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Jul 24, 2016 at 18:54 comment added Masayoshi Kaneda (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 $0=(\lambda_1\tau_1+\cdots+\lambda_n\tau_n)(a)=\lambda_1\tau_1(a)+\cdots+\lambda_n\tau_n(a)=\lambda_1\hat{a}(\tau_1)+\cdots+\lambda_n\hat{a}(\tau_n)=\lambda_i$.
Jul 24, 2016 at 18:53 comment added Masayoshi Kaneda 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)
Jul 24, 2016 at 17:55 vote accept Masayoshi Kaneda
Jul 24, 2016 at 15:47 answer added Nik Weaver timeline score: 10
Jul 24, 2016 at 14:46 comment added Nate Eldredge @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.
Jul 24, 2016 at 13:44 comment added Igor Khavkine 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).
Jul 24, 2016 at 13:40 comment added Nate Eldredge Where is the commutativity of $\mathcal{A}$ used in the proof you already know?
Jul 24, 2016 at 12:44 history asked Masayoshi Kaneda CC BY-SA 3.0