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

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 $A$.

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I haven't got a reference to hand, but I think some of E.C. Lance's early papers on nuclear $C^*$-algebras have some discussion of this, because a key tool used in his proofs is a careful analysis of states on tensor products –  Yemon Choi Jan 29 '10 at 21:31
K are the compact operators? –  Martin Brandenburg Jan 30 '10 at 1:52
@Martin: yes, I think that's fairly standard. It's certainly the usual definition of the stabilization of a $C^*$-algebra. –  Yemon Choi Jan 30 '10 at 5:19
I fixed a typo and removed some pointless LaTeX in the title. –  Harry Gindi Mar 5 '10 at 5:28

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

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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.

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