Principal groupoids with hidden group factors

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026. New original text is public domain (CC0).

A principal groupoid has no nonidentity isotropy arrows. Its measurable structure can nevertheless carry a group coordinate invisible to the unit sigma-field. The preceding binary example gives an abelian random-operator algebra. Here we construct the full countable-group version, classify all its proper transverse functions, and compute a properly infinite factor when the group is the free group on two generators.

The factor example also tests the usual tensor construction of two commuting copies. An explicit controlled translation lies in the relative commutant of one copy and outside the other; the two copies do not generate the tensor-field algebra. This is a failure of that particular construction at the weaker measurable scope. It does not prove that no other endomorphism and symmetry satisfy Connes's Corollary 11. That source assertion remains open at this scope.

We use the fully proved subgroup in Countable generation and isotropy topologies, Theorem 1.2, and its perfect-set lemma. Principal groupoids with extra fibre information supplies the binary predecessor and the random-operator conventions. Ordinary sigma-finite integration, separable Hilbert spaces, the double commutant theorem and Hilbert tensor products are prerequisites. Every measure, kernel, covariance, commutant and factor calculation needed for the present construction is proved below. No group-factor classification, tensor primeness theorem or general modular bridge is imported.

1. A countable partition carrying no ordinary Borel information

Put $X=[0,1]$, with ordinary Borel sigma-field $\mathcal B_X$ and Lebesgue probability $\ell$. Let $D\subset\mathbb R$ be the rational vector subgroup from the cited full construction: $1\notin D$, and every nonempty perfect subset of $\mathbb R$ meets $D$ and $D+1$. In fact every integer translate $D+n$ meets every perfect set, by applying the intersection property to its translate by $-n$. Distinct integer translates are disjoint: a nonzero integer in $D$ would put $1$ in $D$ by rational division.

Let $\Gamma$ be a nonempty countable group, with identity $e$. Choose distinct integers $j(g)$, $g\in\Gamma$. For $g\ne e$ set $A_g=X\cap(D+j(g))$, and set $$A_e=X\setminus\bigcup_{g\ne e}A_g,\qquad a(x)=g\quad(x\in A_g). \tag{1.1}$$ Thus $A_e$ contains $X\cap(D+j(e))$. Every class is nonempty and meets every nonempty perfect subset of $X$.

Lemma 1.1 (all nontrivial unions). Every Borel set disjoint from one $A_g$ is Lebesgue-null. If $\Gamma$ has at least two elements, every nonempty proper union of the classes $A_g$ is neither Borel nor Lebesgue-measurable.

Proof. A positive-measure Borel set contains a nonempty perfect subset by the complete perfect-set lemma in the cited topology lesson. That subset meets $A_g$, proving the first assertion. A nonempty proper union $C$ and its complement each contain some class. A positive-measure Borel subset of either one would contain a perfect set missing a class in the other, which is impossible. Consequently both have inner Lebesgue measure zero. If either were Lebesgue-measurable, so would its complement, and their union would have measure zero instead of one. $\square$

In particular, if a scalar function $q$ on the countable set $\Gamma$ has $q\circ a$ ordinary Borel, or even completion-measurable, then $q$ is constant. Otherwise the inverse image of one of its attained values is a nonempty proper union in Lemma 1.1. This also applies to extended nonnegative values.

Give the underlying set $X$ the enlarged sigma-field $$\mathcal S=\sigma(\mathcal B_X,(A_g)_{g\in\Gamma}).$$ Its sets are exactly $$D_0=\bigcup_{g\in\Gamma}(B_g\cap A_g),\qquad B_g\in\mathcal B_X. \tag{1.2}$$ These sets form a sigma-field containing the generators. For any sigma-finite Borel measure $\beta$ on $X$ which vanishes on every Borel set disjoint from any $A_g$, define $$\eta_\beta(D_0)=\sum_{g\in\Gamma}\beta(B_g). \tag{1.3}$$ The stated null condition makes this well-defined: two presentations have branchwise Borel symmetric differences disjoint from $A_g$. For disjoint $D_n$, their branchwise Borel representatives overlap only in such null sets; disjointizing in their enumeration order proves countable additivity, then summing the nonnegative branch values proves it for $\eta_\beta$. A countable finite-$\beta$ cover, intersected with the countably many $A_g$, proves sigma-finiteness. The measure $\ell$ satisfies the null condition by Lemma 1.1.

For $\beta=\ell$, write $\eta=\eta_\ell$. On every branch, $$\eta(B\cap A_g)=\ell(B)\quad(B\in\mathcal B_X). \tag{1.4}$$ Simple functions and monotone convergence give $$L^2(X,\mathcal S,\eta)\cong K\otimes\ell^2(\Gamma),\qquad K=L^2(X,\ell). \tag{1.5}$$ Explicitly, a vector with branchwise Borel representatives $f_g$ goes to $(f_g)_{g\in\Gamma}$. Formula (1.3) gives its norm; branchwise simple approximation proves both well-definedness and surjectivity. This is an isomorphism of measured vector spaces, not a pointwise bijection of $X$ with $X\times\Gamma$.

2. One pair arrow, with a measurable group label

Take the underlying pair groupoid $G=X\times X$, with arrow $(y,x):x\to y$, inverse $(x,y)$ and product $(z,y)(y,x)=(z,x)$. Adjoin the labels $$d(y,x)=a(y)a(x)^{-1},\qquad \mathcal B_G=\sigma\bigl(\mathcal B(X^2),\,d^{-1}(\{g\}):g\in\Gamma\bigr). \tag{2.1}$$ This sigma-field is separated and countably generated; all arrow singletons are measurable. Its trace on the identity diagonal is exactly $\mathcal B_X$, since $d(x,x)=e$.

Proposition 2.1 (all groupoid operations). The pair groupoid with (2.1) is measurable, principal and has one orbit. Its range-fibre sigma-field, under the source bijection, is $\mathcal S$. If $|\Gamma|>1$, the arrow space is not standard Borel.

Proof. Range and source are measurable through the ordinary product generators. Equality of two unit coordinates is product-Borel, so the composable-pair set is measurable. The identities $$d(x,y)=d(y,x)^{-1},\qquad d(z,x)=d(z,y)d(y,x) \tag{2.2}$$ make inversion and composition measurable on each new generator; the ordinary generators are handled by the ordinary pair operations. Every endpoint pair has exactly one arrow, so isotropy is trivial and there is one orbit.

At fixed $y$, the equation $d(y,x)=g$ is $a(x)=g^{-1}a(y)$. As $g$ varies these are all the classes $A_h$. This proves the fibre assertion. If the arrows were standard Borel, a measurable range fibre would be standard Borel and its injective measurable source map to the standard Borel $X$ would have measurable inverse, by the exact Borel-image prerequisite used in the predecessor lesson. It would make all $A_h$ ordinary Borel, contradicting Lemma 1.1. $\square$

Define $\nu^y$ to be $\eta$ under the source bijection $G^y\to X$. For ordinary Borel $B,C$ and finite $F\subset\Gamma$, $$\nu^y\bigl((C\times B)\cap\{d\in F\}\bigr) =1_C(y)|F|\ell(B). \tag{2.3}$$ Each relative label has mass one. In particular $\nu^y(G^y)=|\Gamma|$, which is infinite when $\Gamma$ is infinite.

Proposition 2.2 (a faithful proper transverse function). The family $\nu$ is a measurable, faithful, proper transverse function, including the uniform translated-cover form of properness.

Proof. The events in (2.3), first with singleton $F$, form a generating pi-system. Choose increasing finite $F_n\uparrow\Gamma$. On the measurable cover $\{d\in F_n\}$, every fibre measure is finite with total $|F_n|$. The finite-measure pi-lambda argument on each cover and monotone convergence therefore prove kernel measurability for every arrow event, without complementing an infinite total mass.

Left translation preserves the actual source coordinate, hence preserves $\eta$. This proves exact transverse invariance; nonzeroness gives faithfulness. The arrow sets $E_n=\{d\in F_n\}$ exhaust $G$. Translation by a label $k$ replaces $F_n$ by a translate of the same finite size. Consequently every translated range-fibre mass is $|F_n|$, uniformly in the arrow used for translation. This proves properness. $\square$

3. Every proper transverse function and the transverse measure

We verify the measure on this example for all proper transverse functions, not just for $\nu$. A transverse function $\kappa$ identifies under source with a single measure $\eta_0$ on $(X,\mathcal S)$, because left translations between every two range fibres preserve source and the measure.

For ordinary Borel $B$, kernel measurability of the event $\{d=e,\ s\in B\}$ says that $$y\longmapsto\eta_0(B\cap A_{a(y)}) \tag{3.1}$$ is ordinary Borel. Lemma 1.1 forces its values to be the same for every group label. Thus there is a Borel measure $\beta$ on $X$ with $$\eta_0(B\cap A_g)=\beta(B)\quad(g\in\Gamma). \tag{3.2}$$ Countable additivity of $\beta$ follows by evaluating disjoint Borel unions on one fixed branch. Any Borel set disjoint from one $A_g$ has $\beta$-measure zero, by (3.2). Countable branch decomposition now shows $\eta_0=\eta_\beta$ as in (1.3).

Properness makes $\beta$ sigma-finite. Indeed, restrict a countable finite-mass properness cover to one fixed range fibre and write each term as $\bigcup_g(B_{n,g}\cap A_g)$. Every $\beta(B_{n,g})$ is finite, and the $B_{n,g}$'s together cover $X$, since their branch intersections cover every actual point. Conversely, given a sigma-finite $\beta$ satisfying the null condition above, choose increasing Borel $C_n\uparrow X$ with $\beta(C_n)<\infty$. The arrow cover $$\{s\in C_n,\ d\in F_n\} \tag{3.3}$$ has translated fibre mass $|F_n|\beta(C_n)$. Its kernel property follows from the finite-stage pi-lambda proof above. It therefore defines a proper transverse function. This classifies all of them.

Theorem 3.1 (the transverse measure). For the proper function corresponding to $\beta$, set $$\Lambda(\kappa)=\beta(X). \tag{3.4}$$ This is a nonzero, sequentially normal, semifinite and sigma-finite transverse measure of modulus $1$. Moreover $\Lambda(\nu)=1$ and $\Lambda_\nu=\ell$.

Proof. Formula (3.2) gives additivity, homogeneity and continuity along increasing transverse functions, by ordinary measure monotone convergence on a fixed branch. We check the convolution axiom explicitly. Let $\pi$ be any measurable probability kernel on range fibres. Write $$\pi_h^u(B)=\pi^u\{(u,v):v\in B,\ d(u,v)=h\}.$$ For fixed $h,B$ this is ordinary Borel in $u$, and $\sum_h\pi_h^u(X)=1$.

If $u\in A_t$, then $v\in A_l$ is equivalent to $d(u,v)=tl^{-1}$. Integrating against the source measure $\eta_\beta$ of $\kappa$, countable branch decomposition and Tonelli give $$\begin{aligned} (\kappa*\pi)^y\{s\in B,\ a(s)=l\} &=\sum_{t\in\Gamma}\int_X\pi_{tl^{-1}}^u(B)\,d\beta(u)\\ &=\int_X\sum_{h\in\Gamma}\pi_h^u(B)\,d\beta(u) =:\beta'(B). \end{aligned} \tag{3.5}$$ The event on the left is interpreted in the source fibre sigma-field; no assertion that $a$ is a unit Borel function is used. The result is independent of $l$. If the convolution is proper, its classifying measure is $\beta'$, and $\beta'(X)=\int 1\,d\beta=\beta(X)$. This is exactly the modulus-one axiom, including infinite values.

For semifiniteness, finite-$\beta$ sets $C_n\uparrow X$ give proper $\kappa_n=(1_{C_n}\circ s)\kappa\le\kappa$ with finite $\Lambda(\kappa_n)=\beta(C_n)\uparrow\beta(X)$. For sigma-finiteness in the transverse sense, the constant sequence $\nu_n=\nu$ has finite transverse value one and faithful supremum. Finally $(f\circ s)\nu$ corresponds to $f\ell$ for every nonnegative ordinary Borel $f$. Hence $\Lambda_\nu(f)=\int f\,d\ell$. $\square$

The integrated arrow measure has, in the relative-label coordinates, ordinary product values $$\int_G F(y,x,d(y,x))\,dm =\sum_{g\in\Gamma}\int_{X^2}F(y,x,g)\,d\ell(y)d\ell(x). \tag{3.6}$$ This follows first for the generating rectangles and then for nonnegative measurable $F$. It is sigma-finite. Inversion interchanges $x,y$ and sends $g$ to $g^{-1}$, so it preserves $m$, in agreement with modulus $1$.

When $\Gamma$ is infinite, every nonzero proper transverse function has infinite fibre total mass. Indeed (3.2) gives total $\sum_g\beta(X)$. It can still have finite positive transverse value (3.4). Fibre mass, transverse value and unit measure are three different quantities.

4. The regular field and its factor

Use $H_y=L^2(G^y,\nu^y)$. Under the measurable relative-label chart, (1.5) identifies every fibre with $$V=K\otimes\ell^2(\Gamma),\qquad U(z,y)=1_K\otimes\lambda_{d(z,y)},\qquad \lambda_h\delta_g=\delta_{hg}. \tag{4.1}$$ The chart uses $d(y,x)$, rather than the non-Borel unit function $a(y)$. If $b_j$ is a countable Borel orthonormal basis of $K$, its measurable fibre vectors are $b_j(s)1_{\{d=g\}}$. They form an orthonormal basis at every unit. Equation (2.2) verifies (4.1) and the representation law.

The representation is square integrable, directly at the coefficient criterion. For the bounded measurable basis section $e_{j,g}=b_j\otimes\delta_g$ and $\zeta\in V$, (2.3) gives $$\int_{G^y}|\langle\zeta,U(\gamma)e_{j,g}\rangle|^2\,d\nu^y(\gamma) =\sum_{h\in\Gamma}|\langle\zeta,b_j\otimes\delta_{hg}\rangle|^2 \le\|\zeta\|^2. \tag{4.2}$$ The countable family is total at every unit. Each associated coefficient map is a contraction; for any fixed $g$, its Gram operators sum to the identity as $j$ ranges over the basis. Thus this is the exact square-integrability notion of the programme, with explicit total sections.

Put $R_\Gamma=\{\lambda_h:h\in\Gamma\}'$ on $\ell^2(\Gamma)$. This notation means the commutant, not an assumed identification theorem for a group von Neumann algebra.

Theorem 4.1 (the entire random-operator algebra). In the chart (4.1), $$M=\operatorname{End}_\Lambda(H) =B(K)\,\overline\otimes\,R_\Gamma . \tag{4.3}$$

Proof. A measurable bounded operator field has ordinary Borel matrix coefficients in the basis above. There are no nonempty saturated negligible sets: the underlying groupoid has one orbit and $\Lambda(\nu)=1$.

Fix a unit $y_0$. Exact covariance makes $$T_y=(1\otimes\lambda_{a(y)})C(1\otimes\lambda_{a(y)})^*, \tag{4.4}$$ where $C=(1\otimes\lambda_{a(y_0)})^*T_{y_0}(1\otimes\lambda_{a(y_0)})$. Each matrix coefficient of this field is $q(a(y))$ for a function $q$ on $\Gamma$. Lemma 1.1 makes every such coefficient constant. Totality of the basis therefore makes all conjugates in (4.4) equal to $C$; thus $C$ commutes with $1\otimes\lambda_h$ for every $h$. Conversely every constant operator in this commutant is measurable and equivariant.

For completeness the commutant is precisely the tensor algebra in (4.3). Matrix coefficients in an orthonormal basis of $K$ of any commuting operator lie in $R_\Gamma$. Its finite $K$-matrix compressions therefore belong to $B(K)\overline\otimes R_\Gamma$; they converge strongly to the operator. The converse commutation holds on elementary tensors and passes to the von Neumann closure. This proves both equality and the normal concrete realization. Completion-measurable coefficients would give the same result, by the completion assertion in Lemma 1.1. $\square$

Now take $\Gamma=F_2=\langle a,b\rangle$, the group of reduced words in $a,a^{-1},b,b^{-1}$.

Lemma 4.2 (the free-group commutant is a finite factor). The infinite-dimensional algebra $R_{F_2}$ is a factor with normal faithful tracial state $$\tau(T)=\langle T\delta_e,\delta_e\rangle. \tag{4.5}$$

Proof. Right translations $\rho_h\delta_g=\delta_{gh^{-1}}$ commute with every left translation, so belong to $R_{F_2}$. If $T$ is central in $R_{F_2}$, it commutes with both left and right translations. Write $c_g=\langle T\delta_e,\delta_g\rangle$. Conjugating $\delta_g$ by the unitary $\lambda_h\rho_h$ shows $c_{hgh^{-1}}=c_g$.

Every nonidentity conjugacy class is infinite. Given a nonempty reduced word $w$, choose a letter $u$ which is neither the inverse of its first letter nor its last letter. At most two of the four letters are forbidden. All words $u^nwu^{-n}$ are reduced with lengths $2n+|w|$, so are distinct. A square-summable family constant on such a class has value zero there. Thus $T\delta_e=c_e\delta_e$; commutation with $\lambda_g$ gives $T\delta_g=c_e\delta_g$ for every $g$. The centre is scalar.

The vector functional (4.5) is normal and positive. For any $T\in R_{F_2}$, its matrix entry in row $g$, column $h$, is $c_{h^{-1}g}$, by commutation with left translations. Consequently $T^*\delta_e$ has coefficients $\overline{c_{g^{-1}}}$, and $$\tau(T^*T)=\|T\delta_e\|^2 =\|T^*\delta_e\|^2=\tau(TT^*). \tag{4.6}$$ Polarizing this equality gives $\tau(ST)=\tau(TS)$ for arbitrary $S,T$ in the algebra. If $T\ge0$ and $\tau(T)=0$, then $T^{1/2}\delta_e=0$; its commutation with all $\lambda_g$ makes it zero on every basis vector. This proves faithfulness. Finally the infinitely many $\rho_h$ are linearly independent, by applying a finite linear combination to $\delta_e$. The algebra is infinite-dimensional. $\square$

Corollary 4.3 (a properly infinite principal random-operator factor). For $\Gamma=F_2$, (4.3) is a nonzero properly infinite semifinite factor. It has a finite faithful trace on every rank-one $K$-corner, and is of type $\mathrm{II}_\infty$.

Proof. A central element first commutes with $B(K)\otimes1$, so is $1\otimes Z$; then Lemma 4.2 makes $Z$ scalar. Split a countable basis of the infinite-dimensional $K$ into two infinite subsets. The corresponding isometries on $K$, tensored with $1$, have orthogonal ranges summing to the identity. Hence the identity is properly infinite.

The usual positive diagonal sum $\operatorname{Tr}_K\otimes\tau$ is a normal faithful semifinite trace: finite $K$-matrix corners have finite trace, and their increasing projections converge strongly to one. A rank-one corner is $R_{F_2}$, a finite infinite-dimensional factor by Lemma 4.2, so is type $\mathrm{II}_1$. Equivalence of the rank-one $K$-corners and their countable sum give type $\mathrm{II}_\infty$. The positive diagonal trace construction and the finite-factor terminology are the stated basic von Neumann prerequisites, not a group-factor classification theorem. $\square$

Trivial set-theoretic isotropy, one orbit, sigma-finite transverse measure and a properly infinite factor therefore do not restore standard Borel fibre density. The missing group coordinate is a measurable phenomenon.

5. A controlled translation obstructs the canonical tensor copies

Retain $\Gamma=F_2$. On $H\otimes H$, the group coordinate is $\ell^2(\Gamma)\otimes\ell^2(\Gamma)$, with diagonal transport $\lambda_l\otimes\lambda_l$. The same coefficient argument as in (4.4) gives $$E=\operatorname{End}_\Lambda(H\otimes H) =\{(1_K\otimes\lambda_l)\otimes(1_K\otimes\lambda_l):l\in\Gamma\}'. \tag{5.1}$$ Here the commutant is on $V\otimes V$, in the constant chart. The tensor field is also square integrable directly: for a product basis section $b_i\otimes\delta_g\otimes b_j\otimes\delta_k$, its coefficient-square integral is the sum of the squared coefficients of a vector along the orthonormal family indexed by $(lg,lk)$, $l\in\Gamma$. This sum is at most that vector's squared norm. All product basis sections are bounded, measurable and total. Put $$P=\{T\otimes1:T\in M\},\qquad Q=\{1\otimes T:T\in M\}. \tag{5.2}$$ They are normal unital copies of $M$ in $E$, commute, and the self-adjoint flip $\Sigma$ belongs to $E$, satisfies $\Sigma^2=1$, and exchanges them.

For $h\in\Gamma$, let $p_k$ be projection onto $\mathbb C\delta_k$ in the second group coordinate, and define, with the $K$ identities suppressed, $$W_h=\sum_{k\in\Gamma}\lambda_{khk^{-1}}\otimes p_k,\qquad W_h(\delta_g\otimes\delta_k)=\delta_{khk^{-1}g}\otimes\delta_k. \tag{5.3}$$ This is a norm-one unitary: on the orthogonal second-coordinate slices it is a unitary left translation, and its inverse is $W_{h^{-1}}$.

Proposition 5.1 (both failures are explicit). For $h=a$, $$W_a\in P'\cap E,\qquad W_a\notin Q,\qquad W_a\notin P\vee Q. \tag{5.4}$$ Thus these tensor copies are not mutual relative commutants and do not generate $E$.

Proof. Simultaneous left translation sends $k$ to $lk$ and conjugates $\lambda_{khk^{-1}}$ to $\lambda_{lkhk^{-1}l^{-1}}$. Reindexing the orthogonal sum proves $$(\lambda_l\otimes\lambda_l)W_h(\lambda_l\otimes\lambda_l)^*=W_h.$$ Thus $W_h\in E$. Every coefficient $\lambda_{khk^{-1}}$ commutes with the first copy of $R_\Gamma$, and $W_h$ is the identity on the first $K$-coordinate. The full tensor commutant calculation (4.3) therefore gives $W_h\in P'$.

Both $P$ and $Q$ commute with the first-coordinate left translation $\lambda_b\otimes1$. So does their generated von Neumann algebra. But on the second-coordinate slice $k=e$, $$\begin{aligned} (\lambda_b\otimes1)W_a(\delta_e\otimes\delta_e) &=\delta_{ba}\otimes\delta_e,\\ W_a(\lambda_b\otimes1)(\delta_e\otimes\delta_e) &=\delta_{ab}\otimes\delta_e. \end{aligned} \tag{5.5}$$ The distinct reduced words $ba,ab$ give orthogonal vectors. Hence $W_a$ fails that commutation and lies in neither $Q$ nor $P\vee Q$. $\square$

Even the equivalence $H\cong H\otimes H$ is available here; it does not repair (5.4). Explicitly the unitary $$C(\delta_g\otimes\delta_k)=\delta_g\otimes\delta_{g^{-1}k} \tag{5.6}$$ turns $\lambda_l\otimes\lambda_l$ into $\lambda_l\otimes1$. Rearrange the Hilbert tensor coordinates and choose a basis bijection between the countably infinite multiplicity spaces $K$ and $K\otimes K\otimes\ell^2(\Gamma)$. This gives a constant measurable unitary field $J:H\to H\otimes H$ intertwining every arrow. Transporting (5.2) and the flip through $J$ gives an endomorphism and symmetry inside $M$, but transports the extra relative-commutant element $W_a$ as well.

This pinpoints the hypothesis used by the positive standard Borel proof: commuting with all global random operators must imply commuting with the full first fibre algebra. Here the first fibre values give $B(K)\overline\otimes R_\Gamma$, whose commutant still contains all $1_K\otimes\lambda_h$. Proposition 5.1 uses precisely that surviving space.

Open diagram at full size

Figure 1. The branch boxes represent measured events, not intervals and not extra arrows. Equations (2.1)–(2.3) give the single pair arrow, label multiplication and equal branch masses. Equation (4.3) and Corollary 4.3 give the properly infinite factor despite trivial isotropy. The lower panel displays the exact action of $W_a$ and the two different words in (5.5); its norm is one, it belongs to $P'\cap E$, and it lies outside $Q$ and $P\vee Q$. Complete proof locators: Theorems 3.1 and 4.1, Corollary 4.3, Proposition 5.1. Compare Connes, Corollary 11, author-hosted PDF 44–45; the actual standard Borel boundary of Claude-SQ, Sections 5 and 8. Full-size diagram.

5A. Joint generation and an internal flip can still leave an extra commutant

We now take a different member of the countable-group family in Sections 1–4. Let $$\Gamma=F(a_0,a_1,a_2,\ldots),\qquad \Delta=F(b_0,b_1,b_2,\ldots). \tag{5A.1}$$ There are two distinct maps between these free groups: $$\pi(a_0)=e,\quad \pi(a_{n+1})=b_n;\qquad \beta(a_n)=b_n\quad(n\ge0). \tag{5A.2}$$ The first is a surjective homomorphism with nontrivial kernel $H=\ker\pi$; the second is an isomorphism. The construction uses both maps for different purposes. Neither identifies the two-generator factor in Section 5 with the factor used here.

For a countable group $G$, let $\lambda_g\delta_x=\delta_{gx}$, $\rho_g\delta_x=\delta_{xg^{-1}}$, and $R_G=\{\lambda_g:g\in G\}'$.

Lemma 5A.1 (the discrete regular commutation theorem, with proof). For every countable group, $$\{\lambda_g:g\in G\}'=\{\rho_g:g\in G\}'',\qquad \{\rho_g:g\in G\}'=\{\lambda_g:g\in G\}''. \tag{5A.3}$$

Proof. Let $T$ commute with all left translations and $U$ with all right translations. If $c=T\delta_e$ and $d=U\delta_e$, their matrix entries, with output index first, are $T_{x,z}=c_{z^{-1}x}$ and $U_{x,z}=d_{xz^{-1}}$. For arbitrary $x,z$, $$\begin{aligned} (TU)_{x,z} &=\sum_y c_{y^{-1}x}d_{yz^{-1}} =\sum_k c_kd_{xk^{-1}z^{-1}},\\ (UT)_{x,z} &=\sum_y d_{xy^{-1}}c_{z^{-1}y} =\sum_k d_{xk^{-1}z^{-1}}c_k . \end{aligned} \tag{5A.4}$$ The first substitution is $k=y^{-1}x$, the second $k=z^{-1}y$. Each sum is absolutely convergent by Cauchy–Schwarz, since both coefficient sequences are in $\ell^2(G)$ and the index maps are bijections. Thus $TU=UT$. Every $T$ in the left commutant therefore belongs to the double commutant of the right translations. The reverse inclusion follows from commutation of the translation generators. Taking commutants gives the second equality. $\square$

Lemma 5A.2 (the kernel removes every nonscalar first coefficient). We have $$R_\Gamma\cap\{\rho_h:h\in H\}'=\mathbb C1. \tag{5A.5}$$ Both $R_\Gamma$ and $R_\Delta$ are finite infinite-dimensional factors with the trace in (4.5).

Proof. Fix a nonidentity reduced word $w\in\Gamma$, and choose $j\ge1$ whose generator $a_j$ does not occur in $w$. Put $h=a_ja_0a_j^{-1}$. Then $h\in H$, and for $m\ge1$, $$h^mwh^{-m}=a_ja_0^ma_j^{-1}w a_ja_0^{-m}a_j^{-1} \tag{5A.6}$$ is reduced, with length $2m+4+|w|$. No cancellation reaches $w$, because $a_j$ does not occur there. These words are distinct, so every nonidentity $H$-conjugacy orbit is infinite.

If $T$ belongs to the left side of (5A.5), it commutes with $\lambda_h\rho_h$ for every $h\in H$. This unitary fixes $\delta_e$ and sends $\delta_w$ to $\delta_{hwh^{-1}}$. The coefficients of $T\delta_e$ are consequently constant on those orbits. Square summability forces every nonidentity coefficient to vanish. Thus $T\delta_e=c_e\delta_e$, and commutation with all $\lambda_g$ gives $T=c_e1$.

A central element of $R_\Gamma$ satisfies this condition because every $\rho_h$ belongs to $R_\Gamma$; hence its centre is scalar. The trace proof in Lemma 4.2 uses only the translation matrix formula: $\tau(T^*T)=\|T\delta_e\|^2=\|T^*\delta_e\|^2=\tau(TT^*)$, and polarization gives traciality. A positive operator of trace zero has square root zero on $\delta_e$ and, by commutation, on every $\delta_g$; it is zero. Thus the vector trace is faithful and normal. The right translations are linearly independent, so the factor is infinite-dimensional. The basis unitary $\delta_g\mapsto\delta_{\beta(g)}$ carries $R_\Gamma$ onto $R_\Delta$, giving the same assertions for the latter. $\square$

On $\mathcal V=\ell^2(\Gamma)\otimes\ell^2(\Delta)$, define $$A=\{\lambda_g\otimes1:g\in\Gamma\}'',\qquad B=\{\rho_g\otimes\rho_{\pi(g)}:g\in\Gamma\}''. \tag{5A.7}$$ The two algebras commute.

Lemma 5A.3 (both copies are normal finite factors). The algebras $A$ and $B$ are normal unital copies of $R_\Gamma$. For $B$, the unitary $$D(\delta_x\otimes\delta_q)= \delta_x\otimes\delta_{q\pi(x)^{-1}} \tag{5A.8}$$ conjugates every $\rho_g\otimes\rho_{\pi(g)}$ to $\rho_g\otimes1$.

Proof. The inverse of the basis permutation in (5A.8) sends $(x,q)$ to $(x,q\pi(x))$. Under the diagonal right action, its second transformed coordinate is $$q\pi(x)\pi(g)^{-1}\pi(xg^{-1})^{-1}=q.$$ This proves the asserted conjugation. Lemma 5A.1 identifies the algebra generated by the $\rho_g$'s with $R_\Gamma$; tensoring its defining representation with an identity is a normal faithful unital representation. For $A$, the inversion unitary $I_\Gamma\delta_x=\delta_{x^{-1}}$ sends $\rho_g$ to $\lambda_g$, so gives a normal isomorphism from $R_\Gamma$ to the left regular algebra. $\square$

Theorem 5A.4 (the entire generated algebra). The join is $$E=A\vee B=B(\ell^2(\Gamma))\overline\otimes R_\Delta. \tag{5A.9}$$ It is a properly infinite semifinite factor of type $\mathrm{II}_\infty$.

Proof. If $T$ commutes with $A$, every $\Delta$-matrix coefficient of $T$ belongs to $R_\Gamma$. If it also commutes with $B$, use $h\in H$: the corresponding generator of $B$ is $\rho_h\otimes1$. Each coefficient then lies in (5A.5) and is scalar. Hence $T=1\otimes C$, with $C$ bounded; this follows by applying $T$ to elementary tensors and its scalar matrix coefficients. Commutation with the remaining $B$-generators, and surjectivity of $\pi$, give $C\in\{\rho_q:q\in\Delta\}'$.

Conversely every such $1\otimes C$ commutes with both sets of generators. Therefore $$E'=1\otimes\{\rho_q:q\in\Delta\}' =1\otimes\{\lambda_q:q\in\Delta\}''. \tag{5A.10}$$ Taking the commutant gives (5A.9). Explicitly, its $\Gamma$-matrix coefficients lie in $R_\Delta$; finite matrix compressions belong to the displayed tensor algebra and converge strongly to the original operator. Lemma 5A.2 and the diagonal trace $\operatorname{Tr}_{\ell^2(\Gamma)}\otimes\tau_\Delta$ give factoriality and semifiniteness exactly as in Corollary 4.3. The rank-one corner is a finite infinite-dimensional factor; splitting the countable $\Gamma$-basis into two infinite subsets gives the two isometries proving proper infiniteness. $\square$

Lemma 5A.5 (an internal symmetry exchanging the generators). The unitary $$J(\delta_x\otimes\delta_q)= \delta_{x^{-1}}\otimes\delta_{q\pi(x)^{-1}} \tag{5A.11}$$ belongs to $E$, satisfies $J=J^*=J^{-1}$, and gives $JAJ=B$.

Proof. Applying the displayed basis permutation twice returns $(x,q)$; thus $J^2=1$ and it is a self-adjoint unitary. It commutes with every $1\otimes\lambda_v$, since replacing $q$ by $vq$ replaces its transformed value by $vq\pi(x)^{-1}$. Equation (5A.10) then places $J$ in $E$. Direct substitution gives $$J(\lambda_g\otimes1)J(\delta_x\otimes\delta_q) =\delta_{xg^{-1}}\otimes\delta_{q\pi(g)^{-1}}. \tag{5A.12}$$ The operator on the right is the generator of $B$ indexed by $g$, proving the asserted equality. $\square$

Proposition 5A.6 (generation has not removed the extra commutant). Put $$Z=1\otimes\rho_{b_0}. \tag{5A.13}$$ This is a norm-one unitary in $A'\cap E$ which does not belong to $B$.

Proof. Membership of $Z$ in $E$ follows from (5A.9); it commutes with $A$ because it acts only on the second coordinate. The closed span $$\mathcal G_\pi= \overline{\operatorname{span}}\{ \delta_x\otimes\delta_{\pi(x)}:x\in\Gamma\} \tag{5A.14}$$ is reducing for every generator of $B$: the graph point $(x,\pi(x))$ is sent to $(xg^{-1},\pi(xg^{-1}))$. Its orthogonal projection therefore commutes with all of $B$. The vector $\delta_e\otimes\delta_e$ lies in this graph subspace, whereas $$Z(\delta_e\otimes\delta_e)= \delta_e\otimes\delta_{b_0^{-1}}\perp\mathcal G_\pi. \tag{5A.15}$$ Thus $Z$ cannot belong to $B$. Conjugation by $J$ also gives an extra element in $B'\cap E\setminus A$. $\square$

The finite factors $A,B$ are not yet copies of the properly infinite $E$. The following step establishes that precise source requirement as well.

Theorem 5A.7 (two generating stable copies with an internal flip). On $\mathcal W=L\otimes L\otimes\mathcal V$, where $L=\ell^2(\mathbb N)$, put $$\begin{aligned} \widetilde M&=B(L)\overline\otimes B(L)\overline\otimes E,\\ P&=B(L)\overline\otimes1_L\overline\otimes A,\qquad Q=1_L\overline\otimes B(L)\overline\otimes B. \end{aligned} \tag{5A.16}$$ There are a normal injective unital endomorphism $\sigma$ of $\widetilde M$ and a self-adjoint unitary $S\in\widetilde M$ with $$\sigma(\widetilde M)=P,\quad S P S=Q,\quad S^2=1,\quad P\vee Q=\widetilde M,\quad Q\subsetneq P'\cap\widetilde M. \tag{5A.17}$$ The reverse relative-commutant equality fails as well. This factor is normally isomorphic to the principal random-operator factor of Sections 1–4 for $\Gamma=F_\infty$.

Proof. By (5A.9), $\widetilde M=B(L\otimes L\otimes\ell^2(\Gamma))\overline\otimes R_\Delta$. Choose a basis unitary from the countably infinite first Hilbert space onto $L$. The group isomorphism $\beta$, followed by inversion on $\Gamma$, gives the normal isomorphism $R_\Delta\to A$ from Lemmas 5A.2–5A.3. Tensoring these maps gives a normal unital isomorphism from $\widetilde M$ onto $P$; composing with $P$'s inclusion defines $\sigma$.

Let $F_L$ exchange the two $L$-coordinates and set $S=F_L\otimes J$. Both factors are self-adjoint unitaries of square one, on distinct coordinates. The first belongs to $B(L\otimes L)$, and Lemma 5A.5 puts the second in $E$, so $S\in\widetilde M$. It sends $P$ onto $Q$. These algebras commute. Their join contains both $L$-matrix algebras and $A\vee B=E$, so is exactly $\widetilde M$.

The operator $\widetilde Z=1_L\otimes1_L\otimes Z$ belongs to $P'\cap\widetilde M$. If it belonged to $Q$, taking a unit-vector matrix coefficient in its second $L$-coordinate would put $Z$ in $B$, contrary to Proposition 5A.6. Thus the inclusion in (5A.17) is strict. Conjugation by $S$ proves strictness in the reverse direction.

Finally Theorem 4.1 for the countable $\Gamma$ gives $M_0=B(K)\overline\otimes R_\Gamma$, with $K$ separable and infinite-dimensional. A basis unitary $K\to L\otimes L\otimes\ell^2(\Gamma)$ and the group isomorphism $\beta$ give a normal isomorphism $M_0\to\widetilde M$. Transporting $\sigma,S,\widetilde Z$ through it retains every assertion. All the faithful proper transverse and square-integrable field hypotheses are supplied by the full constructions of Sections 2–4 for this $\Gamma$. $\square$

Open diagram at full size

Figure 2. The homomorphism $\pi$ kills $a_0$ and shifts the remaining generators; the separate isomorphism $\beta$ matches the finite factors. The kernel-conjugacy argument gives the entire join (5A.9). The internal permutation $J$ is exactly (5A.11), not the plain tensor flip. The graph vector in (5A.15) proves strict relative commutation even after stabilization. Complete proof locators: Lemmas 5A.1–5A.3 and 5A.5, Theorem 5A.4, Proposition 5A.6 and Theorem 5A.7. Source target: Connes, Corollary 11, author-hosted PDF 44–45. This is a generating pair that fails the relative-commutant clause; it does not exclude every possible pair in this factor.

5B. Why closing the relative commutants does not repair this pair

Keep exactly the groups, quotient, representation and operators of Section 5A. The question here is whether replacing a copy by its full relative commutant repairs that construction. The product-group argument below supplies the complete commutant calculation without assuming a general tensor-intersection theorem.

Write $\mathcal L_G=\{\lambda_g:g\in G\}''$, and take all unqualified commutants in $B(\mathcal V)$. Here $D$ denotes the second enlarged algebra; the earlier untwisting unitary in (5A.8) is not used in this section.

Lemma 5B.1 (the full relative commutants). Define $$C=A'\cap E=R_\Gamma\overline\otimes R_\Delta, \qquad D=B'\cap E=JCJ. \tag{5B.1}$$ These are finite factors, with $B\subsetneq C$ and $A\subsetneq D$.

Proof. Equation (5A.10) gives $E=(1\otimes\mathcal L_\Delta)'$. Consequently $A'\cap E$ is the commutant of all left translations in both coordinates. Identify $\mathcal V$ with $\ell^2(\Gamma\times\Delta)$ by its displayed basis. Then $$\lambda_{(g,q)}=\lambda_g\otimes\lambda_q,\qquad \rho_{(g,q)}=\rho_g\otimes\rho_q. \tag{5B.2}$$ Lemma 5A.1, now applied to this countable product group, says that the required commutant is the algebra generated by its right translations. This is precisely the spatial tensor algebra in (5B.1): the generators with $q=e$ and $g=e$ generate the two factors, and every product generator belongs to their join.

For completeness, the product group has infinite conjugacy classes away from the identity. A nonidentity first coordinate has infinitely many conjugates by Lemma 5A.2; if the first coordinate is the identity, use the nonidentity second one. The coefficient and trace proof of that lemma applies to the product group as well. It gives a scalar centre and the faithful normal tracial state at $\delta_{(e,e)}$. Thus $C$ is a finite factor. Conjugation by the internal involution $J$ transports the relative commutant of $A$ to that of $JAJ=B$, proving the formula for $D$ and its finiteness. The strict inclusions are Proposition 5A.6 and its conjugate. $\square$

Proposition 5B.2 (one-sided closure gives mutual commutants and loses every spatial flip at this finite level). We have $$C'\cap E=A,\qquad A\vee C=E. \tag{5B.3}$$ There is no unitary in $B(\mathcal V)$, and hence none in $E$, conjugating $A$ onto $C$. In particular the original $J$ cannot exchange this repaired pair.

Proof. For $T\in C'\cap E$, take matrix coefficients in the first coordinate. Membership in $E=B(\ell^2(\Gamma))\overline\otimes R_\Delta$ puts each coefficient in $R_\Delta$; commutation with $1\otimes R_\Delta\subset C$ puts it in the centre of that factor. All these coefficients are scalar. Thus $T=T_0\otimes1$, with $T_0$ bounded, and commutation with $R_\Gamma\otimes1\subset C$ gives $T_0\in\mathcal L_\Gamma$ by Lemma 5A.1. This proves the first equality; the converse inclusion follows from the definitions. Since $B\subset C$, (5A.9) proves the second.

The ambient commutants, rather than the relative ones, distinguish the two represented algebras: $$A'=R_\Gamma\overline\otimes B(\ell^2(\Delta)), \qquad C'=\mathcal L_\Gamma\overline\otimes\mathcal L_\Delta. \tag{5B.4}$$ The first formula follows by taking second-coordinate matrix coefficients in $A'$: these belong to $R_\Gamma$, and finite matrix compressions in the unrestricted second coordinate lie in the stated tensor algebra and converge strongly. The second follows by applying Lemma 5A.1 to the right translations of the product group in (5B.2).

The first algebra in (5B.4) is properly infinite. Split the countably infinite $\Delta$-basis into two infinite subsets; the associated basis isometries $s_0,s_1$ have orthogonal ranges adding to the identity. Their tensors with $1$ are isometries in $A'$, and neither is unitary. The second algebra has the faithful normal tracial state $$\tau_0(T)= \langle T(\delta_e\otimes\delta_e),\delta_e\otimes\delta_e\rangle. \tag{5B.5}$$ Indeed inversion in the product group carries it to the finite right algebra of Lemma 5B.1 and fixes the identity vector. An isometry in an algebra with this faithful tracial state must be unitary: $\tau_0(1-vv^*)=\tau_0(1-v^*v)=0$, so faithfulness gives $vv^*=1$.

If a unitary $U$ conjugated $A$ onto $C$, it would conjugate their ambient commutants as well. It would send a nonunitary isometry of $A'$ to a nonunitary isometry of $C'$, contradicting the preceding trace calculation. The original flip also fails directly, since $JAJ=B\subsetneq C$. This argument concerns the displayed finite-level representation; it does not assert that every stabilized pair has the same ambient obstruction. $\square$

Proposition 5B.3 (closing both sides produces an explicit nonzero commutator). The algebras $C,D$ are exchanged by $J$, jointly generate $E$, and have scalar intersection. They do not commute. Put $Z=1\otimes\rho_{b_0}$ as before, and $Y=JZJ$. The exact second-coordinate action is $$Y(\delta_x\otimes\delta_q)= \delta_x\otimes \delta_{q\pi(x)^{-1}b_0^{-1}\pi(x)}. \tag{5B.6}$$ For the unit vector $\xi=\delta_{a_2}\otimes\delta_e$, it gives $$\begin{aligned} ZY\xi&=\delta_{a_2}\otimes \delta_{b_1^{-1}b_0^{-1}b_1b_0^{-1}},\\ YZ\xi&=\delta_{a_2}\otimes \delta_{b_0^{-1}b_1^{-1}b_0^{-1}b_1}. \end{aligned} \tag{5B.7}$$ In particular, $$\|[Z,Y]\xi\|=\sqrt2,\qquad C\cap D=\mathbb C1,\qquad C\vee D=E. \tag{5B.8}$$