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

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.

## 1 Answer

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.