Unitary couplings and finite injective factors
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A balanced family can approximately intertwine two tuples of unitaries without containing a single unitary. We will first sample finitely many intertwiners, bound their operator norms without increasing their errors, and pass to an ultrapower. Exact intertwiners there equate two corner dimensions. Projection comparison gives a unitary, and a unitary representative returns the desired approximation to the original algebra.
We use the full balanced coupling construction, the von Neumann multiplier ultrapower, and the complete center-valued trace comparison and finite polar-decomposition proofs. In a faithfully tracial algebra every bounded sequence is a multiplier: . Thus its ultrapower is the quotient of all bounded sequences by those with vanishing ultralimit -norm, and its faithful normal state is a trace: the coordinate traces of and agree before taking the scalar ultralimit. Factoriality of the ultrapower is unnecessary for the proof below. NT2 proves the bounded faithful-state topology and the normal-state countability facts; TL1–3 supplies the complete finite trace Hilbert-space and expectation proofs.
The coupling, lifting, clipping and sampling arguments allow every finite von Neumann algebra with a faithful normal tracial state ; separability is unnecessary. The final application to injectivity uses a factor. Separable predual is assumed only for a generating sequence and the identification with .
The central corner comparison first explains the unitary-conjugacy conclusion. We then establish its unitary lift, bounded clipping and finite sampling mechanisms. Radial projection onto an operator-norm ball proves clipping directly; the complete self-adjoint matrix argument gives a second proof.
Central corner comparison and unitary conjugacy
Theorem 4.1. For a fixed faithfully tracial finite algebra , every integer and , there is such that any -related tuples , in have a unitary with
The modulus need not be specified explicitly. Neither factoriality nor separability is assumed.
Proof. The proof uses bounded clipping in Lemma 2.2 and unitary lifting in Lemma 3.1, both proved below. If the assertion fails, choose and -related tuples , such that
Choose a free ultrafilter and let be the corresponding unitary classes in . In , put
This is a finite von Neumann algebra with the faithful normal restriction of . The coordinate projections belong to .
For each fixed integer , Lemma 2.2 gives bounded in operator norm by , for every . They therefore define classes . Their coupling errors have ultralimit zero, so . The operators
belong to , and (8) gives
For any central projection , traciality and give . Cauchy–Schwarz in (22) consequently yields
The left side is independent of , so it is zero. Testing all central projections, and then the positive and negative spectral projections of their center-valued trace difference, proves . Faithfulness of justifies that last implication. Finite projection comparison in gives a partial isometry with , .
These supports force , with unitary in . Its membership in says . Lemma 3.1 gives unitary representatives , and hence
There are only finitely many , so an -large set satisfies all these errors below . This contradicts (19). Taking the adjoint of the implementing unitary converts the intertwining form into (18).
Equation (22) bounds squared norms. Taking square roots gives (23). The proof uses its decay to zero, not the faster rate that would result from dropping that square root.
Lifting the implementing unitary
Lemma 3.1. Every unitary of has a representative consisting of unitaries of .
Proof. Take a bounded representative . Unitarity in the quotient implies . Extend each polar partial isometry to a unitary , so . This extension is possible in a finite algebra: the two support projections are equivalent, hence have equal center-valued traces, and their complements are equivalent by projection comparison. Add a partial isometry between those complements.
For , . Thus
The sequence represents the same unitary.
Bounded clipping by radial projection
Lemma 2.2. For every , the coupled tuples have operators satisfying
Proof by projection onto an operator-norm ball. The sampling lemma below supplies the 's; we first prove directly why clipping their singular values cannot increase an intertwining error.
For and the polar decomposition , put
The formula gives an element with , even when has a kernel. Its polar support satisfies , and . For any with ,
Indeed , and follows from . Expanding the trace square gives
Thus is the unique nearest element to , in trace -norm, among the operators of norm at most . Faithfulness of the trace proves uniqueness. The explicit formula supplies existence; no additional projection theorem is being assumed.
Apply (RC2) to , with their images , , using and , respectively. Adding the resulting inequalities yields
Dividing when , and treating equality separately, proves
Left and right multiplication by unitaries preserve the operator-norm ball and the trace -distance. Uniqueness in (RC3) therefore gives
Taking and applying (RC4) to , we obtain
We now verify every assertion in (8). For , choose : each squared positive-sum defect equals , and the coupling energies are zero. For , choose from Lemma 1.1 and set
Functional calculus gives , , and . In particular .
For ,
For the left side vanishes. For , subtracting from gives . The trace inequality
does not require operator monotonicity of the positive-part square. To see this, put . For ,
Equality holds for . Hence is the supremum of the left expression over . Replacing by the larger increases each expression, since , proving (RC8).
Since , (RC7)--(RC8) imply
Also . Expanding both squared distances from the identity therefore gives
Applying the same argument to , for which , proves the reversed-product bound. Finally (RC6) preserves each individual intertwining error. Their sum is less than by Lemma 1.1. This proves all of (8), including its original constants, in an arbitrary faithfully tracial finite algebra.
Second proof, by self-adjoint matrix clipping. For , take . Now let , choose the 's from Lemma 1.1 and put
Polar decomposition proves the last equality. Also , , so .
Set and . For scalar ,
For , the difference between the right numerator and is ; for the left side is zero.
We also need trace monotonicity: if , then
To prove it without an operator-monotonicity assumption, use
The expression is at most by and completing the trace square; equality is attained at . Positivity of the trace of a product of positives proves (11).
Apply (10)–(11) to each . Summing gives
Since , : their difference is . Therefore
The reversed products have the same proof.
Finally use the real clipping function . In , with its normalized product trace, set
Functional calculus gives . Direct computation of the two block entries gives
The corresponding identity for has in place of . Equation (6) and covariance of functional calculus under unitary conjugation show that each clipped error is no larger than its original error. Summing proves the final bound in (8).
Trace-Hilbert Lipschitz calculus
Lemma 2.1. If is Lipschitz with constant , then for bounded self-adjoint in a faithfully tracial finite algebra,
Proof. First suppose , have finite spectra. The weights are nonnegative, because they equal . Expanding squares and using both projection partitions gives
Uniform spectral approximations to converge in operator norm, as do their continuous functional calculi on the common bounded spectral interval. Passing to the limit proves (6). The projections from different partitions need not commute.
Finite sampling of a countable coupling
Suppose and are -related, in the terminology of the preceding lesson, with and . Thus
Lemma 1.1. For every integer , there are such that
Proof. Choose with , where
Their scalar traces agree. The continuous notation , with coordinate phases , has a finite realization for all the moments used here: take the independently and uniformly from . Every phase exponent occurring in a second or fourth moment lies between and ; the average of each nonzero such power is zero, just as for uniform circle phases. All expectations below can therefore be finite averages on the phase choices and their finite products. Put . Orthogonality of the phases gives
The fourth moment has an exact noncommuting expansion:
Indeed is nonzero exactly when the two multisets and agree. The two pairings give the first two terms in (5); the all-equal pairing is counted twice and must be subtracted once. The bounds prove the inequality. Exchanging with gives as well.
For independent samples, let . Independence and (5) yield
Consequently . The same estimate holds for the averaged reversed products. Each averaged coupling energy has expectation less than .
Markov's inequality bounds the probabilities of the first two failures in (2) by numbers strictly below each, and the probability of each energy failure by a number strictly below . Their union has probability less than . Choose a sample outside it and put . This gives all the inequalities simultaneously.
Finite and semifinite injective factors
Theorem 5.1. Every injective factor is locally approximately finite dimensional. If its predual is separable, it has a generating increasing dyadic sequence and is isomorphic to the tracial infinite product .
Proof. Fix a finite tuple of unitaries and an error . Theorem 4.1 gives a coupling threshold . Balanced reconstruction supplies a matrix subfactor and unitary models which are -related to . Equation (18) approximates every inside the actual finite-dimensional subalgebra .
Every element is a finite linear combination of at most four unitaries, as shown in the trace-preserving model lesson. Apply the tuple argument to all unitary summands in a finite set with an error budget divided by their coefficient sums. The resulting approximants have an operator-norm bound independent of the shrinking budget. Bounded -norm convergence for a faithful normal trace is -strong* convergence; thus this is local AFD in the von Neumann sense.
For separable predual, the finite-factor exact containment and uniqueness theorem converts local approximation into a generating increasing dyadic matrix sequence. Its compatible trace-GNS identification gives . No uniqueness theorem for injective factors was assumed in obtaining the local approximation.
Corollary 5.2. Every separable injective factor is isomorphic to .
Proof. We first construct for a nonzero finite projection . The type II∞ hypothesis supplies such a projection. Its corner is a finite factor and has no minimal projections, because a minimal projection there would also be minimal in . Thus it is of type , and the full center-trace construction gives its faithful normal normalized trace. Separable predual gives a faithful normal state on , by NT2.
Choose a maximal orthogonal family of projections equivalent to . Zorn's lemma applies because unions of chains remain orthogonal families. The family is countable: under the faithful normal state every member has positive value, while every finite sum of those values is at most one. The finite threshold sets with values at least , , exhaust it. Write and . Factor projection comparison and maximality imply : if , another copy could be added.
There must be infinitely many . Otherwise, connecting the finitely many copies and the residual subprojection of identifies with a corner of a finite matrix algebra over . That matrix algebra is finite: summing its diagonal entries under the faithful corner trace gives a faithful normal finite trace. This would make the identity of finite, contrary to type II∞.
If , put . If , choose equivalent to , and partial isometries with , , for . Put and . Orthogonality of the initial and final supports makes
a strong sum in , with
Thus the residual projection is absorbed without changing the size of a finite corner. The projections sum to one, and each is equivalent to ; the partial isometry identifies with .
For completeness, the countable matrix coordinates give the entire spatial tensor product. Choose with , , and . In a faithful representation of on , the map
is a unitary: the squared coordinate norms sum to , and every finitely supported tuple has preimage . It sends to the scalar matrix units. For , the bounded strong sum becomes . Conversely, every finite block compression of has entries . Those compressions converge strongly to , because their support projections increase to one. This proves equality with , rather than only an embedding. Flipping the tensor legs and identifying the equivalent corners gives the asserted decomposition.
Corner permanence of injectivity now makes injective, and it is a factor with separable predual. Theorem 5.1 identifies it with . The matrix decomposition uses countable projection comparison, independently of the finite injective-to-AFD result.
This also supplies the finite-corner input for the AFD MASA regularity branch: AFD implies injective, an injective finite corner is AFD by Theorem 5.1, and the finite regularity construction applies. The nonfactor construction is treated in Central disintegration and measurable matrix assembly and Injective algebras and separable tracial envelopes. The separate second finite proof develops the finite-rank model approach, using Repairing small matrix corners.
6. Problems with complete solutions
Corner comparison and the AFD conclusion
Exercise 6. Explain why the bound is needed in Theorem 4.1.
Solution. For fixed , it makes each coordinate sequence an element of the bounded sequence algebra defining . A bound only on its trace -norm would not give such an element: operators can have large norms on small projections. Lemma 2.2 retains the small intertwining errors while supplying the required uniform operator bound.
Exercise 7. Recover the normalization in (22).
Solution. The initial defect is block diagonal with upper-left entry and lower-right entry zero. The normalized product trace assigns the upper block weight . Its squared -norm is therefore half the corresponding squared norm in , at most . The final defect is the same computation in the lower block.
Exercise 8. Derive (23) using both defects in (22).
Solution. Each defect has -norm at most . Pairing it with a central projection changes its trace by at most that number, since . The two middle sums have the same pairing with by traciality and centrality. The triangle inequality therefore gives total difference at most , which tends to zero.
Exercise 9. What is needed to turn equality of all central projection pairings into projection equivalence in ?
Solution. The canonical center-valued trace satisfies . Its self-adjoint difference has zero pairing with every central projection. Its positive and negative spectral projections and faithfulness of the trace force the difference to vanish. The finite projection comparison theorem then makes and equivalent inside . Equality of their total scalar traces alone would not suffice in a nonfactor .
Exercise 10. Distinguish the hypotheses of Theorem 5.1's local and sequential conclusions.
Solution. Local AFD follows for every injective factor, using its faithful normalized trace and finite unitary tests. To choose a countable dense list for a generating matrix sequence, the proof assumes separable predual. That sequence has the tracial dyadic product identification with . The local conclusion therefore does not supply a countable sequence or an isomorphism with separable for a factor with nonseparable predual.
Clipping and finite phase averages
Exercise 3. Verify (10) when .
Solution. For , its left side is zero. For , the inequality is . Subtracting from gives . Equality occurs at ; the bound is valid without changing its constant.
Exercise 4. Why may (11) be used although is not assumed operator monotone?
Solution. Equation (12) proves monotonicity of its scalar trace value. If , then for every , because the trace of is nonnegative. Taking suprema gives (11). It asserts no operator inequality between the two functional-calculus values.
Exercise 5. Why are the weights in (7) nonnegative even when do not commute?
Solution. Traciality gives . The sandwich is positive, so its trace is nonnegative. Expanding the two finite spectral squares uses these weights and the trace, without multiplying the projection partitions as if they commuted.
Exercise 1. Explain the subtraction term in the phase moment (5).
Solution. The index conditions and each yield a surviving phase average. When all four indices agree, both conditions count the same term. Subtracting removes the second copy and leaves every surviving term counted exactly once. Noncommuting operator products retain their original order.
Exercise 2. Check why the simultaneous good-sample set has positive measure.
Solution. The two positive-sum failures each have probability strictly below . The energy failures each have probability strictly below . The union bound therefore gives total failure probability strictly below . Its complement has positive measure and contains a sample realizing all inequalities in (2).
References and proof scope
Uffe Haagerup, A new proof of the equivalence of injectivity and hyperfiniteness for factors on a separable Hilbert space, Journal of Functional Analysis 62 (1985), 160–201. Section 4 develops phase sampling, spectral clipping and ultrapower corner comparison; Theorem 5.3 applies coupling to injective finite factors. The sampling construction here uses finite fourth-root phase choices, with all noncommuting moments and simultaneous probability bounds supplied.
The radial clipping proof constructs the nearest point of an operator-norm ball in trace -norm, proves its contraction and two-sided unitary covariance, and gives every squared-defect estimate. The self-adjoint matrix proof is also complete: the trace-Hilbert Lipschitz inequality and scalar trace monotonicity are proved here. The ultrapower proof includes a finite polar unitary lift and the central-pairing bound. Neither factoriality of the ultrapower nor an external Lipschitz or unitary-lifting theorem is needed.
Claire Anantharaman and Sorin Popa, An introduction to II₁ factors, Proposition 4.2.5, p.63, gives finite matrix coordinates. Exercise 8.1, p.135, states the infinite finite-corner decomposition. Corollary 11.2.4, p.188, uses that exercise to classify separable AFD type II∞ factors. The full proof above constructs countably many orthogonal copies, absorbs the residual projection and proves the spatial matrix decomposition before applying the finite injective result. The exercise statement is not used as an unproved prerequisite.
The injective application retains the factor hypothesis of balanced reconstruction. Separable predual enters the generating dyadic sequence and identification with , and the countable II∞ decomposition. The programme lemma Uniqueness of the injective II₁ factor, Lemma 2.3, uses the same polar unitary-lifting mechanism at factor generality; its flip/McDuff route is a different proof. Central disintegration and measurable matrix assembly treats the nonfactor application.