An isomorphism of the free group factors

Proof. The vectors U_g, indexed by g\in\F_d, are orthonormal because \tau(U_h^*U_g)=\tau(U_{h^{-1}g})=\mathbf 1_{g=h}. They span L^2(M,\tau). To see this, let \eta be orthogonal to their span. Choose \eta_m\in M converging to \eta in L^2. The normal functionals f_m(x)=\tau(\eta_m^*x) converge in norm to f(x)=\langle x,\eta\rangle, since \norm{f_m-f}_{M^*}\leq\normtwo{\eta_m-\eta}. The space M_* of normal linear functionals is norm closed in M^*, so f is normal. It vanishes on the ultraweakly dense word algebra and therefore on all of M. Thus \eta=0.

Consequently the map V:\ell^2(\F_d)\to L^2(M,\tau) with V\delta_g=U_g is unitary. It intertwines \lambda(x_j) with left multiplication by U_j. Conjugation by V identifies their double commutants. The double commutant on the right is the faithful normal left representation of W^*(U)=M, so this spatial identification gives the claimed isomorphism. It preserves the trace because V\delta_e=1.□

Proof. Consider a product of centered elements A_{g_j}c_jA_{g_j}^*, with c_j\in W^*(C) and g_j\ne g_{j+1}. Cyclicity of the trace rewrites its trace as

All interior group elements are nonidentity and hence centered. If the last group element is the identity, omit it; otherwise it too is centered. Freeness of W^*(A_1,\ldots,A_n) and W^*(C) makes the displayed trace zero. For a single factor this is simply \tau(c_1)=0. The assertion about the other components follows by grouping free subalgebras.□

Proof. For the first assertion, combine consecutive group symbols and consecutive powers of S. Remove identity group factors. Split each polynomial in S into its scalar trace and a centered part. A term with all alternating factors centered has trace zero by freeness of the group algebra and W^*(S). Removing a scalar factor permits a reduction to fewer alternating factors, without changing the total group label. Induction on the number of factors ends with either a centered alternating product or a single group element whose label remains nonidentity. Both have trace zero.

For the second assertion, successively use A_g S=S_g A_g to move every group factor to the right. The final group factor is exactly the label, and is therefore 1.□

Proof. We prove the more general bound

for a finite free family of subalgebras and centered self-adjoint z_j in the respective components.

Work in the tracial L^2 space of their generated algebra. Freeness identifies this space with the orthogonal sum of the scalar vector and the reduced centered tensor spaces, whose successive component indices are different. To check the identification, take inner products of reduced centered words. If the two letters at the joining point belong to different components, freeness gives zero. If they belong to the same component, split their product into its trace and its centered part; the latter again has zero contribution, while the scalar term shortens both words. Iteration gives orthogonality for different index sequences and the tensor-product inner product for matching sequences. The reduced word span is dense by the argument used in Lemma 2.2.

Let P_j project onto the reduced tensors beginning with component j, and let L_{z_j} denote left multiplication. On (1-P_j)L^2, multiplication prefixes z_j, so

The ranges of the B_j are orthogonal. For every vector \xi this gives

The blocks in the opposite direction are B_j^*, because z_j is self-adjoint. The block (1-P_j)L_{z_j}(1-P_j) is zero. The remaining diagonal blocks P_jL_{z_j}P_j act on mutually orthogonal subspaces; their sum has norm at most \max_j\norm{z_j}. Summing these three parts proves (2.3), since the left representation is faithful.

Apply this bound to z_g=k(g)S_g. Both \normtwo{S_g} and \norm{S_g} are at most \pi, and \max_g\abs{k(g)}\leq\norm{k}_{\ell^2}. This proves (2.2).□

Proof. First take r=1. Give a monomial its group label by ignoring each occurrence of z and multiplying the remaining group symbols in order. Each term of \sigma(k) has label e. The replacement rules (3.5) consequently preserve a monomial’s label term by term. If that label is nonidentity, Lemma 2.4 makes every evaluated derivative term have trace zero.

An identity-label monomial is a polynomial in the z_g. To see this algebraically, repeatedly use a_gz=z_ga_g to move group factors to the right; the final group factor is the identity. Because these are identities in \mathcal P and \delta is a derivation on that algebra, differentiating this rewriting gives the same result as differentiating the original monomial. It is therefore enough to consider p=F(z_g:g\in I), with I\subset\Gamma finite and F a noncommutative polynomial.

For a fixed g\in I, remove the commuting own-variable term from the commutator in (3.6). Its evaluated direction is

The element Y_g is a bounded self-adjoint in B_g=W^*(S_b:b\ne g), and W^*(S_g) is free from B_g by Lemma 2.3.

We use the following direct consequence of freeness. If a unital subalgebra \mathcal A is free from a unital subalgebra \mathcal B and U\in\mathcal B is unitary, then U\mathcal A U^* is free from \mathcal B. Indeed, conjugating an alternating product of centered elements from these two algebras by U^* gives an alternating product of centered elements from \mathcal A and \mathcal B: a fixed centered b\in\mathcal B becomes U^*bU, which is still centered in \mathcal B. Its trace is zero. Conjugation preserves the marginal law of each element of \mathcal A. Since freeness determines mixed moments from the marginal laws, it follows that replacing S_g by US_gU^* preserves its joint law with any fixed collection of elements of B_g.

Apply this observation with U=\exp(\ii uY_g) for real u. The scalar function obtained by replacing every occurrence of S_g in F by \exp(\ii uY_g)S_g\exp(-\ii uY_g) and leaving all other variables fixed has constant trace. Its derivative at zero is the trace of the sum over those occurrences with the direction \ii[Y_g,S_g] inserted. Sum these zero derivatives over g\in I. The product rule and (3.6) identify the sum exactly with \tau(\operatorname{ev}_0(\delta p)), proving the case r=1. Linearity handles arbitrary finite sums of monomials.

The result just proved is the identity \tau\circ\operatorname{ev}_0\circ\delta=0 on all of \mathcal P. Since \delta maps finite polynomials to finite polynomials, applying it to \delta^{r-1}p proves (3.7). In this application every derivative of every coefficient polynomial is included. The auxiliary conjugations above are used only to compute a first variation at the initial tuple; they need neither lift to motions of the a_j nor be jointly integrated.□

Proof. Regard E^n, with its maximum norm, as a real Banach space. Adjunction is bounded and real-linear, so \mathcal V is a finite smooth polynomial map. Its norm and its Lipschitz constant are bounded on every fixed bounded ball. Picard iteration on a sufficiently short interval gives a unique local solution; repeated differentiation makes it smooth.

Each H_j(X) is self-adjoint, even before unitarity is known. Along a local solution,

Thus X_j^*X_j=1. Such an isometry is unitary in M: the projection 1-X_jX_j^* has trace 1-\tau(X_j^*X_j)=0, and faithfulness makes it zero. This equality also holds in E, so \norm{X_j}=1 throughout the local solution.

On this unitary trajectory,

If a maximal interval had a finite endpoint, this bounded derivative would make the tuple norm-Cauchy as time approached that endpoint. Its limit lies in the closed algebra E and is again a unitary tuple. Local existence at the limit continues the solution, a contradiction. The same argument applies at either endpoint, giving a global solution. Uniqueness follows from local uniqueness on overlapping intervals. This continuation uses completeness and bounded derivative, not compactness of a bounded ball.□

Proof. The differential equation and the product rule give, for every integer r\geq0,

We verify that these derivatives give an actual convergent Taylor expansion for the trace.

Choose fixed polynomial representatives for the derivatives of the formal letters in (3.5). There are constants J\geq1 and B_1\geq1 such that replacing any one letter by its derivative increases monomial length by at most J and produces total absolute coefficient sum at most B_1. These constants are finite because the h_j have finite support. If a representative of p has degree at most l, where l\geq1, and total absolute coefficient sum b_0, repeated replacement gives a representative of \delta^r p with degree at most l+rJ and coefficient sum at most

Let R=\max(1,\norm{S}). All the evaluated group letters are unitary, so (3.9) implies, uniformly for real t,

The last inequality uses l+vJ\leq(l+J)(v+1).

Set f_p(t)=\tau(\operatorname{ev}_t(p)) and c_p=B_1R^J(l+J). The integral form of Taylor’s remainder and (3.10) bound the remainder after degree r at increment u by b_0R^l(c_p|u|)^{r+1}. Consequently f_p is real-analytic near every real time, with positive radius at least c_p^{-1}. By Lemma 3.2, all its positive-order derivatives at zero vanish. It is constant near zero and hence, by real-analytic continuation on the connected line \R, constant everywhere.

The constants here may depend on the fixed field and on p. No radius common to all polynomials or to different choices of the h_j is required. The bounds used only unitarity of the solution and boundedness of S; freeness at positive times has not been assumed.□

Proof of Proposition 3.1. For a fixed real t, polynomial evaluation defines a map

This is well defined. If \operatorname{ev}_0(p)=0, then (3.8) applied to p^*p gives

so faithfulness implies \operatorname{ev}_t(p)=0. Thus any additional relations at the evaluated initial tuple are respected as a consequence of actual moment invariance. The map is unital, complex-linear, multiplicative, and adjoint-preserving because formal evaluation has those properties.

For an element b of a finite von Neumann algebra with faithful tracial state,

For completeness, the upper bound is immediate; for the lower bound, every nonzero spectral projection of b^*b sufficiently near its norm has positive trace by faithfulness. Applying (3.8) to (p^*p)^k therefore shows that \varphi_t is isometric. It extends uniquely to a trace-preserving isometric *-homomorphism on B, with range

Indeed, its range is closed and is the norm closure of the polynomials in these time-t generators.

We next prove B_t=B, which establishes surjectivity. Apply Lemma 3.3 inside B_t to the initial tuple A(t), using the same fixed element S and the same polynomial field. Its global solution, viewed in M, must be u\mapsto A(t+u) by uniqueness. At u=-t it recovers the original tuple. Thus every A_j belongs to B_t, and S already belongs to B_t, giving B\subseteq B_t. The forward construction gives the reverse inclusion. Hence \varphi_t is a C^*-automorphism of B fixing S.

Since \exp(\ii S)=C, the von Neumann algebra generated by B is M. Its vectors are therefore dense in L^2(M,\tau), as in the density argument in Lemma 2.2. Trace preservation and surjectivity define a Hilbert-space unitary

on L^2(M,\tau), fixing [1]. If L_b denotes left multiplication, then on the dense set of vectors coming from B,

The faithful normal left representation identifies the von Neumann closure of L(B) with M. Conjugation by U_t therefore gives a normal *-automorphism \beta_t of M extending \varphi_t. It is trace-preserving because U_t fixes [1], and it fixes S. This normal extension is unique by ultraweak density of B.

The maps form a one-parameter group. For fixed s, the path t\mapsto\varphi_s(A(t)) satisfies the same polynomial equation, because \varphi_s is a bounded *-homomorphism fixing S. Its initial tuple is A(s), so uniqueness gives

The equality on the generators and on S yields \varphi_s\varphi_t=\varphi_{s+t} on B, and uniqueness of normal extension gives \beta_s\beta_t=\beta_{s+t} on M. Also \beta_0 is the identity. Norm continuity on polynomials in the generators follows from the solution paths; density and the common isometry bound extend it to point-norm continuity on B. No point-norm continuity on arbitrary elements of M is needed.

It remains to prove the bounds. For each finitely supported real k,

by (2.2). Integrating \norm{A_j'(t)}\leq K\|h_j\|_{\ell^2} between zero and t proves (3.3).

For a fixed w\in\Gamma, evaluation of (3.6) along the flow gives

Moreover s_t(\e)=S. Direct differentiation, with the order of factors retained, yields

The exterior factors are unitary, so its norm is at most K\|D_w-\e\|_{\ell^2}. Integrating between zero and t and multiplying on the left by \exp(\ii tS) proves (3.4).□

Proof. First consider a nonempty reduced word in b_1,\ldots,b_m. Combining consecutive powers with the same index writes it as

Its expansion in x_2,x_3 is

Every displayed x_3 power is nonzero, and each intervening x_2 power is nonzero. This word cannot reduce to the identity. Thus the b_j are a free basis of the subgroup they generate. The free product decomposition

then shows that a,b_1,\ldots,b_m freely generate their subgroup: an alternating product of nonidentity elements from \langle a\rangle and \langle b_1,\ldots,b_m\rangle is also reduced in the displayed free product.

In the abstract free group on a,b_1,\ldots,b_m, the substitution fixing a and sending each b_j to a b_j has inverse given by b_j\mapsto a^{-1}b_j, again fixing a. Consequently

form a free family. On the abstract free group with free basis u_1,\ldots,u_m, define two endomorphisms by

For every j, cancellation gives

and the second identity also holds for j=1. Thus \varphi and \psi are inverse automorphisms. Since p_j=u_1\cdots u_j, this proves the assertion about the p_j.

For the cocycle assertion, the inverse rule D_{x_j^{-1}}=-\lambda(x_j^{-1})D_{x_j} shows that D vanishes on the subgroup generated by x_2,\ldots,x_n. In particular, D_{b_j}=0. The positive occurrences of a in w_m have respective prefixes p_0,p_1,\ldots,p_m, and there are no other occurrences of a or a^{-1}. Applying the cocycle identity successively therefore gives

□

Proof. Write V_j=\lambda(p_j) and V_0=1. Lemma 4.2 implies that a word in V_1,\ldots,V_m and their inverses has trace one if its word in these free symbols reduces to the identity, and trace zero otherwise. This trace rule applies inside the given regular representation. Indeed, for H=\langle p_1,\ldots,p_m\rangle, the spaces \ell^2(Ht), as Ht ranges over the right cosets of H, are reducing copies of the left regular representation of H. The vector \e lies in the copy \ell^2(H). Its vector state therefore agrees with the canonical trace of H on polynomials and on the spectral projections under consideration.

Fix an integer r\geq 1. Expanding Q_m^r gives

Each summand is zero or one. In a summand equal to one, every nonzero index used must occur at least twice, since a generator occurring only once cannot disappear under free reduction. There are therefore at most r distinct nonzero indices.

If exactly r distinct nonzero indices occur, all 2r positions are nonidentity and each index occurs exactly twice. The two occurrences have opposite exponents, as follows already by abelianizing the identity word. Pair the two positions with the same index. This matching is unique for the chosen word. A sequence of adjacent inverse cancellations produces a noncrossing matching: a pair deleted after other pairs can only enclose pairs already deleted, and two deleted pairs cannot cross. Conversely, a noncrossing matching with opposite exponents at its endpoints reduces to the identity by deleting an innermost pair and proceeding by induction. Every noncrossing perfect matching of \{1,\ldots,2r\} has opposite parity at the two endpoints of each pair. To see this, the positions strictly inside that pair must be matched among themselves, so their number is even. The signs in (4.7) alternate; therefore every noncrossing perfect matching has the required opposite signs.

The number of noncrossing perfect matchings is

For completeness, splitting a matching at the partner of its first position gives the recurrence c_0=1 and c_r=\sum_{j=0}^{r-1}c_jc_{r-1-j}. Its formal generating series satisfies C(z)=1+zC(z)^2, so the solution with constant term one is (1-\sqrt{1-4z})/(2z); the binomial expansion gives the displayed coefficient \Cat_r. For each matching, labeling its r pairs with distinct indices from \{1,\ldots,m\} gives

choices, with the value zero when m<r. Ordering the pairs by their first endpoints makes this count unambiguous, and the uniqueness of the equal-index matching prevents double counting. Thus the contribution from exactly r distinct indices is \Cat_r(m)_r.

Every remaining contributing term has at most r-1 distinct nonzero indices. This includes every term with an identity summand: such a term has at most 2r-1 nonidentity positions, while each index it uses occurs at least twice. For fixed r, there are only finitely many patterns specifying the identity positions and the partition of the other positions into equal-index classes. A pattern with k\leq r-1 classes has at most m^k labelings. Summing over those patterns gives an O_r(m^{r-1}) bound, with its constant depending only on r. We have proved

We next justify passage from moments to weak convergence. For r=1, direct expansion gives \tau_\Gamma(Q_m)=m+1, so the first moments of \mu_m are bounded by two. Markov’s inequality gives tightness on [0,\infty). For each fixed integer r\geq 1, the moment convergence in (4.8), applied also at r+1, gives

as R\to\infty. Consequently, if a subsequence converges weakly to a probability measure \mu, its moments pass to the limit. More explicitly, apply weak convergence to the bounded continuous functions \min\{x^r,R^r\} and use the displayed tail bound; the corresponding bound for \mu follows by first applying weak convergence to bounded truncations of x^{r+1}. Letting R\to\infty proves

The density in (4.6) realizes these moments, including the zeroth moment. Write B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\,dt for the Euler beta integral. The substitution x=4t and the beta integral give, for every integer r\geq 0,

The middle equality follows from B(\tfrac12,\tfrac32)=\pi/2 and the recurrence B(a+1,b)=aB(a,b)/(a+b). In particular the density has total mass one.

These moments determine a probability measure on [0,\infty) uniquely. Indeed, \Cat_r\leq 4^r. If a measure satisfying (4.9) had mass c>0 on [4+\epsilon,\infty) for some \epsilon>0, then

for every r, a contradiction for large r. Its support is therefore contained in [0,4]. On this compact interval, uniform polynomial approximation of continuous functions shows that equality of all moments implies equality of the measures. Every subsequential weak limit is consequently \nu. Tightness now gives weak convergence of the full sequence \mu_m to \nu. Finally, \nu has the displayed integrable density and thus has no atom at zero.□

Proof of Lemma 4.1. Use the operators T_m,Q_m and the measures \mu_m above. Since \nu(\{0\})=0, choose a fixed d>0 so small that

The interval [0,d] is closed, so weak convergence and the Portmanteau inequality give

We may therefore choose m sufficiently large that both

hold. Keep this d and this m fixed for the remainder of the construction.

Set

Thus R_m is a bounded positive operator, with \norm{R_m}\leq (dm)^{-1}. The spectral theorem gives

In particular, these operations allow a kernel or arbitrarily small nonzero spectrum of Q_m. Define the coefficient vector

Using Q_m=T_mT_m^* in the indicated order gives

The vector h^{(0)} has real coefficients. To check this directly, let J denote coefficientwise complex conjugation on \ell^2(\Gamma). Each left shift is real in the standard basis, so J commutes with T_m,T_m^* and Q_m, and J\e=\e. Commutation with real polynomials in Q_m extends by uniform approximation to real continuous functions of Q_m. The continuous functions

converge pointwise, with a uniform bound, to 1_{(dm,\infty)}(x) on the spectrum of Q_m. Their functional calculi converge strongly to E_m, hence J also commutes with E_m. Finally,

where g_m is real and continuous on [0,\infty). Thus J commutes with R_m as well, proving Jh^{(0)}=h^{(0)}.

Real finitely supported coefficient vectors are dense in the real subspace of \ell^2(\Gamma). Choose such a vector h with

The elementary bound \norm{T_m}\leq m+1 and (4.13)–(4.14) yield

For this finitely supported h, form the cocycle with h_1=h and all other generator values zero, and take w=w_m. Equation (4.4) gives D_w=T_mh, proving (4.1).

The choices have been made in the order d, then m, then the finite coefficient approximation. The last approximation uses the norm of the already fixed operator T_m; the lemma requires no bound on the support size of h uniform in \eta.□

Proof. Choose 0<\eta<\varepsilon/K, with K=3\pi as in (2.2). Lemma 4.1 gives a finitely supported real coefficient vector h_1, with h_j=0 for j\geq2, and a word w\in\F_n for which

Apply Proposition 3.1 to these coefficients and write \beta=\beta_1. Its displacement estimates (3.3) and (3.4) give

Moreover, \beta fixes S and therefore fixes C=\exp(\ii S).

The free-group basis transformation that fixes the first n symbols and sends the last symbol c to c w is invertible: its inverse fixes the first n symbols and sends c to c w^{-1}. By Lemma 2.2, it induces a trace-preserving normal automorphism \gamma of M satisfying

Set \alpha=\gamma^{-1}\circ\beta. Since \gamma fixes every A_j and hence A_w, we have the exact identities

An automorphism is isometric, so these identities prove (5.1), with the displayed order of the factors. The image under \alpha of the original freely generating Haar tuple is again such a tuple. Finally,

because \beta(C)=C and \gamma(C A_w^*)=C A_w A_w^*=C.□

Proof of Theorem 1.1. Fix n\geq3 and let M=L(\F_{n+1}), with its canonical normalized trace and canonical freely generating Haar tuple

Write A^{(k)}=(A_1^{(k)},\ldots,A_n^{(k)}). Choose a sequence (y_j)_{j\geq1} that is dense in L^2(M,\tau) and consists of elements of the rational complex group word algebra of the initial tuple. Such a sequence exists by Lemma 2.2; coefficients in \mathbb{Q}+\ii\mathbb{Q} make this algebra countable. Put r_0=1.

We construct freely generating Haar tuples (A^{(k)},C^{(k)}) and positive numbers r_k by induction. Suppose the construction has reached stage k-1, where k\geq1. The choices at stage k are made in the following order.

First, for each 1\leq j\leq k, choose a finite linear combination F_{j,k} of group words in n+1 variables such that

This is possible because the current completed tuple generates M and its group word algebra is dense in L^2(M,\tau). Let H_k\geq0 be a common bound for the finitely many constants L(F_{j,k}) in (5.2). Choose

Next apply Proposition 5.1 to (A^{(k-1)},C^{(k-1)}) with tolerance \varepsilon_k. Let \alpha_k and w_k\in\F_n be its automorphism and word, and set

The new completed tuple is freely generating Haar, and

In particular, the last expression is close to the old completion, as needed in (5.3).

Now define word polynomials in n variables by

Since \normtwo{x}\leq\norm{x}, equations (5.2), (5.3), and (5.5) imply

This estimate uses A_{w_k}^{(k)} as a single unitary slot of F_{j,k}. Thus the tolerance chosen in (5.4) does not depend on the length of the word that the perturbation subsequently produces.

Finally, let L_k\geq0 be a common L^2-Lipschitz bound for the finitely many polynomials G_{j,k}, and choose

These constants are finite by (5.2). They are chosen after w_k is known, so any increase in word length is accommodated in the budget for future steps. Both quantities in the minimum are positive, and the induction can continue.

Figure 1 records this order of choices and the role of the budget in controlling all later perturbations.

For \ell>k, the geometric decrease of the budgets gives

Using (5.5), we therefore obtain, for p>k,

Since r_k\leq2^{-k}r_0, each sequence (A_j^{(k)})_k converges in operator norm to a unitary A_j^{(\infty)}, and

The limits are unitaries because the unitary group is norm closed. For every nonidentity reduced word v\in\F_n,

by operator-norm continuity of each fixed word evaluation and of the trace. Thus the limiting tuple has the free Haar word distribution.

It remains to prove that this tuple generates M. For every j\leq k, equations (), (5.7), and (5.8) yield

Fixing j and letting k\to\infty through k\geq j shows that y_j belongs to the L^2 closure of the word algebra of A^{(\infty)}. Since the y_j are dense, this word algebra is dense in all of L^2(M,\tau).

To pass from this density to von Neumann generation, set N=W^*(A_1^{(\infty)},\ldots,A_n^{(\infty)}). The standard finite-trace conditional expectation E_N:M\to N is trace-preserving and extends to the orthogonal projection of L^2(M,\tau) onto L^2(N,\tau). In particular it is L^2-contractive and fixes every element of N. Each G_{j,k}(A^{(\infty)}) belongs to N, so

Hence the projection fixes the dense family (y_j) and is the identity on L^2(M,\tau). For x\in M, this implies E_N(x)=x by faithfulness of \tau, and therefore N=M.

The limiting tuple is now a freely generating Haar n-tuple in M. Lemma 2.2 supplies a unital trace-preserving normal *-isomorphism from L(\F_n) onto M=L(\F_{n+1}), proving the theorem.□

Proof of Theorem 1.2. Apply Theorem 1.1 with n=3 and n=4 to obtain

Let \Theta:N_3\to N_5 be the composition of these normal trace-preserving isomorphisms. Amplifying by \sqrt2 gives a unital normal trace-preserving isomorphism

Formula (6.1) identifies its domain and range as follows:

These identifications and the amplified map compose to give \Phi:L(\F_2)\to L(\F_3). Every map in the composition is a unital normal trace-preserving *-isomorphism. Thus \Phi is bijective and has all the asserted properties.□

Proof. Theorem 1.2 rules out the pairwise-nonisomorphic alternative, so all parameters, including \infty, give the same isomorphism class. For finite r>1 and every t>0, formula (6.1) then gives

Thus \mathcal F(N_r)=\R_{>0}. Finally, N_\infty\cong N_2 and functoriality of amplification give N_\infty^t\cong N_2^t\cong N_2\cong N_\infty for every t>0.□

Proof. Fix an integer n\geq2. Theorem 1.2 and successive applications of Theorem 1.1 provide a unital normal trace-preserving *-isomorphism

for n=2 we may take the identity. If Z is any finite self-adjoint tuple in L(\F_n) and p is a noncommutative *-polynomial, then

Thus coordinatewise transport by \theta_n preserves the joint tracial *-distribution and hence each of the four dimensions. Normality and surjectivity also give W^*(\theta_n(Z))=\theta_n(W^*(Z)).

Let S^{(n)}=(s_1,\ldots,s_n) be a freely independent variance-one semicircular tuple in the semicircular realization of L(\F_n) [15, Remarks 1.10–1.11]. It generates L(\F_n) as a von Neumann algebra, and the standard semicircular computations give

see [17, Section 2.4(c) and (d)] for \delta and [1, Section 5, opening paragraph] for \delta_0. Set X^{(n)}=\theta_n(S^{(n)}). The preceding transport observations prove the required generation and the two microstates values.

For the nonmicrostates dimensions, let u_1,\ldots,u_n be the canonical group unitaries in L(\F_n) and put

The entries are self-adjoint, and u_j=a_j+\ii b_j and u_j^*=a_j-\ii b_j. Hence they generate the algebraic group algebra \C\F_n as a unital *-algebra, and generate L(\F_n) as a von Neumann algebra. Apply Mineyev and Shlyakhtenko’s group-algebra theorem [8, Theorem 4.1] to this tuple, before transporting it. It gives

Here \beta_0^{(2)}(\F_n)=0 and \beta_1^{(2)}(\F_n)=n-1 are the usual L^2-Betti numbers of the free group; see [2, Section 2.1.1]. Define Y^{(n)}=\theta_n(G^{(n)}). Again its joint tracial *-distribution is unchanged and it generates M, proving the remaining assertions. Taking n=2 and n=3 gives different values on generators of the same factor, and letting n grow gives unboundedness.□