irreducible representations and pure states of $M_n(\mathbb{C})$

Question

We consider the $C^\ast$-algebra $A=M_n(\mathbb{C})$, which can be seen as $L(\mathbb{C}^n)$. Prove that:

(1) $id:A\to A$ is a irreducible $\ast$-representation of $A$

(2) all irreducible $\ast$-representations $\pi:A\to L(H_{\pi})$ of $A$ are eqivalent to $id$, i.e. there is an isometric isomorphism $V:\mathbb{C}^n\to H$ such that $V^*\pi(a)=id(a)V$ for all $a\in A$.

(3) Let $R(A)$ be the set of all pure states of $A$, endowed with the weak$*$-topology, let $S^{2n-1}=\{v\in\mathbb{C}^n: \|v\|_2=1\}$ and for all $u,w \in S^{2n-1}$, $u\sim w :\iff $ there is a $\lambda\in \mathbb{C}$ such that $\lambda u=w$. Prove that $R(A)$ is homeomorphic to $S^{2n-1}/\sim$.

Number (1) I've already done. For (2): There is a hint on the worksheet: Consider for $i,j=1,..,n$ the matrices $E_{ij}$ with 1 at the $i=j-th$ position and $0$ elsewhere. First of all prove that $\pi(E_{11})\neq 0$ for all irreducible representations $\pi:A\to L(H_{\pi})$ and choose a $v_1\in \pi(E_{11})H$ such that $\|v_1\|=1$. Then consider $v_i=\pi(E_{i1})v_1$. I have done the instruction so far, but then I'm stuck. I think $B=\{v_1,..,v_n\}$ has to be an orthonormal basis of $H$, but is $B$ linearly independent? Let $k_1v_1+...+k_nv_n=0$, i.e. $k_1\pi(E_{11})v_1+..+k_n\pi(E_{n1})v_1=(k_1\pi(E_{11})+..+k_n\pi(E_{n1}))v_1=0$ and $v_1\neq 0$, then it must be $k_1\pi(E_{11})+..+k_n\pi(E_{n1})=0$. It is $\pi(E_{j1})\neq 0$ for all $j$, but why should follow that $k_j=0$ for all $j=1,..,n$ (a littlebit awkward this question..)?

And do I have to prove that $\overline{\pi(M_n(\mathbb{C}))span\{v_1,..,v_n\}}=H$?.

Then I defined $U:\mathbb{C}^n\to H, e_k\mapsto v_k$, $e_k=(0,..,1,..0)$ with 1 at the k-th position. Then $U$ is unitary and I otain $U^*\pi(\enspace)U=id(\enspace)$.

For (3):From lecture I know that there is a bijection between R(A) and $Irr(A)/\sim$, where $Irr(A)$ is the set of all $(\pi,x)$, $\pi$ cyclic $*$-representation of $A$ which is irreducible with cyclic vector $x$, and $\sim$ is the equivalence from (2). From (2) we know that all irreducible $*$-representations $(\pi,x)$ are equivalent to $(id,y)$, $x$ and $y$ are clyclic vectors. But I don't know how to continue.

I appreciate your help. Regards

Answer 1

(2): Note that any nonzero $*$-homomorphism $\pi:M_n (\mathbb C)\to B (H) $ is one-to-one, since its kernel is an ideal.

Now you have $$ 0=k_1\pi(E_{11})+\cdots+k_n\pi(E_{n1})=\pi(k_1E_{11}+\cdots+k_nE_{n1}). $$ As $\pi$ is one-to-one, $$ 0=k_1E_{11}+\cdots+k_nE_{n1}=\begin{bmatrix}k_1&0&\cdots&0\\ \vdots&&\ddots\\ k_n&0&\cdots&0, \end{bmatrix} $$ so $k_1=\cdots=k_n=0$.

The equality $\overline{\pi(M_n(\mathbb{C}))span\{v_1,..,v_n\}}=H$ makes no sense; it is $\overline{\pi(M_n(\mathbb{C}))span\{v_1,..,v_n\}}=H_\pi$, which is obvious since each $v_j$ is constructed as $\pi(E_{j1})v_1$.

(3) If you do GNS for a pure state $\varphi$, you get an irreducible representation. So $$ \varphi(A)=\langle \pi_\varphi(A)\xi_\varphi,\xi_\varphi\rangle =\langle V^*AV\xi_\varphi,\xi_\varphi\rangle=\langle A\eta,\eta\rangle $$ with $\eta=V\xi_\varphi$. Also $\|\varphi\|=\varphi(I)=\|\eta\|^2$. So the pure states are determined by the set $\{\eta\in\mathbb C^n:\ \|\eta\|=1\}$, modulo those vectors that produce the same state.

If $\nu=\lambda\eta$ for some $\lambda\in\mathbb C$, then $|\lambda|=1$ and the states $\langle\cdot\nu,\nu\rangle$ and $\langle\cdot\eta,\eta\rangle$ are equal. And if $\nu$ and $\eta$ are linearly independent, the range projection $P$ onto $\mathbb C\eta$ satisfies $P\eta=\eta$ and $\|P\nu\|<1$, which shows that $\langle P\eta,\eta\rangle\ne\langle P\nu,\nu\rangle$. This shows that the map $\varphi\to[\eta]$ is a bijection. It remains to test the bicontinuity.