# product states on the tensor product *-algebra

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 $$A⊗B$$?

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 $$\psi(\eta_{\varphi}(x)) = (\varphi \otimes \psi)(x)$$.
• better to read the details of my question: $x\neq0$ implies $\omega\times\varphi(x^*x)>0$, this is usually meant for "separate points" in operator algebras. By the way, I asked for that and the answer didn't respond about that question. – francesco fidaleo Oct 8 '18 at 20:26