# Isomorphism $SO(3)\cong SU(2)/\{+I,-I\}$: is it just a group isomorphism?

Reading the proof that $SO(3)\cong SU(2)/\{+I,-I\}$, I see that it hinges on the isomorphism theorem for groups. Since I am studying this in a Lie groups course, I would expect these two groups to be actually lie group isomorphic, and not just group isomorphic.

So my question is: is there any reason why the group isomorphism coming from the isomorphism theorem should be smooth?

• The formulas you get will manifestly consist of smooth functions. – Lord Shark the Unknown Jun 17 '18 at 8:46

## 1 Answer

It is clearly a Lie group homomorphism, that is, it is a differentiable map. You just consider the set $H=\{ai+bj+ck\,|\,a,b,c\in\mathbb{R}\}\subset\mathbb H$ and see $SU(2)$ as the quaternions with norm $1$. Now, each $q\in SU(2)$ acts on $H$ by $h\mapsto qhq^{-1}$ and it is easy to prove that the action of each $q$ is an element of $SO(3,\mathbb{R})$. This is cleary a differentiable map. Since $\{\pm\operatorname{Id}\}$ is a discrete subgroup of $SU(2)$, it induces a differentiable map from $SU(2)/\{\pm\operatorname{Id}\}$ into $SO(3,\mathbb{R})$. The fact that we use the isomorphism theorem to prove that this induces an isomorphism between $SU(2)/\{\pm\operatorname{Id}\}$ and $SO(3,\mathbb{R})$ changes nothing about that

• But the map that you described does not need to be the same map coming from the isomorphism theorem, or does it? – Soap Jun 17 '18 at 9:22
• @Soap I am not sure I understand your question. The isomorphism theorem is used to prove that the map that I discribed is an isomorphism. It does not provide a map. – José Carlos Santos Jun 17 '18 at 9:25
• Not explicitly, but isn't it the unique map that composed with the natural map $G\rightarrow G/\ker\phi$ gives the epimorphism $\phi$ we started with? – Soap Jun 17 '18 at 9:35
• @Soap Yes. To be more precise: if $\phi\colon G\longrightarrow H$ is an epimorphism, then the map$$\begin{array}{ccc}G/\ker\phi&\longrightarrow&H\\g\ker\phi&\mapsto&\phi(g)\end{array}$$is an isomorphism. As you see, the isomorphism is totally determind by $\phi$, which is given from the start. – José Carlos Santos Jun 17 '18 at 9:40