Finite corners, unitary displacement and the tensor tail

Programme exposition: CC0-1.0 to the extent of rights held. Human-source terms remain separate.

The fullness and hypercentrality lesson needs three constructions beyond the centralizing quotient: division of a finite algebra with its center retained, a unitary moving a trace-zero element, and an exact map onto a tensor tail. Its nonfactor example also needs a proof of its center and type, rather than an appeal to unspecified fibers.

We use the complete retained CF calculus and Hilbert arguments, SC00–07 measure and Lp proofs, CP preduals, BD bounded density, PC0–7/FCT0–7, NT1–2, FG2's complete dyadic factor model, and CQ2's trace representation. Inner products below are linear in the second variable; conjugating SC07's convention gives this convention with the same Hilbert norm.

FH0. A normalized tracial state on a finite factor is normal

Let $N$ be a nonzero finite factor, and let $\sigma$ be any positive normalized tracial functional on $N$. Normality is not assumed. FCT0–7 constructs the canonical faithful normal center-valued trace. Since $Z(N)=\mathbb C1$, write it as $T(x)=\tau_0(x)1$; its scalar $\tau_0$ is a faithful normal normalized trace.

If $N$ has a minimal projection $p$, choose a maximal orthogonal family of copies of $p$. It fills $1$: a nonzero residual has a polar bridge to $p$ by PC2, whose nonzero initial projection must equal $p$, supplying another copy. The family is finite by FCT2's prohibition of infinitely many orthogonal copies of one nonzero projection in a finite algebra. Write it as $p_1,\ldots,p_d$. The corners $p_iNp_i$ are scalar by the bounded spectral calculus and minimality. Partial isometries between the copies give matrix units $e_{ij}$, and every $x\in N$ is their finite linear combination, because each corner $p_i x p_j$ has one scalar coefficient after transport into $pNp$. Traciality gives $\sigma(p_i)=1/d$, and $\sigma(e_{ij})=\sigma(p_i e_{ij})=\sigma(e_{ij}p_i)=0$ for $i\ne j$. Thus $\sigma=\tau_0$ on this entire finite matrix algebra.

If $N$ has no minimal projection, it has no abelian projection: PC4 makes the center of each nonzero factor corner scalar, so an abelian corner would be one-dimensional and its unit minimal. FH1's finite type criterion therefore makes $N$ a type $\mathrm{II}_1$ factor. AF2's full trace-cut construction, based on FCT3's halving and FCT7 comparison, cuts any projection to any prescribed $\tau_0$-mass between zero and its mass. It constructs a partition of $1$ into $m$ equivalent projections, for each integer $m\geq1$. Traciality and normalization give $\sigma$-mass $1/m$ to each cell. FCT7 shows that any projection of $\tau_0$-mass $a/m$, $0\leq a\leq m$, is equivalent to a sum of $a$ such cells and therefore has $\sigma$-mass $a/m$.

For any projection $q$, put $\alpha=\tau_0(q)$. Choose integers $a_m=\lfloor m\alpha\rfloor$ and $b_m=\lceil m\alpha\rceil$. Cut inside $q$ to obtain $q_m^-\leq q$ of mass $a_m/m$, and add a cut of its complement to obtain $q_m^+\geq q$ of mass $b_m/m$. The added mass is at most $\tau_0(1-q)$, so the construction covers the endpoint cases as well. Positivity gives

Both endpoints tend to $\alpha$, so $\sigma(q)=\tau_0(q)$. Uniform bounded spectral step approximation gives equality on self-adjoint elements: positive normalized functionals are bounded of norm one by CF, so equality on projection steps passes to the norm limit. Real and imaginary parts give equality everywhere. Hence $\sigma=\tau_0$ and is normal. No separable-predual assumption or normality assertion about an arbitrary trace on a nonfactor algebra was used.

FH1. The finite type criterion and halving without a factor assumption

Let $N$ be a finite von Neumann algebra. An abelian projection $p$ means a projection whose corner $pNp$ is commutative. The central type I part is precisely the part generated by the central supports of these projections. The full construction of this part is in FCT3: put

Every nonzero projection below $z_{\mathrm I}$ has a nonzero abelian subprojection. A polar bridge to an abelian corner supplies one; absence of all such bridges would make its central support orthogonal to every support in the displayed join, a contradiction by PC2. FCT3 constructs the finite homogeneous families on this part by central comparison, with all central pieces retained. Thus this description does not presume a type I factor classification.

If $N\ne0$ has no nonzero abelian projection, $z_{\mathrm I}=0$. Finiteness excludes the infinite type II and type III parts. The resulting finite algebra with no abelian part is type $\mathrm{II}_1$; this statement does not say that its center is scalar. A faithful normalized trace, when specified, remains its finite trace, independent of whether its predual is separable. For the zero algebra the projection split is $0=0+0$, with no nonzero type or factor conclusion.

Here is the required division directly. In any nonzero projection corner $eNe$, absence of an abelian projection implies noncommutativity. Some projection $a$ in the corner is not central: otherwise the complete bounded spectral step approximation would make every self-adjoint element central, and real and imaginary parts would make the whole corner commutative. A noncentral $a$ has a nonzero off-diagonal element $ax(e-a)$ or its adjoint. PC1 polar decomposition gives two nonzero orthogonal equivalent projections, one below $a$, the other below $e-a$.

Choose a maximal collection of such pairs inside $e$, with all their initial and final projections mutually orthogonal. Unions of chains remain collections of this kind, so PC0's choice argument applies. If the complementary remainder were nonzero, its corner would supply another pair by the preceding construction. The remainder must therefore be zero. PC1's arbitrary orthogonal strong* sums add the pair projections and implementing partial isometries. They give

For an implementing $v$ with $v^*v=p,\ vv^*=q$, the elements $e_{11}=p,\ e_{22}=q,\ e_{21}=v,\ e_{12}=v^*$ are a full two-by-two matrix system with support $e$. Orthogonality gives its four-index products exactly. No countability of the maximal family, predual separability, factoriality, or scalar trace was assumed. Taking $e=1$ gives the unital system needed by the absorption lesson.

FH2. Compact convex orbits and the explicit displacement bound

Let $N$ be a nonzero finite factor with faithful normalized trace $\tau$, and $b\in N$ with $\tau(b)=0$. FH0 proves that this trace is normal, even if normality was not part of its specified hypotheses. Then some unitary $u\in N$ satisfies

The zero element is included.

The unitary orbit is contained in the operator-norm ball of radius $\|b\|$. CP/CF's full compact-ball construction makes that ball ultraweakly compact. Let $K$ be the ultraweakly closed convex hull of the orbit, taken inside this ball. It is nonempty, compact and convex; addition and scalar multiplication are ultraweakly continuous. Unitary conjugation is an ultraweak homeomorphism by the fixed-multiplier CP test and preserves the orbit, hence $K$.

The trace norm has the exact scalar representation

Scalar Schwarz gives the upper bound. For $x\ne0$, choose $d=x/\|x\|_2$, an element of $N$, to attain equality; at zero both sides vanish. Each scalar pairing in (FH2.2) is normal by NT1. A supremum of their continuous absolute values is ultraweakly lower semicontinuous. More explicitly its sublevel set at $c\geq0$ is the intersection of all the closed scalar sublevel sets at $c$.

Let $m=\inf_{x\in K}\|x\|_2$. The closed nonempty sets $\{x\in K:\|x\|_2\leq m+1/j\}$ are nested in compact $K$; their intersection contains a minimizer. It is unique. If $x,y$ were distinct minimizers, faithfulness makes $\|x-y\|_2>0$, and the Hilbert parallelogram identity gives

contradicting convexity and minimality. Conjugating by any unitary gives another minimizer, so the unique one commutes with every unitary. The complete CF four-unitary decomposition then makes it commute with every element of $N$. Since $N$ is a factor it is scalar. Trace normality gives trace zero at every point of $K$, so the scalar minimizer is zero.

For $b\ne0$, suppose (FH2.1) failed for every unitary. The orbit would lie in the closed trace ball $\{x:\|x-b\|_2\leq\|b\|_2/2\}$. This ball is convex by the trace triangle inequality and ultraweakly closed by (FH2.2). It would contain $K$, hence zero. But zero has distance $\|b\|_2$ from $b$, a contradiction. For $b=0$, the identity unitary satisfies the desired bound. This proof uses the whole normal topology, not just an assumed weak Hilbert compactness of the orbit.

FH3. A finite matrix formula gives the entire tensor tail

Let $R$ be FG2's dyadic factor with normalized trace $\tau$, and $D_r\cong M_d(\mathbb C)$, $d=2^r$, its initial $r$ tensor legs. Fix matrix units $(e_{ij})$ of $D_r$ with diagonal sum $1$. Define on every $x\in R$

This is a finite sum; there is no compact-group measure or convergence assumption in its definition.

It is normal by NT1's fixed-multiplier predual maps. Put $a_{ij}=e_{ji}/\sqrt d$. Formula (FH3.1) is $\sum a_{ij}^*x a_{ij}$, with $\sum a_{ij}^*a_{ij}=1$. The same formula on operator matrices proves complete positivity, by compressing a positive matrix with each fixed multiplier and summing. It is unital. NT1's complete two-by-two Schwarz proof therefore gives

Positive order gives norm contraction, and applying the trace gives trace $2$-norm contraction. Trace preservation itself follows directly:

Multiplying (FH3.1) by $e_{ab}$ on either side gives the same sum $d^{-1}\sum_j e_{aj}x e_{jb}$. Thus the image commutes with $D_r$. Conversely an $x$ commuting with every $e_{ij}$ is fixed by the formula, since $\sum_{i,j}e_{ij}e_{ji}=d1$. Its range is exactly $D_r'\cap R$.

We identify that range with the full remaining tensor factor. On an elementary finite tensor in a stage $D_s$, $s\geq r$, matrix multiplication in the first $r$ legs shows that (FH3.1) replaces that initial matrix by its normalized trace times its identity, leaving all subsequent legs intact. Finite sums give the same conclusion for every local tensor polynomial. BD supplies uniformly bounded strong approximation of any $x\in R$ by such polynomials: apply its complete C* density theorem to the norm closure of their union. Fixed multiplication is strongly continuous on bounded sets, so $\mathcal E_r(x)$ is in the strongly closed algebra generated by the remaining legs. The reverse inclusion follows because each remaining leg commutes with $D_r$. Hence

The right side is normally trace-preservingly isomorphic to $R$. On the local trace Hilbert vectors, shifting each leg index by $r$ preserves their inner products, since those products are the finite products of normalized matrix traces in FG2. It therefore extends to a unitary from FG2's whole completed trace Hilbert space onto the tail's trace Hilbert space. It intertwines every local multiplication and, by BD and CP, their complete von Neumann closures. Conjugation by the unitary is normal in both directions. To identify that tail Hilbert closure with the given tail algebra, apply the complete STC coefficient proof for an already normal state to the restriction of the vector trace, and ST1–2's faithful normal image and inverse proof. Its state is already normal and faithful by restriction, so these inputs have their full hypotheses. CQ2 gives its trace Hilbert model. FG2 consequently supplies its factor property, type $\mathrm{II}_1$, trace, and separable predual. No uniqueness theorem for arbitrary hyperfinite factors is a premise.

Finally,

Each fixed-multiplier term tends to zero in trace $2$-norm for an ordinary or ultrafilter central sequence, by NT2. The finite sum gives the corresponding convergence of $\mathcal E_r(x_n)-x_n$, for each fixed $r$. The complete preceding finite-algebra criterion equates centrality and predual centralizing here. This argument supplies the quantitative tail approximation used in both parts of Theorem 5.2.

Finite Pauli witnesses. Put

All four matrices are unitary. For $b=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$, direct multiplication gives

Adding and dividing by four therefore gives $4^{-1}\sum_{u\in\mathcal P}ubu^*=\operatorname{tr}_2(b)1$, for every complex matrix, without a self-adjointness assumption.

For $s>r$, let $F_{r,s}=\bigotimes_{j=r+1}^{s}M_2\subset N_r$ and let $\Gamma_{r,s}$ consist of the $4^{s-r}$ tensor products of matrices from $\mathcal P$ in those legs, with identity on the initial $r$ legs. Applying the one-leg formula successively to an elementary tensor gives its product trace times the identity. Elementary tensors span $F_{r,s}$, so linearity proves

Unitary invariance of the trace norm and expansion of the square now give

In particular, if $\tau(b)=0$, at least one member of this finite family moves $b$ by at least $\sqrt2\|b\|_2$.

Write $\Gamma_r=\bigcup_{s>r}\Gamma_{r,s}$. For any trace-zero $a\in N_r$, this family contains a unitary moving $a$ by at least $\|a\|_2/2$. At $a=0$, use the identity. Otherwise the tail Hilbert identification proved above makes finite tail polynomials dense in trace $2$-norm. Choose $c\in F_{r,s}$ with $\|c-a\|_2<\varepsilon/2$ and set $b=c-\tau(c)1$. Trace Schwarz gives $|\tau(c)|\leq\|c-a\|_2$, so $\tau(b)=0$ and $\|b-a\|_2<\varepsilon$. A unitary supplied by (FH3.6) satisfies

Taking $\varepsilon=\|a\|_2/8$ gives a lower bound at least $5\|a\|_2/8>\|a\|_2/2$, since $\sqrt2\geq1$. Thus the witness can always be a finite Pauli word in the specified tail.

For any $x\in R$, put $b=\mathcal E_r(x)-\tau(x)1\in N_r$, which has trace zero. Choose the preceding unitary for $b$. If $b=0$, the identity has the same half-displacement property. With $\varepsilon_r(x)=\|x-\mathcal E_r(x)\|_2$ and $\Delta(x)=\|x-\tau(x)1\|_2$, the triangle inequality gives

This is a finite-matrix witness estimate followed by trace-norm approximation. It applies to every bounded element of $R$, and its unitary commutes with the whole prescribed initial matrix stage.

FH4. The full $L^\infty([0,1])\bar\otimes R$ example

The example uses the complete SC00–07 Lebesgue and integration proofs, not an unnamed measurable-field theorem. Restrict SC02's complete real-line Lebesgue measure to $[0,1]$, where its mass is one; write it as $\lambda$. SC05–07 proves the complex integral, scalar Schwarz, completeness of $L^2(\lambda)$, and density of bounded simple functions.

Let $H_A=L^2(\lambda)$ and let $M_f$ multiply by a class $f\in L^\infty(\lambda)$. It is bounded with norm exactly $\|f\|_\infty$: the upper estimate follows by integration of $|f h|^2\leq\|f\|_\infty^2|h|^2$; for each number below the essential supremum, the set where $|f|$ exceeds it has positive measure, and its normalized characteristic vector proves the lower estimate. Adjoints are $M_{\bar f}$, and products are ordinary products. These operators form the algebra $A$.

In fact $A'=A$. If $T$ commutes with $A$, put $g=T1\in L^2$. For a bounded simple function $h$, commutation gives $Th=hg$. For each measurable $E$,

Consequently $|g|\leq\|T\|$ almost everywhere: applying the squared inequality to any positive-measure set where $|g|\geq\|T\|+\varepsilon$ would contradict it. These level sets for a countable positive sequence of $\varepsilon$'s cover the alleged exceptional set. Thus $g\in L^\infty$, and $T=M_g$ by simple-function density. Conversely multiplication operators commute. Therefore $A''=A$, proving its von Neumann closure. Its vector state at $1$ is the integral; it is normalized and normal by CP, and faithful because a positive measurable function of zero integral is zero almost everywhere by SC04.

Use the trace Hilbert representation $H_R$ of $R$, with cyclic vector $\Omega_R$. Form the Hilbert tensor completion $H_A\otimes H_R$. Its elementary inner product is the product of the two leg inner products. Positivity and nondegeneracy on any finite tensor sum follow by choosing an orthonormal basis of the finite span of its first-leg vectors and summing the squared norms of the resulting second-leg coefficients; Hilbert completion is CF's complete construction. This computation also proves boundedness of each $S\otimes T$, with norm at most $\|S\|\|T\|$, by first applying either leg to the finite orthogonal coefficient expansion and then completing.

Let $N=(A\odot R)''$ on this space. The vector $\Omega=1\otimes\Omega_R$ is cyclic because the bounded-simple and $R$-local cyclic vectors are dense. It is separating: the right multipliers of $R$ commute with $R$, and $A$ commutes with itself, so $A\odot R_{\rm right}$ commutes with $N$; its orbit of $\Omega$ is also dense. If $x\Omega=0$, commutation kills that dense orbit and boundedness makes $x=0$.

The vector state on the local tensor algebra has the product formula

Finite expansion and the leg trace laws prove traciality for every pair of local tensor polynomials. It extends to their norm closure by vector-state continuity. The two normal-functional extensions in CQ2, using BD, then prove the trace law on all of $N$. The separating-vector argument proves faithfulness; CP proves normality. Its value at the unit is one, and the isometry trace argument of CQ2 proves that $N$ is finite. This is precisely the concrete spatial product $L^\infty([0,1])\bar\otimes R$, with the specified integral-times-trace state.

Its center is exactly $A\otimes1$, as can be proved without a generic tensor commutation theorem. Take $x\in Z(N)$. For $h,h'\in H_A$, mixed compression between the maps $\xi\mapsto h\otimes\xi$ and $\xi\mapsto h'\otimes\xi$ produces a bounded operator $x_{h',h}$ on $H_R$. It commutes with $R$, because $x$ is central, and with the entire commutant $R'$, because each local tensor and hence $N$ commutes with $1\otimes R'$. Thus

The corresponding bounded scalar sesquilinear form in $h,h'$ is represented by $S\in B(H_A)$, by the full CF Hilbert Riesz argument. Coefficients on elementary vectors give $x=S\otimes1$. Commutation with $A\otimes1$ makes $S\in A'=A$. Conversely $A\otimes1$ commutes with every local tensor and its closure, so

This center is nontrivial: the characteristic function of $[0,1/2]$ gives a central projection of trace $1/2$.

To verify its type, rather than referring to type II fibers, suppose $p\in N$ were a nonzero abelian projection. FCT3's full finite homogeneous-family construction gives a nonzero central $w\leq c(p)$ and a finite integer $k\geq1$ such that $w$ is a sum of $k$ orthogonal copies of $pw$. FCT0–7 constructs the normalized faithful center-valued trace $T_N$, so

Choose a dyadic matrix stage $D_r\subset R$ of size $d=2^r>k$. Its diagonal projections, compressed by $w$, are equivalent and sum to $w$. Each therefore has center-valued trace $w/d$. FCT7 comparison puts a copy $q$ of one of them under $pw$, since $w/d\leq w/k$.

But PC4 identifies the center of the abelian corner $pwNpw$ with $Z(N)pw$, so every subprojection of $pw$, including $q$, is $pwt$ for a central projection $t\leq w$. Center linearity of the trace gives

A nonzero scalar multiple $k/d\in(0,1)$ of the nonzero projection $w$ is not a projection: its square is a different scalar multiple. This contradicts the choice of $t$. Thus $N$ has no nonzero abelian projection. FH1 proves it is type $\mathrm{II}_1$, with its nontrivial center retained. This proves the whole example and does not assert that this particular $N$ is a central sequence algebra.

FH5. Strong*-zero perturbations against varying centralizing sequences

Let $M$ be a von Neumann algebra with a faithful normal state $\varphi$. Let $(y_n)$ be bounded and centralizing, either ordinarily or along $\omega$, and let $(z_n)$ be bounded and strongly*-zero in that same mode of convergence. Then both $y_nz_n$ and $z_ny_n$ are strongly*-zero in that mode. This gives a direct proof without moving a varying bounded multiplier through strong* convergence.

Put $\sup_n\|y_n\|\leq K$, $\sup_n\|z_n\|\leq L$. The immediate initial-square estimate is

For the final square, first move the last $y_n^*$ to the front using its state commutator; next move the last $z_n^*$ to the front using its state commutator. NT1 bounds each scalar error by the norm of the remaining product times that commutator norm. The resulting expression has a direct positive order bound:

All terms on the right tend to zero. Apply the same statement to the pair $y_n^*,z_n^*$, which satisfies the same hypotheses, to get strong*-zero convergence of $y_n^*z_n^*$. Taking adjoints gives convergence of $z_ny_n$. NT2's two-seminorm criterion completes the assertion.

In particular scalar-trivial sequences are hypercentral. Writing $x_n=\lambda_n1+z_n$ with bounded $\lambda_n$ and strongly*-zero $z_n$, their commutator with any bounded centralizing $y_n$ is $z_ny_n-y_nz_n$, which the preceding estimates make strongly*-zero. The proof treats ordinary convergence and ultrafilter convergence with the same finite inequalities.

Complete input boundaries

FH0 uses full finite comparison and cancellation, arbitrary scalar cuts and bounded spectral approximation, proving the trace normality needed by FH2. FH1 uses the full center-sensitive PC/FCT route. FH2 uses compact predual balls, trace Schwarz, normal multiplier tests and the Hilbert parallelogram identity. FH3 uses the already constructed dyadic factor and the explicit finite matrix formula. FH4 uses the complete SC00–07 measure and Lp construction, CF Hilbert and compactness proofs, CP/BD, CQ2, the dyadic matrix stages, and full PC/FCT comparison. No compact Haar average, direct integral, type I factor classification or unspecified measurable-fiber theorem is imported.

FH5 uses the complete positive-functional Schwarz, adjoint module identities and faithful-state topology proofs in NT1–2. In particular its products do not presume continuity of multiplication by a varying bounded operator.