Proof. The vectors \(U_g\), indexed by \(g\in\mathbb F_d\), 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 \(\left\lVert f_m-f\right\rVert_{M^*}\leq\left\lVert \eta_m-\eta\right\rVert_2\). 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 \(V:\ell^2(\mathbb F_d)\to L^2(M,\tau)\) 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 \[\tau\bigl(c_1 A_{g_1^{-1}g_2}c_2\cdots A_{g_{r-1}^{-1}g_r}c_r A_{g_r^{-1}g_1}\bigr).\] 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 \[ \left\lVert \sum_j z_j\right\rVert \leq 2\Bigl(\sum_j\left\lVert z_j\right\rVert_2^2\Bigr)^{1/2} +\max_j\left\lVert z_j\right\rVert \tag{3}\] 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 4.

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 \[B_j=P_jL_{z_j}(1-P_j),\qquad \left\lVert B_j\right\rVert\leq\left\lVert z_j\right\rVert_2.\] The ranges of the \(B_j\) are orthogonal. For every vector \(\xi\) this gives \[\left\lVert \sum_jB_j\xi\right\rVert^2 =\sum_j\left\lVert B_j\xi\right\rVert^2 \leq\Bigl(\sum_j\left\lVert z_j\right\rVert_2^2\Bigr)\left\lVert \xi\right\rVert^2.\] 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 \(\max_j\left\lVert z_j\right\rVert\). Summing these three parts proves (3), since the left representation is faithful.

Apply this bound to \(z_g=k(g)S_g\). Both \(\left\lVert S_g\right\rVert_2\) and \(\left\lVert S_g\right\rVert\) are at most \(\pi\), and \(\max_g\left\lvert k(g)\right\rvert\leq\left\lVert k\right\rVert_{\ell^2}\). This proves (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 (8) consequently preserve a monomial’s label term by term. If that label is nonidentity, Lemma 6 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 (9). Its evaluated direction is \[\operatorname{ev}_0(\delta z_g)=\mathrm i[Y_g,S_g],\qquad Y_g=\sum_{b\ne g}D_g(b)S_b.\] 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 5.

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 \(U=\exp(\mathrm iuY_g)\) for real \(u\). The scalar function obtained by replacing every occurrence of \(S_g\) in \(F\) by \(\exp(\mathrm iuY_g)S_g\exp(-\mathrm iuY_g)\) 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 \(\mathrm i[Y_g,S_g]\) inserted. Sum these zero derivatives over \(g\in I\). The product rule and (9) 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 (10). 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, \[\frac{d}{dt}(X_j^*X_j) =-\mathrm iX_j^*H_j(X)X_j+\mathrm iX_j^*H_j(X)X_j=0.\] 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 \(\left\lVert X_j\right\rVert=1\) throughout the local solution.

On this unitary trajectory, \[\left\lVert X_j'\right\rVert\leq \left\lVert S\right\rVert\sum_g|h_j(g)|.\] 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 \(r\geq0\), \[ \frac{d^r}{dt^r}\operatorname{ev}_t(p) =\operatorname{ev}_t(\delta^r p). \tag{12}\] 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 (8). 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 \[b_0 B_1^r\prod_{v=0}^{r-1}(l+vJ).\] Let \(R=\max(1,\left\lVert S\right\rVert)\). All the evaluated group letters are unitary, so (12) implies, uniformly for real \(t\), \[\begin{align*} \frac{1}{r!} \left\|\frac{d^r}{dt^r}\operatorname{ev}_t(p)\right\| &\leq b_0R^l(B_1R^J)^r \frac{\prod_{v=0}^{r-1}(l+vJ)}{r!} \\ &\leq b_0R^l\bigl[B_1R^J(l+J)\bigr]^r. \tag{13}\end{align*}\] 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 (13) 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 9, all its positive-order derivatives at zero vanish. It is constant near zero and hence, by real-analytic continuation on the connected line \(\mathbb R\), 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 8. For a fixed real \(t\), polynomial evaluation defines a map \[\varphi_t:\operatorname{ev}_0(\mathcal P) \longrightarrow B, \qquad \varphi_t\bigl(\operatorname{ev}_0(p)\bigr) =\operatorname{ev}_t(p).\] This is well defined. If \(\operatorname{ev}_0(p)=0\), then (11) applied to \(p^*p\) gives \[\tau\bigl(\operatorname{ev}_t(p)^* \operatorname{ev}_t(p)\bigr)=0,\] 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, \[\left\lVert b\right\rVert =\lim_{k\to\infty}\tau((b^*b)^k)^{1/(2k)}.\] 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 (11) to \((p^*p)^k\) therefore shows that \(\varphi_t\) is isometric. It extends uniquely to a trace-preserving isometric \(*\)-homomorphism on \(B\), with range \[B_t=C^*(A_1(t),\ldots,A_n(t),S).\] 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 10 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 \(\exp(\mathrm iS)=C\), 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 4. Trace preservation and surjectivity define a Hilbert-space unitary \[U_t[b]=[\varphi_t(b)]\qquad(b\in B)\] on \(L^2(M,\tau)\), fixing \([1]\). If \(L_b\) denotes left multiplication, then on the dense set of vectors coming from \(B\), \[U_t L_b U_t^*=L_{\varphi_t(b)}\qquad(b\in B).\] 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 \[\varphi_s(A_j(t))=A_j(s+t).\] 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 \(k\), \[s_t(k)=\beta_t(s(k)),\qquad \left\lVert s_t(k)\right\rVert=\left\lVert s(k)\right\rVert\leq K\|k\|_{\ell^2}\] by (2). Integrating \(\left\lVert A_j'(t)\right\rVert\leq K\|h_j\|_{\ell^2}\) between zero and \(t\) proves (6).

For a fixed \(w\in\Gamma\), evaluation of (9) along the flow gives \[\frac{d}{dt}A_w(t)=\mathrm is_t(D_w)A_w(t).\] Moreover \(s_t(\delta_e)=S\). Direct differentiation, with the order of factors retained, yields \[\frac{d}{dt}\bigl(\exp(-\mathrm itS)A_w(t)\bigr) =\mathrm i\exp(-\mathrm itS)\bigl(s_t(D_w)-S\bigr)A_w(t).\] The exterior factors are unitary, so its norm is at most \(K\|D_w-\delta_e\|_{\ell^2}\). Integrating between zero and \(t\) and multiplying on the left by \(\exp(\mathrm itS)\) proves (7). ◻

Proof. First consider a nonempty reduced word in \(b_1,\ldots,b_m\). Combining consecutive powers with the same index writes it as \[b_{j_1}^{q_1}\cdots b_{j_l}^{q_l}, \qquad q_i\in\mathbb Z\setminus\{0\}, \qquad j_i\ne j_{i+1}.\] Its expansion in \(x_2,x_3\) is \[x_2^{j_1}x_3^{q_1} x_2^{j_2-j_1}x_3^{q_2}\cdots x_2^{j_l-j_{l-1}}x_3^{q_l}x_2^{-j_l}.\] 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 \[\Gamma=\langle a\rangle *\langle x_2,\ldots,x_n\rangle\] 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 \[u_j=a b_j,\qquad 1\leq j\leq m,\] form a free family. On the abstract free group with free basis \(u_1,\ldots,u_m\), define two endomorphisms by \[\varphi(u_j)=u_1\cdots u_j, \qquad \psi(u_1)=u_1, \qquad \psi(u_j)=u_{j-1}^{-1}u_j\quad (j\geq 2).\] For every \(j\), cancellation gives \[\begin{align*} \psi\varphi(u_j) &=u_1(u_1^{-1}u_2)\cdots(u_{j-1}^{-1}u_j)=u_j,\\ \varphi\psi(u_j) &=(u_1\cdots u_{j-1})^{-1}(u_1\cdots u_j)=u_j \quad (j\geq 2), \end{align*}\] 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 \[D_{w_m}=\sum_{j=0}^m\lambda(p_j)h=T_mh.\] ◻

Proof. Write \(V_j=\lambda(p_j)\) and \(V_0=1\). Lemma 13 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 \(\delta_e\) 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 \[ \tau_\Gamma(Q_m^r) =\sum_{i_1,\ldots,i_{2r}\in\{0,\ldots,m\}} \tau_\Gamma\bigl( V_{i_1}V_{i_2}^*\cdots V_{i_{2r-1}}V_{i_{2r}}^* \bigr). \tag{20}\] 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 (20) alternate; therefore every noncrossing perfect matching has the required opposite signs.

The number of noncrossing perfect matchings is \[\mathop{\mathrm{Cat}}_r=\frac{1}{r+1}\binom{2r}{r}.\] 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 \(\mathop{\mathrm{Cat}}_r\). For each matching, labeling its \(r\) pairs with distinct indices from \(\{1,\ldots,m\}\) gives \[(m)_r=m(m-1)\cdots(m-r+1)\] 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 \(\mathop{\mathrm{Cat}}_r(m)_r\).

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 \[ \tau_\Gamma(Q_m^r) =\mathop{\mathrm{Cat}}_r(m)_r+O_r(m^{r-1}), \qquad \int x^r\,d\mu_m(x)\longrightarrow\mathop{\mathrm{Cat}}_r. \tag{21}\]

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 (21), applied also at \(r+1\), gives \[\sup_m\int x^{r+1}\,d\mu_m(x)<\infty, \qquad \sup_m\int_{(R,\infty)}x^r\,d\mu_m(x) \leq \frac{1}{R}\sup_m\int x^{r+1}\,d\mu_m(x) \longrightarrow 0\] 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 \[ \int x^r\,d\mu(x)=\mathop{\mathrm{Cat}}_r\qquad(r\geq 0). \tag{22}\]

The density in (19) 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 \(r\geq 0\), \[\begin{align*} \frac{1}{2\pi}\int_0^4 x^{r-1/2}(4-x)^{1/2}\,dx &=\frac{4^{r+1}}{2\pi} B\left(r+\frac12,\frac32\right)\\ &=\frac{(2r)!}{r!(r+1)!} =\mathop{\mathrm{Cat}}_r. \end{align*}\] 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, \(\mathop{\mathrm{Cat}}_r\leq 4^r\). If a measure satisfying (22) had mass \(c>0\) on \([4+\epsilon,\infty)\) for some \(\epsilon>0\), then \[4^r\geq\mathop{\mathrm{Cat}}_r\geq c(4+\epsilon)^r\] 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 12. 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 \[\nu([0,d])<\frac{\eta^2}{16}.\] The interval \([0,d]\) is closed, so weak convergence and the Portmanteau inequality give \[\limsup_{m\to\infty}\mu_m([0,d]) \leq\nu([0,d])<\frac{\eta^2}{16}.\] We may therefore choose \(m\) sufficiently large that both \[ \mu_m([0,d])<\frac{\eta^2}{8}, \qquad \frac{1}{dm}<\frac{\eta^2}{16} \tag{23}\] hold. Keep this \(d\) and this \(m\) fixed for the remainder of the construction.

Set \[E_m=1_{(dm,\infty)}(Q_m), \qquad R_m=f_m(Q_m), \qquad f_m(x)= \begin{cases} 0,&0\leq x\leq dm,\\ x^{-1},&x>dm. \end{cases}\] Thus \(R_m\) is a bounded positive operator, with \(\left\lVert R_m\right\rVert\leq (dm)^{-1}\). The spectral theorem gives \[ Q_m R_m=E_m, \qquad R_m Q_m R_m=R_m, \qquad \tau_\Gamma(1-E_m)=\mu_m([0,d]). \tag{24}\] In particular, these operations allow a kernel or arbitrarily small nonzero spectrum of \(Q_m\). Define the coefficient vector \[h^{(0)}=T_m^*R_m\delta_e\in\ell^2(\Gamma).\] Using \(Q_m=T_mT_m^*\) in the indicated order gives \[\begin{align*} T_mh^{(0)}&=E_m\delta_e,\tag{25}\\ \|T_mh^{(0)}-\delta_e\|_{\ell^2}^2 &=\tau_\Gamma(1-E_m)<\frac{\eta^2}{8},\tag{26}\\ \|h^{(0)}\|_{\ell^2}^2 &=\langle R_mQ_mR_m\delta_e,\delta_e\rangle =\tau_\Gamma(R_m) \leq\frac{1}{dm}<\frac{\eta^2}{16}. \tag{27}\end{align*}\]

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 \(J\delta_e=\delta_e\). Commutation with real polynomials in \(Q_m\) extends by uniform approximation to real continuous functions of \(Q_m\). The continuous functions \[x\longmapsto \min\{1,\max\{0,l(x-dm)\}\}, \qquad l=1,2,\ldots,\] 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, \[R_m=E_m g_m(Q_m), \qquad g_m(x)=\frac{1}{\max\{dm,x\}},\] 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 \[\|h-h^{(0)}\|_{\ell^2} <\frac{\eta}{4(m+1)}.\] The elementary bound \(\left\lVert T_m\right\rVert\leq m+1\) and (26)–(27) yield \[\begin{align*} \|h\|_{\ell^2} &<\frac{\eta}{4}+\frac{\eta}{4(m+1)}<\eta,\\ \|T_mh-\delta_e\|_{\ell^2} &<\frac{\eta}{\sqrt{8}}+\frac{\eta}{4}<\eta. \end{align*}\] For this finitely supported \(h\), form the cocycle with \(h_1=h\) and all other generator values zero, and take \(w=w_m\). Equation (17) gives \(D_w=T_mh\), proving (14).

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). Lemma 12 gives a finitely supported real coefficient vector \(h_1\), with \(h_j=0\) for \(j\geq2\), and a word \(w\in\mathbb F_n\) for which \[\|h_1\|_{\ell^2}<\eta, \qquad \|D_w-\delta_e\|_{\ell^2}<\eta.\] Apply Proposition 8 to these coefficients and write \(\beta=\beta_1\). Its displacement estimates (6) and (7) give \[\max_j\left\lVert \beta(A_j)-A_j\right\rVert<\varepsilon, \qquad \left\lVert \beta(A_w)-C A_w\right\rVert<\varepsilon.\] Moreover, \(\beta\) fixes \(S\) and therefore fixes \(C=\exp(\mathrm iS)\).

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 4, it induces a trace-preserving normal automorphism \(\gamma\) of \(M\) satisfying \[\gamma(A_j)=A_j \quad(1\leq j\leq n), \qquad \gamma(C)=C A_w.\] Set \(\alpha=\gamma^{-1}\circ\beta\). Since \(\gamma\) fixes every \(A_j\) and hence \(A_w\), we have the exact identities \[\begin{align*} \alpha(A_j)-A_j &=\gamma^{-1}\bigl(\beta(A_j)-A_j\bigr),\\ \alpha(A_w)-C &=\gamma^{-1}\bigl(\beta(A_w)-C A_w\bigr). \end{align*}\] An automorphism is isometric, so these identities prove (28), with the displayed order of the factors. The image under \(\alpha\) of the original freely generating Haar tuple is again such a tuple. Finally, \[C'=\alpha(C)=\gamma^{-1}(C)=C A_w^*,\] because \(\beta(C)=C\) and \(\gamma(C A_w^*)=C A_w A_w^*=C\). ◻

Proof of Theorem 1. Fix \(n\geq3\) and let \(M=L(\mathbb F_{n+1})\), with its canonical normalized trace and canonical freely generating Haar tuple \[(A_1^{(0)},\ldots,A_n^{(0)},C^{(0)}).\] 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 4; coefficients in \(\mathbb{Q}+\mathrm i\mathbb{Q}\) 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 \[ \left\lVert F_{j,k}(A^{(k-1)},C^{(k-1)})-y_j \right\rVert_2<2^{-k}. \tag{30}\] 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 (29). Choose \[ 0<\varepsilon_k< \min\left\{\frac{r_{k-1}}2,\, \frac{2^{-k}}{1+H_k}\right\}. \tag{31}\]

Next apply Proposition 15 to \((A^{(k-1)},C^{(k-1)})\) with tolerance \(\varepsilon_k\). Let \(\alpha_k\) and \(w_k\in\mathbb F_n\) be its automorphism and word, and set \[A_j^{(k)}=\alpha_k(A_j^{(k-1)}), \qquad C^{(k)}=\alpha_k(C^{(k-1)}).\] The new completed tuple is freely generating Haar, and \[ \max_j\left\lVert A_j^{(k)}-A_j^{(k-1)}\right\rVert<\varepsilon_k, \qquad \left\lVert A_{w_k}^{(k)}-C^{(k-1)}\right\rVert<\varepsilon_k. \tag{32}\] In particular, the last expression is close to the old completion, as needed in (30).

Now define word polynomials in \(n\) variables by \[G_{j,k}(X)=F_{j,k}(X,X_{w_k}) \qquad(1\leq j\leq k).\] Since \(\left\lVert x\right\rVert_2\leq\left\lVert x\right\rVert\), equations (29), (30), and (32) imply \[\begin{align*} \left\lVert G_{j,k}(A^{(k)})-y_j\right\rVert_2 &\leq \left\lVert F_{j,k}(A^{(k)},A_{w_k}^{(k)}) -F_{j,k}(A^{(k-1)},C^{(k-1)})\right\rVert_2 \\ &\qquad +\left\lVert F_{j,k}(A^{(k-1)},C^{(k-1)})-y_j\right\rVert_2 \\ &<H_k\varepsilon_k+2^{-k}<2\cdot2^{-k}. \tag{33}\end{align*}\] This estimate uses \(A_{w_k}^{(k)}\) as a single unitary slot of \(F_{j,k}\). Thus the tolerance chosen in (31) 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 \[ 0<r_k\leq \min\left\{\frac{r_{k-1}}2,\, \frac{2^{-k}}{1+L_k}\right\}. \tag{34}\] These constants are finite by (29). 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 \[\varepsilon_\ell<\frac{r_{\ell-1}}2 \leq \frac{r_k}{2^{\ell-k}}.\] Using (32), we therefore obtain, for \(p>k\), \[\max_j\left\lVert A_j^{(p)}-A_j^{(k)}\right\rVert \leq\sum_{\ell=k+1}^{p}\varepsilon_\ell \leq r_k.\] Since \(r_k\leq2^{-k}r_0\), each sequence \((A_j^{(k)})_k\) converges in operator norm to a unitary \(A_j^{(\infty)}\), and \[ \max_j\left\lVert A_j^{(\infty)}-A_j^{(k)}\right\rVert\leq r_k. \tag{35}\] The limits are unitaries because the unitary group is norm closed. For every nonidentity reduced word \(v\in\mathbb F_n\), \[\tau(A_v^{(\infty)}) =\lim_{k\to\infty}\tau(A_v^{(k)})=0,\] 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 [eq:stage-new-approximation], (34), and (35) yield \[\begin{align*} \left\lVert G_{j,k}(A^{(\infty)})-y_j\right\rVert_2 &\leq \left\lVert G_{j,k}(A^{(\infty)})-G_{j,k}(A^{(k)})\right\rVert_2 +\left\lVert G_{j,k}(A^{(k)})-y_j\right\rVert_2 \\ &\leq L_k r_k+2\cdot2^{-k} \leq3\cdot2^{-k}. \tag{36}\end{align*}\] 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 \[\left\lVert E_N(y_j)-y_j\right\rVert_2 \leq 2\left\lVert G_{j,k}(A^{(\infty)})-y_j\right\rVert_2 \longrightarrow0.\] 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 4 supplies a unital trace-preserving normal \(*\)-isomorphism from \(L(\mathbb F_n)\) onto \(M=L(\mathbb F_{n+1})\), proving the theorem. ◻