup$ – YCor Dec 10 '16 at 17:45
  • $\begingroup$ As I explained as a comment in the link, assuming that $G$ is Lie with finitely many components, or that $G$ is compact Lie, is no restriction. Indeed, once a $(k+2)$-tuple $(n_1,n_2,m_1,\dots,m_k)$ is achieved, you can pass to the Zariski closure of its image in $V_1\oplus V_2$ (which is reductive), and then to a maximal compact subgroup (which is Zariski-dense), yielding the same decomposition on the tensor product. $\endgroup$ – YCor Dec 10 '16 at 17:49
  • $\begingroup$ @VictorProtsak: Dear Victor, could you provide a reference to your multiplicity-free case answer to my question? $\endgroup$ – Adam Dec 10 '16 at 18:36

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