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?



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


