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Why is $SO(3, \mathbb{C}) \cong PSL(2, \mathbb{C})$? I can't seem to be able to construct an explicit isomorphism between them.

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@Geoff : I think you mean $Lie(SL(2,C))$ instead of $SL(2,C)$. –  user10676 Apr 4 '12 at 18:29

1 Answer 1

Let $SL(2,\mathbb C)$ act on the $2\times 2$ matrices of trace $0$ by conjugation. This is a 3-dimensional $\mathbb C$-vectorspace, and for the trace (euclidian inner product) we have $$ \text{trace}(S^{-1}ASS^{-1}BS)=\text{trace}(AB) $$ This gives you a homomorphism $SL(2,\mathbb C)\rightarrow SO(3,\mathbb C)$, which is surjective etc.

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Could please explain how $PSL(2,\mathbb{C}) \cong SL(2,\mathbb{C})/\{\pm\}$ comes into play? And what is $S$? –  draks ... Apr 4 '12 at 21:35
    
above $S$ would be any matrix in $SL(2,\mathbb C)$. Think what happens in this operation if you replace $S$ with $-S$, what is the kernel of the homomorphism $SL(2,\mathbb C)\rightarrow SO(3,\mathbb C)$? –  Blah Apr 5 '12 at 7:59
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How do we have the homomorphism mapping into $SO(3, \mathbb{C})$? –  HJK Apr 6 '12 at 6:24
    
Because every $S \in SL(2,\mathbb C)$ is an orthogonal linear endomorphism of a three-dimensional $\mathbb C$ inner product space (in this case the $2\times 2$ matrices of trace 0) –  Blah Apr 6 '12 at 15:21

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