What is the resulting group of $(\operatorname{U}(1)\times \operatorname{U}(1)) / \mathbb{Z}_2 $

Question

I found this similar question to mine, however in my case I have the direct product of two copies of the group. My intuition tells me I should get $\operatorname{U}(1)$, but intuition is not proof. This question arises due to having a field theory where a field transforms under $\operatorname{U}(1)$ two different ways. As such we need to modulo out by any simultaneous commuting $\operatorname{U}(1)$ transforms which may be identified with doing nothing. If both $\operatorname{U}(1)$'s act by $-1$, it is clear this group element should be identified with the identity. Any help or hints are appreciated

Answer 1

Apply the first isomorphism theorem to the morphism $$\varphi:U(1)\times U(1)\to U(1)$$ $$(u,v)\mapsto uv^{-1}.$$ The quotient group you are describing is $$U(1)=\operatorname{im}\varphi\simeq(U(1)\times U(1))/\ker\varphi,$$ but $$\ker\varphi=\{(u,u)\mid u\in U(1)\}\simeq U(1)$$is bigger than the subgroup generated by $(-1,-1)$, which is $$\{(u,u)\mid u=\pm1\} \simeq\Bbb Z_2.$$

Answer 2

In general, $\dim(G/H)=\dim(G)-\dim(H)$. Since $\dim S^1=1$ and $\dim S^0=0$, we expect the quotient group $S^1\times_{S^0}S^1$ to have dimension $1\!+\!1\!-\!0\,=\,2$, so the quotient can't be $S^1$.

(Note $S^1\cong\mathrm{U}(1)$ and $S^0\cong\mathbb{Z}_2$.)

Hint. Apply $1$st Iso. Thm. to the map $(u,v)\mapsto(uv,uv^{-1})$ from $T^2=S^1\times S^1$ to itself.