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
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?

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    $\begingroup$ 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. $\endgroup$ Commented Jul 30, 2019 at 13:34
  • $\begingroup$ 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. $\endgroup$ Commented Jul 30, 2019 at 21:31
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    $\begingroup$ @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? $\endgroup$
    – Emanuele
    Commented Jul 31, 2019 at 7:39
  • $\begingroup$ @TorstenSchoeneberg: yes sorry. I re-edited the last paragraph. Is it now clear? $\endgroup$
    – Emanuele
    Commented Jul 31, 2019 at 7:55
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    $\begingroup$ 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. $\endgroup$
    – Emanuele
    Commented Aug 1, 2019 at 18:55

1 Answer 1


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.


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