Is $(SU(2)\times U(1))/\{\pm 1\}\cong SO(3)\times U(1)$?

Asked Jun 04 '25 at 09:56

Active Jun 04 '25 at 11:14

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This question arises from the classification of lie subgroups of $SO(4)$. I wonder if the Lie group $$(SU(2)\times U(1))/\{\pm 1\}$$ is isomorphic to $SO(3)\times U(1)$. I believe there exists some “twisting” in this group. If not, how to think of this group?

Added: What if replacing $SU(2)$ by $U(1)$? i.e. how to think of the group $(U(1)\times U(1))/\{\pm 1\}$? I think this should be easier, but still it's unclear to me.

Any suggestions would be appreciated. Thanks!

edited Jun 04 '25 at 11:14

asked Jun 04 '25 at 09:56

Yunfeng Gong

We have $SU(2)/{\pm 1}\cong SO(3)$, see here. Use this for the direct product. – Dietrich Burde Jun 04 '25 at 10:01
@DietrichBurde In my question, $1$ means the identity element of the product group, i.e. $\pm 1$ means $\pm (I,1)$. – Yunfeng Gong Jun 04 '25 at 10:03
1

But the natural $SU(2)\times U(1)\to SO(3)\times U(1)$ cannot induce a isomorphism in my question. The kernel is ${\pm 1,1}$. – Yunfeng Gong Jun 04 '25 at 10:07
No, the kernel is ${1}\times {\pm 1}\cong {\pm 1}$, because $C_1\times C_2\cong C_2$ for cyclic groups. – Dietrich Burde Jun 04 '25 at 10:08
You are wrong. For example, $\mathbb{Z}/\mathbb{Z}\not\cong\mathbb{Z}/2\mathbb{Z}$ but $\mathbb{Z}\cong 2\mathbb{Z}$. – Yunfeng Gong Jun 04 '25 at 10:15
I am only saying that the kernel is isomorphic to $C_2$. – Dietrich Burde Jun 04 '25 at 10:19
Let us continue this discussion in chat. – Yunfeng Gong Jun 04 '25 at 10:21

Is $(SU(2)\times U(1))/\{\pm 1\}\cong SO(3)\times U(1)$?

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