# Precise intuition for the exact isomorphism $SO(3) \times SO(3)$, $\frac{SU(2) \times SU(2)}{\mathbb{Z}_2}$ and $SO(4)$

Am I being precise to claim the following group isomorphisms: $${SO(3) \times SO(3)}{}=\frac{SO(4)}{\mathbb{Z}_2}$$ $$\frac{SU(2) \times SU(2)}{\mathbb{Z}_2 \times \mathbb{Z}_2 }=\frac{SO(4)}{\mathbb{Z}_2}$$ $$\frac{SU(2) \times SU(2)}{\mathbb{Z}_2}={SO(4)}$$

What are the precise intuitions for the exact isomorphism?

p.s. It seems that some of the previous posts are not very intuitive nor precise:1. recovering-the-two-su2-matrices-from-so4-matrix and 2. Why is SO(3)×SO(3) isomorphic to SO(4)?

• What do you find imprecise about the posts you link to? The second one gives a detailed explanation of the first isomorphism on your list, and the first one gives a detailed explanation of the third isomorphism on your list. If you find something about those posts unclear, your best bet is probably to post a comment on the answers asking for clarification about whatever specific point is confusing. The second isomorphism on your list follows from the first, together with the isomorphism $SU(2)/\{\pm 1\}\cong SO(3)$, which is a standard fact you can find discussed in many places. Oct 16, 2017 at 16:40
• To be specific about the isomorphism $SU(2)/\{\pm 1\} \cong SO(3)$: view $SU(2)$ as the unit quaternion group, and $SO(3)$ as the group of rotations of $S^2$. Each quaternion $q$ acts by conjugation on the quaternions, yielding a rotation $R$ of the 2-sphere in the purely imaginary quaternions (this is a computation). The map $q\mapsto R$ is a homomorphism $SU(2) \to SO(3)$. Only 1 and -1 act trivially, and so the kernel is $\{\pm 1\} \cong \mathbb{Z}/2\mathbb{Z}$. These ideas are spelled out in my article math.iupui.edu/%7Edramras/double-tip.html and references therein. Oct 16, 2017 at 16:41

The classification of connected Lie groups goes like this: every connected Lie group $G$ has a Lie algebra $\mathfrak{g}$ which determines the isomorphism class of its universal cover $\widetilde{G}$. There is a covering map

$$1 \to \pi_1(G) \to \widetilde{G} \to G \to 1$$

exhibiting $G$ as a quotient of $\widetilde{G}$ by the fundamental group $\pi_1(G)$, which turns out to be a discrete subgroup of the center $Z(\widetilde{G})$ of $\widetilde{G}$. Thus $G$ is classified by the pair consisting of the simply connected Lie group $\widetilde{G}$, together with a particular choice of discrete subgroup $\pi_1(G)$ of the center $Z(\widetilde{G})$.

In this case, all of the Lie groups you describe have the same universal cover, namely $Spin(4) \cong SU(2) \times SU(2)$. The center of this group is $\mathbb{Z}_2 \times \mathbb{Z}_2$ (one $\mathbb{Z}_2$ sitting in each copy of $SU(2)$) and so we can classify all Lie groups with this universal cover by classifying all subgroups of this center (which are all discrete). They are as follows:

• The trivial subgroup $1$. The quotient by this is just $Spin(4) \cong SU(2) \times SU(2)$.
• $\mathbb{Z}_2 \times 1$. The quotient by this is $SO(3) \times SU(2)$.
• $1 \times \mathbb{Z}_2$. The quotient by this is $SU(2) \times SO(3)$.
• The diagonal copy of $\mathbb{Z}_2$. The quotient by this is $SO(4)$.
• $\mathbb{Z}_2 \times \mathbb{Z}_2$. The quotient by this is $SO(3) \times SO(3)$.

So, everything you wrote down is basically fine except that in the third line it's ambiguous what you mean by $\mathbb{Z}_2$. The $\mathbb{Z}_2$ you want is the diagonal one, not the "left" or "right" ones above.

• This is all by way of explaining what the right statements are. To explain how to prove them requires a bit more work; one place to start is with the quaternions. Oct 17, 2017 at 20:53
• Thanks -- do you know this as well:... math.stackexchange.com/questions/2549783 thanks you "_" Dec 3, 2017 at 23:37
• @ Qiaochu Yuan, you said the negative answer for the new post. But what if we consider $\frac{SU(m) \times SU(m) \times U(1)}{(\mathbb{Z}_m)^2}=\frac{SU(m) \times U(m)}{(\mathbb{Z}_m)^2}$ case instead, do we have simpler expression for $m=2,3,4,5,..?$ Thanks! Dec 4, 2017 at 16:10