Proof of The Amalgamated Free Product of Two Noncommutative Laws over a Common Marginal: Representations, Amalgamation, Freeness and the Tracial Vacuum State

Preliminaries. Throughout, $\varepsilon\in\{1,2\}$; $\langle\cdot,\cdot\rangle$ and $\lVert\cdot\rVert$ are the inner product and norm of $\mathcal{H}$, which is a complex Hilbert space by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §hilbert, and $J:\mathcal{F}\to\mathcal{H}$ is the canonical map of The Amalgamated Free Product Space of Two Noncommutative Laws over a Common Marginal, Its Vacuum Vector and Vacuum State §space. 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, sums, scalar multiples and composites of elements of $\mathcal{L}(\mathcal{H})$ lie in $\mathcal{L}(\mathcal{H})$, which is a complex vector space in which composition distributes over sums; composition of maps is associative, so finite products of operators are written left to right without brackets. By The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots §star-algebra, $N$ and $A_{\varepsilon}$ contain $I$ and are closed under sums, scalar multiples, products and adjoints; such memberships are used without comment. 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, for linear maps that have adjoints, $(T^{*})^{*}=T$ (i.e. $T$ is the adjoint of $T^{*}$), $I^{*}=I$, $(cT)^{*}=\overline{c}\,T^{*}$ (so $0^{*}=(0\,I)^{*}=0$) and $(RT)^{*}=T^{*}R^{*}$.

(P1) Embeddings and centred parts. By Marginals of a Noncommutative Law: the Isometry of GNS Spaces, the Trace-Preserving Embedding of Tracial Algebras and the Conditional Expectation §homomorphism, $\pi_{\varepsilon}$ is linear (so $\pi_{\varepsilon}(0)=0$), $\pi_{\varepsilon}(I)=I$, $\pi_{\varepsilon}(xy)=\pi_{\varepsilon}(x)\pi_{\varepsilon}(y)$ and $\tau_{\gamma_{\varepsilon}}(\pi_{\varepsilon}(x))=\tau_{\mu}(x)$ for $x,y\in N$. By Marginals of a Noncommutative Law: the Isometry of GNS Spaces, the Trace-Preserving Embedding of Tracial Algebras and the Conditional Expectation §expectation, $E_{\varepsilon}$ is linear, $E_{\varepsilon}(b^{*})=E_{\varepsilon}(b)^{*}$, $E_{\varepsilon}(\pi_{\varepsilon}(x))=x$ and $\tau_{\mu}(E_{\varepsilon}(b))=\tau_{\gamma_{\varepsilon}}(b)$ for $b\in A_{\varepsilon}$ and $x\in N$. With the centred parts of The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §centred this gives: for $c\in A_{\varepsilon}$, $c^{\circ}\in A_{\varepsilon}^{\circ}$ and $c=c^{\circ}+\pi_{\varepsilon}(E_{\varepsilon}(c))$; for $a\in A_{\varepsilon}^{\circ}$, $a^{\circ}=a-\pi_{\varepsilon}(0)=a$, and $a^{*}\in A_{\varepsilon}^{\circ}$ because $E_{\varepsilon}(a^{*})=E_{\varepsilon}(a)^{*}=0^{*}=0$; for $a\in A_{\varepsilon}^{\circ}$ and $y\in N$, $\pi_{\varepsilon}(y)\,a$ and $a\,\pi_{\varepsilon}(y)$ lie in $A_{\varepsilon}^{\circ}$ (same clause); and for $y\in N$, $\pi_{\varepsilon}(y)^{\circ}=\pi_{\varepsilon}(y)-\pi_{\varepsilon}(y)=0$.

(P2) The canonical map. $J$ is complex-linear with $\langle J\xi,J\eta\rangle=h(\xi,\eta)$ for $\xi,\eta\in\mathcal{F}$ by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §isometry, and $h(\delta_{s},\delta_{t})=h_{0}(s,t)$ by The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §form; hence $\langle J\delta_{s},J\delta_{t}\rangle=h_{0}(s,t)$ for $s,t\in\mathcal{T}$. By The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §actions and The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §actions, for $b\in A_{\varepsilon}$ and $u\in\mathcal{T}$,

By The Amalgamated Free Product Space of Two Noncommutative Laws over a Common Marginal, Its Vacuum Vector and Vacuum State §vectors, $\Omega=J\delta^{N}_{I}$, $\Xi_{N}(x)=J\delta^{N}_{x}$ and $\Xi(t)=J\delta_{t}$. By The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §multilinear, $\Xi_{N}(0)=\Xi_{N}(0\cdot I)=0\cdot\Xi_{N}(I)=0$, and $\Xi(a_{1},\dots,a_{k})=0$ whenever some entry $a_{j}$ is $0$.

(P3) Density. For every $w\in\mathcal{H}$ and real $r>0$ there is $\eta\in\mathcal{F}$ with $\lVert w-J\eta\rVert<r$. Indeed $J(\mathcal{F})$ is dense in $\mathcal{H}$ by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §dense, i.e. its closure is $\mathcal{H}$ (Dense Subset of a Topological Space) for the topology of the metric $(u,u')\mapsto\lVert u-u'\rVert$ (Complex Hilbert Spaces and Bounded Linear Maps: Standing Notation §spaces), and claim 3 of Characterization of the Closure in a Metric Space by Open Balls provides an element $J\eta$ of $J(\mathcal{F})$ at distance less than $r$ from $w$.

(P4) Uniqueness. If $S,S'\in\mathcal{L}(\mathcal{H})$ and $SJ\xi=S'J\xi$ for every $\xi\in\mathcal{F}$, then $S=S'$. Indeed, $R=S-S'\in\mathcal{L}(\mathcal{H})$ satisfies $RJ\xi=0$ for every $\xi$. Let $w\in\mathcal{H}$. For real $r>0$ choose $\eta$ by (P3); since $R$ is linear and $RJ\eta=0$, and $\lVert R\rVert_{\mathrm{op}}\ge0$ is a bound for $R$ (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),

If $Rw\neq0$, then the nonnegative real number $\lVert Rw\rVert^{2}=\langle Rw,Rw\rangle$ is nonzero by claim 4 of Elementary Properties of a Complex Inner Product, so $\lVert Rw\rVert>0$, and $r=\lVert Rw\rVert/(\lVert R\rVert_{\mathrm{op}}+1)$ gives $\lVert Rw\rVert\le\lVert R\rVert_{\mathrm{op}}\lVert Rw\rVert/(\lVert R\rVert_{\mathrm{op}}+1)<\lVert Rw\rVert$, which is impossible. So $Rw=0$, i.e. $Sw=S'w$.

(P5) Separation. If $w\in\mathcal{H}$ and $\langle J\eta,w\rangle=0$ for every $\eta\in\mathcal{F}$, then $w=0$. Indeed, for real $r>0$ choose $\eta$ by (P3); by claim 1 of Elementary Properties of a Complex Inner Product, $\langle w,w\rangle=\langle w-J\eta,w\rangle+\langle J\eta,w\rangle=\langle w-J\eta,w\rangle$, so Cauchy-Schwarz Inequality in a Complex Inner Product Space gives

If $w\neq0$, then $\lVert w\rVert>0$ (claim 4 of Elementary Properties of a Complex Inner Product) and $r=\lVert w\rVert/2$ gives $\lVert w\rVert^{2}\le\lVert w\rVert^{2}/2$, which is impossible.

Claim 1 (Representations). Let $b,b'\in A_{\varepsilon}$, $c\in\mathbb{C}$ and $\xi\in\mathcal{F}$. By The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §actions and The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §relations,

All operators involved lie in $\mathcal{L}(\mathcal{H})$, so (P4) gives $\Lambda_{\varepsilon}(b+b')=\Lambda_{\varepsilon}(b)+\Lambda_{\varepsilon}(b')$, $\Lambda_{\varepsilon}(cb)=c\,\Lambda_{\varepsilon}(b)$, $\Lambda_{\varepsilon}(I)=I$ and $\Lambda_{\varepsilon}(bb')=\Lambda_{\varepsilon}(b)\Lambda_{\varepsilon}(b')$. 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, $\Lambda_{\varepsilon}(b)$ has an adjoint $\Lambda_{\varepsilon}(b)^{*}\in\mathcal{L}(\mathcal{H})$, and $\Lambda_{\varepsilon}(b)$ is the adjoint of $\Lambda_{\varepsilon}(b)^{*}$, i.e. $\langle\Lambda_{\varepsilon}(b)v,v'\rangle=\langle v,\Lambda_{\varepsilon}(b)^{*}v'\rangle$ for $v,v'\in\mathcal{H}$. For $\xi,\eta\in\mathcal{F}$, by (P2), The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §actions and the adjoint relation of The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §actions,

Since the inner product is linear in its second argument, $w=\Lambda_{\varepsilon}(b^{*})J\xi-\Lambda_{\varepsilon}(b)^{*}J\xi$ satisfies $\langle J\eta,w\rangle=0$ for all $\eta$, so $w=0$ by (P5); as $\xi$ was arbitrary, (P4) gives $\Lambda_{\varepsilon}(b^{*})=\Lambda_{\varepsilon}(b)^{*}$. The bound $\lVert\Lambda_{\varepsilon}(b)\rVert_{\mathrm{op}}\le\lVert b\rVert_{\mathrm{op}}$ is part of The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §actions.

Claim 2 (Amalgamation). For $x\in N$ and $\xi\in\mathcal{F}$, The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §actions and the last relation of The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras §relations give $\Lambda_{1}(\pi_{1}(x))J\xi=J\ell_{1}(\pi_{1}(x))\xi=J\ell_{2}(\pi_{2}(x))\xi=\Lambda_{2}(\pi_{2}(x))J\xi$, so $\Lambda_{1}(\pi_{1}(x))=\Lambda_{2}(\pi_{2}(x))$ by (P4).

Claim 3 (Freeness). Let $x\in N$. By (1) and case (a) of The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §actions with $b=\pi_{\varepsilon}(x)$ and $u=(0,I)$, using $\pi_{\varepsilon}(x)\pi_{\varepsilon}(I)=\pi_{\varepsilon}(x)$, $\pi_{\varepsilon}(x)^{\circ}=0$, $E_{\varepsilon}(\pi_{\varepsilon}(x))=x$ (P1), the linearity of $J$ and (P2),

Next let $\mathcal{K}$ be the set of $k\in\mathbb{N}$ such that $\Lambda_{e_{1}}(a_{1})\cdots\Lambda_{e_{k}}(a_{k})\Omega=\Xi(a_{1},\dots,a_{k})$ for every alternating tuple $(a_{1},\dots,a_{k})$ of length $k$ and type $(e_{1},\dots,e_{k})$; we show $\mathcal{K}=\mathbb{N}$ by Principle of Induction for the Natural Numbers. For $k=1$: by (1) and case (a) with $b=a_{1}$, $u=(0,I)$, using $a_{1}\pi_{e_{1}}(I)=a_{1}$, $a_{1}^{\circ}=a_{1}$ and $E_{e_{1}}(a_{1})=0$,

so $1\in\mathcal{K}$. Let $k\in\mathcal{K}$ and let $(a_{1},\dots,a_{k+1})$ be alternating of type $(e_{1},\dots,e_{k+1})$. Then $t'=(a_{2},\dots,a_{k+1})$ is alternating of length $k$ and type $(e_{2},\dots,e_{k+1})$, with $e_{2}=\bar{e}_{1}$ since $e_{2}\neq e_{1}$. By $k\in\mathcal{K}$, (1) and case (b) of The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §actions with $\varepsilon=e_{1}$ and $b=a_{1}$, using $a_{1}^{\circ}=a_{1}$ and $\pi_{e_{2}}(E_{e_{1}}(a_{1}))a_{2}=\pi_{e_{2}}(0)a_{2}=0$,

where for $k=1$ the middle tuple is $(0)$; the last step is (P2). So $k+1\in\mathcal{K}$, and $\mathcal{K}=\mathbb{N}$. Finally, for an alternating tuple $t$ of length $k$, (P2) and The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §form give $\langle\Omega,\Xi(t)\rangle=h_{0}((0,I),(k,t))=0$, since this pair is not of either of the two special forms.

Claim 4 (Marginals). By (P2), The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §form and The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots §trace, $\lVert\Omega\rVert^{2}=\langle\Omega,\Omega\rangle=h_{0}((0,I),(0,I))=\tau_{\mu}(I^{*}I)=\tau_{\mu}(I)=1$, and $\lVert\Omega\rVert\ge0$, so $\lVert\Omega\rVert=1$. Let $b\in A_{\varepsilon}$. By (1) and case (a) with $u=(0,I)$, using $b\pi_{\varepsilon}(I)=b$,

by The Amalgamated Free Product Space of Two Noncommutative Laws over a Common Marginal, Its Vacuum Vector and Vacuum State §state, Claim 3 (applied to the alternating tuple $((\varepsilon,b^{\circ}))$), (P2), The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §form and (P1).

Claim 5 (Traciality). Notation. For $y\in N$ put $\Pi(y)=\Lambda_{1}(\pi_{1}(y))$, which equals $\Lambda_{\varepsilon}(\pi_{\varepsilon}(y))$ for both $\varepsilon$ by Claim 2. By Claim 1 and (P1), for $y,y'\in N$ and $c\in A_{\varepsilon}$,

For an alternating tuple $t=(a_{1},\dots,a_{k})$ of type $(e_{1},\dots,e_{k})$ put $\Lambda(t)=\Lambda_{e_{1}}(a_{1})\cdots\Lambda_{e_{k}}(a_{k})$. By Claim 3, $\Lambda(t)\Omega=\Xi(t)$ and $\varphi(\Lambda(t))=\langle\Omega,\Xi(t)\rangle=0$. Let $W\subseteq\mathcal{L}(\mathcal{H})$ consist of the operators $\Pi(y)$, $y\in N$, and $\Lambda(t)$, $t$ an alternating tuple. Let $\mathcal{V}$ be the intersection of all subsets $U\subseteq\mathcal{L}(\mathcal{H})$ with $W\subseteq U$ and $P+Q\in U$ for all $P,Q\in U$ (the family is nonempty, as it contains $\mathcal{L}(\mathcal{H})$). Then $W\subseteq\mathcal{V}$, $\mathcal{V}$ is closed under sums, and $\mathcal{V}\subseteq U$ for every $U$ in the family; in particular, a subset of $\mathcal{V}$ that contains $W$ and is closed under sums equals $\mathcal{V}$. For $c\in A_{\varepsilon}$, (P1), Claim 1 and Claim 2 give

since $\Lambda_{\varepsilon}(c^{\circ})=\Lambda(((\varepsilon,c^{\circ})))\in W$ and $\Pi(E_{\varepsilon}(c))\in W$.

Step A (left multiplication). Let $\mathcal{A}$ be the set of $P\in\mathcal{V}$ with $\Lambda_{f}(c)P\in\mathcal{V}$ for every $f\in\{1,2\}$ and $c\in A_{f}$. If $P,Q\in\mathcal{A}$ then $\Lambda_{f}(c)(P+Q)=\Lambda_{f}(c)P+\Lambda_{f}(c)Q\in\mathcal{V}$, so $\mathcal{A}$ is closed under sums. We show $W\subseteq\mathcal{A}$; then $\mathcal{A}=\mathcal{V}$. Fix $f$ and $c\in A_{f}$. For $w=\Pi(y)$: $\Lambda_{f}(c)w=\Lambda_{f}(c\,\pi_{f}(y))\in\mathcal{V}$ by (2) and (3). For $w=\Lambda(t)$, $t=(a_{1},\dots,a_{k})$ of type $(e_{1},\dots,e_{k})$: if $f\neq e_{1}$, then by (3) (with $f$ in place of $\varepsilon$), distributivity and (2),

(for $k=1$ the second term is $\Lambda_{e_{1}}(\pi_{e_{1}}(E_{f}(c))a_{1})$); the first term is $\Lambda(c^{\circ},a_{1},\dots,a_{k})$, the tuple being alternating as $c^{\circ}\in A_{f}^{\circ}$ and $f\neq e_{1}$, and the second is $\Lambda(\pi_{e_{1}}(E_{f}(c))a_{1},a_{2},\dots,a_{k})$, alternating by (P1). Both lie in $W$, so $\Lambda_{f}(c)w\in\mathcal{V}$. If $f=e_{1}$, put $y=E_{f}(ca_{1})$; by Claim 1 and (3), $\Lambda_{f}(c)\Lambda_{f}(a_{1})=\Lambda_{f}(ca_{1})=\Lambda_{f}((ca_{1})^{\circ})+\Pi(y)$. For $k=1$ this is $\Lambda(((f,(ca_{1})^{\circ})))+\Pi(y)\in\mathcal{V}$. For $k\ge2$, distributivity and (2) give

(for $k=2$ the second term is $\Lambda_{e_{2}}(\pi_{e_{2}}(y)a_{2})$), which is $\Lambda((ca_{1})^{\circ},a_{2},\dots,a_{k})+\Lambda(\pi_{e_{2}}(y)a_{2},a_{3},\dots,a_{k})$, both tuples alternating since $e_{2}\neq e_{1}=f$ and by (P1). So $\Lambda_{f}(c)w\in\mathcal{V}$ in all cases.

Step B (products). Let $\mathcal{R}$ be the set of $r\in\mathbb{N}$ such that $\Lambda_{f_{1}}(c_{1})\cdots\Lambda_{f_{r}}(c_{r})\in\mathcal{V}$ for all $f_{1},\dots,f_{r}\in\{1,2\}$ and $c_{j}\in A_{f_{j}}$. Then $1\in\mathcal{R}$ by (3), and if $r\in\mathcal{R}$, a product of $r+1$ factors is $\Lambda_{f_{1}}(c_{1})P$ with $P=\Lambda_{f_{2}}(c_{2})\cdots\Lambda_{f_{r+1}}(c_{r+1})\in\mathcal{V}$, hence lies in $\mathcal{V}$ by Step A. So $\mathcal{R}=\mathbb{N}$ by Principle of Induction for the Natural Numbers.

Step C (traciality on $W$). For $S,T\in\mathcal{L}(\mathcal{H})$, the defining relation of the adjoint $S^{*}$ (which exists 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) gives $\varphi(ST)=\langle\Omega,S(T\Omega)\rangle=\langle S^{*}\Omega,T\Omega\rangle$. Also $(S_{1}\cdots S_{k})^{*}=S_{k}^{*}\cdots S_{1}^{*}$ for $k\in\mathbb{N}$ and $S_{1},\dots,S_{k}\in\mathcal{L}(\mathcal{H})$: the set of $k$ for which this holds for all such $S_{j}$ contains $1$, and contains $k+1$ whenever it contains $k$ because $(S_{1}S_{2}\cdots S_{k+1})^{*}=(S_{2}\cdots S_{k+1})^{*}S_{1}^{*}$; so it is $\mathbb{N}$ by Principle of Induction for the Natural Numbers. We show $\varphi(ww')=\varphi(w'w)$ for all $w,w'\in W$; the statement is symmetric in $w,w'$, so three cases suffice.

(i) $w=\Pi(x)$, $w'=\Pi(y)$. By (2), Claim 4 and (P1), $\varphi(ww')=\varphi(\Lambda_{1}(\pi_{1}(xy)))=\tau_{\gamma_{1}}(\pi_{1}(xy))=\tau_{\mu}(xy)$, likewise $\varphi(w'w)=\tau_{\mu}(yx)$, and $\tau_{\mu}(xy)=\tau_{\mu}(yx)$ by The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots §trace.

(ii) $w=\Pi(x)$, $w'=\Lambda(t)$, $t=(a_{1},\dots,a_{k})$ of type $(e_{1},\dots,e_{k})$. By (2), $ww'=\Lambda(\pi_{e_{1}}(x)a_{1},a_{2},\dots,a_{k})$ and $w'w=\Lambda(a_{1},\dots,a_{k-1},a_{k}\pi_{e_{k}}(x))$, both tuples being alternating by (P1). Hence $\varphi(ww')=0=\varphi(w'w)$ by Claim 3.

(iii) $w=\Lambda(s)$, $s=(a_{1},\dots,a_{k})$ of type $(e_{1},\dots,e_{k})$, and $w'=\Lambda(t)$, $t=(b_{1},\dots,b_{l})$ of type $(f_{1},\dots,f_{l})$. By the product rule for adjoints and Claim 1, $w^{*}=\Lambda_{e_{k}}(a_{k}^{*})\cdots\Lambda_{e_{1}}(a_{1}^{*})=\Lambda(s^{*})$, where $s^{*}=(a_{k}^{*},\dots,a_{1}^{*})$ is alternating of type $(e_{k},\dots,e_{1})$ by (P1); similarly $w'^{*}=\Lambda(t^{*})$ with $t^{*}=(b_{l}^{*},\dots,b_{1}^{*})$ of type $(f_{l},\dots,f_{1})$. By Claim 3 and (P2),

By The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §form, the first is $0$ unless $l=k$ and $f_{i}=e_{k+1-i}$ for all $i\in[k]$, and the second is $0$ unless $l=k$ and $e_{i}=f_{k+1-i}$ for all $i\in[k]$; since $i\mapsto k+1-i$ is a bijection of $[k]$, these conditions coincide. If they fail, both sides are $0$. If they hold, then by The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §nested, The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §form and $(T^{*})^{*}=T$, $\varphi(ww')=\tau_{\mu}(X_{k})$ and $\varphi(w'w)=\tau_{\mu}(Y_{k})$, where

for $j\in[k]$, $j\ge2$. Apply Cyclic Symmetry of Nested Conditional Expectations over a Common Marginal §cyclic with the letters $f_{1},\dots,f_{k}$, $u_{j}=a_{k+1-j}\in A_{e_{k+1-j}}=A_{f_{j}}$, $v_{j}=b_{j}\in A_{f_{j}}$ and $x=y=I$. Since $\pi_{f_{1}}(I)=I$, $G_{1}=X_{1}$, and $G_{j}$, $X_{j}$ obey the same recursion; so the set of $j\in\mathbb{N}$ such that $j\le k$ implies $G_{j}=X_{j}$ contains $1$ and contains $j+1$ whenever it contains $j$, and by Principle of Induction for the Natural Numbers $G_{k}=X_{k}$. Likewise, as $f_{k+1-j}=e_{j}$, $v_{k+1-j}=b_{k+1-j}$ and $u_{k+1-j}=a_{j}$ for $j\in[k]$, we get $G_{1}'=E_{e_{1}}(b_{k}a_{1})=Y_{1}$ and the recursion of $G_{j}'$ is that of $Y_{j}$, so $G_{k}'=Y_{k}$ by the same induction. Hence $\tau_{\mu}(X_{k})=\tau_{\mu}(I\,G_{k})=\tau_{\mu}(G_{k}'\,I)=\tau_{\mu}(Y_{k})$.

Step D (conclusion). $\varphi$ is additive, since $\varphi(S+T)=\langle\Omega,S\Omega+T\Omega\rangle=\varphi(S)+\varphi(T)$ by linearity of the inner product in its second argument (Complex Hilbert Spaces and Bounded Linear Maps: Standing Notation §spaces). Let $\mathcal{U}$ be the set of $P\in\mathcal{V}$ with $\varphi(Pw')=\varphi(w'P)$ for all $w'\in W$. It contains $W$ by Step C, and for $P,Q\in\mathcal{U}$, distributivity and additivity give $\varphi((P+Q)w')=\varphi(Pw')+\varphi(Qw')=\varphi(w'P)+\varphi(w'Q)=\varphi(w'(P+Q))$; so $\mathcal{U}=\mathcal{V}$. Next let $\mathcal{U}'$ be the set of $Q\in\mathcal{V}$ with $\varphi(PQ)=\varphi(QP)$ for all $P\in\mathcal{V}$; it contains $W$ because $\mathcal{U}=\mathcal{V}$, and is closed under sums by the same computation; so $\mathcal{U}'=\mathcal{V}$. Now let $r\ge2$, $f_{j}$ and $c_{j}$ be as in the claim. By Step B, $P=\Lambda_{f_{1}}(c_{1})\in\mathcal{V}$ and $Q=\Lambda_{f_{2}}(c_{2})\cdots\Lambda_{f_{r}}(c_{r})\in\mathcal{V}$ (a product of $r-1\in\mathbb{N}$ factors), and $\varphi(PQ)=\varphi(QP)$ is the asserted identity.