A tunnel may generate a proper factor through its relative commutants. If that factor has finite index, its basis can be moved arbitrarily far down the tunnel. This makes it possible to test generation by conditional expectations of bounded vectors. The test is stronger than mere nonzero approximation: it must have one positive lower bound for every sufficiently late level and every vector in a specified norm ball.
Let be separable II₁ factors of finite index and finite depth. The identity inclusion is already classified separately. Write
Theorem 14.4 and reflection make a factor. Unless stated otherwise, the tunnel is arbitrary. Expectations preserve normalized traces.
Corners and downward finite depth
Lemma 15.1. Every consecutive inclusion in the tunnel has finite depth. For every , the algebra is a factor.
Proof. A common corner by a nonzero projection in the smaller factor preserves the standard invariant and finite depth. Here are the needed details. If , the relative-commutant map is . It is faithful on , since has full support in the factor . It is onto : choose finitely many partial isometries with and orthogonal summing to one, and extend an element of the latter commutant by . Commutation with the corner matrix coefficients proves that this extension commutes with and compresses back to . The same construction applies at every level of the common-corner Jones tower. Jones projections commute with , and their full-support criterion is preserved. This proves the claim about corners.
Now consider a downward triple . Its next upward factor is . The common corner inclusion
is isomorphic to . Indeed, . The adjoint pull-down identity gives ; the map identifies this right -module with . Since , its corner is , and becomes . If has finite depth, its dual has finite depth by Theorem 14.4. The corner claim therefore gives finite depth of . Induction moves this conclusion down the tunnel.
For , the expectation onto preserves , since . Consequently
To justify equality, approximate an element of by expectations onto the increasing finite algebras , then apply . The resulting elements lie in the union on the right and converge in . This is the downward relative-commutant closure for the finite-depth adjacent pair , so Theorems 14.3–14.4 make it a factor.
Expectations commute with later relative commutants
Proposition 15.2. If , then
The same commutation holds with replaced by . Its product is , where
Taking increasing limits gives the commuting square
Proof. Conjugation by on has fixed-vector space . The projection onto commutes with that action because . It thus commutes with the projection onto the fixed vectors. Their common range is , proving (15.3).
For clarity, that fixed-vector projection can be obtained without any group amenability assumption. The norm-closed convex hull of the conjugates of a bounded element has a unique vector of least -norm. Invariance of the hull makes this vector fixed. Its pairing with every fixed vector agrees with that of the original element, so it is the orthogonal fixed-vector projection, namely the relative-commutant expectation. The hull is uniformly bounded, hence its limits are elements of .
The subspace is also invariant, so the same projection argument applies to . Alternatively, for and ,
Such products span , since the commuting algebra is finite dimensional. This proves (15.4). Finally in . Pass to the limit in (15.3) and use (15.2). The resulting product of expectations is , proving (15.5).
The same index is visible at every tunnel level
Assume now that . This is a substantive hypothesis; finite depth alone has not yet supplied its existence in the present argument.
Theorem 15.3. For every ,
A partial orthonormal basis for is also such a basis for .
Proof. Put ; this scalar is distinct from the algebra . By (15.5) and the positive-operator index inequality,
The variational characterization of index therefore gives
.
Apply Proposition 14.8 to the skipped-level triple
Its Jones projection satisfies . It commutes with , so . Equation (15.5) gives . The optimal positive-operator bound for , evaluated on the nonzero projection , yields
Multiplicativity now forces equality:
Computing through proves (15.6).
Let be a partial orthonormal basis over , with support projections . Commutation of expectations gives
In the basic construction of , the operators are mutually orthogonal projections. Their sum has normalized trace
Faithfulness makes that sum the identity. Multiplying by and using the compression identity gives
Thus the basis is also a basis over .
The fractional support in this basis measures , not . More precisely, with , one can choose full entries and one additional entry with support of trace . If is an integer, the additional entry is zero. All traces here are normalized; the inclusions preserve that trace. Takesaki's proof of Lemma 4.26(iii) puts in this support formula. Exercise 15.5 shows that this expression can exceed one in an actual generating tunnel.
The basis uses right coefficients as in (15.7): the coefficient follows . Partial orthonormality does not justify reversing that order. The proof above obtains the correctly ordered expansion directly from the basic-construction identity.
A bounded vector detects a proper inclusion
Lemma 15.4. If are II₁ factors with finite index , there is such that
Proof. A proper finite-index inclusion has , by the allowed-index theorem. In its basic construction the Jones projection has normalized trace . Projection comparison provides a unitary with . Pull down to , so and . Orthogonality gives , hence . Also
The corner map is faithful, giving , and its trace gives .
A uniform orbital test forces generation
Theorem 15.5. Suppose the tunnel in (15.1) has . Suppose further that there are and such that for every and every with
some satisfies
Then admits a generating tunnel.
Proof. Choose a countable -dense sequence in the displayed norm-bounded unit sphere, and a countable -dense sequence in the positive unit ball of . Separability supplies both sequences.
We build finite prefixes . Any already chosen finite prefix can be matched with the original one by a unitary : use one-step tunnel uniqueness successively; a correcting unitary at the next level lies in the previous smaller factor and preserves every earlier level.
At stage , take such . The positive-vector index inequality, with its squared -norm convention, gives
Choose sufficiently large that, for ,
This uses the increasing-limit convergence . Apply (15.9) to , obtaining . Increasing-limit convergence then supplies such that
Extend the prefix by through level . Because , it preserves all with , and it fixes pointwise. Thus the old prefix is preserved, as are all earlier detection bounds. Equation (15.10) is preserved too. This completes the induction.
Let . Lemma 15.1 makes it a factor. Passing to limits and using density gives
for every in the stated unit sphere. The positive-vector variational characterization gives . If , Lemma 15.4 supplies a vector of that sphere with , a contradiction. Hence .
Finally . The expectations commute with those onto , by Proposition 15.2 with the fixed initial level. Since the latter converge to the identity on , their restrictions approximate every element of by . Thus the constructed tunnel generates both endpoints.
The quantifiers in (15.9) cannot be replaced by a separate lower bound for each vector or by detection at only one level. The induction uses later-level freedom while preserving all earlier choices.
The two normalization requirements in this argument remain fixed throughout the induction. A dense sequence of orbital test vectors must belong to the same set
that occurs in the hypothesis; it is dense in that set for the tracial metric. The bounded missed vector from Lemma 15.4 belongs to this set when . For positive vectors, the squared variational bound gives the nonsquared norm bound , as used in (15.10). A nonsquared bound alone would only imply , which does not put that missed vector under the required cutoff. These are the needed corrections to the dense-vector cutoff and the norm coefficient in the proof of Takesaki's Lemma 4.25; the generating-tunnel conclusion is retained with the corrected proof above.
Relative commutants of varying algebras in an ultrapower
Let be a free ultrafilter and the tracial ultrapower. A bounded sequence represents zero precisely when . For subalgebras , let .
Lemma 15.6. There is equality
Proof. The right side plainly commutes with . Conversely, put . Coordinatewise expectations define the trace-preserving expectation onto : they are uniformly bounded and contract , so the definition is independent of representatives. An element represented by is outside this subalgebra exactly when
By the least-norm orbit argument in Proposition 15.2, lies in the closed convex hull of the -unitary conjugates of . Hence
Choose realizing at least half of that lower bound whenever it is nonzero. The sequence defines a unitary , and . Thus . This proves the reverse inclusion. No uniform choice of a single unitary in all was required.
The remaining existence argument must establish a finite-index tunnel and the uniform orbital test (15.9). The lemmas above give their consequences without assuming either conclusion in their proofs.
Exercises
Exercise 15.1 — introductory. If , what constant and unitary work in (15.9)?
Solution. Here , every expectation onto is the identity, and every tested vector has -norm one. Take and any .
Exercise 15.2 — intermediate. In Theorem 15.3, explain why one ordinary negative Jones projection cannot simply be declared to be the skipped-level projection .
Solution. The skipped inclusion has index , so its downward Jones projection must have expectation onto . A projection for one adjacent inclusion has the one-step normalization at its own level, and may even belong to . Proposition 14.8 constructs the appropriate projection for the composite inclusion; its commutation with then puts it in .
Exercise 15.3 — intermediate. Suppose a family obeys partial orthonormality over , and . Prove that it is a complete basis.
Solution. The projections in the basic construction are orthogonal. Their sum has normalized trace one by the basic-construction trace formula. Thus it is the identity. Apply it to , use compression, and use faithfulness of to obtain (15.7).
Exercise 15.4 — advanced. In Lemma 15.6, why would testing only constant sequences of unitaries be insufficient?
Solution. Commutation with the diagonal copy of one fixed algebra tests for each constant . The obstruction at coordinate can depend on , and for varying there may be no common unitary detecting it. The proof chooses separately and tests the resulting unitary of . The full ultraproduct relative commutant quantifies over all such bounded sequences.
Exercise 15.5 — advanced. Take the actual index-six hyperfinite group inclusion of Theorem 41.4, and choose a generating tunnel using Theorem 17.5. For , compute the fractional support trace in the basis for . Compare it with .
Solution. The group example is a proper finite-depth inclusion of separable hyperfinite II₁ factors, so Theorem 17.5 applies. Generation makes , whence and . The basis is the single element , with an optional zero second entry. Its fractional support trace is . On the other hand, , so . The compared expression is , which cannot be the normalized trace of a projection. This verifies the erroneous support formula inside the hypotheses of the source lemma, while Theorem 15.3 and its correctly normalized basis remain valid.