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Consider a compact Lie group $G$ of dimension strictly greater than 0. There is a theorem saying that $G$ admits a finite-dimensional faithful representation. Therefore, we can pick one of lowest dimension. Question:

is a lowest-dimension faithful representation of $G$ always irreducible? If not, what is a counterexample and are there some conditions we can add (like $G$ to be simple for instance) that can make the statement become true?

At page 20, Section 3.6 of the file you can find at
https://ir.canterbury.ac.nz/bitstream/handle/10092/5943/joyce_thesis.pdf?sequence=1
the author says:

"For simple and semi-simple groups the primitive (i.e. lowest-dimension faithful) representations are irreducible"

but he doesn't explain nor give any reference.

On the other side, at https://mathoverflow.net/questions/328138/non-faithful-irreducible-representations-of-simple-lie-groups?rq=1 they speak about irreducible representations of Lie algebras that induce non-faithful ones at the group level, and at a certain point they claim that the Lie groups $D_{2l}$ ($l\geq 2$) have the property that all irreducible representations are non-faithful, the center of these groups being non-cyclic. If so, is then the first source I have mentioned in the first link wrong or am I missing something about what these guys are doing in this last link?

Emanuele
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    If $G$ is simple, decompose a primitive representation in irreducible subrepresentations : one of them is nontrivial irreducible, so it is faithful by simplicity; by minimality it must be the whole represnetation. – Maxime Ramzi Jul 30 '19 at 13:34
  • There's a logical glitch in your last paragraph. Earlier, you are asking whether certain faithful representations are necessarily irreducible. The fact that many irreducible representations are not faithful can never contradict that. – Torsten Schoeneberg Jul 30 '19 at 21:31
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    @Max: thank you for the idea. I have a question on this process, that maybe relies on the definition of simple Lie group. Say we define $G$ to be a simple Lie group when it is connected, non-Abelian and it has no connected closed normal subgroups. Therefore any normal subgroup of $G$ is discrete, since the identity component is trivial by simplicity of $G$. Hence we know that the kernel of an irreducible representation of $G$ (being closed and normal) is discrete: is it also connected so to actually conclude that the considered irrep is faithful as you are saying? – Emanuele Jul 31 '19 at 07:39
  • @TorstenSchoeneberg: yes sorry. I re-edited the last paragraph. Is it now clear? – Emanuele Jul 31 '19 at 07:55
  • Ah ! I was using the definition of simple groups from "non-Lie" group theory. A priori it is not clear (to me - I'm not an expert at all) that the kernel is connected. – Maxime Ramzi Jul 31 '19 at 11:37
  • No, it still mixes up the implications. Even if a certain type has no faithful representation at all, that does not contradict the statement that "all faithful rep's of least dimension are irreducible". It's just an empty statement for the simply connected (!) groups of type $D_{even}$, they have no faithful rep's. So what? Of course they still have tons of ("other") irreducible ones. – Torsten Schoeneberg Jul 31 '19 at 15:38
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    I am saying that for $D_{\text{even}}$ all irreducible representations are non-faithful. This implies (and in classical logic is equivalent) that any faithful representation is reducible. Therefore, it doesn't exist a faithful irreducible representation, contradicting the claim in the first source I mentioned. – Emanuele Aug 01 '19 at 18:55
  • You're right, sorry; I was under the impression that those groups have no faithful representations at all (irreducible or not), in which case there would be no contradiction; but they do. I now believe that the first source is wrong in that case, a first counterexample would be a minimal faithful rep of $Spin(8)$, which is probably a sum of two distinct half-spin irreps mentioned here: https://math.stackexchange.com/a/3296484/96384 – Torsten Schoeneberg Aug 01 '19 at 23:31

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You definitely want some sort of simplicity condition: Consider $G =S^1 \times S^1$. It is abelian so any irreducible representation is one dimensional, but none of the one dimensional representations are faithful.

Nate
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