# Other than SU(3), SO(4), SU(2)xU(1), are there compact semisimple Lie groups which exactly two 3-dimensional representations that are dual to each other?

In my original question, I asked which compact Lie groups $G$ have a certain property. Jim and Dan showed that this property is equivalent to $G$ having exactly two irreducible 3-dimensional representations over the complex numbers which are dual to each other.

There are at least three compact Lie groups that have such a property, namely $SU(3)$, $SO(4)$ and $SU(2)\times U(1)$. Let us go through these examples. There are exactly two irreducible 3-dimensional represenations of $SU(3)$ which are the standard representation and its dual. The irreducible 3-dimensional representations of $SO(4)$ descend from its universal cover ${\mathrm{Spin}}(4)=SU(2)\times SU(2)$ which has exactly two dual irreducible 3-dimensional representations $S^2(V)\otimes {\mathbf{1}}$ and ${\mathbf{1}}\otimes S^2(V)$ (here $V$ is the standard representation of $SU(2)$ and ${\mathbf{1}}$ is the trivial representation). Finally, the irreducible 3-dimensional representations of $SU(2)\times U(1)$ are $S^2(V)\otimes W$ and $S^2(V)\otimes W^{\vee}$ where $V$ and $W$ are the standard representations of $SU(2)$ and $U(1)$ respectively.

Okay. That was quite a lot. There are any other examples of semisimple compact Lie groups having exactly two irreducible 3-dimensional representations which are dual to each other? From Jim's answer, it is likely that such Lie groups have to be of low rank.

The original question is given below.

Suppose that $G$ is a compact Lie group with at least two distinct irreducible 3-dimensional representations.

Can one classify those $G$ with the following two properties?

1. For any irreducible 3-dimensional representations $\pi$, the multiplicity of $\pi\otimes\pi$ at the trivial representation is 0.
2. For any two distinct irreducible 3-dimensional representations $\pi$ and $\pi'$, the multiplicity of $\pi\otimes \pi'$ at the trivial representation is 1.

EDIT: To make this question more concrete, let us assume that $G$ is semisimple of rank 2. From Jim's answer, this assumption of rank 2 is probably most relevant because we are looking at 3-dimensional representations.

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@Colin: By now the question has shifted quite a bit, which gets confusing. But your new header is not helpful, since there aren't really as many examples as you list. –  Jim Humphreys Jul 17 '12 at 18:11

[EDITED] At least when the group is simple (or semisimple), finite dimensional irreducible representations are classified precisely by their highest weights and classical formulas for dimensions and tensor product multiplicities apply. (The group can't be a torus, whose irreducible representations are 1-dimensional; in the remaining reductive cases one has to factor in a power of det or other such character to get all irreducible representations.)

Concerning property 1, you have to have a group whose 3-dimensional irreducible representations all fail to be self-dual (ruling out for instance $\mathrm{SU}(2)$): the multiplicitiy of the trivial module can be computed by the dimension of the Hom space from it to the tensor product in question, which is easy to analyze (see Dan's comment below).

It helps to give one or more starting examples where both properties hold. One of these would be $\mathrm{SU}(3)$, which has just a single irreducible 3-dimensional representation (the natural one) and its non-isomorphic dual. (Allowing a reductive group with a nontrivial torus as center will complicate things a bit.) Beyond rank 2 the 3-dimensional irreducibles seem to be very sparse (meaning non-existent, though I haven't checked this rigorously). So I don't immediately see any further examples among the simple or semisimple groups. Weyl's dimension formula is a natural tool here. For instance, in the rank 2 case $\mathrm{SU}(3)$, highest weights are given by ordered pairs of non-negative integers $(r,s)$ and the corresponding dimension is $(r+1)(s+1)(r+s+2)/2$, which takes value 3 only for the dual weights $(1,0), (0,1)$. Duality here comes from the Dynkin diagram automorphism reversing nodes.

P.S. I'm not sure about the exact setting of the question (or its motivation). For instance I'm taking for granted that irreducible representations are over $\mathbb{C}$.

[ADDED] Maybe I'm having trouble translating your compact group language into the representation framework for semisimple Lie algebras over $\mathbb{C}$, but in any case I don't see why your example $\mathrm{SU}(2) \times \mathrm{SU}(2)$ should yield representations which are dual to each other. Instead they look self-dual to me. At least in the Lie algebra setting, the approach I outlined involving Weyl group orbits of weights should rule out further simple examples: for this you'd look case-by-case at the "minuscule" highest weights and the resulting dimensions. Your third compact group example doesn't involve a semisimple group, by the way, which further complicates matters.

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But the key point is that the Lie group must have exactly two irreducible three-dimensional representations, comprising a representation and its dual representation, no? $SL_3$ should work, too. –  Will Sawin Jul 6 '12 at 16:26
@Will: Yes, the question here reduces to the fact that the Hom space between two irreducible representations has dimension 1 or 0 depending on whether they are isomorphic or not. But the question is formulated just for compact Lie groups, though I'm tempted to rephrase it in terms of corresponding groups or Lie algebras over $\mathbb{C}$ and their representations. Your $\mathrm{SL}_3$ has the compact real form I wrote as $\mathrm{SU}(3)$. –  Jim Humphreys Jul 6 '12 at 19:21
@Jim: Thanks for your answer. This question actually comes from physics. It attempts to abstract a property of $SU(3)$. By the way, how do you show that such a Lie group must have exactly two irreducible three-dimensional representations? Also, why do you find it more natural to rephrase this in terms of corresponding groups or Lie algebras? As for the conventions, yes, I am taking the irreducible representations to be over the complex numbers. –  Colin Tan Jul 7 '12 at 2:47
Here's why the properties asked for are equivalent to having exactly two distinct 3-dimensional irreps which are duals of each other. If $\pi$ and $\pi'$ are irreducible representations, note that the multiplicity of the trivial representation in $\pi \otimes \pi'$ is $$\dim \mathrm{Hom}_G(\mathbf 1, \pi \otimes \pi') = \dim \mathrm{Hom}_G(\mathbf 1 \otimes \pi^\vee, \pi') = \dim \mathrm{Hom}_G(\pi^\vee, \pi') =\begin{cases}1 & \pi^\vee \cong \pi' \\\\ 0 & \text{otherwise}.\end{cases}$$ Here $\pi^\vee$ is the dual of the representation $\pi$, and the last step is just Schur's lemma. –  Dan Petersen Jul 7 '12 at 12:11
@Colin: Though it's often more natural in physics to work with compact groups, the finite dimensional representation theory over the complex field translates (by Cartan-Weyl) into parallel questions for complex reductive Lie or algebraic groups, or their Lie algebras. The semisimple case is most critical here, where representations are classified by dominant highest weights, etc. There are many treatments in math and physics textbooks, which differ in notation and language. But your question boils down to a purely algebraic question. See Dan's comment for more elaboration. –  Jim Humphreys Jul 7 '12 at 14:35