3 added 11 characters in body edited Apr 12 '17 at 6:11 Valter Moretti 38.1k44 gold badges7373 silver badges145145 bronze badges Since $$SO(3)$$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation of $$SO(3)$$ in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations, which, in turn, are finite dimensional representations of $$SU(2)$$. So, once you know all finite dimensional of $$SU(2)$$ representations you know everything. Since $$SO(3)$$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation of $$SO(3)$$ in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations, which in turn, are finite dimensional representations of $$SU(2)$$. So, once you know all finite dimensional representations you know everything. Since $$SO(3)$$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation of $$SO(3)$$ in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations which, in turn, are finite dimensional representations of $$SU(2)$$. So, once you know all finite dimensional of $$SU(2)$$ representations you know everything. 2 added 11 characters in body edited Apr 11 '17 at 20:26 Valter Moretti 38.1k44 gold badges7373 silver badges145145 bronze badges Since $$SO(3)$$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation of $$SO(3)$$ in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations, which in turn, are finite dimensional representations of $$SU(2)$$. So, once you know all finite dimensional representations you know everything. Since $$SO(3)$$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations, which in turn, are finite dimensional representations of $$SU(2)$$. So, once you know all finite dimensional representations you know everything. Since $$SO(3)$$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation of $$SO(3)$$ in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations, which in turn, are finite dimensional representations of $$SU(2)$$. So, once you know all finite dimensional representations you know everything. 1 answered Apr 11 '17 at 20:25 Valter Moretti 38.1k44 gold badges7373 silver badges145145 bronze badges Since $$SO(3)$$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations, which in turn, are finite dimensional representations of $$SU(2)$$. So, once you know all finite dimensional representations you know everything.