Let be an irreducible representation of a finite group Say that is tensor indecomposable if any isomorphism implies that either or is a character.
In conversations with Matt and Toby (which permeate the rest of this post as well), the following problem came up:
Question: Let be a finite group. Let be an irreducible representation of Is there a unique decomposition
of as a tensor product of tensor indecomposable representations (up to re-ordering and twist)? I don’t think this can be too hard, but I confess I don’t see how to do it. (Since we didn’t really need this, we didn’t think about it too hard.)
(edit: when I say I don’t think this can be too hard, I don’t mean to imply that I think it is true; just that I think either a proof or counterexample should not be too hard to find — hopefully not both.)
One can ask an analogous problem for Lie groups. Actually, the problem for Lie algebras is actually quite simple (and the answer is positive). It is related to the following:
Let and be irreducible non-trivial representations of a simple Lie group Then is reducible.
Proof: Assume that is irreducible. In particular, it is determined by its highest weight. Let the highest weights of and be and respectively. Then the highest weight of must be But now, by the Weyl character formula, we deduce that
The product term can also be written as:
In particular, since the pairing is non-negative between positive roots and highest weights, we deduce a contradiction unless
for all The assumption that is irreducible, however, is equivalent to saying that has a maximal root and for such a maximal root, we have
Note that this lemma is actually a special case of a theorem of Rajan, who proved that, for simple the factors of a (not necessarily irreducible) tensor product are determined by the representation. In particular, the tensor product of two non-trivial irreducible representations cannot be irreducible.
The problem with the initial question is that it’s hard to construct tensor products of irreducible representations which are irreducible. Or rather, it is easy, simply by taking where is an irreducible representation of and is an irreducible representation of and the tensor is irreducible for Yet the interesting case is something closer to assuming that and are faithful. Actually, this motivates the following question:
Question: Do there exist irreducible non-trivial representations and of a finite simple non-abelian group such that is irreducible?
The argument above for Lie groups suggests that this may not happen for Chevelley groups (although it certainly doesn’t prove this). It also suggests (relating the representation theory of and to ) that it doesn’t happen for the alternating groups either. It almost surely doesn’t happen for the sporadic groups either. So my guess that the answer to the problem above is no, and that this is probably known, and probably requires classification. (Please comment if you know the answer.) Actually, this also reminds me of a similar problem which I think is open.
Question: Fix N. Does there exist a non-trivial representation of a finite group of dimension N such that (of dimension ) is irreducible?
This question came up in my paper with Barry, where I was surprised to find very few examples. I seem to remember that the Mathieu group has an -dimensional representation whose corresponding 120 dimensional adjoint is irreducible. Can one classify all such examples coming from simple groups?