# Linear independency and compactness of the set of pure states of a $C^*$-algebra

Let $\mathcal{A}$ be a noncommutative $C^*$-algebra and $PS(\mathcal{A})$ be the set of its pure states.

Question 1. Is $PS(\mathcal{A})$ linearly independent (as vectors over $\mathbb{R}$)? (If $\mathcal{A}$ is commutative, $PS(\mathcal{A})$ is easily seen to be linearly independent (I learned this from Bernard Russo).)

Question 2. If $\mathcal{A}$ is unital, is $PS(\mathcal{A})$ always compact in the weak* topology? If not, is there any topology that makes $PS(\mathcal{A})$ compact?

• Where is the commutativity of $\mathcal{A}$ used in the proof you already know? – Nate Eldredge Jul 24 '16 at 13:40
• In what sense should $PS(\mathcal{A})$ should be "linearly indpendent"? It's not even a linear space. For example, for $\mathcal{A} = M_2(\mathbb{C})$, the pure states constitute a 2-sphere (the Bloch sphere). – Igor Khavkine Jul 24 '16 at 13:44
• @IgorKhavkine: I assume we mean "linearly independent as a subset of the dual space $\mathcal{A}^*$." So your example seems to show that this is false. – Nate Eldredge Jul 24 '16 at 14:46
• Thank you for your comments and a partial answer, Igor and Nate. In the commutative case, the set of pure states is the character space (= maximal ideal space) $\Omega(\mathcal{A})$ which is a locally compact Hausdorff space in the weak* topology, and $\mathcal{A}\cong C_0(\Omega(\mathcal{A}))$ $*$-isomorphically. Let $\lambda_1\tau_1+\cdots+\lambda_n\tau_n=0$, where $\lambda_1,\dots,\lambda_n\in\mathbb{C}$ and $\tau_1,\dots,\tau_n\in\Omega(\mathcal{A})$. (Continued below) – Masayoshi Kaneda Jul 24 '16 at 18:53
• (Continuation) For any $i\in\{1,\dots,n\}$, by the complete regularity of $\Omega(\mathcal{A})$, $\exists\hat{a}\in C_0(\Omega(\mathcal{A}))$ such that $\hat{a}(\tau_i)=1$ and $\hat{a}(\tau_j)=0$ for $j\ne i$. Thus $0=(\lambda_1\tau_1+\cdots+\lambda_n\tau_n)(a)=\lambda_1\tau_1(a)+\cdots+\lambda_n\tau_n(a)=\lambda_1\hat{a}(\tau_1)+\cdots+\lambda_n\hat{a}(\tau_n)=\lambda_i$. – Masayoshi Kaneda Jul 24 '16 at 18:54

As pointed out in the comments, the answer to question 1 is "no". More explicitly, if $v$ is any unit vector in $\mathbb{C}^2$ then the map $A \mapsto \langle Av,v\rangle$ is a pure state on $M_2$, and the pure states arising from two vectors are distinct unless one vector is a scalar (of modulus 1) multiple of the other. Find five unit vectors, no two of which are scalar multiples of each other; the resulting pure states are distinct but cannot be linearly independent, since they all live in $M_2^*$ which is four dimensional.

The answer to question 2 is also "no". I think of compactness of the pure states as a commutative/finite dimensional phenomenon. A famous theorem of Glimm states that if $A$ is any C*-algebra which has an irreducible representation that contains no compact operators, then the pure states of $A$ are weak* dense in the state space of $A$. (Believe it or not.) There's no obvious topology that would make the set of pure states compact.