Tensor Products

Let W be an irreducible representation of a finite group G. Say that W is tensor indecomposable if any isomorphism W = U \otimes V implies that either U or V 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 G be a finite group. Let V be an irreducible representation of G. Is there a unique decomposition

V = V_1 \otimes V_2 \ldots \otimes V_k

of V 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 V and W be irreducible non-trivial representations of a simple Lie group \mathfrak{g}. Then V \otimes W is reducible.

Proof: Assume that V \otimes W is irreducible. In particular, it is determined by its highest weight. Let the highest weights of V and W be \lambda and \mu respectively. Then the highest weight of V \otimes W must be \lambda + \mu. But now, by the Weyl character formula, we deduce that

\displaystyle{1 = \frac{\dim(V) \dim(W)}{\dim(V \otimes W)} = \prod_{\Phi^{+}} \frac{ \langle \rho,\alpha \rangle \langle \rho + \lambda + \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.

The product term can also be written as:

\displaystyle{1 + \frac{ \langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.

In particular, since the pairing is non-negative between positive roots and highest weights, we deduce a contradiction unless

\langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle = 0

for all \alpha \in \Phi^{+}. The assumption that \mathfrak{g} is irreducible, however, is equivalent to saying that \Phi^{+} has a maximal root \beta, and for such a maximal root, we have

\langle \lambda, \beta \rangle = 0 \Rightarrow \langle \lambda, \alpha \rangle = 0, \ \forall \alpha \in \Phi^{+} \Leftrightarrow \lambda = 0.

Note that this lemma is actually a special case of a theorem of Rajan, who proved that, for simple \mathfrak{g}, 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 U \otimes V where U is an irreducible representation of H and V is an irreducible representation of G and the tensor is irreducible for H \times G. Yet the interesting case is something closer to assuming that U and V are faithful. Actually, this motivates the following question:

Question: Do there exist irreducible non-trivial representations U and V of a finite simple non-abelian group G such that U \otimes V 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 A_n and S_n to \mathrm{GL}_n) 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 V of a finite group G of dimension N such that \mathrm{Hom}^0(V,V) (of dimension N^2 - 1) 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 M_{12} has an 11-dimensional representation whose corresponding 120 dimensional adjoint is irreducible. Can one classify all such examples coming from simple groups?

This entry was posted in Mathematics and tagged , , , , , , . Bookmark the permalink.

7 Responses to Tensor Products

  1. Emmanuel Kowalski says:

    About the last question, it corresponds to the representation having “fourth moment” equal to $2$ in the terminology of Katz in his “Larsen alternative” paper. In Section 1.5 of that paper, he gives examples arising from the Weil representation (and says that he got them from Deligne).In Section 1.6, he gives examples from the Atlas (including some cases for M_{23}, M_{24} and J_4).

  2. vytas says:

    If you look at mod p reps then it is very often that irreducible reps are tensor product of other irreducible reps. e.g. $G=\SL_2(\mathbb F_q)$ then every irreducible $\overline{\mathbb F}_p$ rep is a tensor product of some symmetric powers of the standard rep twisted by Frobenius.

  3. AV says:

    It is related to another nice thing in the Lie group case, which I’ve always wanted to understand better: the “PRV conjecture” (proved by S. Kumar in 1989) says that in fact $V(\lambda + w \mu)$ occurs at least once inside $V(\lambda) \otimes V(\mu)$ for each $w$ in the Weyl group.

  4. KD says:

    Apologies in advance, as my grasp on the material is very shaky…

    I think the answer to the first question is negative. Here is an example. Let V be a vector space over a field F = F_p, where p is an odd prime (let’s say, dim V = 4). Then we have a canonical 2-step nilpotent p-group G which is a central extension

    1-> \wedge^2 V -> G -> V -> 1.

    Fix a primitive additive character \psi of F. Given a skew 2-form f on V of rank 2r, we have a unique irreducible representations W(f) of G of dimension p^r where the center \wedge^2 V acts through \psi(f) and Ker(f) acts trivially. Let \omega be a symplectic form on V. For any decomposition

    \omega = f_1 + f_2

    of \omega (which has rank 4) into a sum of rank 2 forms, we have a maximal factorization

    W(\omega) = W(f_1) \otimes W(f_2).

    So if we have another splitting

    \omega = f_1′ + f_2′,

    we also have

    W(\omega) = W(f_1′) \otimes W(f_2′).

    These factorizations are equivalent by an element of Sp(4, F) which maps f_1 to f_1′ and f_2 to f_2′. However, let us choose the two splittings randomly, and let U \subset \wedge^2 V be the common kernel of f_1, f_1′, f_2, f_2′ as linear functions on \wedge^2 V. Then U is 3-dimensional (since there is only one linear relation between f_1, f_2, f_1′, f_2′) and all the representations in question are representations of G’ = G/U. Also I think that in general G’ probably does not have an automorphism f_1, f_2 to f_1′, f_2′ (where would it map f_1′, f_2′?). Also I think one can do this in the case of larger dim V. E.g. for dim V = 6 we can take

    \omega = f_1 + f_2 + f_3 = f_1′ + f_2′,

    where f_2′ has rank 4 and others have rank 2. Then we will even get two maximal factorizations with different number of factors.

    • KD says:

      I think the answer to the second question is also negative. Let k be the field of order p^2, p > 5, G’ = Sl(2, V) where V is a 2-dimensional k-vector space G = G’/\pm 1. Then G is a simple group and the map Fr: k -> k, a -> a^p induces an automorphism Fr: G -> G. Let r: G-> Gl(Sym^2(V)) be the natural action, r’ = r \circ Fr. Then r \otimes r’ is irreducible.

  5. Pingback: Mathieu Magic | Persiflage

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s