# Why is $\operatorname{Spin}(3) \cong \operatorname{SU}(2)$?

Why is $\operatorname{Spin}(3) \cong \operatorname{SU}(2)$? I can't seem to realize $\operatorname{Spin}(3)$ explicitly as a matrix group and so I'm having issues constructing an isomorphism, of course, I could also use the fact that $1 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow \operatorname{Spin}(3) \rightarrow \operatorname{SO}(3) \rightarrow 1$ is a short exact sequence, but I can't see how this would be useful for $\operatorname{SU}(2)$.

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What's your definition of $Spin(3)$? –  Jason DeVito Apr 6 '12 at 4:17

The exact sequence you mentions shows that $Spin(3)$ is the universal cover of $SO(3)$, i.e. the simply connected Lie group with Lie algebra $so(3)$. Therefore to show that $Spin(3)$ is isomorphic to $SU(2)$ it is sufficient (by uniqueness of universal covering groups) to show that $SU(2)$ also fits into such a short exact sequence. There is a well-known double cover from $SU(2)$ to $SO(3)$. One way to see it is by viewing $SU(2) = S^3$ as the unit quaternions and have them act on $\mathbb R^3$ (viewed as purely imaginary quaternions) by conjugation. This will be an isometry and it is easily checked the kernel is $\pm 1$.
Alternatively, $SU(2)$ acts orthogonally via the adjoint representation on its Lie algebra (viewed as a euclidean space with respect to the Killing form). This representation also gives a double cover $SU(2) \to SO(3)$.