Just in view of the **double** universal covering provided by $SU(2)$, $SO(3)$ must a quotient of $SU(2)$ with respect to a central discrete normal subgroup with **two** elements. This is consequence of a general property of universal covering Lie groups:

*If $\pi: \tilde{G} \to G$ is the universal covering Lie-group homomorphism, the kernel $H$ of $\pi$ is a discrete normal central subgroup of the universal covering $\tilde{G}$ of $G= \tilde{G}/H$, and $H$ is isomorphic to the fundamental group of $G$, i.e. $\pi_1(G)$ (wich, for Lie groups, is Abelian)* .

One element of that subgroup must be $I$ (since a group includes the neutral element). The other, $J$, must verify $JJ=I$ and thus $J=J^{-1}= J^\dagger$. By direct inspection one sees that in $SU(2)$ it is only possible for $J= -I$. So $SO(3) = SU(2)/\{I,-I\}$.

Notice that $\{I,-I\} = \{e^{i4\pi \vec{n}\cdot \vec{\sigma}/2 }, e^{i2\pi \vec{n}\cdot \vec{\sigma}/2 }\}$ stays in the **center** of $SU(2)$, namely the elements of this subgroup commute with all of the elements of $SU(2)$. Moreover $\{I,-I\}=: \mathbb Z_2$ is just the first homotopy group of $SO(3)$ as it must be in view of the general statement I quoted above.

A unitary representations of $SO(3)$ is also a representation of $SU(2)$ through the projection Lie group homomorphism $\pi: SU(2) \to SU(2)/\{I,-I\} = SO(3)$. So, studying unitary reps of $SU(2)$ covers the whole class of unitary reps of $SO(3)$. Let us study those reps.

Consider a unitary representation $U$ of $SU(2)$ in the Hilbert space $H$. The central subgroup $\{I,-I\}$ must be represented by $U(I)= I_H$ and $U(-I)= J_H$, but $J_HJ_H= I_H$ so, as before, $J_H= J_H^{-1}= J_H^\dagger$.

As $J_H$ is unitary and self-adjoint simultaneously, its spectrum has to be included in $\mathbb R \cap \{\lambda \in \mathbb C \:|\: |\lambda|=1\}$. So (a) **it is made of $\pm 1$ at most** and (b) **the spectrum is a pure point spectrum** and so only **proper** eigenspeces arise in its spectral decomposition.

If $-1$ is not present in the spectrum, the only eigenvalue is $1$ and thus $U(-I)= I_H$. If only the eigenvalue $-1$ is present, instead, $U(-I)= -I_H$.

If the representation is **irreducible** $\pm 1$ cannot be simultaneously eigenvalues. Otherwise $H$ would be split into the orthogonal direct sum of eigenspaces $H_{+1}\oplus H_{-1}$. As $U(-1)=J_H$ commutes with all $U(g)$ (because $-I$ is in the center of $SU(2)$ and $U$ is a representation), $H_{+1}$ and $H_{-1}$ would be **invariant** subspaces for all the representation and it is forbidden as $U$ is **irreducible**.

We conclude that,

*if $U$ is an irreducible unitary representation of $SU(2)$, the discrete normal subgroup $\{I,-I\}$ can only be represented by either $\{I_H\}$ or $\{I_H, -I_H\}$.*

Moreover:

*Since $SO(3) = SU(2)/\{I,-I\}$, in the former case $U$ is ***also** a representation of $SO(3)$. It means that $I = e^{i 4\pi \vec{n}\cdot \vec{\sigma} }$ and $e^{i 2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ are both transformed into $I_H$ by $U$.

*In the latter case, instead, $U$ is not a true representation of $SO(3)$, just in view of a sign appearing after $2\pi$, because $e^{i 2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ is transformed into $-I_H$ and only $I = e^{i 4\pi \vec{n}\cdot \vec{\sigma}/2 }$ is transformed into $I$ by $U$.*

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti