Ytukyg

Let $A_{theta}$ be the rotation C$^{*}$-algebra with rotation $theta$. I.e., $A_{theta}=C^{*}(u,v)$, where $vu=e^{2pi i theta}uv$. Suppose that $theta=p/q$, where $p$ and $q$ are non-zero positive integers that are relatively prime.

I am trying to show that $A_{theta}$ is not simple.

Consider the C$^{*}$-algebra $M_{q}(mathbb{C})$ and the two unitary matrices
$$
U = begin{pmatrix}0 & 1 & 0 & cdots & 0\
0 & 0 & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots\
0 & 0 & 0 & cdots & 1\
1 & 0 & 0 & cdots & 0
end{pmatrix}
quadtext{and}quad V=begin{pmatrix}
1 & 0 & 0 & cdots & 0\
0 & e^{2pi i theta} & 0 & cdots & 0\
0 & 0 & e^{4pi i theta} & cdots & 0\
vdots & vdots & vdots & ddots & vdots\
0 & 0 & 0 & cdots & e^{2(q-1)pi itheta}
end{pmatrix}
$$
in $M_{q}(mathbb{C})$. It is easy to check that $VU=e^{2pi i theta}UV$. It follows by the universal property of $A_{theta}$ that there is a $*$-homomorphism $picolon A_{theta}to M_{q}(mathbb{C})$ such that $pi(u)=U$ and $pi(v)=V$. If $pi$ is not injective, then we are finished, since $kerpi$ is a non-zero proper ideal in $A_{theta}$. So, we may suppose $pi$ is inejctive. Now note that $pi$ is an irreducible representation. Otherwise, it would decompose into a direct sum of smaller irreducible ones, which is impossible. But then it must be that $pi(A_{theta})=M_{q}(mathbb{C})$ since irreducible representations acting on a finite-dimensional space are surjective.

Thus, we can conclude that $A_{theta}=C^{*}(u,v)=C^{*}(U,V)=M_{q}(mathbb{C})$. I am looking for a straight-forward way to get a contradiction here.

As you can read in the first paragraph of Chapter VI in Davidson, you can realize unitaries $u,v$ with $uv=e^{2pi i theta}vu$ by taking $u,vin B(L^2(mathbb T))$ where $u$ is multiplication by $z$, i.e. the bilateral shift.

The C$^*$-algebra generated by $u$ is $C^*(u)=C(mathbb T)$, so by restricting your $pi$ to $C^*(u)$ you get a $*$-homomorphism $$pi:C(mathbb T)to M_n(mathbb C).$$ If $pi$ were injective, you would have an infinite-dimensional subalgebra of $M_n(mathbb C)$.