product states on the tensor product *-algebra

Question

Let A and B be two unital $C^∗$-algebras, and $x∈A⊗B$ (the algebraic tensor product $*$-algebra), different from $0$. Is there states $ω_x∈A^∗_+$ and $φ_x∈B^∗_+$ such that, for the product state $ω_x×φ_x$, $(ω_x×φ_x)(x^∗x)>0$? In other words, do the product states separate the points of the tensor product $A⊗B$?

Answer 1

Yes. Lets write $$ 0 \neq x = \sum_{i=1}^n a_i \otimes b_i, $$ so that the $b_i$ are linearly independent. For the moment, fix a state $\varphi$ on $A$. This gives rise to the so-called slice map $$ \eta_\varphi \colon A \otimes B \to B : a \otimes b \mapsto \varphi(a)b. $$ For this $\varphi$, we get $$ \eta_\varphi(x) = \sum_{i=1}^n\varphi(a_i)b_i. $$ Clearly, one can choose $\varphi$ so that $\eta_\varphi(x) \neq 0$, since otherwise, by the linear independence of the $b_i$, $\varphi(a_i) = 0$, for any state $\varphi$ on $A$ and $i \in \{1,2,\cdots,n\}$. That would imply that $a_1=a_2=\cdots=a_n=0$ and therefore $x= 0$. Now, we may take a state $\psi$ on $B$ such that $$ \psi(\eta_{\varphi}(x)) > 0. $$ But $\psi(\eta_{\varphi}(x)) = (\varphi \otimes \psi)(x)$.