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?
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.