Two different positivity arguments control the index. Positive traces on a sequence of recursively constructed projections exclude most numbers below four. At infinite index, averaging in the semifinite basic construction makes the variational constant collapse to zero.
Quantum integers and recursively constructed projections
Suppose has finite index . Set and, using the Jones projections in its tower, put . Shift the generator labels in this section so that the first generator is . The relations become
Define real numbers
The brackets here denote numbers, not subfactor indices. Let be the algebra generated by , inside the tracial tower. Whenever the denominators are nonzero, define
These are the Jones–Wenzl projections before the first zero denominator.
Lemma 7.1. For , if are nonzero, then is a self-adjoint projection, lies in , and satisfies
Proof. We prove the assertions inductively. The base case is , with trace . For , gives , a projection annihilating .
For the general step, put . Write
The operator commutes with , whose generators have indices at most . Using and the adjacent relation gives
and, by the same expansion without the leftmost ,
When , these identities make
a self-adjoint projection, annihilating . The earlier annihilate both and , so they also annihilate . Multiplication by fixes it.
The Markov trace gives
because the newest has expectation onto the preceding factor. Thus
This proves the induction. Notice that the projection calculation itself is algebraic; the trace is what forces its trace value to be nonnegative.
The restriction below four
Theorem 7.2 — Jones index restriction. The index of a II₁ subfactor belongs to
Proof. Infinite index already belongs to the displayed extended interval. Suppose . Write with . The sine addition formula gives
If is not an integer, let be the smallest positive integer with . Then are positive, while . Lemma 7.1 constructs the projection using only the positive denominators , but assigns it trace
This contradicts positivity of the trace. Therefore for an integer , giving the discrete values.
This proves a restriction. Existence at each discrete value needs a separate construction; the theorem alone does not supply it. The diagonal-corner example of the second lesson already realizes every finite value at least four.
Corollary 7.3. A proper II₁ inclusion has index at least two. At a discrete value , the projection vanishes in the tracial Jones tower.
Proof. The first discrete values are for and for ; index one is equality of the two factors. For the second assertion all the denominators needed for are positive, and its trace is . Faithfulness of the trace forces the projection to vanish.
An averaging lemma in infinite trace
We now address the infinite-index part of the positive-operator characterization. The argument uses an inclusion where is a finite factor, is semifinite with a faithful normal trace , and
For a finite partition in , write
.
Lemma 7.4 — small pinching. If , then for every there is such a partition with
Proof. The closed convex hull in of has a unique vector of least norm. All its elements remain bounded positive operators of norm at most . To justify this for a Hilbert-space limit, choose an ultraweakly convergent subnet in the bounded positive ball and test with , which is dense in . Trace pairings identify the ultraweak limit with the Hilbert-space limit. The least-norm vector is invariant under all these conjugations, so belongs to . No nonzero scalar lies in , because . Thus the least-norm vector is zero.
Consequently some satisfies
Indeed, otherwise every finite convex combination would have its pairing with larger than that fixed positive bound and could not approach zero. Approximate in operator norm by a finite-spectrum unitary . Its approximation can be chosen to give
using . Let be the spectral partition of . The corresponding pinching is the orthogonal projection onto the block-diagonal subspace. The diagonal blocks cancel in , so
The previous pairing bound gives . Hence
The same argument applies inside each nonzero corner . Its relative commutant is : a full-corner comparison transfers the commutant from the factor inclusion. Explicitly, choose a finite family of partial isometries , including , with and orthogonal summing to one. An element commuting with extends by to an element commuting with : check its matrix blocks using for . The extension is scalar, and its -corner is . Each such corner identity has infinite -trace by hypothesis.
Refine every block by its new spectral partition. The squared -norms of the orthogonal diagonal blocks add, so each refinement reduces their total by a factor at most . After finitely many refinements the factor is smaller than .
Infinite index gives variational constant zero
Theorem 7.5. For any II₁ inclusion, including infinite index,
where .
Proof. The finite-index case is Corollary 4.6. Assume the index is infinite. If is infinite dimensional, it has nonzero projections of arbitrarily small -trace. Since commutes with , bimodularity gives , and trace preservation gives . Therefore
Suppose next that . In the basic construction , we have
The relative-commutant assertion follows by conjugating with the tracial . Every nonzero projection of has infinite -trace by projection comparison, as in Exercise 1.5. Lemma 7.4, applied to , gives a partition in with
Trace cyclicity and the compression identity give
Since , at least one nonzero has ratio .
Finally, if is finite dimensional but not scalar, choose a partition into minimal projections. The local-index partition formula at infinite index forces some corner to have infinite index. Its relative commutant is scalar, so the preceding argument applies there. Put , let be , and let preserve the normalized corner trace. Trace pairings show that, for ,
Thus the global squared-norm ratio for positive is times the corner ratio, and again has infimum zero.
If for all positive , trace pairing with yields , exactly as in Theorem 3.5. Since that infimum is zero, no positive is possible. The constant zero always works.
Exercises
Exercise 7.1 — introductory. Compute as polynomials in , and the first three values of .
Solution. The recurrence gives , , . The values for are .
Exercise 7.2 — intermediate. Show directly why cannot be a II₁ subfactor index.
Solution. Here , , while . The projection is defined using positive denominators , but its trace would be , which is impossible.
Exercise 7.3 — intermediate. For a positive matrix block operator , explain the factor four in the pinching estimate in Lemma 7.4.
Solution. On a block , conjugation by the finite-spectrum unitary multiplies it by , where . The difference multiplier has absolute value at most two and hence squared absolute value at most four. The diagonal blocks have difference multiplier zero. Orthogonality of the blocks in tracial gives the estimate.
Exercise 7.4 — advanced. An inclusion has infinite index and scalar relative commutant. Why is a small value of not the reason the variational infimum vanishes?
Solution. That trace is fixed at one. The vanishing comes from conjugating and pinching this fixed finite-trace projection by elements of , whose nonzero projections have infinite basic-construction trace. Irreducibility makes the Hilbert-space orbit average zero. The resulting partition contains a projection with arbitrarily small expectation-to-vector norm ratio.
Vaughan F. R. Jones, Index for subfactors, Inventiones Mathematicae 72 (1983), 1–25.
Hans Wenzl, On sequences of projections, Comptes Rendus Mathématiques de l'Académie des Sciences, Canada 9 (1987), 5–9. The recursive projections used above are called Jones–Wenzl projections.
Mihai Pimsner and Sorin Popa, Entropy and index for subfactors, Annales scientifiques de l'École normale supérieure 19 (1986), 57–106.
Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, September 2026. Self-checked by the writing AI. Public domain (CC0).