Relative generation and the generator problem for finite factors

Proof. Let b\in\mathbf Q'\cap\mathbf M have a uniformly bounded representative (b_n), and put

Then \lim_{n\to\omega}\delta_n=0. Otherwise there would be an \varepsilon>0 and a set belonging to \omega on which \delta_n>\varepsilon. Choosing on this set a unitary w_n\in P with \|w_nb_nw_n^*-b_n\|_2>\varepsilon, and setting w_n=1 elsewhere, would give a unitary in \mathbf Q that does not commute with b.

For each n, let K_n be the closed convex hull, in L^2(M,\tau), of the elements wb_nw^*, w\in\mathcal U(P). The unique element c_n\in K_n of smallest 2-norm is fixed by conjugation by every unitary of P: each such conjugation preserves K_n and its Hilbert space norm. Moreover, c_n belongs to M and \|c_n\|\leq\|b_n\|. Indeed, convex combinations from this orbit have this operator norm bound, and an operator-norm-bounded 2-norm limit lies in the same bounded ball of M. To see the latter fact, left multiplication by such a convergent sequence converges strongly on the dense subspace M\subset L^2(M), since

and hence converges strongly on all of L^2(M) by the uniform operator norm bound. The strong limit belongs to M, and its value at 1 is the prescribed 2-norm limit.

Thus c_n\in P'\cap M=\mathbb C1. The trace is constant on K_n and continuous in 2-norm, so c_n=\tau(b_n)1. Every element of K_n is at 2-distance at most \delta_n from b_n. Consequently

The bounded scalar sequence (\tau(b_n)) has a scalar ultralimit, which represents b. This proves the assertion.□

Proof. The subspace X=D\cap\ker\tau is norm separable, hence 2-norm separable. For example, applying d\mapsto d-\tau(d)1 to a countable norm-dense subset of D gives a countable norm-dense subset of X. Apply Theorem 2.2 to obtain a diffuse abelian algebra A\subset\mathbf Q.

We first choose a Haar unitary in A. Diffuseness gives arbitrarily small nonzero subprojections of every nonzero projection: repeatedly split and retain a piece of at most half the trace. Normality of the trace and maximality under inclusion then give subprojections of every prescribed smaller trace. We may therefore choose recursively partitions of 1 by projections

with p_{m,k}=p_{m+1,2k}+p_{m+1,2k+1}. The self-adjoint elements

satisfy \|h_{m+1}-h_m\|\leq2^{-(m+1)} and converge in norm to an element h\in A. For each continuous function f on [0,1],

It follows that t=\exp(2\pi i h) is Haar.

We next remove the centering requirement at the two endpoints, so that (2.2) can be applied to words with arbitrary outside coefficients. Expand each endpoint into its scalar and centered parts. A term with a centered last endpoint vanishes by (2.1), since every nonzero power of t belongs to A and is centered. A term with a centered first endpoint and scalar last endpoint vanishes after moving the first endpoint to the end by traciality. It remains to consider the word with both endpoints omitted. When j=1 this is the Haar property. When j\geq2, rotate the final power t^{n_j} next to the first power t^{n_1}. If n_j+n_1\ne0, the resulting word is covered by (2.1). If n_j+n_1=0, the two powers cancel and leave a word starting and ending with centered elements of D. This word is again covered by (2.1), except when j=2, in which case it is simply the centered element d_1. This proves (2.2).

Laurent polynomials in t are norm dense in C^*(t). As t is Haar, the centered Laurent polynomials are precisely the linear combinations of its nonzero powers. Expanding an alternating centered word, and using (2.2) followed by norm approximation, proves that t is free from D.□

Proof. For any Haar unitary free from D, formula (2.2) follows immediately by expanding the endpoints into scalar and centered parts and applying freeness to each resulting alternating word. Replacing t by t^* merely negates the exponents, so we concentrate on v=atb.

Consider a word

with nonzero integer exponents and centered internal factors d_1,\ldots,d_{j-1}. Expand each power of v into individual letters t and t^{-1}, with elements of D between them. Within one power of v all these letters have the same sign. A gap between opposite signs can therefore occur only between two successive powers of v. Such a gap equals

It is centered by traciality. Every other gap joins equal signs. Expand each of these latter gaps into its scalar and centered parts. In a resulting term, removing a scalar gap only joins letters of equal sign. Thus each block joined across scalar gaps is a nonzero power of t, and consecutive blocks are separated by centered elements of D. The endpoint factors remain arbitrary elements of D. Formula (2.2) shows that every resulting term has trace zero.

We have proved (2.2) for v. Taking j=1 and d_0=d_1=1 proves that v is Haar. The general formula and norm approximation by Laurent polynomials then prove its freeness from D.□

Proof. For a letter a, let B_a be the orthogonal projection onto the closed span of reduced words beginning with a. Split

The first summand has range in the range of B_\gamma. These ranges are mutually orthogonal, so for \eta\in\ell^2(F),

For the second summands, the triangle inequality and Cauchy–Schwarz give

because the projections B_{\gamma^{-1}} are mutually orthogonal. Adding the two estimates proves the lemma.□

Proof. Fix \ell and work first in \mathbf M=M^\omega, with \mathbf Q=P^\omega. We use \tau also for the ultrapower trace.

Free variables and prefixes. Set D_0=C^*(1,u,z,\xi). By Lemma 2.3, choose successively t_j\in\mathcal U(\mathbf Q) Haar and free from D_j, and put D_{j+1}=C^*(D_j,t_j), for 0\le j\le\ell. Each of these C^*-algebras is norm separable. Define

and

Thus g_k is the prefix immediately preceding the kth occurrence of u in W. Since D_j=C^*(D_0,s_1,\ldots,s_j), Lemma 2.4 shows that s_{j+1} is Haar and free from D_j. In particular, s_1,\ldots,s_\ell are free Haar unitaries. Every nonidentity reduced word in them has trace zero, so the group they generate can be identified with the free group G on these generators. Since W=(g_\ell u)t_\ell and g_\ell u\in\mathcal U(D_\ell), the same lemma shows that W is Haar and free from D_\ell. Applying it to W^* and z\in\mathcal U(D_\ell) also shows that

is Haar and free from D_\ell.

For p\in G set Y_p=pyp^{-1}. These form a free family of Haar unitaries. Indeed, if a_1,\ldots,a_r are nonzero integers and successive labels p_1,\ldots,p_r are distinct, then

Each internal ratio is a nonidentity element of G and hence is centered in D_\ell. Equation (2.2) gives zero trace. It follows, by applying this to the reduction of v^*w, that distinct reduced words in the Y_p are orthonormal in L^2(\mathbf M,\tau), including the empty word.

A left perturbation of u at its kth occurrence in W is conjugated by g_k when moved to the front of the word. We therefore use the conjugates g_j^{-1}yg_j: conjugating the jth one by g_j returns y. Taking imaginary parts will give a self-adjoint perturbation, while freeness makes their average small in 2-norm. Put \operatorname{Im}(b)=(b-b^*)/(2i) and define

The g_j^{-1} are distinct in G, so the summands defining h are orthogonal centered imaginary parts of free Haar unitaries. Each has squared 2-norm 1/2. Consequently,

Write H_k=g_khg_k^{-1}, p_{kj}=g_kg_j^{-1}, and c=\alpha/(2\ell). Cancelling consecutive prefixes gives

For the first identity, use e^{iH_k}=g_ke^{ih}g_k^{-1} and g_k^{-1}g_{k+1}=ut_k for 1\le k<\ell; the product order is therefore essential.

Let S=\{p_{kj}:1\le k,j\le\ell\} and let n_p count the ordered pairs (k,j) with p_{kj}=p. For the identity e of G, one has n_e=\ell. If k\ne j, the word

is reduced, since its joining letters have different indices. Its initial positive segment and its final negative segment determine k and j. All off-diagonal pairs therefore give distinct nonidentity words, and

The first-order correlation. Put C_1=\sum_{k=1}^{\ell}iH_k. Each diagonal term cY_{p_{kk}}W equals cz, because p_{kk}=e and Y_eW=z. We show that all other linear terms vanish when paired with \xi. Equation (2.2) for W and D_\ell gives \tau(\xi^*W)=0. For p\ne e, the element p^{-1} is centered in D_\ell, so

For every p\in G, one also has

Indeed, split the internal factor z^*p^{-1}\in D_\ell into its centered and scalar parts. The first contribution vanishes by (2.2); the second is a scalar multiple of \tau(\xi^*pW^2) and vanishes by the same formula. It follows that

The uniform remainder estimate. It remains to bound the contribution of \prod_k e^{iH_k}-1-C_1 uniformly in \ell. Group the ordered product of exponentials by its original degree:

The term C_0 is 1, and the degree-one term is the C_1 above. For fixed \ell, the expansion is absolutely convergent in operator norm, because

The Cauchy product may thus be grouped in this way, and its sum is \prod_k e^{iH_k}. We will prove the bound

which, unlike the preceding norm-convergence bound, is uniform in \ell.

Let F_S be the formal free group with generators T_p, p\in S. In particular T_e is a generator, not the identity of F_S. Let \Phi:\mathbb C[F_S]\to\mathbf M be the homomorphism \Phi(T_p)=Y_p. By the orthonormality proved above, for every finite group-ring sum f=\sum_{w\in F_S}f(w)w one has

Define

and

For a group-ring element f, write |f| for its coefficientwise absolute value, and compare nonnegative coefficient functions coefficientwise. Group multiplication gives

This comparison already includes all free reductions and all contributions collected at the same resulting word. Thus the formal element

satisfies

For the last inequality, each tuple (m_1,\ldots,m_\ell) corresponds to one nondecreasing sequence of row indices in the distributive expansion of (A_1+\cdots+A_\ell)^d. Its factorial weight is at most one, and every other summand has nonnegative coefficients. No commutation of the A_k is used. The index d records the degree before reduction; resulting words may have shorter length, including zero.

Let \lambda now denote the left regular representation of F_S, and let \delta_{\mathrm{id}} be the unit vector at its identity. By (3.5) and (3.6),

This uses only the coefficient 2-norm identity (3.5), with no comparison of operator norms between the two representations. Lemma 3.1 and (3.3) give

Consequently \norm{C_d}_2\le(2\alpha)^d. The operator-norm convergence established above implies convergence of the same partial sums in 2-norm. Taking their tails and using the triangle inequality therefore yields

In particular, orthogonality between different degrees is not required. This proves (3.4).

Combining the first-order correlation with (3.2), (3.4), and Cauchy–Schwarz gives

Lifting for the fixed word length. Each t_j has a representative consisting of unitaries in P. To see this, start with a bounded representative (b_n) of any unitary in P^\omega. The polar part of b_n extends to a unitary w_n\in P, since its initial and final complementary projections have equal trace in the finite factor P. Then b_n=w_n|b_n| and functional calculus gives

Here the inequality follows from |\sqrt t-1|\le|t-1| for t\ge0.

Choose unitary representatives t_{j,n}\in P for the finitely many t_j. Using the constant representatives of u,z, form g_{k,n},W_n,y_n,h_n,u_n',W_n' by exactly the formulas above. For this fixed \ell these are bounded sequences, with h_n=h_n^* and \norm{h_n}\le\alpha. Bounded algebraic operations and continuous functional calculus in the quotient show that they represent the corresponding ultrapower elements. In particular,

Moreover \norm{u_n'-u}_2\le\norm{h_n}_2, by the scalar inequality |e^{it}-1|\le|t| and functional calculus. Choose one index n for which

Both conditions hold on sets belonging to \omega, so their intersection is nonempty. Set \widetilde u=u_n' and \widetilde t_j=t_{j,n}. Equation (3.7) proves (3.1). This choice is made separately for each \ell, as the proposition permits.□

Proof. The unit ball of M is compact and metrizable in the weak-star topology, so it has a countable weak-star dense subset S. For each \xi\in L^2(M,\tau), the functional a\mapsto\langle a,\xi\rangle on M is normal. Indeed, normal functionals form a norm-closed subspace of M^*, and approximation of \xi by elements of M in L^2 gives norm approximation of this functional by normal functionals, since

If \xi is orthogonal to S, normality implies that it is orthogonal to the unit ball, and hence to M. Thus S has dense linear span in L^2(M,\tau), proving separability.

Let (u_n) be a Cauchy sequence in \mathcal U(M) for d. The adjoints are also Cauchy, because \norm{u_n^*-u_m^*}_2=\norm{u_n-u_m}_2. For a\in M,

The left multiplication operators associated with u_n therefore converge strongly on the dense subspace M\subset L^2(M,\tau). Their uniform operator norm bound extends this convergence to the whole Hilbert space. The same holds for the left multipliers associated with u_n^*. Strong closedness of the left regular representation of M gives limits u,v\in M, and passage to inner products gives v=u^*. Multiplication is jointly strongly continuous on bounded sets, so uv=vu=1. Finally, convergence on the vector 1 gives \norm{u_n-u}_2\to0.□

Proof. Let \mathcal P_P denote the family of finite *-polynomials in a unitary variable with coefficients in P. For each fixed u, the values b(u), b\in\mathcal P_P, are dense in L^2(N_u,\tau). To see this, take the operator norm closure of the polynomial algebra. Its von Neumann closure is N_u, and Kaplansky density approximates each element of N_u strongly by bounded elements of this C^*-algebra. Evaluation on 1\in L^2(M,\tau) gives 2-norm approximation, which can then be followed by operator norm approximation by polynomials. The density of N_u in L^2(N_u,\tau) finishes the assertion.

For fixed b\in\mathcal P_P, the map u\mapsto b(u) is continuous in 2-norm. This follows by telescoping products, using the fixed operator norm bounds of the coefficients, the fact that unitaries have norm one, and the identity \norm{u^*-v^*}_2=\norm{u-v}_2. Consequently

is upper semicontinuous: an infimum of any family of continuous real functions is upper semicontinuous. Orthogonality gives

so f_\xi is lower semicontinuous and 0\leq f_\xi(u)\leq\norm{\xi}_2.

For completeness, a bounded lower semicontinuous real function f on a complete metric space is continuous on a dense G_\delta set. For \varepsilon>0, let G_\varepsilon be the union of all open sets on which the oscillation of f is less than \varepsilon. This union is dense. Indeed, in any nonempty open set O, write s=\sup_O f and choose x\in O with f(x)>s-\varepsilon/3. Lower semicontinuity gives an open neighborhood V\subset O of x on which f>f(x)-\varepsilon/3. The oscillation on V is less than \varepsilon. Baire’s theorem now shows that \bigcap_{n\geq1}G_{1/n} is dense, and f is continuous at each of its points.

Apply this construction to f_\eta for \eta in a countable dense subset of L^2(M,\tau) and intersect the resulting dense G_\delta sets. By Baire’s theorem, their intersection \mathcal C is again a dense G_\delta set. For arbitrary \xi,\eta\in L^2(M,\tau), contractivity of e_u gives

Thus, for u\in\mathcal C and v\in\mathcal U(M),

Approximating \xi by an \eta from the countable dense subset, and then letting v\to u, proves continuity of every f_\xi at u.□

Proof of Theorem 1.1. Take u\in\mathcal C from Lemma 4.2, and suppose that N_u\ne M. There is a unitary z\in M\setminus N_u, since the unitaries linearly span M. Let E_{N_u}:M\to N_u be the trace-preserving conditional expectation, and put

The conditional expectation is the restriction to M of e_u. Therefore

Fix

For every integer \ell\geq1, Proposition 3.2 provides a unitary u_\ell\in M and unitaries t_{0,\ell},\ldots,t_{\ell,\ell}\in P such that, with

one has

The fixed choice (4.2) gives

It follows from (4.4) that

for all sufficiently large \ell. On the other hand, \norm{W_\ell}_2=1 and W_\ell\in L^2(N_{u_\ell},\tau), so

By (4.3), u_\ell\to u in 2-norm. Continuity of f_\xi at u gives f_\xi(u_\ell)\to f_\xi(u)=0, a contradiction. Thus M=W^*(P,u) for every u\in\mathcal C.

To see that the full set of relative generators is G_\delta, choose (\eta_j)_{j\geq1} dense in L^2(M,\tau). With d_{\eta_j} as in Lemma 4.2, this set is

Each set in the intersection is open by upper semicontinuity. Vanishing of every distance is equivalent to L^2(N_v,\tau)=L^2(M,\tau), hence to N_v=M by the trace-preserving conditional expectation. The set is dense because it contains \mathcal C.□

Proof. By uniqueness of the hyperfinite \mathrm{II}_1 factor with separable predual [15, Section 1.3], we may use its standard tensor model

On the mth tensor site, let

with the identity on the other sites. On that site, 2d_mq_m(1-d_m) is the matrix unit e_{12}, so d_m,q_m generate the full matrix algebra. Each of the two families (d_m) and (q_m) consists of commuting projections.

For any countable family of commuting projections (e_m), the operator norm convergent series

recovers every e_m by spectral calculus. Inductively, after recovering e_1,\ldots,e_{m-1}, form

The restriction of r_m to 1-e_m lies between 0 and 3^{-m}, whereas its restriction to e_m lies between 2\,3^{-m} and 3\,3^{-m}. Since e_m commutes with r_m,

Applying this encoding separately to (d_m) and (q_m) gives two self-adjoint elements whose generated von Neumann algebra contains every tensor site, and hence is all of R.□

Proof. Let M be such a factor. By Theorem 5.1, applied with N=M, it contains a hyperfinite \mathrm{II}_1 subfactor P with P'\cap M=\mathbb C1. The trace-preserving conditional expectation onto P embeds P_* isometrically into M_* by precomposition, so P has separable predual. Lemma 5.2 gives self-adjoint a,b\in P with P=W^*(a,b), and Theorem 1.1 gives u\in\mathcal U(M) with M=W^*(P,u). Hence

□

Proof. Write N=W^*(a_1,a_2,a_3,a_4) with a_j=a_j^*, and choose C>\norm{a_1}+\norm{a_2}. Put

Let E_{ij} be the standard matrix units of M_2(N). The first diagonal block of A has spectrum in [-\norm{a_1},\norm{a_1}], and the second has spectrum in [C-\norm{a_2},C+\norm{a_2}]. For \norm{a_1}<\lambda<C-\norm{a_2}, spectral calculus therefore gives

The off-diagonal cuts of B give E_{12}=E_{11}BE_{22}\in Q and E_{21}=E_{12}^*\in Q. The first corner then contains

Thus NE_{11}\subset Q. Multiplying this corner by the matrix units yields NE_{ij}\subset Q for every i,j, and hence Q=M_2(N).□

Proof of Theorem 1.2. Let M be a \mathrm{II}_1 factor with separable predual and normalized trace \tau. Choose a projection p\in M with \tau(p)=1/2. Set N=pMp, with normalized trace \tau_N(a)=2\tau(a). This corner is a \mathrm{II}_1 factor with separable predual: precomposition with the normal compression a\mapsto pap embeds N_* isometrically into M_*. Since p and 1-p have the same trace, there is a partial isometry v\in M with v^*v=p and vv^*=1-p. The corresponding matrix units give a normal unital *-isomorphism

Proposition 5.3, applied to N, and Lemma 5.4 provide self-adjoint A,B\in M_2(N) with W^*(A,B)=M_2(N). Define x=\Phi(A+iB)\in M. Since the real and imaginary parts of x are \Phi(A) and \Phi(B), respectively,

□

Proof. For a bounded self-adjoint d-tuple Z, one has \delta(Z)\leq d and \delta_0(Z)\leq d. Both dimensions equal n for a freely independent variance-one semicircular n-tuple; see [21, Section 2.4(a), (c), and (d)] for \delta and [2, Section 5] for \delta_0. The standard semicircular n-tuple generates L(\mathbb F_n) as a von Neumann algebra. A self-adjoint pair supplied by Theorem 1.2 generates the same factor and has each dimension at most 2<n. These two tuples give the assertion.□