Timeline for 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?
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| Jul 16, 2012 at 22:18 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Jul 16, 2012 at 18:03 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Jul 8, 2012 at 8:01 | vote | accept | CommunityBot | ||
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| Jul 7, 2012 at 15:31 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Jul 7, 2012 at 14:35 | comment | added | Jim Humphreys | @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. | |
| Jul 7, 2012 at 12:11 | comment | added | Dan Petersen | 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. | |
| Jul 7, 2012 at 2:47 | comment | added | user2529 | @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. | |
| Jul 6, 2012 at 19:21 | comment | added | Jim Humphreys |
@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 $\mathrm{SU}(3)$.
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| Jul 6, 2012 at 16:26 | comment | added | Will Sawin | 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. | |
| Jul 6, 2012 at 15:51 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Jul 6, 2012 at 14:34 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Jul 6, 2012 at 13:51 | history | answered | Jim Humphreys | CC BY-SA 3.0 |