Proof of Cyclic Symmetry of Nested Conditional Expectations over a Common Marginal

Each result cited below is used for the data in its own statement. For $\varepsilon\in\{1,2\}$, the maps $\pi_{\varepsilon}$ and $E_{\varepsilon}$ are those of Two Noncommutative Laws with a Common Marginal: Standing Notation for Their Amalgamated Free Product §embeddings. By Marginals of a Noncommutative Law: the Isometry of GNS Spaces, the Trace-Preserving Embedding of Tracial Algebras and the Conditional Expectation §expectation, for $S,T\in N$ and $c\in A_{\varepsilon}$,

by Marginals of a Noncommutative Law: the Isometry of GNS Spaces, the Trace-Preserving Embedding of Tracial Algebras and the Conditional Expectation §homomorphism, $\pi_{\varepsilon}(I)=I$; 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 (for the law $\gamma_{\varepsilon}$), $\tau_{\gamma_{\varepsilon}}(cc')=\tau_{\gamma_{\varepsilon}}(c'c)$ for $c,c'\in A_{\varepsilon}$; and all products below lie in the algebra indicated 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. Taking $S=I$ or $T=I$ in the first identity gives $x\,E_{\varepsilon}(c)=E_{\varepsilon}(\pi_{\varepsilon}(x)\,c)$ and $E_{\varepsilon}(c)\,z=E_{\varepsilon}(c\,\pi_{\varepsilon}(z))$ for $x,z\in N$.

Step 1 (one layer). Let $\varepsilon\in\{1,2\}$, $u,v\in A_{\varepsilon}$ and $x,z\in N$. Then

Indeed, by the identities above the left side equals $\tau_{\mu}(E_{\varepsilon}(\pi_{\varepsilon}(x)u\pi_{\varepsilon}(z)v))=\tau_{\gamma_{\varepsilon}}(\pi_{\varepsilon}(x)u\pi_{\varepsilon}(z)\,v)$, which by the trace property (with $c=\pi_{\varepsilon}(x)u\pi_{\varepsilon}(z)$ and $c'=v$) equals $\tau_{\gamma_{\varepsilon}}(v\pi_{\varepsilon}(x)u\,\pi_{\varepsilon}(z))=\tau_{\mu}(E_{\varepsilon}(v\pi_{\varepsilon}(x)u\,\pi_{\varepsilon}(z)))=\tau_{\mu}(E_{\varepsilon}(v\pi_{\varepsilon}(x)u)\,z)$.

Step 2 (induction on $k$). For $k\in\mathbb{N}$ let $P(k)$ be the assertion: for all $f_{1},\dots,f_{k}\in\{1,2\}$, all $u_{j},v_{j}\in A_{f_{j}}$ ($j\in[k]$) and all $x,y\in N$, the elements $G_{k}$ and $G_{k}'$ defined in the statement satisfy $\tau_{\mu}(xG_{k})=\tau_{\mu}(G_{k}'y)$. Let $\mathcal{K}=\{k\in\mathbb{N}:P(k)\}$; we show that $1\in\mathcal{K}$ and that $k+1\in\mathcal{K}$ whenever $k\in\mathcal{K}$, so that $\mathcal{K}=\mathbb{N}$ by Principle of Induction for the Natural Numbers.

For $k=1$: $G_{1}=E_{f_{1}}(u_{1}\pi_{f_{1}}(y)v_{1})$ and $G_{1}'=E_{f_{1}}(v_{1}\pi_{f_{1}}(x)u_{1})$, and $P(1)$ is (1) with $\varepsilon=f_{1}$, $u=u_{1}$, $v=v_{1}$, $z=y$.

Let $k\in\mathcal{K}$ and let data $f_{1},\dots,f_{k+1}$, $u_{j},v_{j}$, $x,y$ be given, with $G_{1},\dots,G_{k+1}$ and $G_{1}',\dots,G_{k+1}'$ as in the statement. The first $k$ elements $G_{1},\dots,G_{k}$ are also the elements $G_{1},\dots,G_{k}$ of the truncated data $(f_{j},u_{j},v_{j})_{j\in[k]}$ with the same $y$, since both are given by the same first term and the same recursion for the indices $j\le k$ (induction on $j$). Put

Since $G_{k+1}=E_{f_{k+1}}(u_{k+1}\pi_{f_{k+1}}(G_{k})v_{k+1})$, (1) with $\varepsilon=f_{k+1}$ and $z=G_{k}$ gives $\tau_{\mu}(xG_{k+1})=\tau_{\mu}(x'G_{k})$. By $P(k)$ for the truncated data, with $x'$ in place of $x$ and the same $y$,

For the data of length $k+1$ the statement defines $G_{1}'=x'$ and $G_{j}'=E_{f_{k+2-j}}(v_{k+2-j}\pi_{f_{k+2-j}}(G_{j-1}')u_{k+2-j})$ for $2\le j\le k+1$. We claim $G_{j+1}'=H_{j}$ for every $j\in[k]$, by induction on $j$: for $j=1$, $G_{2}'=E_{f_{k}}(v_{k}\pi_{f_{k}}(G_{1}')u_{k})=E_{f_{k}}(v_{k}\pi_{f_{k}}(x')u_{k})=H_{1}$; and if $j<k$ and $G_{j+1}'=H_{j}$, then, since $k+2-(j+2)=k+1-(j+1)=k-j$,

Hence $G_{k+1}'=H_{k}$, and $\tau_{\mu}(xG_{k+1})=\tau_{\mu}(x'G_{k})=\tau_{\mu}(G_{k+1}'y)$. This is $P(k+1)$, so $k+1\in\mathcal{K}$.

Therefore $P(k)$ holds for every $k\in\mathbb{N}$, which is the claim.