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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 $\pi:A\rightarrow B(H)$ 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 $\pi:A\rightarrow B(H)$ 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

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

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 $\pi:A\rightarrow B(H)$ 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 $\varphi(a\otimes K)={\rm tr}(S^{\ast}\pi(a) SK)$ 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 $$\langle a\otimes\xi,b\otimes\eta\rangle_{\varphi}:=\langle T(b^{\ast}a)\xi,\eta\rangle_{H},$$ and set $N_{\varphi}={\rm span}${$a\otimes\xi\in A\odot H\mid\langle T(a^{\ast}a)\xi,\xi\rangle_{H}=0$}. Set $H_{\varphi}:=\overline{A\odot H/N_{\varphi}}$. Then we can define a representation $\pi:A\rightarrow B(H)$ 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

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 $\pi:A\rightarrow B(H)$ 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 $\pi:A\rightarrow B(H)$ 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

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 $\pi:A\rightarrow B(H)$ 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 $\varphi(a\otimes K)={\rm tr}(S^{\ast}\pi(a) SK)$ for all $a\in A$ and $K\in\mathcal{K}(H)$.

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 $\varphi(a\otimes K)={\rm tr}(T(a)K)$.

(to be continued)

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 $\pi:A\rightarrow B(H)$ 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 $\varphi(a\otimes K)={\rm tr}(S^{\ast}\pi(a) SK)$ 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 $$\langle a\otimes\xi,b\otimes\eta\rangle_{\varphi}:=\langle T(b^{\ast}a)\xi,\eta\rangle_{H},$$ and set $N_{\varphi}={\rm span}${$a\otimes\xi\in A\odot H\mid\langle T(a^{\ast}a)\xi,\xi\rangle_{H}=0$}. Set $H_{\varphi}:=\overline{A\odot H/N_{\varphi}}$. Then we can define a representation $\pi:A\rightarrow B(H)$ 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