Chapter 3. Operator squares and completeness of exact lines

An exact line is a one-dimensional space spanned by a function of modulus one that can be multiplied by modulus-one functions at the other positions in the same row or column to give a constant phase.

This chapter follows Definition 4.1 and Lemmas 4.2–4.7 of R (pp. 14–21). Fix a parameter square satisfying the local kernel identities for grouped rectangles from Chapter 2. The argument decomposes the space visible from both the row and the column into exact lines. The steps establish intertwining of projections, fixed positive parts, an abelian unitary group, and the reverse inclusion needed for completeness.

Fixed source, from page 14

1. The conclusion to be proved

Fix a full parameter square $E$ satisfying Proposition 3.5. All subsequent probability laws are conditioned on this $E$. Partition the rows into three nonempty groups $P,A,B$ and the columns into three nonempty groups $r,s,t$. A rectangular cell $Y_{i,l}$ is the complete state vector in that rectangle.

Set $H_{i,l}=L^{2}(\mu _{i,l})$, where $\mu _{i,l}$ is the marginal law of the rectangular cell. The inner product is linear in the second variable:

Every span below means a closed linear span. The row-active space $V$ of a cell is the closed span of the conditional expectations, onto that cell, of bounded functions of the other two cells in its row. The column-active space $W$ is defined using the other two cells in its column. Both contain the constants and are invariant under complex conjugation.

The aim is to show that $D=V\cap W$ is spanned by modulus-one functions satisfying both the row and the column product relations. At the corner cell $(P,r)$, the proof gives the more specific description

Here, $h(z)$ takes values in an abelian unitary group constructed in the proof. Finite monomials in the group's characters, evaluated at $h(z)$, satisfy both exact product relations and span this $L^{2}$ space.

Completeness is the remaining issue: constructing exact lines gives only $L^{2}(\sigma (h))\subseteq D$. The reverse inclusion needs a separate argument; $D$ cannot yet be treated as an algebra of functions.

2. The precise probabilistic input

Section 3 of R gives the following statements for every ordering of the row groups and column groups:

1. The three rectangular cells in each grouped row or column are pairwise independent.
2. Given the complete grouped row $P$, any two grouped columns are independent, and each column's conditional law depends only on its entries in $P$.
3. Given row group $P$ and the two whole columns other than $t$, the conditional law of $Y_{A,t}$ is its own row kernel given $Y_{A,r},Y_{A,s}$.

The input does not assert joint independence of all three rectangular cells, symmetry under interchanging the axes, or mixing of the auxiliary state system.

Write

Both $(z,c,d)$ and $(z,x,y)$ are pairwise independent triples. By condition 2, given $(z,c,d)$, the conditional kernel of $(x,y)$ depends only on $z$. Hence $(c,d)$ and $(x,y)$ are conditionally independent given $z$. Since $x$ and $z$ are independent, for any bounded $f,g$,

Thus $x$ is independent of the whole pair $(c,d)$. The same holds for $y$. Conversely, each of $c$ and $d$ is independent of the whole pair $(x,y)$. Combining this with independence of $c$ and $d$, and of $x$ and $y$, gives the four triple product laws needed later:

No joint independence of all four variables $(x,y,c,d)$ is claimed. The group structure will use the fourth-order correlation that these hypotheses still allow.

3. Two basic facts about exact tuples

Definition 4.1 and Lemma 4.2 appear on pp. 14–15. For an $m$-coordinate law $\lambda$, an exact tuple satisfies $|\chi _i|=1$ and $\prod \chi _i(Y_i)=c$, where $c$ is a fixed constant of modulus one. The product identity holds almost everywhere under $\lambda$. Representatives $\chi _i$ that differ by a constant phase define the same line.

First, an exact line belongs to the active space of its coordinate. Indeed,

is already a bounded function of the other coordinates. Conditioning on $Y_i$ leaves the right-hand side equal to $\chi _i$.

Second, if $m\ge 3$ and every subfamily of $m-1$ coordinates is independent, distinct exact lines are orthogonal and their partner lines are unique. The partners are the lines at the other coordinates that form an exact tuple with the given line. To prove this, put $a_i=|\int \chi _i \,\mathrm{d}\mu _i|$. Solve for $\chi _i$ and integrate over the other coordinates to obtain

If some $a_i=0$, the product identity forces at least one other $a_j=0$. Every right-hand side then contains a zero, so all $a_j$ vanish. If they are all positive, set $b_i=-\log a_i\ge 0$. Then $2b_i=\sum _jb_j$, and summing gives $(m-2)\sum _jb_j=0$. Thus all $a_i=1$. The conditions $|\chi _i|=1$ and $|\mathbb E[\chi_i]|=1$ imply that $\chi _i$ is almost everywhere constant. An exact tuple therefore consists either entirely of constants or entirely of mean-zero components.

The coordinatewise quotient of two exact tuples is again exact. If the two lines differ at one cell, the quotient there is nonconstant. All quotient functions then have mean zero, so corresponding representatives are orthogonal at every cell. If the lines agree at one cell, that quotient is constant, forcing every quotient to be constant; this proves uniqueness of the partner lines. Multiplication and conjugation give the group operations. A separable Hilbert space contains at most countably many nonzero mutually orthogonal lines.

The completeness proof below mainly uses the inclusion of exact lines in active spaces and invariance under conjugation. Uniqueness of partners does not replace the operator argument.

4. Four edge operators and two rectangle identities

For Lemma 4.3 (pp. 16–17), arrows indicate the direction in which an operator sends functions:

For example, in the law of row $A$,

The variable $x$ is independent of each individual cell $s$ and $t$ in that row. Given $x$, their marginals remain the fixed laws $\mu _{A,s}$ and $\mu _{A,t}$. Jensen's inequality gives

Thus the operator is a contraction between fixed Hilbert spaces. Its adjoint is reverse conditional expectation for the same conditional joint kernel. The other three edges are treated in the same way. Testing matrix coefficients on countable dense sets gives parameter measurability from the regular conditional kernels, and hence strong measurability on each vector.

Now fix the five cells $(z,c,d,x,y)$. Given row group $P$, column $s$ is independent of column $r$, and column $t$ is independent of column $r$. Further conditioning on $x,y$ therefore leaves the marginal of the remaining two cells in column $s$ equal to its column kernel given $c$; column $t$ has the corresponding kernel given $d$. Denote the joint $L^{2}$ spaces of these pairs by $\mathcal H_s$ and $\mathcal H_t$, and let $Q:\mathcal H_t\to \mathcal H_s$ be their conditional-expectation operator given the five cells.

Embed each single-cell function space into the corresponding two-cell space, with isometries $\iota _{A,s}$, $\iota _{B,s}$, $\iota _{A,t}$, and $\iota _{B,t}$. Condition 3 from Section 2 gives exactly

The same condition applies with $s$ and $t$ reversed, giving

Also, $J_s=\iota _{A,s}^*\iota _{B,s}$ and $J_t=\iota _{A,t}^*\iota _{B,t}$. Consequently,

The reverse identity is

Together these give equation (23) of the source. Both are needed: the first alone does not give the two directions of projection intertwining used below. Countable dense vector tests and boundedness make them identities of whole operators on a common conull set for the five-cell law.

5. Intertwining before projection onto the intersection

For Lemma 4.4 (pp. 17–18), commutation of the orthogonal projections onto $V$ and $W$ must first be proved.

5.1 What the edge ranges span

The essential closed span of the ranges of $L_A(x)$ is $V_{A,s}$. To see this, condition a product test $w(x)h(Y_{A,t})$ from the other two cells onto $Y_{A,s}$. The result is the Bochner integral, that is, the norm integral in the Hilbert space,

The other two cells have a product marginal, so finite sums of product tests are dense in their $L^{2}$ space. If $v$ is orthogonal to every displayed integral, fix $h$ from a countable dense family and vary bounded $w$. This gives $\langle v,L_A(x)h\rangle =0$ for almost every $x$. Density and the uniform contraction bound extend the conclusion to every $h$. Thus orthogonality to all integrals is exactly orthogonality to the ranges of almost all edge operators. The converse follows by integration.

The argument applies to every edge and reverse edge. Each edge operator therefore has range in the target active space for its own direction, and annihilates the orthogonal complement of the corresponding source active space. Here and below, range arguments are first made on countable dense sets of vectors and then placed on a common conull set.

5.2 Both rectangle identities are needed for intertwining

From $L_AJ_t=J_sL_B$, the range of $J_s(c)L_B(y)$ lies in $V_{A,s}$. The pair $(c,y)$ has a product law. Fubini therefore gives, for almost every fixed $c$, this range inclusion for almost every $y$. The ranges of $L_B(y)$ essentially span $V_{B,s}$, so

Taking the adjoint of the second rectangle identity gives $J_s(c)^*L_A(x)=L_B(y)J_t(d)^*$. Independence of $(c,x)$ and the essential span of the ranges of $L_A$ similarly give

This means that $J_s(c)$ sends the orthogonal complement of $V_{B,s}$ into the orthogonal complement of $V_{A,s}$. Only after combining the two inclusions do we obtain

Likewise, the first rectangle identity and the product law of $(x,d)$ give $L_A(x)W_{A,t}\subseteq W_{A,s}$. The second identity and the product law of $(x,c)$ give the adjoint inclusion. Hence

The same argument applies to every edge and every permitted ordering of the groups.

5.3 Commutation and density of the restricted ranges

By (25), $P^V_{A,s}$ preserves the essential closed span of the ranges of $J_s(c)$, namely $W_{A,s}$. As a self-adjoint projection, it also preserves $W_{A,s}^\perp$. Thus $P^V$ commutes with $P^W$, and we may now write

The space $D$ remains invariant under conjugation. Combining (26) with the row-active restriction of $L_A$ itself gives

The same holds for the other edges. Finally, project the range-span identity for $L_A$ onto $D_{A,s}$ and use the preceding identity. The ranges of the restricted family $L_A|_D$ then essentially span the target $D$. The reverse family has the same property. This is the density used in the cancellation argument below.

6. Why the positive parts are fixed

For Lemma 4.5 (pp. 18–19), work on the four $D$ spaces and write

Set $H_{1}=D_{A,s}$, $H_{2}=D_{A,t}$, $H_{3}=D_{B,s}$, and $H_{4}=D_{B,t}$. The operator types are $F:H_{2}\to H_{1}$, $G:H_{4}\to H_{3}$, $K:H_{3}\to H_{1}$, and $N:H_{4}\to H_{2}$. The source and target of every composition can now be checked explicitly.

The two rectangle identities are equation (27) of the source:

Combining these with their adjoints gives

For the first identity, for example, the adjoint of the second rectangle identity gives $K^*F=GN^*$. Multiply on the left by $K$ and substitute $KG=FN$. The other three identities likewise use only well-defined compositions, without inverses.

The parameters $(c,x,d)$ in the first identity of (28) have a product law. Take an independent copy $c'$, compare the identities at common $x,d$, and subtract:

This holds for almost every $(c,c',x)$. The ranges of $F(x)$ essentially span $H_{1}$, so the operator in brackets vanishes. Thus $KK^*$ is independent of $c$. Take the adjoint of the same identity and compare $d,d'$. The essential span of the ranges of $F^*$ is $H_{2}$, giving a fixed $NN^*$.

The second identity of (28) has product parameters $(c,y,d)$, giving fixed $K^*K$ and $N^*N$ in the same way. In the first identity of (29), the product parameters $(x,d,y)$ give fixed $F^*F$ and $G^*G$. In its second identity, the product parameters $(x,c,y)$ give fixed $FF^*$ and $GG^*$. These comparisons use the triple product laws proved in Section 2. Joint independence of all four parameters is unnecessary.

If $v\in \operatorname{Ker}(F^*F)$, the fixed source square gives $F(x)v=0$ for almost every $x$. Thus $v$ is orthogonal to the ranges of almost all $F(x)^*$, whose essential span is $H_{2}$. Hence $v=0$. Every source and target square is injective by the same argument.

For typical $x$, the operator $F(x)$ has zero kernel and dense range, the latter following from $\operatorname{Ker} F(x)^*=0$. The polar decomposition $F(x)=U_F(x)|F|$ consequently has full initial and final spaces, so $U_F(x)$ is unitary between the corresponding $D$ spaces.

6.1 Preserving the rectangle identity without a bounded inverse

Put $A=|F|$ and $B=|G|$. These are fixed positive injective contractions. The identity $F^*FN=NG^*G$ gives $A^{2}N=NB^{2}$. Uniform polynomial approximation to the square-root function on $[0,1]$ then yields $AN=NB$.

Substituting $F=U_FA$ and $G=U_GB$ into $FN=KG$ gives

We cannot multiply by a purported bounded $B^{-1}$. Positivity and injectivity of $B$ only give density of $\operatorname{Ran} B$, since $(\operatorname{Ran} B)^\perp =\operatorname{Ker} B=0$. The bounded operator in brackets vanishes on the dense range $\operatorname{Ran} B$ and therefore on all of $H_{4}$.

Thus $U_FN=KU_G$. The two sides have polar decompositions

Both right-hand factors are positive, and both preceding products are unitary. Uniqueness of polar decomposition gives

For example, on $\ell ^{2}$ the operator $Be_n=n^{-1}e_n$ for $n\ge1$ is injective with dense range, but its inverse has no bounded extension. The cancellation argument also applies to this $B$. A uniform lower bound on the positive parts is therefore unnecessary.

Strong measurability of the polar parts on each vector follows from

Each approximating operator has norm at most one. The fixed source square is injective, and spectral calculus gives the strong limit. A countable sequence of $\varepsilon$ suffices for measurability.

7. From the polar rectangle relation to an abelian group

In Lemma 4.6 (pp. 19–20), denote the four polar parts by $a(x)$, $b(y)$, $c_{0}(c)$, and $d_{0}(d)$, respectively. They satisfy equation (30) of the source. At this point they are still unitary maps between different $H_{i}$ spaces; a common unitary group has yet to be identified.

7.1 Reference identifications at one typical $z_{0}$

Fix a typical $z_{0}$. Given $z_{0}$, the pairs $(a,b)$ and $(c_{0},d_{0})$ are independent. Each individual conditional marginal equals its unconditional marginal, because $z$ is independent of each of $x,y,c,d$.

The strong operator topology on unitary maps is separable and metrizable; multiplication and inversion are continuous. The identity $a d_{0}=c_{0} b$ is a closed condition. Since the conditional joint law is the product of the two pair laws, the identity holds on the full Cartesian product of their conditional supports. Otherwise continuity would give an open neighborhood of positive measure where it fails, contradicting the almost-everywhere identity.

Choose one point $(a_*,b_*)$ and $(c_*,d_*)$ from the two supports. They satisfy $a_*d_*=c_*b_*$. Take $\mathcal H=H_4$ and fix the four identifications

Replacing each edge by $J_{\mathrm{target}}^* UJ_{\mathrm{source}}$ makes all four reference edges equal to $\operatorname{id}$ and preserves the rectangle relation. The maps $J_{i}$ are chosen once and thereafter do not depend on $x,y,c,d$ or $z$.

Both conditional pair supports now contain $(\operatorname{id},\operatorname{id})$. Fixing the second pair at $(\operatorname{id},\operatorname{id})$ forces $a=b$ on the first support; reversing the roles forces $c_{0}=d_{0}$. Thus $a$ and $b$ have the same unconditional marginal, denoted by $\rho$, while $c_{0}$ and $d_{0}$ have the same marginal, denoted by $\tau$. Both supports contain $\operatorname{id}$.

In the two conditional supports at $z_{0}$, substitute $(u,u)$ and $(v,v)$ to obtain $uv=vu$. Thus elements of $\operatorname{supp} \rho$ commute with elements of $\operatorname{supp} \tau$. This cross-commutation does not yet say that either support is a group.

7.2 Translation invariance from general $z$

For a general typical $z$, fix a pair $(c_{0},d_{0})=(u,v)$. Conditional independence leaves the marginals of $a$ and $b$ equal to $\rho$. The rectangle relation $a=u b v^{-1}$ therefore gives

Here $u\rho v^{-1}$ is the image of $\rho$ under $w\mapsto uwv^{-1}$. The identity holds for almost every actual pair $(c_{0},d_{0})$. Since $c$ and $d$ are independent, the two polar variables have unconditional joint law $\tau \otimes \tau$.

For a bounded continuous test $f$, the map

is continuous: the group operations are continuous, and bounded convergence passes pointwise convergence through the integral. The measure identity therefore extends from $\tau \otimes \tau$-almost every pair to every point of $\operatorname{supp} \tau \times \operatorname{supp} \tau$. Interchanging the two pairs likewise gives $\tau =u\tau v^{-1}$ for all $u,v\in \operatorname{supp} \rho$.

Write $S=\operatorname{supp} \rho$ and $T=\operatorname{supp} \tau$. Taking $v=\operatorname{id}$ shows that every $u\in T$ satisfies $uS=S$. Since $\operatorname{id}\in S$, we have $u\in S$, hence $T\subseteq S$. The reverse argument gives $S\subseteq T$. Therefore $S=T=\mathcal G$.

The set $\mathcal G$ is closed in the ambient unitary group. For $g\in \mathcal G$, the identity $g\mathcal G =\mathcal G$ makes it closed under multiplication, and $\operatorname{id}\in g\mathcal G$ gives $g^{-1}\in \mathcal G$. Cross-commutation now becomes commutation within the group. Thus $\mathcal G$ is a closed abelian unitary subgroup, and $\rho$ is invariant under all its left translations.

7.3 Why the ratio depends only on $z$

Since $\mathcal G$ is abelian, the rectangle relation becomes

Given $z$, the two sides are independent random elements and are almost everywhere equal. Their common conditional law must be a point mass. This can be checked on a countable determining class: for an event $A$ with conditional probability $p$, the probability that the first element belongs to $A$ and the second does not belong to $A$ is $p(1-p)$. Equality makes this zero, forcing a point mass for the conditional kernel on the standard Borel space.

The regular conditional law is measurable in $z$, and a Dirac probability kernel determines a measurable point. Hence there is a measurable $h(z)\in \mathcal G$ such that

This measurable point is recovered with $E$ fixed. Neither the reference identifications nor the later basis vectors are yet claimed measurable in the full parameter $E$.

8. Establishing a common eigenbasis

Commutation of unitary operators alone does not give a common eigenbasis. For example, the abelian unitary group generated by the bilateral shift on $\ell ^{2}(\mathbb Z )$ has no nonzero eigenvector: the eigenvalue equation forces every coordinate to have the same modulus, which is incompatible with square summability.

The additional input here is the left-invariant probability law $\rho$ just constructed. In $\mathcal H$, choose an orthonormal basis $(e_n^0)$, take $q_n>0$ and $\sum q_n<\infty$, and define $Q e_n^0=q_n e_n^0$. This operator is positive, injective, and trace class. Form the Bochner integral in trace norm

The integral is well defined. For finite-rank $Q$, strong unitary convergence $U_m\to U$ implies trace-norm convergence of $U_mQU_m^*$. For general $Q$, approximate by finite-rank operators and use $\|U(Q-Q_N)U^*\|_{1}=\|Q-Q_N\|_{1}$ to control the tails uniformly. The integrand is therefore trace-norm continuous, with constant norm $\|Q\|_{1}$, so its integral exists and is a positive trace-class operator.

For every $V\in \mathcal G$, left invariance of $\rho$ gives

Thus $Q_{0}$ commutes with all of $\mathcal G$. Moreover, for every $v\ne 0$,

Since $U^*v\ne 0$ and $Q$ is injective, every integrand is strictly positive. No uniform positive lower bound is required: a strictly positive measurable function has strictly positive integral against a probability measure. Thus $Q_{0}$ is injective.

By the spectral theorem for positive compact operators, each nonzero eigenspace is finite dimensional. Their orthogonal sum spans the orthogonal complement of $\operatorname{Ker} Q_{0}$, which is all of $\mathcal H$. Since $Q_{0}$ commutes with $\mathcal G$, every such finite-dimensional eigenspace is invariant under every group element. The resulting commuting unitary matrices can be simultaneously diagonalized. Combining their common eigenbases gives, in $\mathcal H$, an orthonormal eigenbasis $(e_n)$.

The common eigenbasis thus comes from the positive injective compact operator $Q_{0}$ in the commutant. It does not follow from an unrestricted spectral assertion for abelian groups. The proof neither assumes $\mathcal G$ compact nor appeals to a Haar measure before constructing an invariant measure.

Write $U e_n=\gamma _n(U)e_n$. Each $\gamma _n$ is a continuous circle-valued character, preserving group multiplication. For diagonal unitaries, coordinatewise convergence is equivalent to strong convergence. Given $v=\sum v_ne_n$, first retain finitely many coordinates so that $\sum_{n>N}|v_n|^2$ is small, then control those coordinates by their convergence. The converse follows by applying the operators to each $e_n$. These coordinates therefore generate the Borel structure of $\mathcal G$.

For $h(z)$, trigonometric polynomials in finitely many coordinates are dense in $L^{2}$ for its actual law. Indeed, trigonometric polynomials are uniformly dense in continuous functions on a finite-dimensional torus, and those functions are dense in $L^{2}$ for a finite Borel probability measure. Increasing conditional expectations onto the finite-coordinate $\sigma$-algebras give the remaining density argument. Consider all finite character monomials

Here, $J$ is a finite subset of the common eigenbasis index set and $m_n$ is an integer for each $n\in J$.

These monomials span $L^{2}(\sigma (h(z)))$. The law of $h(z)$ need not first be identified with Haar measure.

By (31),

In column $r$, the three representatives $\gamma (h(z))$, $\overline{\gamma (a(x))}$, and $\gamma (b(y))$ have product one. In row $P$, the corresponding partners are $\overline{\gamma (c_{0}(c))}$ and $\gamma (d_{0}(d))$. Thus $\gamma (h(z))$ satisfies both exact product relations and belongs to $D_{P,r}$. This proves $L^{2}(\sigma (h(z)))\subseteq D_{P,r}$; the reverse inclusion remains.

9. The reverse inclusion without an algebra assumption on $D$

For Lemma 4.7 (p. 21), take any three-cell row with positions $r,s,t$. Let $P_r$, $P_s$, and $P_t$ be the already constructed orthogonal projections onto the intersections of the row-active and column-active spaces. Define the bilinear contraction

Since $(Y_s,Y_t)$ has a product marginal,

Contractivity of conditional expectation gives equation (35) of the source:

This estimate holds for all $g,h\in L^{2}$. The projected functions $P_s g$ and $P_t h$ need not remain bounded.

9.1 Two incoming edges give (33)

Let $L_{r\leftarrow t}(v)$ be the row kernel given $Y_s=v$, acting from $t$ to $r$. For bounded tests, disintegrate over $Y_s$ and use independence of $Y_s$ and $Y_r$ to get

The established intertwining relation $P_rL_{r\leftarrow t}(v)=L_{r\leftarrow t}(v)P_t$ allows the bounded linear projection to pass inside the Bochner integral, giving

Alternatively, condition on $Y_t$ and use the edge $L_{r\leftarrow s}$. The same argument gives

These Bochner integrals extend to all $L^{2}$ inputs, because

By (35) and dense approximation in $L^{2}$, both identities hold for every $L^{2}$ input, including projected inputs that may be unbounded. Idempotence then gives

This is equation (33) of the source:

Thus if $g\in D_s$ and $h\in D_t$, then $C_r(g,h)$ is fixed by $P_r$ and belongs to $D_r$. These contractions in fact span $D_r$. The unrestricted product contractions span the row-active space $V_r$, and $P_rV_r=D_r$. By (33), every projected contraction is a contraction with both inputs restricted to $D$. The same argument works along columns.

This uses neither closure of $D$ under multiplication nor a spanning conclusion from a single edge. Both incoming edges are required.

9.2 First use of spanning: the $A$ and $B$ cells depend on their polar variables

In row $A$, take $g\in D_{A,s}$ and $h\in D_{A,t}$. Contracting onto $x=Y_{A,r}$ and using the fixed marginals gives

The conjugation is required by the convention that the inner product is linear in the second variable. Since $D$ is invariant under conjugation, the formula can be computed entirely on the restricted $D$ spaces.

With the fixed source positive part $A_{0}=|L_A|$, we have $L_A(x)|_D=a(x)A_{0}$. Under the fixed Hilbert-space identifications in Section 7, this positive part is conjugated by a fixed unitary and remains independent of $x$. The coefficient above is therefore a measurable function of $a(x)$. The restricted spanning identity from the previous subsection and closedness of $L^{2}(\sigma (a(x)))$ give

Similarly, $D_{B,r}\subseteq L^{2}(\sigma (b(y)))$. Only inclusion is needed here; equality has not been asserted.

9.3 Second use of spanning: every corner-cell $D$ function depends on $h(z)$

Use the restricted spanning identity in column $r$. Its inputs lie in $D_{A,r}$ and $D_{B,r}$ and can therefore be written as $f(a(x))$ and $g(b(y))$, respectively, as $L^{2}$ functions.

First take $f,g$ bounded. Given $z$, the law of $b$ is still the fixed law $\rho$, while $a=h(z)b$. Hence

The right-hand side depends only on $h(z)$ and thus belongs to $L^{2}(\sigma (h(z)))$.

If $f(a)$ and $g(b)$ are only in $L^{2}$, approximate them by bounded truncations in their own $\sigma (a)$ and $\sigma (b)$. The truncations need not remain in $D$: their only purpose is to show that the limiting contraction is measurable with respect to $\sigma (h)$, and the restricted spanning assertion is not applied to the truncations themselves. The product marginal of $x$ and $y$, together with (35), gives

The target space $L^{2}(\sigma (h))$ is closed, so every restricted contraction belongs to it. These contractions span $D_{P,r}$, proving $D_{P,r}\subseteq L^{2}(\sigma (h(z)))$.

Combining this with the inclusion proved in Section 8 gives (34). Only now can the multiplicative structure be identified. By the permitted reorderings of row groups and column groups, any rectangular cell can be moved to the corner. Exact lines therefore span the intersection of the row-active and column-active spaces at every cell.

10. Quantifiers and common conull sets

Versions of the original probability kernels are first chosen on countable determining classes. Edge identities are first checked on countable dense pairs of vectors and extended to whole operators by contractivity. The product marginals justify each particular Fubini comparison and the choice of two independent parameter values. No uncountable intersection of exceptional sets for all bounded functions is required.

The choice of $z_{0}$ in Lemma 4.6, the reference unitary identifications, and the common eigenbasis are all made with a typical $E$ fixed. The conclusion concerns only the active spaces and exact lines of the original conditional laws and is therefore intrinsic. Measurability of the intermediate choices in $E$ has not been established. Section 5 of R must construct measurable label families separately.