Finite bases, bounded vectors and a positive-operator inequality

A finite-index inclusion behaves like a finite-dimensional vector space with coefficients in an algebra. The last coordinate may occupy only a projection corner. This is why a nonintegral index is possible, and why an ordinary basis of unitaries is too restrictive.

We use the basic-construction identities and trace normalization proved in A projection that remembers an inclusion, Theorems 1.2 and 1.4 and Corollary 1.5. The finite-trace supporting reading, M1–M6, M9 and M10, supplies the normal trace-preserving expectations, positivity and finite tracial commutation. Projection comparison and halving are proved in Projections and types of von Neumann algebras, Theorem 5.5 and Proposition 13.3. We explain their precise trace consequences below. The final example constructs its tensor factors and computes the index through a supported six-by-six corner. The finite basis itself proves the bounded-vector comparison directly in every normal representation.

Basic references are [Anantharaman–Popa], [Pimsner–Popa] and [Jones]. For a freely readable comparison, see Claire Anantharaman and Sorin Popa, An introduction to II₁ factors, Section 9.4, especially Proposition 9.4.5 and Proposition 9.4.8. The arguments below also give an explicit fractional final support and the uniform norm bound for the basis vectors. Except in the section explicitly treating every index, are II₁ factors of finite index . Put , , and . All expectations below preserve their normalized traces.

Recovering a coefficient from an operator

Lemma 3.1 — pull-down. Let be the trace-preserving expectation. Then

The normal linear map

satisfies

In particular, .

Proof. For , the trace formula in the first lesson gives

Faithfulness gives the first identity. For ,

Thus on . This algebra is ultraweakly dense in , by M7/(M11) of the finite-trace supporting reading: its ultraweak closure is the closed ideal generated by the full projection . Both sides are ultraweakly continuous in , so the equality holds on . Normality of follows from its formula.

Now . Applying the positive expectation gives

Taking norms proves the bound. Finally, , and the equality proves the reverse inclusion.

The coefficient is unique: if with , evaluation on gives .

A basis with a fractional final coordinate

Here are the trace consequences of the projection prerequisites. In a finite factor, comparison says that one of two projections is equivalent to a subprojection of the other. Equivalence preserves a faithful trace; if the two traces agree, the unused complement has trace zero and is zero. Thus equal trace gives equivalence, and smaller trace gives subequivalence.

In a II factor, Proposition 13.3 halves every projection into two equivalent orthogonal projections. Repeating this in each corner gives a nested dyadic partition of the identity. For any , let be the sum of the first pieces at level , ordered so that each piece's children are consecutive. Then and . The supremum has trace by normality. Applying the same construction with the normalized trace in a nonzero corner gives every trace between zero and that corner's trace. In particular, any finite list of nonnegative trace values summing to one is realized by orthogonal projections summing to one: choose each projection in the unused corner, and use the final remainder for the last entry.

Write , where and . The preceding construction gives a projection with . If , take . Let

Theorem 3.2 — partial orthonormal basis. There are such that

They satisfy

The expansion is unique among expressions with . Each has norm at most . If , the last element is zero and can be omitted.

Conversely, any family satisfying the displayed inner-product identities, with these support projections and traces, satisfies all the expansion identities.

Proof. Since , choose orthogonal projections summing to one with

Projection comparison in the II₁ factor gives partial isometries with

Put . Lemma 3.1 gives and . Orthogonality of the gives for . Hence

Faithfulness of on gives the inner-product identities. Also , so . Summing the final projections gives .

Applying proves . Multiplying the projection identity by gives

The injectivity of on proves the expansion. Its coefficients belong to , since . Applying proves uniqueness.

For the converse, put . The inner-product identities give , so each is a partial isometry with initial projection , and their ranges are mutually orthogonal. Thus ; injectivity of on gives . Their range sum is a projection with normalized trace

Faithfulness forces that sum to equal one. Applying and multiplying by give the scalar sum and expansion exactly as above; applying recovers each coefficient in , proving uniqueness. Pull-down uniqueness gives , and gives . If , then .

The theorem uses a right module convention: coefficients stand on the right, and are recovered as . Taking adjoints gives the corresponding left-coordinate formula

Switching the placement of coefficients without taking adjoints would generally be wrong.

For a direct test, let , with a II₁ factor, and use the one-element basis

It is an orthonormal right basis at index one. For , the prescribed right coefficient is . Thus , whereas

The correct left coefficient for the adjoint basis is , as in the preceding formula.

This also diagnoses the coefficient order in Theory of Operator Algebras III, Chapter XIX, Theorem 2.27(ii)(g), printed page 436: with coefficients determined by , the expansion is . The calculation in its proof on printed page 438 uses this right order. The displayed left order in the statement fails even in the identity inclusion above.

Proposition 3.3 — change of basis. If and are two such bases with the same support projection , then

defines a unitary in the corner

Here a unitary in the corner means .

Proof. The coefficient formula gives , so . Using the expansions and the inner-product identities gives

Interchanging the bases gives .

Corollary 3.4. The map

is a unitary of right -modules. The algebraic right module is isomorphic to , and is finite projective.

Proof. For bounded coefficients, the squared norm of the image is

The expansion proves dense range. Thus the isometry extends to a unitary. The bounded-coordinate expansion and uniqueness prove the algebraic assertion; each is a direct summand of the free module .

The inequality supplied by a finite basis

Theorem 3.5 — positive-operator bound. For every ,

Consequently,

Proof. Let , write , and form the row . Theorem 3.2 gives , so . Therefore

Every positive is for , proving the first assertion. Multiplying by on both sides and taking the trace gives

Bimodularity and trace preservation identify the left side with .

Corollary 3.5a — the exact constant. The largest constant such that for every is . There is a projection with

Proof. Choose a projection with , using the trace-range construction above. Since and , comparison in the II₁ factor gives a partial isometry with

Its initial projection gives . Pull-down supplies with . Jones compression and injectivity on give

Applying to , using bimodularity and , gives

Set , so . The equality implies , hence . Therefore

This is a nonzero projection with , and

If holds for every positive , apply it to and compress by :

Since , this forces . Theorem 3.5 proves that works. Finally,

The argument includes , when may be chosen.

A positive bound , with , also forces finite index, as the following argument proves. The downward basic construction provides a further structural explanation in the later tower theory.

The optimal constants at every index

Let be II₁ factors with normalized trace , let , , , and , where for . Put when . Define

Theorem 3.5b — both optimal constants at every index. For every such inclusion,

If , a nonzero projection attains both norm infima. If , there is a sequence of nonzero bounded positive elements for which both norm ratios tend to zero. The finite proof is Theorem 3.5 and Corollary 3.5a, (PP1)–(PP4): projection comparison and pull-down construct with and , giving optimality and equality. We supply the infinite case in full.

Small ambient supports with one standard coordinate

Assume . The basic-construction reading proves that is a semifinite II∞ factor and that

The second assertion is its Exercise 1.5, equation BC8: finite matrix amplification of the given possibly infinite trace and factor comparison transfer the infinite identity trace to every nonzero projection of the finite subfactor .

Choose any and a projection of trace , by the written dyadic trace-range construction. Factor projection comparison in makes either subequivalent to or subequivalent to . The latter would contradict (PI3), because a trace preserves equivalence and is monotone on projections. Therefore there is with

Let . Then , , and the map from to is the restriction of to , hence has norm at most one. Indeed , so commutes with every bounded right -action. No bounded element of is assumed to represent .

A positive lower bound would make that vector bounded

Suppose for all positive , for some . The positive normal functional

is well defined by the normal right action in the finite-trace reading M10. Although that action is antimultiplicative, it is linear, positive and normal; those are the properties used here. For ,

Thus positivity, the assumed inequality and trace preservation give

For , antimultiplicativity gives , so

Consequently extends to a bounded operator on of norm at most . For , , and density gives commutation on the completed space. The finite tracial commutation theorem M10.1 therefore gives for a bounded , with , and . Since , trace faithfulness gives . Now

Choosing (and ) is a contradiction. Therefore . This proves the implication from an operator lower bound to finite index directly, without assuming that an arbitrary vector is an affiliated operator.

Bounded positive witnesses for the norm constants

Let . It is a *-algebra: products reduce by . The basic-construction reading Theorem 1.2 proves that its closure acts nondegenerately and is strongly dense in . Kaplansky's density theorem, Theorem 7.1(1), applicable to a possibly nonunital and nonclosed *-algebra, therefore supplies contractions such that

Only this one-vector approximation is required, so no separability of is used. Since , write

For an explicit formula, if , take . The norm of is at most one, and compression gives

The implication from the compressed inequality uses the faithful standard action of on . Also , so

Put . For large , this is a nonzero bounded positive element. Bounded polar decomposition, Proposition 7.2, applied to inside , shows that the support of is equivalent to the range support of , which is at most . Thus . Trace Cauchy–Schwarz, applied to , gives

Equation (PI9) gives , and therefore

For each integer , choose , construct the corresponding , and choose an approximant with . The resulting nonzero bounded positive satisfies . Hence the squared infimum is zero, and taking square roots gives the unsquared infimum zero. In the finite case PP4 gives the unsquared constant . This proves (PI2) at finite index, infinite index and the index-one endpoint.

Theorem 3.5b supplies the infinite-index converse and both sharp constants. Its norm convention also clarifies Takesaki, Theory of Operator Algebras III, XIX.4.14, printed pages 475–477: the operator bound is , while the best nonsquared bound is . The projection in (PP3) tests the distinction for every finite . With squared norms, (PI2) gives , including zero at infinite index. The later tower lesson retains a separate downward-construction route.

When an apparently unbounded vector is bounded

A vector may be represented by an unbounded affiliated operator. Call it right -bounded if the map

extends to a bounded operator .

Theorem 3.6. If is right -bounded, then for an element , and

The operator belongs to , has norm , and satisfies

Proof. The map intertwines the right -actions on its domain and range. Thus commutes with the right -action and belongs to . As , its norm is exactly . Also . Pull-down gives ; put . Evaluating on gives . The remaining identities follow from Lemma 3.1.