# Lowest-dimensional faithful representations of compact Lie groups

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?

• 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. Jul 30, 2019 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. Jul 30, 2019 at 21:31
• @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? Jul 31, 2019 at 7:39
• @TorstenSchoeneberg: yes sorry. I re-edited the last paragraph. Is it now clear? Jul 31, 2019 at 7:55
• 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. Aug 1, 2019 at 18:55

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.