why the irrational rotation algebra $A_{\theta}$ is $C(T^{2})$ when $\theta =0$

Question

since the irrational rotation algebra $A_{\theta}$ is commutative when $\theta =0$, it has the form $C(X)$ for some space $X$ and by universal property of $A_{\theta}$, there is a homomorphism from $A_{\theta}$ to $C(T^{2})$ , so there must be a continous map from $T^{2}$ to $X$ which induce the homomorphism, I have difficulty to show it is a homeomorphism.

Answer 1

When $\theta=0$, $A_0$ is the universal C$^*$-algebra generated by two commuting unitaries $u,v$. Your $X$ is the space of characters. Any multiplicative $\phi:C^*(u,v)\to\mathbb C$ is determined by the pair $(\phi(u),\phi(v))\in\mathbb T^2$. And conversely, given $(\lambda,\mu)\in\mathbb T^2$, we may define a character by $\phi(u)=\lambda$, $\phi(v)=\mu$. Indeed, we may see $\lambda$ and $\mu$ as two unitaries, so by universality there exists a $*$-homormophism $\phi:C^*(u,v)\to C^*(\lambda,\mu)\subset\mathbb C$, with $\phi(u)=\lambda$, $\phi(v)=\mu$. So the map $\gamma:\phi\longmapsto (\phi(u),\phi(v))$ is bijective.

In $X$ we consider the weak$^*$-topology. If $\phi_j\to\phi$, this means that $\phi_j(x)\to\phi(x)$ for all $x\in C^*(u,v)$. In particular we may take $x=u$ and $x=v$ to get that $\phi_j(u)\to\phi(u)$ and $\phi_j(v)\to\phi(v)$. So $\gamma$ is continuous. As $X$ is compact and $\mathbb T^2$ is Hausdorff, $\gamma^{-1}$ is also continuous. Thus $\gamma$ is a homeomorphism.