The basic construction can be repeated. Going upward is canonical once the inclusion is given. Going downward requires a choice, but all choices at one step differ by a unitary in the smaller factor. The downward step also produces an element that makes the positive-operator index bound sharp.
where is the projection of onto . Although these projections are first constructed on different Hilbert spaces, each is an element of ; the compatible algebra inclusions put all of them in a common algebraic tower.
Theorem 4.1. Every is a II₁ factor, every consecutive inclusion has index , and the normalized traces agree on overlaps. If preserves the trace, then
Proof. For the first step, represent on . The first lesson gives . The commutant-index formula in the second lesson gives . Conjugation by carries to , preserving index, so . Here conjugation is a conjugate-linear isomorphism; standard Hilbert spaces and their real positive traces preserve the same dimension under it. The trace and compression assertions are the first lesson's trace formula. Applying the same argument recursively proves every assertion.
Theorem 4.2 — projection relations. In this tower,
Proof. Each commutes with , while for . This proves distant commutation. The compression identity gives
For the other adjacent identity it suffices to consider , because the same reasoning applies at every level. On , . For ,
For ,
These vectors span a dense subspace of . The bounded operators therefore agree everywhere.
Proposition 4.3. The union , completed in its common trace representation, generates a II₁ factor .
Proof. The compatible normalized traces give a faithful trace on the algebraic union and a finite von Neumann algebra in its GNS representation. Let . Trace-preserving expectations onto the increasing satisfy in , because their range union is dense in the completed Hilbert space. Bimodularity gives , hence . Thus is scalar. The algebra contains the diffuse factor , so it has type II₁.
This construction of uses the common trace. It does not assert that the whole infinite tower acts normally on the original space .
A downward step from a module of dimension one
Theorem 4.4 — downward construction. There is a II₁ subfactor such that
where the isomorphism fixes . Its image of satisfies
Proof. Choose a projection with , and let . Left multiplication makes it an -module with dimension . Its restriction to has dimension . Finite-module classification therefore identifies it, as a left -module, with .
Let be the transported left -action on , and let be the tracial conjugation there. Since , its commutant lies in . Define
This is a II₁ factor. The reciprocal dimension formula gives
Conjugation by changes this to the left -dimension of the standard -space. Thus . The basic construction of is
Transporting back to gives the desired isomorphism. The upward trace formula for says that the trace-preserving expectation from its basic construction to sends to . Hence .
Which projections can be downward Jones projections?
The scalar-expectation condition completely characterizes them.
Theorem 4.5 — recognition and uniqueness. Let be a projection with . Then
is a downward basic construction. If is another such projection, there is a unitary with . Conversely, every -unitary conjugate of has the same scalar expectation.
Proof. Since , the left -dimension of is one. The map
is isometric, because
Its range is a closed -submodule of dimension one, and so is all of . For , there is therefore a unique with . For ,
The tracial bounded-multiplier criterion shows that , with . To see this criterion directly, represent as an affiliated -operator. Testing the inequality on spectral projections of forces ; hence it is bounded. Thus . Taking adjoints gives . The linear span is consequently a two-sided ideal of . Its weak closure is nonzero and, since is a factor, equals . This proves .
Transport the left -action to through , and write . The operator commutes with . Indeed, its matrix coefficients are
and tracial cyclicity gives exactly the same coefficient for . Since the transported algebra is , its conjugated commutant is
The same reciprocal-dimension calculation as in Theorem 4.4 gives and identifies the transported algebra with the basic construction of . Moreover . Because commutes with , its range contains . Its range has -dimension , the same as . Faithfulness of the dimension trace forces equality. Thus , proving recognition with the actual Jones projection.
For uniqueness, , so there is a unitary with . The coefficient argument above gives with . Then . Applying gives . In a finite algebra a coisometry is unitary, so . The converse follows from bimodularity.
The factor in the last proof is identified inside before the basic-construction assertion. A projection merely having the right trace would not suffice; its full expectation must be scalar.
The sharp constant at finite index
Corollary 4.6. For a finite-index II₁ inclusion, the largest constant such that for every is . Also
Proof. The third lesson proves the lower bounds. Choose the projection from Theorem 4.4. If , compression by gives , so . Also
Thus both bounds are attained.
The corresponding nonsquared norm ratio has a different constant:
Indeed, all the ratios are nonnegative, so taking square roots in Corollary 4.6 gives the lower bound. The same projection , with and , attains it: its two norms are and . Thus the operator-order constant and the squared-norm constant are , while the nonsquared-norm constant is .
This distinction corrects the displayed norm formula in Takesaki's Theorem XIX.4.14. Its equation (14′) uses nonsquared -norms but puts on the left. At finite nontrivial index that constant is ; equivalently, both norms must be squared to obtain the inverse index. The operator-order formula is retained. Positivity restricts the index, Theorem 7.5, proves the squared formula also at infinite index, where both norm constants are zero. At index one all three constants are one.
Continuing downward
Apply Theorem 4.4 to , then repeat. This gives a tunnel
in which every consecutive triple is a basic construction. Its Jones projections satisfy the same adjacent and distant relations as the upward projections. At any stage the next choice is unique up to a unitary in the current smaller factor, by Theorem 4.5. The tunnel is therefore an additional choice, whereas the upward tower is fixed by the original inclusion.
Proposition 4.7 — the same convention at every integer level. For every integer , the projection implements , and
Here the compression identity applies to , and every expectation preserves the compatible normalized trace. The adjacent and distant relations of Theorem 4.2 hold for all integer indices.
Proof. Every consecutive triple in the downward construction is a basic construction by Theorems 4.4–4.5. The upward construction has the same property by Theorem 4.1. Applying the basic-construction compression and Markov identities to each such triple gives (4.2). The normalized traces agree: downward levels inherit the trace of , and upward traces restrict to that same trace. For distant indices, the later projection commutes with the earlier projection's containing factor. For adjacent indices, apply the calculation of Theorem 4.2 in their consecutive triple; no step depends on an index being nonnegative. This proves all the assertions.
In particular, is the projection for , and . The already reserved is the projection for , while is the projection for . Naming the projection by the middle level avoids shifting it when the tunnel is introduced.
In later lessons we will ask for a tunnel whose finite-dimensional relative commutants generate the original factors. The existence of a tunnel alone says nothing about that stronger approximation property.
Exercises
Exercise 4.1 — introductory. If , compute , , and .
Solution. There are four consecutive inclusions from to , so multiplicativity gives . The Markov trace gives , and the projection relation gives .
Exercise 4.2 — intermediate. For , construct a projection with expectation .
Solution. Choose matrix units for a unital -subfactor of , which exists by diffuse projection comparison. Put
Then , and multiplication of matrix units gives . Partial trace gives
The index is four, so Theorem 4.5 recognizes this as a downward Jones projection. It also witnesses the optimal inequality constant , even though scalar averaging on alone has optimal constant .
Exercise 4.3 — intermediate. Why does a projection of trace fail to satisfy the hypothesis of Theorem 4.5 automatically?
Solution. Trace preservation only gives ; it does not force to be scalar. In the tensor inclusion , choose a projection of trace and set . Then , but .
Exercise 4.4 — advanced. For index , find a partial isometry between adjacent Jones projections and its initial and final projections.
Solution. Set . The adjacent relations give
Thus neighboring projections are equivalent inside the algebra they generate. This equivalence will later propagate central-support stabilization.
Exercise 4.5 — intermediate. For the actual index-four tensor inclusion and projection of Exercise 4.2, compute both expectation-to-vector norm ratios. Determine the optimal constants for the operator-order and nonsquared-norm inequalities.
Solution. Since is a projection of trace and ,
Theorem 3.5 gives the universal lower bounds, and attains both. The optimal nonsquared-norm constant is therefore . The optimal operator-order constant is : compressing by forces . In particular the larger norm constant cannot be used in that operator-order inequality.
Exercise 4.6 — intermediate. Which Jones projection implements ? Give its containing factor, its expectation onto the middle level and its generated basic construction. At nontrivial index, can it belong to the middle factor?
Solution. It is , and
If it belonged to , this expectation would instead fix it, giving . At , the scalar is not a projection. Thus the projection lies in the next factor and not in the middle factor. At index one the projection is one and every tower and tunnel level agrees.