Seeing $PSL_2(\mathbb{C}) \cong SO_3(\mathbb{C})$ How can I see the isomorphism between projective special linear group (order 2) and the special orthogonal group (order 3)?
I know only this setting $PSL_2(\mathbb{C}) = SL_2(\mathbb{C})/Z(SL_2(\mathbb{C})$, but it has not helped me a lot.
 A: Let $V$ be the natural $2$-dimensional representation of $\operatorname{SL}_2(\mathbb{C})$. Then the symmetric square $S^2(V)$ with the natural action is a $3$-dimensional representation of $\operatorname{SL}_2(\mathbb{C})$ with an $\operatorname{SL}_2(\mathbb{C})$-invariant symmetric form. The representation $S^2(V)$ has $Z(\operatorname{SL}_2(\mathbb{C}))$ as its kernel, so this gives you an isomorphism $\operatorname{PSL}_2(\mathbb{C}) \cong \operatorname{SO}_3(\mathbb{C})$.
A: Quick and dirty argument is that the real special unitary group $SU(2)/\{1,-1\}$ is isomorphic to the real rotation group $SO(3)$. Therefore their corresponding complexifications $PSL_2(\mathbb{C})$ and $SO_3(\mathbb{C})$ should also be isomorphic. 
A: $PSL_2(\mathbb{C})$ is the image of $GL_2(\mathbb{C})$ in its action by conjugation on the space of $2\times 2$ matrices over $\mathbf{C}$. The subspace $\mathfrak{l}$ of matrices of trace $0$ is invariant under this action and has a nondegenerate symmetric bilinear form $(A,B) = {\rm trace}(AB)$ which is invariant under conjugation. $\mathfrak{l}$ has dimension $3$, which provides the desired homomorphism from $PSL_2(\mathbb{C})$ to $SO_3(\mathbb{C})$. The map is onto from dimensional considerations: the both group, seen as real Lie groups, have real dimension $6$.
Actually, $\mathfrak{l} = \mathfrak{sl}_2$ is the Lie algebra of $PSL_2(\mathbb{C})$, the action by conjugation is the adjoint action, and the bilinear symmetric form ${\rm trace}(AB)$ is (proportional to) the Killing form.
A: Here is another viewpoint: $SO_3(\mathbb{C})$ is exactly the set of automorphisms of $\mathbb{P}^2_{\mathbb{C}}$ preserving a quadric curve (the one corresponding to the bilinear form $SO_3$ is defined by).  Any quadric curve in $\mathbb{P}^2_{\mathbb{C}}$ is isomorphic to $\mathbb{P}^1_{\mathbb{C}}$ which has automorphism group $PSL_2(\mathbb{C})$ giving a map $SO_{3}(\mathbb{C})\to PSL_2(\mathbb{C})$.  This is the desired isomorphism.
A: You will find the rest in the paragraph "Connection between SO(3) and SU(2)" of (https://en.wikipedia.org/wiki/Rotation_group_SO(3)).
