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Dec 22, 2020 at 22:50 comment added Qiaochu Yuan @Piet: yes, that's right (for finite-dimensional representations).
Dec 22, 2020 at 22:19 comment added LSpice @PietBongers, assuming that $\pi(G)$ means $\pi_1(G)$, then this is so; Wikipedia calls it the homomorphisms theorem, although I am not sure if this terminology is standard.
Dec 22, 2020 at 21:00 comment added Piet Bongers @QiaochuYuan: Thanks a lot for the extra comments. So if I have understood correctly, for a Lie group $G$, will the irreps of $G$ coincide with the irreps of $\frak{g}$ when $\pi(G)$ is trivial?
Dec 22, 2020 at 20:06 history edited Qiaochu Yuan CC BY-SA 4.0
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Dec 22, 2020 at 20:06 vote accept Piet Bongers
Dec 22, 2020 at 20:03 comment added Qiaochu Yuan @Piet: it's exactly as LSpice says, you get a double cover so again there are "twice" as many irreducibles.
Dec 22, 2020 at 20:02 comment added LSpice @PietBongers, addressed in a comment on the main question (although maybe @‍QiaochuYuan will have more to say).
Dec 22, 2020 at 19:57 comment added Piet Bongers So the example I am thinking about is $U_2$ and $SU_2 \times U_1$. Does $SU_2 \times U_1$ have more irreducible representations than $U_2$?
Dec 22, 2020 at 19:53 vote accept Piet Bongers
Dec 22, 2020 at 19:55
Dec 22, 2020 at 19:41 history answered Qiaochu Yuan CC BY-SA 4.0