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I have just read that all irreducible representations of $SO(3)$ are in bijective correspondance with the irreducible representations $V_n$ of $SU(2)$ in which $-E$ acts as the identity.

Can someone help me how to understand that this indeed holds.

As far as I know, there is only an epimorphism $\pi : SU(2) \to SO(3)$, having kernel {$E,-E$} (diagonal matrices.)

But not sure how this helps to get bijection between their irreducible representations nor what exactly is meant when saying "in which $-E$ acts as the identity".

Many thanks for any help

edward_scissorhands
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  • The action is a $SU(2)\times V\to V$ which can be written as $SU(2)\to GL(V)$. Now it is meant that $-E$ is mapped to the identity. – Berci May 13 '17 at 17:48

1 Answers1

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This a consequence of the universal property of quotients and the first isomorphism theorem.

A representation of $SU(2)$ is the same thing as a continuous group homomorphism $$\rho :SU(2)\to GL(V),$$ (with $V$ a vector space) and representations with $-E$ acting as the identity are those where $\rho(-E)$ is the identity on $V$, i.e. the homomorphisms $\rho$ whose kernel contains the normal subgroup $\{E,-E\}$. By the universal property of quotient groups, these are the same thing as continuous homomorphisms $$SU(2)/\{E,-E\}\to GL(V)$$ (where $SU(2)/\{E,-E\}$ is endowed with the quotient topology). Now $SU(2)/\{E,-E\}$ is isomorphic to $SO(3)$ by the first isomorphism theorem, and the isomorphism is continuous, hence an homeormorphism (for a more detailed explanation about this isomorphism, see this question). Thus the continuous homomorphisms $SU(2)/\{E,-E\}\to GL(V)$ are exactly the representations of $SO(3)$.

This shows that representations of $SO(3)$ are in bijection with those of $SU(2)$ such that $-E$ acts trivially. Now irreducible representations are those where $V$ has no proper invariant subspace. Since this only depends on the image of the homomorphism $\rho $ in $GL(V)$, the bijection described above restricts to a bijection between irreducible representations.

Arnaud D.
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  • Thanks! Just one thing to make clear, when you use the fact that a representation is, in general, the same thing as a group homomorphism is this the definition you use or a proposition derived from the standard definition of representation(i.e. the one which demands a continuous group action together with linear translations)? – edward_scissorhands May 13 '17 at 18:08
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    To be honest I'm not really used to talk about representation of Lie Groups, so I hadn't taken continuity into account. This shouldn't change the answer too much, I'll edit some details. – Arnaud D. May 13 '17 at 18:16
  • Thank you very much, I just don't get why this isomorphism must necessarily be continuous? – edward_scissorhands May 13 '17 at 19:02
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    If we use the quotient topology on $SU(2)/{E,-E}$, then the isomorphism is continuous if and only if the original epimorphism $SU(2)\to SO(3)$ is (this is a consequence of the definition of the quotient topology). – Arnaud D. May 13 '17 at 19:41