Proof of A Trace-Preserving Unital -Homomorphism between Tracial W-Probability Spaces is Implemented by an Isometry

Each result cited below is universally quantified over the data in its own statement.

Both $(H_{0},M_{0},\Omega_{0})$ and $(H,M,\Omega)$ are cyclic tracial operator algebras by Tracial W*-Probability Spaces §space, so The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra applies to each, and $J_{0}$, $J$ are conjugations with $J_{0}(S\Omega_{0})=S^{*}\Omega_{0}$ for $S\in M_{0}$ and $J(A\Omega)=A^{*}\Omega$ for $A\in M$ by The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §conjugation. By Tracial W*-Probability Spaces §trace and Cyclic Tracial Operator Algebras and Their Traces §trace, $\tau_{0}(S)=\langle\Omega_{0},S\Omega_{0}\rangle$ and $\tau(A)=\langle\Omega,A\Omega\rangle$. Adjoints are used in the form $\langle T^{*}w,v\rangle=\langle w,Tv\rangle$ of Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint, equivalently (by conjugate symmetry) $\langle v,T^{*}w\rangle=\langle Tv,w\rangle$; every bounded linear map between Hilbert spaces has a bounded adjoint by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint, and $(T^{*})^{*}=T$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus. The maps $\pi(S)-\pi(T)$ and $\pi(S-T)$ agree, since $\pi(S+(-1)T)=\pi(S)+(-1)\pi(T)$.

Claim 1 (Isometry). Let $D=M_{0}\Omega_{0}=\{S\Omega_{0}:\ S\in M_{0}\}$. Since $S\Omega_{0}+T\Omega_{0}=(S+T)\Omega_{0}$, $c(S\Omega_{0})=(cS)\Omega_{0}$ and $0=(0\cdot I)\Omega_{0}$, with $S+T$, $cS$, $0\cdot I\in M_{0}$ by Cyclic Tracial Operator Algebras and Their Traces §star-algebra, $D$ is a linear subspace of $H_{0}$, and it is dense by Cyclic Tracial Operator Algebras and Their Traces §cyclic.

For $R\in M_{0}$ we have $\pi(R)\in M$, and by The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §trace applied to $(H,M,\Omega)$ and to $(H_{0},M_{0},\Omega_{0})$,

If $S\Omega_{0}=T\Omega_{0}$ for $S,T\in M_{0}$, then with $R=S-T$ this gives $\lVert\pi(S)\Omega-\pi(T)\Omega\rVert=\lVert\pi(R)\Omega\rVert=\lVert R\Omega_{0}\rVert=0$, so $\pi(S)\Omega=\pi(T)\Omega$. Hence $V_{0}:D\to H$, $V_{0}(S\Omega_{0})=\pi(S)\Omega$, is well defined, and $\lVert V_{0}\xi\rVert=\lVert\xi\rVert$ for $\xi\in D$. It is linear: $V_{0}(S\Omega_{0}+T\Omega_{0})=\pi(S+T)\Omega=\pi(S)\Omega+\pi(T)\Omega$ and $V_{0}(c\,S\Omega_{0})=\pi(cS)\Omega=c\,\pi(S)\Omega$. By Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §linear (with $C=1$) there is exactly one $V\in\mathcal{L}(H_{0},H)$ extending $V_{0}$, and $\lVert V\xi\rVert=\lVert\xi\rVert$ for every $\xi\in H_{0}$. Thus $V$ satisfies $VS\Omega_{0}=\pi(S)\Omega$ for all $S\in M_{0}$; any $V'\in\mathcal{L}(H_{0},H)$ with this property agrees with $V$ on $D$, so $V'=V$ by Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality. Moreover $V\Omega_{0}=VI\Omega_{0}=\pi(I)\Omega=\Omega$.

For $S,T\in M_{0}$, using that $\pi(S)^{*}$ and $S^{*}$ are the adjoints of $\pi(S)$ and $S$,

So $\langle\eta,V^{*}V\xi\rangle=\langle\eta,I_{H_{0}}\xi\rangle$ for all $\xi,\eta\in D$; since $V^{*}V,I_{H_{0}}\in\mathcal{L}(H_{0})$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations, Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality (with $E=D$) gives $V^{*}V=I_{H_{0}}$.

Claim 2 (Intertwining). Let $S\in M_{0}$. The maps $\pi(S)V$ and $VS$ belong to $\mathcal{L}(H_{0},H)$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations, and for $T\in M_{0}$, since $ST\in M_{0}$,

So they agree on $D$, and $\pi(S)V=VS$ by Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality. Applying this to $S^{*}\in M_{0}$ gives $\pi(S)^{*}V=\pi(S^{*})V=VS^{*}$; taking adjoints with Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $(\pi(S)^{*}V)^{*}=V^{*}\pi(S)$ and $(VS^{*})^{*}=SV^{*}$, so $V^{*}\pi(S)=SV^{*}$.

Next, $VJ_{0}=JV$. For $x,y\in H$ one has $\lVert Jx\rVert^{2}=\langle Jx,Jx\rangle=\langle x,x\rangle=\lVert x\rVert^{2}$ and $Jx-Jy=J(x-y)$, by Conjugation of a Complex Hilbert Space §conjugation (additivity and $J((-1)y)=-Jy$); likewise for $J_{0}$. Hence for $\xi,\eta\in H_{0}$, $\lVert JV\xi-JV\eta\rVert=\lVert V(\xi-\eta)\rVert=\lVert\xi-\eta\rVert$ and $\lVert VJ_{0}\xi-VJ_{0}\eta\rVert=\lVert J_{0}(\xi-\eta)\rVert=\lVert\xi-\eta\rVert$, so $JV$ and $VJ_{0}$ are continuous maps $H_{0}\to H$. For $S\in M_{0}$,

Let $S_{D}:D\to H$ be the restriction of $JV$ to $D$. It is additive, satisfies $S_{D}(c\xi)=\overline{c}\,S_{D}\xi$ (as $V$ is linear and $J$ conjugate-linear) and $\lVert S_{D}\xi\rVert=\lVert\xi\rVert$. By Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §conjugate-linear (with $C=1$) there is exactly one continuous map $H_{0}\to H$ extending $S_{D}$; both $JV$ and $VJ_{0}$ are such maps by the display, so $VJ_{0}=JV$.

Finally, $V^{*}J=J_{0}V^{*}$. Let $\xi\in H_{0}$ and $\eta\in H$. Using $\langle Jx,Jy\rangle=\langle y,x\rangle$ and $JJx=x$ (and the same for $J_{0}$),

Taking $\xi=V^{*}J\eta-J_{0}V^{*}\eta$ gives $\langle\xi,\xi\rangle=0$, so $V^{*}J\eta=J_{0}V^{*}\eta$ by claim 4 of Elementary Properties of a Complex Inner Product (definiteness).

Claim 3 (Injectivity). Let $S\in M_{0}$ with $\pi(S)=0$. Then $VS\Omega_{0}=\pi(S)\Omega=0$, so $\lVert S\Omega_{0}\rVert=\lVert VS\Omega_{0}\rVert=0$ by Claim 1, and $S=0$ by The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §separating applied to $(H_{0},M_{0},\Omega_{0})$. If $\pi(S)=\pi(T)$ for $S,T\in M_{0}$, then $\pi(S-T)=\pi(S)-\pi(T)=0$, so $S-T=0$ and $S=T$; thus $\pi$ is injective. For $S\in M_{0}$ and $\xi\in H_{0}$, by Claims 1 and 2 and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound,

So $\lVert\pi(S)\rVert_{\mathrm{op}}$ is a bound for $S$, and $\lVert S\rVert_{\mathrm{op}}\le\lVert\pi(S)\rVert_{\mathrm{op}}$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound.