The state space of the stabilization of a C*-algebra - MathOverflow

The state space of the stabilization of a C*-algebra

2010-01-29T21:09:47Z

Given a $C^*$-algebra $A$, I wonder up to what extent we can describe the state space of the stabilization $A\otimes K$ of $A$ in terms of the state space of $A$. Of course, the "tensor-product" states are the most obvious ones in general. But this seems to be far from the whole state space of the stabilization of $A$.

Answer by Kamran Reihani for The state space of the stabilization of a C*-algebra

2010-02-03T01:42:44Z

Thanks, Yemon! I haven't found exactly what I am looking for in the papers you mentioned or some others yet. Part of the problem is how often we can find a separable state on the stabilization of $A$. Here, a separable state means one which can be expressed as a convex combination of the tensor-product states on $A\otimes{\cal K}$. Any other state is called entangled. (Apparently, this terminology comes from Quantum Information Theory.) The study of separable states is still under progress for the tensor product of matrix algebras (cf. [1]).

[1]: E. Alfsen F. Shultz, Unique decompositions, faces, and automorphisms of separable states, http://arxiv.org/abs/0906.1761v3.

Answer by Kamran Reihani for The state space of the stabilization of a C*-algebra

2010-02-03T05:25:33Z

From the general point of view of $C^*$-algebra theory, Bill Paschke mentioned to me today an interesting classification of states (or even, positive functionals) on the stabilization of $A$ as follows:

Proposition. Let $A$ be unital $C^{\ast}$-algebra, $H$ be a Hilbert space and $\varphi$ be a positive linear functional on $A\otimes\mathcal{K}(H)$. Let ${\cal L}^1(H)$ denote the ideal of trace-class operators on $H$. Then there exist a Hilbert space $H_{\varphi}$, a representation $\pi:A\rightarrow B(H_\varphi)$ and an operator $S:H\rightarrow H_{\varphi}$ such that $S^*S\in{\cal L}^1(H)$ (so $S^{\ast}\pi(a)\,S\in\mathcal{L}^1(H)$ for $a\in A$), and $\varphi(a\otimes K)={\rm tr}(S^{\ast}\pi(a)\,S\,K)$ for all $a\in A$ and $K\in\mathcal{K}(H)$. Moreover, $\varphi$ is a state iff ${\rm tr}(S^{\ast}S)=1$.

Proof. Since ${\cal K}(H)^{\ast}\cong{\cal L}^1(H)$, any positive linear functional $\varphi$ can be regarded as a completely positive map $T:A\rightarrow{\cal L}^1(H)$ with $\varphi(a\otimes K)={\rm tr}(T(a)K)$. Define a sesquilinear form on the algebraic tensor product $A\odot H$ by $$\langle a\otimes\xi,b\otimes\eta\rangle_{\varphi}:=\langle T(b^{\ast}a)\xi,\eta\rangle_{H},$$ set $N_{\varphi}={\rm span}\{a\otimes\xi\in A\odot H\mid\langle T(a^{\ast}a)\xi,\xi\rangle_{H}=0\}$ and $H_{\varphi}:=\overline{A\odot H/N_{\varphi}}$. Then we can define a representation $\pi:A\rightarrow B(H_\varphi)$ by $$\pi(a)(b\otimes\eta+N_{\varphi}):=ab\otimes\eta+N_{\varphi}.$$ Now, let $S:H\rightarrow H_{\varphi}$ be the operator defined by $$S\xi:=1\otimes\xi+N_{\varphi}.$$ Then $S^{\ast}:H_{\varphi}\rightarrow H$ is given by $$S^{\ast}(b\otimes\eta+N_{\varphi})=T(b)\eta,$$ and we have $S^{\ast}S=T(1)\in{\cal L}^1(H)$. Moreover, $S^{\ast}\pi(a)\,S=T(a)$ , which is what we need. It is easy to see that $||\varphi||={\rm tr}(S^{\ast}S)$. Q.E.D