Proof. We first check closure under sequential inverse limits. For a sequence of extensions, take the local state and parameter spaces to be the spaces of compatible paths, with their consistent probability laws. They are standard Borel spaces. At every array level the consistent finite-stage laws determine a law on compatible array paths. Time acts coordinatewise. Projection to \(X\), consistency under fixing any axis, and unconditional last-axis independence pass to the limit by testing finite-stage functions. Independence of the local \(X\)-state and the limiting parameter follows in the same way.

For the conditional requirement, fix a proper family \(J\) of last-axis planes and bounded functions \(F_j\) on those planes, all pulled from one finite stage. Let \(\mathcal P_n^{(s)}\) denote the full parameter array at stage \(s\), and \(\mathcal P_{n-1,j}^{(s)}\) its restriction to plane \(j\). For every sufficiently large \(s\), the defining property gives \[\mathbb E\left[\prod_{j\in J}F_j\,\middle|\,\mathcal P_n^{(s)}\right] =\prod_{j\in J}\mathbb E[F_j\mid\mathcal P_{n-1,j}^{(s)}].\] Increasing martingale convergence on both sides gives the same identity for the limiting parameter fields. The factors are uniformly bounded, so their products converge in \(L^2\). This proves both independence and dependence of each plane law only on its own parameters, first for finite-stage tests and then, by density, for all bounded tests.

For a fixed test and data type, the norm of the indicated conditional expectation is nondecreasing under extension and bounded above by the norm of the test. At each constructed stage, choose countable dense families of bounded tests in every \(L^2(m_n)\), and list all pairs of a test and one of the finitely many data types at that level. Construct successive extensions as follows. At round \(q\), process the first \(q\) listed pairs from each of the first \(q\) stages already constructed. For each pair choose an extension whose projection norm is within \(2^{-q}\) of the supremum over extensions of the current stage. There is no requirement to attain the supremum. Append these finitely many extensions and continue. Every listed pair is processed infinitely often with error tending to zero. Taking the sequential inverse limit gives an object \(M_\infty\) in the class.

Let \(Z\) be any extension of \(M_\infty\), and fix a test listed at some finite stage. At any sufficiently late processing step, \(Z\) is also an extension of the current stage. Its projection norm is consequently at most that step’s supremum, which is at most the chosen next-stage norm plus \(2^{-q}\). The next-stage norm is at most the limiting norm. Letting \(q\to\infty\) proves that \(Z\) cannot increase the norm for this test. Finite-stage functions are dense in the inverse-limit \(L^2\) spaces, and conditional expectation is a contraction. The same conclusion follows for every \(F\in L^2(m_n)\) of \(M_\infty\).

Finally, the pullback of \(\mathcal D\) is contained in \(\mathcal D'\). The Pythagorean identity for these nested orthogonal projections is \[\begin{align*} &\lVert \mathbb E[F\circ\theta_n\mid\mathcal D']\rVert_2^2 -\lVert \mathbb E[F\mid\mathcal D]\circ\theta_n\rVert_2^2\\ &\hspace{2cm}= \lVert \mathbb E[F\circ\theta_n\mid\mathcal D'] -\mathbb E[F\mid\mathcal D]\circ\theta_n\rVert_2^2. \end{align*}\] The equality of norms therefore gives (4). ◻

Proof. The tower property shows that \(K_j(E_j,\cdot)\) is also the conditional law of \(O_j\) given \(E_j\). Let \(M_j(x,\cdot)\) be its image under \(o\mapsto a_j(x,o)\). Disintegration gives, for almost every \(x\) with respect to the law of \(E_j\), \[K_j(x,\,\mathrm do)\delta_{a_j(x,o)}(\,\mathrm d\alpha) =M_j(x,\,\mathrm d\alpha)Q_j(x,\alpha;\,\mathrm do).\] Substituting \(x=e_j(E)\) preserves these almost-sure identities. The joint law of \((E,O,A,B)\) thus factors as \[\mathop{\mathrm{Law}}(E)(\,\mathrm de) \left[\bigotimes_{j=1}^sM_j(e_j(e),\,\mathrm d\alpha_j)\right] L(e,\alpha;\,\mathrm db) \left[\bigotimes_{j=1}^sQ_j(e_j(e),\alpha_j;\,\mathrm do_j)\right].\] The first three factors give the law of \((E,A,B)\), proving (5). ◻

Proof. We prove the statement by induction on \(|R|\), simultaneously for all array levels and all projection slots \(a\notin R\). Saturation will be used only for extensions already supplied by smaller parent sets. The resulting screening identity and Lemma 5 will then verify the conditional product law for the larger parent set. Write an entry of the old level \(n+1\) as \(y_{c,w}\), where \(c\in I\) is the internal axis and \(w\in I^n\) comprises the remaining axes. The proposed projection retains \(y_{a,w}\). It preserves every array law by axis consistency, preserves the \(X\)-state through \(\kappa(y_a)\), and recovers the old parameter from (6). Time invariance, consistency, and unconditional last-axis independence for the candidate follow directly from the old level \(n+1\).

At the local level, condition on \((u(y_c))_{c\in I}\) in \(m_1\). The set \(R\cup\{a\}\) is proper. Its states are conditionally independent, with the law at \(a\) conditioned only on \(u(y_a)\). Since the \(X\)-image of that conditional law is \(\mu_X\), the selected \(X\)-state still has law \(\mu_X\) after conditioning further on \((y_c)_{c\in R}\). This verifies the required independence from the enlarged parameter. If \(R=\varnothing\), the remaining conditional last-axis requirement is exactly the old condition at level \(n+1\), so the base case is proved.

Now suppose \(R\ne\varnothing\) and that the lemma holds for smaller parent sets. Fix \(n\geq1\) and a missing last-axis slot \(j_0\). In the old law \(m_{n+1}\) let \[E=(u(y_{c,w}))_{c,w},\qquad O=(y_{c,w}:w_n\ne j_0).\] For \(j\ne j_0\), let \(O_j\) be the full plane at last slot \(j\), \(E_j\) its old parameter array, and \(A_j\) its states at internal indices \(c\in R\). Finally, let \[B=(y_{c,w}:c\in R,\ w_n=j_0)\] be the parent states in the missing plane. Conditional on \(E\), the \(O_j\) are independent, with each plane law depending only on \(E_j\). We claim that \[ \mathop{\mathrm{Law}}(B\mid E,O) \text{ depends on }O\text{ only through }(A_j)_{j\ne j_0}. \tag{7}\]

Enumerate \(R\) and reveal its missing-plane slices one internal index \(c\) at a time. Let \(R'\) be the set of indices already revealed. It has smaller cardinality than \(R\). The induction hypothesis supplies a valid extension with parent set \(R'\) and projection slot \(c\). Apply saturation to a bounded test \(F\) of the missing-plane slice \((y_{c,w}:w_n=j_0)\), viewed as a function on the projected old level \(n\). For the data type \(\mathcal G_{n,j_0}\), the old data are the parameters of the row at \(c\) and its states outside the missing plane; the enlarged data are \(E\), all of \(O\), and the already revealed missing parent slices. More explicitly, if \(B_{R'}\) denotes those already revealed missing slices, write \(A_j^{R'}\) for the states of \(O_j\) at internal indices in \(R'\). This smaller-parent extension has \(\mathcal P'_n=\sigma(E,(A_j^{R'})_{j\ne j_0},B_{R'})\), \(\mathcal O'_{n,j_0}=\sigma(O)\), and \(\mathcal G'_{n,j_0}=\sigma(E,O,B_{R'})\). Hence (4) says exactly that \[ \mathbb E[F\mid E,O,(y_{l,w})_{l\in R',w\in I^n}] =\mathbb E[F\mid (u(y_{c,w}))_{w\in I^n}, (y_{c,w}:w_n\ne j_0)]. \tag{8}\] The right side is the old \(m_n\) conditional kernel at the fixed internal index \(c\). It involves only \(E\) and the part of \(O\) belonging to that same \(c\), and it does not depend on previously revealed missing states. Multiplication of these successive kernels proves (7).

Apply Lemma 5 to the proper family of planes \(O_j\). Conditioning on \(E\), all the \(A_j\), and \(B\) preserves their independence and conditions each plane only on \((E_j,A_j)\). These are precisely the complete enlarged parameters and each plane’s own enlarged parameters, respectively. The law of such a plane is the old \(m_n\), now viewed as the candidate level \(n-1\); its conditional law given \((E_j,A_j)\) is therefore exactly the one required by Definition 3. We have proved the requirement for \(k-1\) planes. Marginalization proves it for every proper family.

There are only countably many levels and finitely many choices of parents and slots at each induction step. Testing on countable determining classes makes all the kernel equalities simultaneous. The induction is complete. ◻

Proof. Column independence in (1) is the defining conditional independence at level two. To obtain rows, use the extension in Lemma 6 with parent set \(R\) and projection slot \(a\notin R\). For a bounded test \(F\) of the entire projected row, saturation for the full parameter data \(\mathcal P_1\) gives \[ \mathbb E[F(Y_{a,I})\mid E,Y_{R,I}] =\int F\,\mathrm d\lambda_{p_a}. \tag{14}\] Reveal the members of any proper row family in turn. The previous members number at most \(k-2\), so (14) applies at every step. Their joint conditional law is the product of the indicated row laws.

Assertion (2) is the conditional last-axis requirement at level one of the extension with parents \(R\). Its local law is \(m_1\) on each column, and its own parameter there is \((q_j,Y_{R,j})\), giving precisely the stated marginal kernel. For (3), use the same extension and saturation for \(\mathcal G_{1,j_0}\). The projected old data are \(p_a\) and the other entries of row \(a\), while the extension data are \(E\), the parents, and all columns except \(j_0\). This gives (12).

We prove the grouped assertions without interchanging array axes. Each row group is a proper set of rows, so its full law is the product of its individual row laws. Within each individual row, any proper set of column coordinates is independent. Consequently any two grouped cells in that grouped row are independent. The identical conclusion for a grouped column follows by applying the corresponding column assertions. Since \(|P|\leq k-2\), (2) applies to the parents \(P\). The union of any two column groups is proper, which gives the claimed conditional independence of grouped columns.

It remains to prove (13). We first handle a single missing column \(j_0\). Given \(E\), \(Y_{P,I}\), and all other columns, reveal \(y_{a,j_0}\) successively for \(a\in A\). At each step the parents consist of \(P\) and the previously revealed rows in \(A\). Their number is at most \[|P|+|A|-1=k-|B|-1\leq k-2.\] All their remaining entries are already known from the observed columns. Equation (12) therefore gives the kernel of the individual row at that step. These kernels do not depend on the previously revealed entries in column \(j_0\), and hence \[ \mathop{\mathrm{Law}}(Y_{A,j_0}\mid E,Y_{P,I},Y_{I,I\setminus\{j_0\}}) =\bigotimes_{a\in A} \lambda_{p_a}(\,\mathrm dy_{a,j_0}\mid Y_{a,I\setminus\{j_0\}}). \tag{15}\]

Now choose \(j_0\in t\), put \(J=t\setminus\{j_0\}\) and \(C=r\cup s\), and retain \(W=Y_{A,J}\) as part of the output. The set \(C\cup J\) consists of exactly \(k-1\) columns. By (2), conditional on \(E\) and \(Y_{P,I}\), those columns are independent. In each of them, \(P\cup A\) is a proper row set. Thus the \(A\)-entries, even after conditioning on its \(P\)-entries, have the plain product marginal \(\bigotimes_{a\in A}\eta_{u_{a,j}}\). It follows that \[ \mathop{\mathrm{Law}}(W\mid E,Y_{P,I},Y_{I,C}) =\bigotimes_{a\in A,\,j\in J}\eta_{u_{a,j}}. \tag{16}\] Use (15) and then integrate away the additional states in columns \(J\), retaining \(W\). Its right side depends on those columns only through \(W\), so the resulting joint kernel for \((W,Y_{A,j_0})\) is the measure in (16), followed by the product of the individual row kernels in (15).

Under the row-group law \(\Lambda_A\), the coordinates in \(C\cup J\) are also independent with exactly those marginals: the rows in \(A\) are independent and each row has independent proper coordinates. Consequently the just obtained kernel is exactly the conditional law of \(Y_{A,t}\) under \(\Lambda_A\) given \(Y_{A,C}\). This proves (13), including the case \(J=\varnothing\). Only cardinalities of the groups were used, so every ordering of the groups is allowed. All choices are finite; the earlier convention on kernel versions makes the assertions simultaneous. ◻

Proof. It is enough to prove (18) for \(J=I\setminus\{j_0\}\), and then take marginals. Write \[Z=Y_{a,I},\qquad O=(Y_{I,j})_{j\in J},\qquad S=(y_{a,j})_{j\in J}.\] Use the extension with no parents and projection slot \(a\). Saturation for the data \(\mathcal O_{1,j_0}\) says, for every bounded row test \(F\), \[ \mathbb E[F(Z)\mid O]=\mathbb E[F(Z)\mid S]. \tag{19}\] Since \(S\) is a function of both \(Z\) and \(O\), this is conditional independence of \(Z\) and \(O\) given \(S\). One direct verification is to multiply (19) by a bounded function of \(O\) and condition on \(S\); the resulting factorization is the conditional-independence identity. It also gives the reversed screening statement \(\mathop{\mathrm{Law}}(O\mid Z)=\mathop{\mathrm{Law}}(O\mid S)\).

Unconditionally the columns in \(O\) are independent, each with law \(m_1\). Conditioning on their separate shared states therefore gives \[\mathop{\mathrm{Law}}(O\mid Z) =\bigotimes_{j\in J}m_1(\,\mathrm dY_{I,j}\mid y_{a,j}).\] We identify the parameter image of each factor. In a single column, the joint law of its parameter tuple \(q\) and its state \(y_a\) at slot \(a\) is \[ \pi(\,\mathrm dq)\eta_{q[a]}(\,\mathrm dy_a). \tag{20}\] Disintegrating \(\pi\) at \(q[a]\) rewrites this as \[\nu(\,\mathrm du)K_a(u,\,\mathrm dq)\eta_u(\,\mathrm dy_a).\] Since \(u(y_a)=u\) almost surely under \(\eta_u\), the conditional law of \(q\) given the shared state \(y_a\) is exactly \(K_a(u(y_a),\,\mathrm dq)\). Thus \[\mathop{\mathrm{Law}}((q_j)_{j\in J}\mid Z) =\bigotimes_{j\in J}K_a(u(y_{a,j}),\,\mathrm dq_j).\] The right side is a function only of \(p_a\), which itself is a function of \(Z\). Conditioning down to \(p_a\) proves (18).

For the last assertion, take \(a=i\) and retain just \(j=i\). The tuple \(p_i\) has marginal \(\pi\), and its coordinate at \(i\) is \(u_{i,i}\). Given \(p_i\), the law of \(q_i\) is \(K_i(p_i[i],\cdot)\). Disintegrating once more over that coordinate shows that \[\mathop{\mathrm{Law}}(p_i,q_i\mid u_{i,i}=u) =K_i(u,\,\mathrm dp_i)K_i(u,\,\mathrm dq_i),\] as required. ◻

Proof. For an exact tuple put \(a_i=\lvert \int\chi_i\,\mathrm d\mu_i\rvert\). Solving the product relation for \(\chi_i\) and using independence of the remaining coordinates gives \[a_i=\prod_{j\ne i}a_j,\qquad 0\le a_i\le1.\] If some \(a_i=0\), at least one other \(a_j\) is zero, and the same equations then force all \(a_j\) to be zero. Otherwise put \(b_i=-\log a_i\). The equations give \(2b_i=\sum_j b_j\) for every \(i\); since \(m\ge3\), all \(b_i\) vanish. Thus either all means have modulus one, in which case all the functions are constant almost surely, or all means are zero.

The quotient of two exact tuples is again exact. If their lines differ at one coordinate, its quotient function is nonconstant, so all quotient means vanish. In particular the two representatives are orthogonal at that coordinate. If their lines agree at one coordinate, the quotient there is constant, forcing constant quotients everywhere. This proves uniqueness of partners. Products and conjugates preserve exact tuples; uniqueness makes multiplication of lines well defined and makes each coordinate projection bijective. A family of nonzero mutually orthogonal lines in a separable Hilbert space is countable. Comparing any exact tuple with the constant tuple gives the final assertion. ◻

Proof. Condition on \(z,c,d,x,y\). The marginal law of the two remaining entries in column \(s\) is the column law given \(c\), and similarly column \(t\) has its law given \(d\). Indeed, given row \(P\), each of these columns is independent of column \(r\). Denote the corresponding two joint \(L^2\) spaces by \(\mathcal H_s\) and \(\mathcal H_t\). Let \(Q:\mathcal H_t\to\mathcal H_s\) be the conditional Markov operator between them under this five-cell conditioning, and let \(\iota_{A,s},\iota_{B,s},\iota_{A,t},\iota_{B,t}\) be the isometric embeddings of the individual cell spaces into the appropriate joint spaces.

The own-row kernel assertion gives \[Q\iota_{A,t}=\iota_{A,s}L_A,\qquad Q\iota_{B,t}=\iota_{B,s}L_B.\] The same assertion with \(s\) and \(t\) reversed gives \[Q^*\iota_{A,s}=\iota_{A,t}L_A^*,\qquad Q^*\iota_{B,s}=\iota_{B,t}L_B^*.\] Also \(J_s=\iota_{A,s}^*\iota_{B,s}\) and \(J_t=\iota_{A,t}^*\iota_{B,t}\). Consequently \[L_AJ_t=\iota_{A,s}^*Q\iota_{B,t}=J_sL_B, \qquad L_A^*J_s=\iota_{A,t}^*Q^*\iota_{B,s}=J_tL_B^*.\] Countable dense sets of tests make these operator equalities simultaneous on a conull set. The grouped locality assertions hold for all the indicated orderings, so the same proof applies to each of them. ◻

Proof. First, the essential closed range span of \(L_A(x)\) is \(V_{A,s}\). Conditional expectations of product tests onto \(Y_{A,s}\) have the form \[ \int w(x)L_A(x)h\,\mathrm d\mu_{A,r}(x), \tag{24}\] where \(w,h\) are bounded. Such tests span the active space by density of product functions in the \(L^2\) space of the other two coordinates. A vector orthogonal to every integral in (24) is orthogonal to \(L_A(x)h\) almost everywhere, by varying \(w\). A countable dense set of \(h\) then makes it orthogonal to the whole range for almost every \(x\). The converse is immediate. This proves the assertion, and the same argument applies to every incident operator in either direction. In particular each such operator has its range in its own-direction active space and annihilates the orthogonal complement of the source active space.

By the first rectangle identity, \(J_s(c)L_B(y)\) has range in \(V_{A,s}\). The pair \((c,y)\) is independent. Fubini and the essential range span of \(L_B\) therefore give \(J_s(c)V_{B,s}\subseteq V_{A,s}\) for almost every \(c\). The adjoint of the second identity is \(J_s(c)^*L_A(x)=L_B(y)J_t(d)^*\). Independence of \((c,x)\) gives the reverse inclusion for the adjoint. Together these say \[ P^V_{A,s}J_s(c)=J_s(c)P^V_{B,s}. \tag{25}\] Likewise, the first rectangle identity and independence of \((x,d)\) show that \(L_A(x)W_{A,t}\subseteq W_{A,s}\); the second identity and independence of \((x,c)\) give the adjoint assertion. Thus \[ P^W_{A,s}L_A(x)=L_A(x)P^W_{A,t}. \tag{26}\] The same reasoning works at every edge.

Equation (25) shows that \(P^V_{A,s}\) preserves the essential range span of \(J_s\), which is \(W_{A,s}\). As an orthogonal projection it also preserves \(W_{A,s}^{\perp}\), and therefore commutes with \(P^W_{A,s}\). This proves commutation in every cell. Combining (26) with the own-row active-space reduction gives \(P^D_{A,s}L_A=L_AP^D_{A,t}\), and similarly for all other edges. Finally, projecting the essential range span \(V_{A,s}\) onto \(D_{A,s}\) shows that the restricted ranges of \(L_A\) span \(D_{A,s}\): \(P^D_{A,s}\mathop{\mathrm{Ran}}L_A=\mathop{\mathrm{Ran}}(L_AP^D_{A,t})\). The corresponding argument handles each edge and its adjoint. ◻

Proof. The two rectangle equations and their adjoints imply \[\begin{align*} KK^*F&=FNN^*,& K^*KG&=GN^*N,\tag{28}\\ F^*FN&=NG^*G,& FF^*K&=KGG^*. \tag{29}\end{align*}\] The parameter triples in these four equations are, respectively, \((c,x,d)\), \((c,y,d)\), \((x,d,y)\) and \((x,c,y)\). Each has product law by the independence observations preceding Lemma 11.

For the first equation, compare two independent values \(c,c'\) using the same \(x,d\). Fubini gives \[\bigl(K(c)K(c)^*-K(c')K(c')^*\bigr)F(x)=0\] for almost every \((c,c',x)\). The essential range span of \(F\) is its whole target \(D\) space, so the two target squares agree. The adjoint equation and the essential range span of \(F^*\) similarly make \(NN^*\) fixed. The second equation in (28) makes \(K^*K\) and \(N^*N\) fixed. The two equations in (29), using the range spans of \(N,K\) and their adjoints, make \(F^*F,G^*G,FF^*,GG^*\) fixed. These comparisons are operator comparisons: countable dense vector tests first give a common conull set, after which boundedness gives equality on every vector.

If \(v\) is in the kernel of the fixed square \(F^*F\), then \(F(x)v=0\) for almost every \(x\). Hence \(v\) is orthogonal to the ranges of almost every \(F(x)^*\), whose essential span is the whole source \(D\). Thus \(v=0\). The same reasoning proves injectivity of all source and target squares. Each operator consequently has zero kernel and dense range, so its polar part has full initial and final spaces and is unitary.

To preserve the rectangle identity, put \(A=|F|\) and \(B=|G|\). From \(F^*FN=NG^*G\) we obtain \(A^2N=NB^2\). Polynomial approximation to the square-root function on \([0,1]\) gives \(AN=NB\). Write \(F=U_FA\) and \(G=U_GB\). The equation \(FN=KG\) becomes \[(U_FN-KU_G)B=0.\] The range of the injective self-adjoint \(B\) is dense, so the bounded operator in parentheses vanishes. If \(N=U_N|N|\) and \(K=U_K|K|\), then \[U_FN=(U_FU_N)|N|,\qquad KU_G=(U_KU_G)(U_G^*|K|U_G)\] are polar decompositions of the same operator. Uniqueness yields \(U_FU_N=U_KU_G\). This argument uses no inverse bound for the positive parts. Finally, polar parts are strongly measurable on vectors, for example by the bounded approximations \(F(F^*F+\varepsilon\mathop{\mathrm{id}})^{-1/2}\) as \(\varepsilon\downarrow0\). ◻

Proof. Give the unitary spaces the strong operator topology. On unitaries, inversion and multiplication are continuous, and these spaces are separable metrizable. Conditional on a typical fixed value \(z_0\), the pair \((a,b)\) and the pair \((c_0,d_0)\) are independent, while each individual marginal equals its unconditional marginal. The continuous identity (30) therefore holds throughout the product of the two conditional pair supports. Choose one point \((a_*,b_*)\) from the first support and one point \((c_*,d_*)\) from the second. They satisfy \(a_*d_*=c_*b_*\). Identify the four vertex spaces by these reference unitaries, starting at the lower right vertex. The two possible identifications of the upper left vertex agree by this equation, so all four reference edge values become identity. Use these fixed identifications from now on.

The conditional pair supports at \(z_0\) now contain \((\mathop{\mathrm{id}},\mathop{\mathrm{id}})\). Fixing the column pair at this point in (30) forces \(a=b\) throughout the other pair support. Fixing the row pair similarly forces \(c_0=d_0\). The two row marginals therefore have a common law \(\rho\) and the two column marginals a common law \(\tau\); these are also their unconditional laws. The supports of \(\rho\) and \(\tau\) both contain identity. Substitution of diagonal pair values into (30) shows that these two supports commute elementwise.

For general \(z\), conditional independence of the two pairs and their fixed individual marginals imply \[ \rho=u\rho v^{-1} \quad\text{for almost every actual pair }(u,v)=(c_0,d_0). \tag{32}\] Indeed, after fixing \(z,c_0,d_0\), the equation says \(a=c_0bd_0^{-1}\), and \(a,b\) each have law \(\rho\). Unconditionally, \(c_0,d_0\) are independent, so their pair law is \(\tau\otimes\tau\). The map \((u,v)\mapsto u\rho v^{-1}\) is continuous for weak convergence of probability measures: apply strong-topology continuity and dominated convergence to each bounded continuous test. Thus (32) holds for every \(u,v\in\mathop{\mathrm{supp}}\tau\). The symmetric argument gives \(\tau=u\tau v^{-1}\) for every \(u,v\in\mathop{\mathrm{supp}}\rho\).

Taking \(v=\mathop{\mathrm{id}}\) shows that each support is preserved by left translation by elements of the other. Because identity belongs to both, they contain one another. Write their common closed support as \(\mathcal G\). For \(g\in\mathcal G\) we have \(g\mathcal G=\mathcal G\). This proves closure under products, and \(\mathop{\mathrm{id}}\in g\mathcal G\) supplies \(g^{-1}\in\mathcal G\). Thus \(\mathcal G\) is a group. It is abelian by the earlier commutation, and (32) gives left invariance of \(\rho\). Equation (30) now gives equality of the two ratios in (31). Given \(z\), the ratios are independent and equal almost surely, hence have a common point-mass conditional law. A countable determining family in the standard unitary group makes its point measurable in \(z\), giving \(h\).

Choose a positive injective trace-class operator \(Q\) on \(\mathcal H\) and form the trace-norm integral \[Q_0=\int_{\mathcal G}UQU^*\,\mathrm d\rho(U).\] The integrand is trace-norm continuous under strong-unitary convergence, as follows first for finite-rank \(Q\) and then by trace-norm approximation. The integral is positive and trace-class, and left invariance of \(\rho\) makes it commute with every element of \(\mathcal G\). For \(v\ne0\) each quadratic form \(\langle v,UQU^*v\rangle\) is positive; hence so is \(\langle v,Q_0v\rangle\). Thus \(Q_0\) is injective. Its finite-dimensional positive eigenspaces span \(\mathcal H\). Each is invariant under the commuting unitaries, so simultaneous diagonalization in each gives a full orthonormal eigenbasis \((e_n)\).

Write \(Ue_n=\gamma_n(U)e_n\), with \(\gamma_n(U)\in\mathbb T\). These coordinates are continuous characters. On diagonal unitaries, coordinatewise convergence is equivalent to strong convergence, by approximation of any vector by its finite basis truncations. The coordinates therefore generate the Borel information of \(\mathcal G\). Trigonometric polynomials on every finite coordinate torus are dense in \(L^2\) of its actual probability law. Increasing conditional expectations onto the finite-coordinate fields then show that the functions \[\gamma(h(z)),\qquad \gamma=\prod_{n\in J}\gamma_n^{m_n},\quad J\text{ finite},\quad m_n\in\mathbb Z,\] span \(L^2(\sigma(h(z)))\). No identification of the law of \(h\) as Haar measure is needed for this density statement. Finally, (31) gives \[\gamma(h(z)) =\gamma(a(x))\overline{\gamma(b(y))} =\gamma(c_0(c))\overline{\gamma(d_0(d))}.\] Each factor is unimodular. Multiplying by the conjugate factors shows that \(\gamma(h(z))\) is exact both in column \(r\) and in row \(P\). ◻

Proof. Since \((Y_s,Y_t)\) has product marginal, the bilinear contraction is well defined on \(H_s\times H_t\) and satisfies \[ \lVert C_r(g,h)\rVert_2\le\lVert g\rVert_2\lVert h\rVert_2. \tag{35}\] For bounded inputs let \(L_{r\leftarrow t}(v)\) be the row Markov operator from \(t\) to \(r\) given \(Y_s=v\). Pairwise independence gives the vector-valued integral formula \[C_r(g,h)=\int g(v)L_{r\leftarrow t}(v)h\,\mathrm d\mu_s(v).\] Lemma 12 says \(P_rL_{r\leftarrow t}(v)=L_{r\leftarrow t}(v)P_t\). Moving the projection through the integral proves \(P_rC_r(g,h)=C_r(g,P_th)\). Using instead the edge from \(s\) to \(r\) conditioned on \(Y_t\) proves \(P_rC_r(g,h)=C_r(P_sg,h)\). The bound (35) extends both identities to all \(L^2\) inputs, including unbounded projected tests. Idempotence then gives \[P_rC_r(g,h)=P_rC_r(g,P_th)=C_r(P_sg,P_th),\] proving (33). Restricted contractions are fixed by \(P_r\) and hence lie in \(D_r\). Unrestricted product contractions span the row-active space; projecting that space onto \(D_r\) proves the restricted spanning assertion. The column proof is identical.

Apply this result first in row \(A\). For \(g\in D_{A,s}\) and \(h\in D_{A,t}\), its contraction onto \(x=Y_{A,r}\) is \[C_{A,r}(g,h)(x)=\langle \overline g,L_A(x)h\rangle.\] Conjugation preserves \(D_{A,s}\). On the restricted spaces, \(L_A(x)=a(x)A_0\) for a fixed positive operator \(A_0\) by Lemma 13. Thus every displayed coefficient is a function of \(a(x)\) alone. The space \(L^2(\sigma(a(x)))\) is closed, being the range of a conditional-expectation projection, so the spanning assertion gives \[D_{A,r}\subseteq L^2(\sigma(a(x))),\qquad D_{B,r}\subseteq L^2(\sigma(b(y))).\] Apply the column version of the spanning assertion in column \(r\). Given \(z\), we have \(a=h(z)b\) and \(b\) has its fixed marginal law \(\rho\). For bounded measurable \(f,g\) the contraction of \(f(a)\) and \(g(b)\) onto \(z\) is therefore \[\int_{\mathcal G} f(h(z)u)g(u)\,\mathrm d\rho(u),\] a function of \(h(z)\). The product marginal of \((x,y)\), the bound (35), and bounded approximation extend this conclusion to the two \(D\) spaces and their contraction span. Closure therefore yields \(D_{P,r}\subseteq L^2(\sigma(h(z)))\). In the other direction, the monomials of Lemma 14 are exact for both laws, hence lie in \(D_{P,r}\), and they span \(L^2(\sigma(h(z)))\). This proves (34). We have used no assumption that \(D\) was already closed under multiplication. Permuting row groups and column groups places any grouped cell at the corner and proves the last assertion. ◻

Proof. Isolate row \(a\) and column \(j\) as singleton groups, and split each of their complements into two nonempty groups. At the isolated cell, the row-active and column-active spaces do not depend on the chosen splits: the other two grouped cells still contain exactly all the other entries of that row or column. Lemma 12 gives commutation, and Lemma 15 gives a spanning family of simultaneous exact lines for the two split three-cell laws. We take this family to contain all such lines: each is in the intersection, and orthogonality leaves no additional line outside the spanning family.

Fix the row-group partition and vary the column-group partition. Every resulting family is a subset of the mutually orthogonal exact lines of the same fixed split column law, by Lemma 10. All have the same closed span \(V_{a,j}\cap W_{a,j}\). Two subsets of one orthogonal family with the same closed span must coincide: a line belonging only to one would be orthogonal to the other span. Now fix the initial column grouping and vary the row grouping. The same argument, using the fixed split row law, shows that the initial family is exact for every split in both directions.

For \(k=3\) the grouped laws already have individual coordinates. For \(k>3\), fix a unimodular representative \(\chi_j\) of one of the lines and consider its row completions. For each split \(S\) of \(I\setminus\{j\}\), there are unimodular functions \(a_S,b_S\) and \(c_S\in\mathbb T\) with \[\chi_j(Y_{a,j})a_S(Y_{a,S})b_S(Y_{a,I\setminus(S\cup\{j\})})=c_S.\] Absorb \(c_S^{-1}\) into \(a_S\). All the resulting completion functions then equal \(\overline{\chi_j(Y_{a,j})}\) under the row law, so they define the same function \(\phi\) of the other coordinates. Those \(k-1\) coordinates have product marginal. For each singleton split, \(\phi\) lies in a one-dimensional line in that coordinate tensored with the remaining \(L^2\) spaces. The orthogonal projections to these individual line factors commute and all fix \(\phi\). Their product has the full one-dimensional tensor product as range, so \(\phi\) is a pure product of its unimodular individual factors, up to one scalar. This proves exactness for the full row law. The column argument is the same. Conversely, every line exact for both full laws is active for both and belongs to their intersection. This proves part (1).

For part (2), make the chosen row a singleton group, the chosen columns \(s,t\) singleton groups, and put all remaining columns in the third group. Split the remaining rows into two nonempty groups. The intertwining in Lemma 12 is exactly the stated identity for this grouping. To obtain the column assertion, isolate the two chosen rows instead and apply the already proved column intertwining; there is again no interchange of axes.

Finally, solving the exact product relation for \(\chi_t\) and conditioning gives (36). Its scalar has modulus one. The projections \(P^W\) commute with conjugation. Thus the intertwining in part (2) carries membership of \(\mathbb C\chi_t\) in \(W_{a,t}\) to membership of \(\mathbb C\chi_s\) in \(W_{a,s}\). The reverse row Markov operator gives the converse. The same proof applies to column partners. ◻

Proof. Fix \(i\). By Lemma 8, the row and column tuples meeting at cell \((i,i)\) are conditionally independent with law \[\pi_i^u:=\pi(\,\mathrm dp\mid p_i=u)\] given their common entry \(u\). Proposition 16 therefore gives, for almost every such pair \((p,q)\), \[ \Pi_i(p)\Pi_i(q)=\Pi_i(q)\Pi_i(p),\qquad \mathop{\mathrm{Ran}}\Pi_i(p)\cap\mathop{\mathrm{Ran}}\Pi_i(q)\subseteq C_i(p). \tag{38}\]

Here the almost-everywhere assertion does not presuppose a measurable description of \(C_i\). Indeed, choose a measurable conull set of parameter squares on which the conditional kernel hypotheses of Proposition 16 hold on countable determining classes. That Proposition gives its pointwise conclusion on every square in this set. Disintegrate this set over the two tuples meeting at \((i,i)\). On a measurable conull set of pairs the conditional measure assigns probability one to admissible squares, and is concentrated on the prescribed pair. Each such pair has an admissible completion. The conclusions in (38) depend only on the two tuple laws, so they hold for that pair. No measurable choice of a completion or of an exact line is required. Disintegration over \(u\) then gives conull sections for the product law \(\pi_i^u\otimes\pi_i^u\).

For a typical \(u\) define the positive contraction on \(H_u\) \[ B_{i,u}=\int \Pi_i(q)\,\pi_i^u(\,\mathrm dq), \qquad Q_{i,u}=\text{the projection onto }\mathop{\mathrm{Ker}}B_{i,u}. \tag{39}\] The weak operator integral is well defined, and its matrix entries are measurable in \(u\). Since \(0\le B_{i,u}\le\mathop{\mathrm{id}}\), \(Q_{i,u}\) is the strong limit of \((\mathop{\mathrm{id}}-B_{i,u})^n\) and is measurable too. In the fixed ambient \(\ell^2\) coordinates, use the projection of each standard basis vector \(e_j\) into \(H_u\) and extend the operators by zero on \(H_u^\perp\). Then \[\int\lVert \Pi_i(p)Q_{i,u}e_j\rVert^2\,\pi_i^u(\,\mathrm dp) =\langle Q_{i,u}e_j,B_{i,u}Q_{i,u}e_j\rangle=0.\] After a countable intersection, \(\Pi_i(p)Q_{i,u}=0\) for almost every \(p\). Taking adjoints gives \[ \mathop{\mathrm{Ran}}\Pi_i(p)\subseteq(\mathop{\mathrm{Ker}}B_{i,u})^\perp. \tag{40}\] This range containment is an operator assertion on a measurable conull set: it is tested by \(Q_{i,u}\Pi_i(p)e_j=0\) for all \(j\).

Fix \(p\) satisfying (40) and having a conull \(q\)-section in (38). If \(v\in\mathop{\mathrm{Ran}}\Pi_i(p)\) is orthogonal to \(C_i(p)\), commutation implies that \(\Pi_i(q)v\) belongs to the intersection in (38). Consequently \[\lVert \Pi_i(q)v\rVert^2=\langle v,\Pi_i(q)v\rangle=0 \quad\text{for }\pi_i^u\text{-almost every }q.\] Thus \(\langle v,B_{i,u}v\rangle=0\), and \(v\in\mathop{\mathrm{Ker}}B_{i,u}\). Equation (40) forces \(v=0\). The conull \(q\)-section used here gives whole-subspace statements, so it works for every such \(v\); a measurable selection of \(v\) is unnecessary. Every exact line is active, by its defining product relation. This proves (37). There are finitely many slots \(i\). ◻

Proof. We first enumerate the exact lines over tuple parameters, then show that these enumerations describe a family depending only on a local entry. The distinction is necessary: Lemma 17 alone gives no such independence from the other parameters.

Proof. For a nonnegative measurable function \(F\) on \(Z^I\), define \[ \int F\,\mathrm d\widetilde\pi =\int\sum_{\ell\in\mathcal L(p_s)} F\bigl((p_t,\Phi_{st}(p)\ell)_{t\in I}\bigr)\,\mathrm d\pi(p). \tag{48}\] Unique exact matching makes this independent of the chosen source slot \(s\). Define the square measure in the same way, counting its unique matched arrays from one chosen cell. The last assertion of Proposition 18 makes this independent of the chosen cell. Counting from a coordinate, row, or column gives the claimed marginals. The preimages of the first \(N\) labels of any one coordinate have measure at most \(N\) and exhaust the lifted spaces. Thus the measures are sigma-finite.

Disintegrating (48) first over \(p_s=u\) gives (46). For each prescribed source label there is exactly one completion over every compatible tuple; fixing that label introduces no weight into the parameter kernel. This proves directly that the conditional kernels are probability kernels even when the total lifted measure is infinite. Similarly, given a labelled row of the square, its label has exactly one extension over every compatible parameter square. Its conditional parameter-square law is therefore the original law given that row’s parameter tuple. Lemma 8 gives independent proper columns; their labels are then determined separately by their shared labels and their own parameter tuples. This proves the last assertion of the Lemma.

Write \(\widetilde\pi_{st}\) for the \((s,t)\) marginal. Its two marginals are \(\xi\), and its kernels in the two directions are \(K_{st}\) and \(K_{ts}\). Jensen’s inequality and the marginal identities give \(L^2\) contraction and \(M_{st}^*=M_{ts}\). Time equivariance of exact matching preserves both lifted measures and gives \(VM_{st}=M_{st}V\).

Finally, test the \((s,t)\) pair in row \(b\) of the lifted square with \(f,g\in L^2(\xi)\). Conditional on labelled row \(a\), its two column transitions are independent applications of \(M_{ab}\), since two columns form a proper family when \(k\ge3\). Hence \[\begin{align*} \langle f,M_{st}g\rangle &=\int\overline{f(z_{b,s})}\,g(z_{b,t}) \,\mathrm d\widetilde\vartheta\\ &=\int\overline{(M_{ab}f)(z_{a,s})}\, (M_{ab}g)(z_{a,t})\,\mathrm d\widetilde\vartheta\\ &=\langle M_{ab}f,M_{st}M_{ab}g\rangle =\langle f,M_{ab}^*M_{st}M_{ab}g\rangle. \end{align*}\] These integrals are absolutely integrable by Cauchy–Schwarz with the pair marginals \(\xi\). Alternatively one may first use bounded tests of finite-measure support and then pass in \(L^2\). The calculation proves (47) without normalizing the sigma-finite measures. ◻

Proof. Fix \(s\ne t\) and write \(M=M_{st}\). Taking \((a,b)=(s,t)\) and \((t,s)\) in (47) gives \[ M=M^*MM=MMM^*. \tag{50}\] Let \(\mathcal H_1=\overline{\mathop{\mathrm{Ran}}M}\) and \(\mathcal H_2=\overline{\mathop{\mathrm{Ran}}M^*}\). The first identity puts \(\mathcal H_1\) in the fixed space of \(M^*M\), which lies in \(\mathcal H_2\). The adjoint of the second identity puts \(\mathcal H_2\) in the fixed space of \(MM^*\), which lies in \(\mathcal H_1\). They are therefore equal to a closed space \(\mathcal H\). On it \(M^*M=MM^*=\mathop{\mathrm{id}}\), while \(M\) vanishes on \(\mathcal H^\perp=\mathop{\mathrm{Ker}}M\). Thus \(M\) is unitary on \(\mathcal H\), with inverse \(M^*\) there. For every \(F\in\mathcal H\), equality in the conditional-expectation norm gives \[ F(z_t)=(MF)(z_s) \quad\widetilde\pi_{st}\text{-almost everywhere}. \tag{51}\] Indeed, the integral of the squared difference is \(\lVert F\rVert_2^2-\lVert MF\rVert_2^2=0\).

Choose a measurable enumeration \(\ell_n(u)\) without repetition, and set \(h(u,\ell_n(u))=2^{-n}\) on its domain, starting at \(n=1\). Then \(h>0\) and \(\lVert h\rVert_2^2\le\sum_{n\ge1}4^{-n}<\infty\). The probability kernel sends \(h\) to a strictly positive function \(v=Mh\) almost everywhere: the kernel avoids every \(\xi\)-null set for almost every source point, by its marginal identity, and the integral of a positive function under a probability law is positive. Also \(v\in\mathcal H\).

The time operator \(V\) is unitary and commutes with \(M\) in both directions, so it preserves \(\mathcal H\). Define the measurable countable vector \[ \psi(z)=\bigl((M^nV^r v)(z)\bigr)_{(n,r)\in\mathbb Z^2} \in\mathbb C^{\mathbb Z^2}. \tag{52}\] Negative powers of \(M\) are taken on \(\mathcal H\). All its coordinates have finite versions outside one null set, and its \((0,0)\) coordinate is positive there. On \(\mathbb C^{\mathbb Z^2}\) define \[(D_0c)_{n,r}=c_{n+1,r},\qquad (S_0c)_{n,r}=c_{n,r+1}.\] These are commuting invertible coordinate reindexings. Using (51) for the countably many coordinates gives \[ \psi(z_t)=D_0\psi(z_s),\qquad \psi(\tau z)=S_0\psi(z) \tag{53}\] almost everywhere in the respective measures.

We claim that \(\psi\) has countable essential image. Given a source point \(z=(u,\ell)\), the parameter of its \(K_{st}\) target has law \(\nu\), since \(p_s,p_t\) are independent under \(\pi\) and fixing a source label does not tilt that law. Consider the occurrence relation \[R(z,u')\quad\Longleftrightarrow\quad D_0\psi(z)=\psi(u',\ell_n(u')) \text{ for some defined }\ell_n(u').\] It is measurable: existence is a countable union and equality of vectors is a countable conjunction of coordinate equalities. The first identity in (53) implies \(\nu\{u':R(z,u')\}=1\) for \(\xi\)-almost every \(z\). Tonelli’s theorem for the sigma-finite product \(\xi\otimes\nu\) selects one \(u'\) for which \(R(z,u')\) holds for \(\xi\)-almost every \(z\). That target fibre has only countably many \(\psi\)-values. Invertibility of \(D_0\) proves the claim.

Let \(\mathcal B\) be the positive-mass values in this countable essential image. For \(b\in\mathcal B\) put \[E_b=\{z:\psi(z)=b\},\qquad B_b(u)=\{\ell:(u,\ell)\in E_b\}.\] Every \(b\) has positive \((0,0)\) coordinate \(b_{0,0}\), and \[ 0<\xi(E_b)\le \frac{\lVert v\rVert_2^2}{|b_{0,0}|^2}<\infty. \tag{54}\] Equal marginals in the first identity of (53) and time invariance in the second show that \(D_0\) and \(S_0\) permute \(\mathcal B\), preserving each level’s mass along their orbits. After discarding a \(\nu\)-null set, all labels in every remaining fibre belong to these levels, and each \(B_b(u)\) is finite. Countability of \(\mathcal B\) makes these assertions simultaneous.

Counting the labels in (53) shows that, for \(\pi\)-almost every \(p\), its matching isomorphism \(\Phi_{st}(p)\) carries every block \(b\) onto block \(D_0b\). Because each fibre’s label list and \(\mathcal B\) are countable, intersecting the corresponding conull sets upgrades the transported label identity to every label in every block for such a tuple. Disintegrate the measurable conull set of such tuples over \((p_s,p_t)\). Since its marginal is \(\nu\otimes\nu\), there is a measurable conull set of pairs \((u,w)\) admitting a tuple with this property. Fubini chooses a common source \(u_*\) for which almost every target is in this set. For each such target \(u\), choose one resulting group isomorphism \(I_u:\mathcal L(u_*)\to\mathcal L(u)\). Each \(I_u\) takes block \(b\) to block \(D_0b\) simultaneously for every \(b\). The choices are used only pointwise.

Fix one typical target \(u_0\), set \(\mathcal L=\mathcal L(u_0)\), and define \(J_u=I_uI_{u_0}^{-1}\). These group isomorphisms preserve every block index. Thus all \(|B_b(u)|\) are constant almost everywhere. Equation (54) makes each constant finite, and its positive left side makes it nonzero. Let \(B_b=B_b(u_0)\) and \(m_b=|B_b|\).

The second identity of (53) gives the asserted time permutation of blocks. Restrict the local parameter set to the intersection of all integer translates of the conull sets just used and of the time-equivariance set. Choose \(u\) in this invariant conull set. The automorphism \[A=J_{T_Uu}^{-1}\tau_uJ_u\] of \(\mathcal L\) maps every \(B_b\) to \(B_{S_0b}\). Take \(G=A^{-1}\). Iteration gives (49). This last step fixes one automorphism realizing the permutation of blocks; it makes no assertion about the maps inside a block at other parameters or later times. ◻

Proof. Condition the nonzero product integral at level \(1\) on \(p\). For any chosen slot \(i\), the conditional expectation of the product of the other bounded inputs onto that slot belongs to its active space. If \(f_i(x)\) were orthogonal to that space for almost every \(p\), the product integral would vanish. In using complex inner products here one can conjugate the other product; the active space is conjugation invariant. Lemma 17 and Proposition 18 therefore give a choice \(f=f_i\) with a nonzero coefficient on some local label on a set of positive \(\nu\)-measure. The identity-label coefficient is zero: \[\int f(x(y))\,\mathrm d\eta_u(y)=\int f\,\mathrm d\mu_X=0.\] Thus nonidentity labels supply the nonzero coefficient.

For a block \(b\in\mathcal B\) define the measurable function \[c_b(u)=\sum_{\ell\in B_b(u)} \left|\int f(x(y))\overline{\chi_{u,\ell}(y)}\,\mathrm d\eta_u(y)\right|.\] The countably many blocks cover all labels, so some fixed block \(b\) has \(\int c_b\,\mathrm d\nu>0\). Its cardinality \(m_b\) is fixed and finite, and hence \(c_b\le m_b\lVert f\rVert_\infty\). Set \(B=B_b\subset\mathcal L\) using Lemma 20, and \(c=\int c_b\,\mathrm d\nu\).

For each typical \(u\) choose the block-preserving identification \(J_u\) from that Lemma and put \(\chi_\ell^u=\chi_{u,J_u\ell}\). Proposition 18 gives (55). By (49), the sum in (56) equals the intrinsic measurable sum \[ \sum_{\ell'\in B_{S_0^{-r}b}(u)} \left|\int f(T_X^r x(y)) \overline{\chi_{u,\ell'}(y)}\,\mathrm d\eta_u(y)\right|. \tag{58}\] Only the set of lines is relevant; changing representative phases or permuting labels inside this set changes no summand magnitude. This proves measurability without a measurable choice of \(J_u\).

The conditional laws are equivariant: \((T_Y^r)_*\eta_u=\eta_{T_U^r u}\). Time carries \(B_{S_0^{-r}b}(u)\) bijectively onto \(B_b(T_U^r u)\), taking representatives to representatives up to scalar phase. The change of variables \(y'=T_Y^r y\) in (58) consequently gives \[ s_r(u)=c_b(T_U^r u). \tag{59}\] Invariance of \(\nu\) proves its integral is \(c\) for every \(r\). Finally, \(G^r\) bijects \(B\) onto \(B_{S_0^{-r}b}\), so every moving block has the same cardinality \(|B|=m_b\). Each coefficient is at most \(\lVert f\rVert_\infty\). This proves the bound in (57) and completes the Proposition. ◻

Proof. For deterministic complex numbers \(w_m\) with \(|w_m|\le1\) and \(0<\theta\le1\), the elementary inequality \(e^t\le1+t+2t^2\) for \(|t|\le1\) gives \[\exp\bigl(\theta\Re(w_m\overline{\chi_{l_m}})\bigr) \le 1+\theta\Re(w_m\overline{\chi_{l_m}})+2\theta^2.\] The right side is nonnegative. Multiply these inequalities over \(m<n\) and integrate. On expanding the product, a nonconstant term contains one factor \(\chi_{l_m}\) or \(\overline{\chi_{l_m}}\) at each of a nonempty set of distinct indices. By (60), it is a constant multiple of the representative of a signed sum of those labels. Dissociation makes that label nonzero, so its integral vanishes. Hence \[ \int\exp\left(\theta\sum_{m<n} \Re(w_m\overline{\chi_{l_m}})\right)\,\mathrm d\eta \le(1+2\theta^2)^n\le e^{2\theta^2 n}. \tag{62}\] For \(0<\varepsilon\le1\), choose \(\theta=\varepsilon/4\) and apply Markov’s inequality. Uniformly over the deterministic coefficients, \[ \eta\left\{\frac1n\sum_{m<n} \Re(w_m\overline{\chi_{l_m}})>\varepsilon\right\} \le e^{-\varepsilon^2 n/8}. \tag{63}\]

We next allow the coefficients to be the random word \((F_m)\). Put \[a_n=\sqrt{H_n/n}+n^{-1/2}.\] Call a word likely if its probability is at least \(e^{-a_n n}\). There are at most \(e^{a_n n}\) likely words. For the random word, the mean of minus the logarithm of its probability is \(H_n\), so the total probability of the other words is at most \(H_n/(a_n n)\). Both \(a_n\) and this last bound tend to zero.

Fix arbitrary deterministic phases \(\omega_m\in\mathbb T\). On the event that \((F_m)_{m<n}\) equals a particular likely word \((v_m)\), the sum \[Y_n=\frac1n\sum_{m<n} \Re(\omega_m F_m\overline{\chi_{l_m}})\] is the sum in (63) with \(w_m=\omega_m v_m\). We do not condition the character law on that word: the event in question is simply a subset of the corresponding event for deterministic coefficients. A union bound therefore gives \[ \eta\{Y_n>\varepsilon\} \le \frac{H_n}{a_n n} +\exp\left(a_n n-\frac{\varepsilon^2 n}{8}\right) \longrightarrow0. \tag{64}\] This explains why arbitrary coupling is allowed.

Finally choose each \(\omega_m\) to make \(\omega_m\int F_m\overline{\chi_{l_m}}\,\mathrm d\eta\) nonnegative real, choosing any phase if that integral is zero. These choices are deterministic. Since \(Y_n\le1\), we have \[0\le\frac1n\sum_{m<n} \left|\int F_m\overline{\chi_{l_m}}\,\mathrm d\eta\right| =\int Y_n\,\mathrm d\eta \le\varepsilon+\eta\{Y_n>\varepsilon\}.\] First let \(n\) tend to infinity and then let \(\varepsilon\) tend to zero. ◻

Proof. Suppose the first alternative fails. Remove any initial zero coefficients from a nonempty signed relation and apply a power of \(G^{-1}\). We obtain \[b_0l+b_1Gl+\cdots+b_qG^q l=0, \qquad b_0,b_q\in\{-1,1\},\quad b_a\in\{-1,0,1\}.\] If \(q=0\), then \(l=0\) and the third alternative holds with \(p=d=1\). For \(q\ge1\), the unit coefficient at the upper end expresses \(G^ql\) as an integral combination of the preceding \(q\) labels. The coefficient at the lower end does the same in the other direction. Applying the relation to all translates shows that \(K\) is generated by \(l,Gl,\ldots,G^{q-1}l\). In particular, its torsion subgroup \(K_{\rm tor}\) is finite and its quotient \(\overline K\) is a finite-rank lattice.

The induced map \(\overline G\) is a lattice automorphism. If it has an eigenvalue \(\lambda\) with \(|\lambda|>1\), choose a nonzero complex left eigenfunctional \(\varphi\) on \(\overline K\otimes_\mathbb Z\mathbb C\). The two-sided orbit of \(\overline l\) spans this vector space. Thus \(\varphi(\overline l)\ne0\), since otherwise \(\varphi(\overline G^r\overline l)=\lambda^r \varphi(\overline l)=0\) for every integer \(r\). Choose \(p\) with \(R=|\lambda|^p>2\). On each residue class \(j\), the absolute values of the projected labels are \(|\lambda^j\varphi(\overline l)|R^m\). In any finite signed relation on distinct indices, the term with largest index strictly dominates the sum of all earlier terms, because \(\sum_{a<m}R^a<R^m\). There can be no such relation in \(K\). This proves the second alternative.

It remains to consider the case with no expanding eigenvalue. If the lattice has positive rank, \(\det\overline G=\pm1\) implies that every eigenvalue has modulus one. We use the bounded-coefficient argument in Kronecker’s root-of-unity criterion (Kronecker 1857, Theorem I); see also (Damianou 2015, Theorem 1 and its proof). The characteristic polynomial of \(\overline G^n\) has integer coefficients, each bounded independently of \(n\): its coefficients are elementary symmetric functions of a fixed number of numbers of modulus one. Only finitely many such polynomials exist. For each eigenvalue \(\lambda\), all the powers \(\lambda^n\) are roots of these finitely many polynomials. Hence two powers agree and \(\lambda\) is a root of unity.

Choose \(p\) such that \(\overline G^p\) is unipotent and \(G^p\) is the identity on \(K_{\rm tor}\). The latter is possible because the automorphism group of a finite group is finite. This argument includes the rank-zero case. For some \(d_0\ge1\), the map \(N=G^p-\mathop{\mathrm{id}}\) satisfies \(N^{d_0}K\subseteq K_{\rm tor}\), and \(NK_{\rm tor}=0\). Consequently \(N^{d_0+1}K=0\). The additional application of \(N\) accounts for possible maps from a free part into torsion; an invariant splitting of \(K\) has not been assumed. Taking \(d=d_0+1\) and expanding \((\mathop{\mathrm{id}}+N)^m\) proves (65). ◻

Proof. We induct on the degree. If \(P(m)=l_0\) is constant, condition \(\overline{\chi_{l_0}}\) onto \(x\). Its conditional expectation is a bounded measurable function \(g\) on \(X\), and \[\int F(T_X^{pm}x)\overline{\chi_{l_0}}\,\mathrm d\eta =\int F(T_X^{pm}x)g(x)\,\mathrm d\mu_X(x)\longrightarrow0\] by mixing and centering. Mixing for bounded functions follows from the set formulation by simple-function approximation. This proves the base case, including \(l_0=0\).

For the inductive step, choose deterministic phases \(\omega_m\) aligning the integrals in the assertion. In the fixed Hilbert space \(L^2(\eta)\) put \[v_m=\omega_m F(T_X^{pm}x)\overline{\chi_{P(m)}}.\] Their norms are bounded by \(\lVert F\rVert_\infty\). At a fixed positive lag \(h\), the absolute value of \(\langle v_m,v_{m+h}\rangle\) equals \[\left|\int F_h(T_X^{pm}x) \overline{\chi_{P(m+h)-P(m)}}\,\mathrm d\eta\right|, \qquad F_h=(F\circ T_X^{ph})\overline F.\] The constant phases from alignment and from (60) disappear inside this absolute value. Apply the induction hypothesis to the centered function \(F_h-\int F_h\,\mathrm d\mu_X\) and the polynomial of lower degree. The remaining constant contribution has absolute value at most \(|\int F_h\,\mathrm d\mu_X|\). Therefore \[ \limsup_{n\to\infty}\frac1n\sum_{m<n} |\langle v_m,v_{m+h}\rangle| \le\left|\int F_h\,\mathrm d\mu_X\right|. \tag{66}\]

The following averaging step is the Hilbert-space van der Corput argument; see (Bergelson and Håland Knutson 2008, Theorem 4.1) for a real Hilbert-space formulation, using the real part of the inner product in the complex case. For completeness, fix \(H\ge1\). Replacing each \(v_m\) in its mean by \(H^{-1}\sum_{a=0}^{H-1}v_{m+a}\) changes the mean by a vector of norm at most \(2H\lVert F\rVert_\infty/n\). Convexity of the squared norm bounds the squared norm of this new mean by the mean of the squared norms of its summands. Expanding those squares, grouping terms by the lag, and using (66) gives \[\limsup_{n\to\infty} \left\lVert\frac1n\sum_{m<n}v_m\right\rVert^2 \le\frac{\lVert F\rVert_\infty^2}{H} +\frac2H\sum_{h=1}^{H-1} \left(1-\frac hH\right) \left|\int F_h\,\mathrm d\mu_X\right|.\] The harmless shift of the averaging range by at most \(H\) indices does not change any of these limiting bounds. Mixing and centering imply \(\int F_h\,\mathrm d\mu_X\to0\) as \(h\to\infty\). Letting \(H\to\infty\) shows that the Hilbert-space means tend to zero. Testing them against the constant function \(1\) yields exactly the mean of the aligned absolute integrals, proving the induction. ◻

Proof of Proposition 22. Apply Lemma 25. In the first alternative, Lemma 24 applies directly. In the second, apply it on each of the finitely many residue classes modulo \(p\), using the progression-word entropy estimate after that lemma. In the third, (65) gives a polynomial on every residue class. Apply Lemma 26 there with \(F=f\circ T_X^j\). This function is centered by invariance of \(\mu_X\). In each case the finitely many progression averages tend to zero and combine to give the full average in (61). ◻