The method of topology is very clear.Then there's a question asking to use adjoint representation of lie group $SU(2)$ $(\operatorname{adj}:SU(2)\to GL(su(2)))$to prove this. I can't solve this .
$SU(2)$ is a covering space of $SO(3)$.
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differential-geometry
lie-algebras
lie-groups
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4Dear henry, The group $SU(2)$ acts on its Lie algebra via the adjoint rep'n (more concretely, by conjugation). This Lie algebra is 3-dimensional, and so this gives a representation of $SU(2)$ by $3\times 3$ matrices. Now check that these matrices lie in $SO(3)$ (perhaps by finiding a quadratic form on the Lie algebra $su(2)$ that this action preserves). Regards, – 2012-03-14