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How does the map $\phi:SU(2) \times SU(2) \rightarrow SO(4)$ induce a double covering.

I understand that we can use the quaternions here by defining the action of $SU(2) \times SU(2)$ on $\Bbb V$ by $(g, h) \cdot z := g z h^* .$

I see the action is a homomorphism and has kernel { \pm 1 } But I don't see how this will induce the double covering of $\phi$.

Thanks in advance!

Toasted_Brain
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  • Well, it is a $2$-to-$1$ map, so $SU(2) \times SU(2)$ covers $SO(4)$ twice. – F. Conrad Feb 04 '23 at 23:44
  • Why is that a 2-1 map? I don't really see that. Thanks! – Toasted_Brain Feb 04 '23 at 23:47
  • You can find it on this site: https://math.stackexchange.com/questions/459675/prove-that-the-map-phis3-times-s3-to-bf-gl4-bbb-r-defined-via-quatern

    Also, note that $SU(2) \cong S^3$. You can probably find this homomorphism explicitly on the internet.

     – F. Conrad Feb 05 '23 at 00:38
  • Thanks for the info. But I still don't see why the homomorphism is a 2-1 covering map, i.e. each point has two pre-images... Maybe I'm missing something that's obvious :( – Toasted_Brain Feb 05 '23 at 00:58
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    If the kernel is ${ \pm 1}$, and $x$ is in the preimage of $y$, then the full preimage of $y$ is $\pm x$. – Torsten Schoeneberg Feb 05 '23 at 01:33

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For a group homomorphism $\phi\colon G\rightarrow H$, let $K=\{1=k_1,k_2,\dots,k_n\}$ be the kernel consisting of $n$ elements. Then, if $h=\phi (g)$ is in its image, we can see that $$\begin{split} \phi^{-1}(h)&=\{g,gk_2,\dots,gk_n\}\\ &=\{g,k_2g,\dots,k_ng\}, \end{split}$$ therefore $\phi$ is exactly $n$-to-$1$ onto the image (set-theoretically).

Moreover, in your case of $\phi\colon \mathrm{SU}(2)\times \mathrm{SU}(2)\rightarrow \mathrm{SO}(4)$, the surjectivity is automatic, as follows. Let $G'=\mathrm{Im}(\phi)$ be the image. Since $\mathrm{SU}(2)\times \mathrm{SU}(2)$ is compact, $G'$ is a closed subgroup, hence a Lie subgroup. If $\dim G'<\dim \mathrm{SO}(4)=\dim\mathrm{SU}(2)\times\mathrm{SU}(2)$, the implicit function theorem at regular points show that $\phi^{-1}(g')$ is infinity, a contradiction. Therefore $\dim G'=\dim \mathrm{SO}(4)$ and the connectedness of the latter implies $G'=\mathrm{SO}(4)$. Note that, this argument　also shows that $\phi$ is a 2-to-1 covering map topologically as well, as $\phi$ has everywhere regular points by translations.

Ayaka
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