# Free independence and Haar tuples *Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).* This course proves OpenAI's theorem that the free group factors \(L(\mathbb F_n)\), \(2\le n<\infty\), are all isomorphic to one another [OAI]. The first four lessons construct an isomorphism \(L(\mathbb F_n)\cong L(\mathbb F_{n+1})\) for \(n\ge3\). The idea is to find, inside \(L(\mathbb F_{n+1})\), \(n\) unitaries that are free Haar unitaries and generate the whole factor; such a tuple identifies the factor with \(L(\mathbb F_n)\) (Lemma 2.4). The last four lessons compute, after Voiculescu, the corner of \(L(\mathbb F_n)\) cut down by a projection of trace \(\frac12\), which is \(L(\mathbb F_{4n-3})\); with the rank step this gives \(L(\mathbb F_2)\cong L(\mathbb F_3)\). ([OAI] uses Dykema's amplification formula for interpolated free group factors for this last step.) Shlyakhtenko later recast the argument as an automorphism approximation property of inclusions and proved the corresponding isomorphism for the C*-algebras generated by free semicircular families [Sh 2026]; Section 4 of the lesson [Corners of free group factors](corners-of-free-group-factors.md) states his results. This first lesson collects the tools. Section 1 recalls free independence in a tracial von Neumann algebra and proves the facts we need: freeness passes to generated von Neumann algebras and to groupings, mixed moments of free algebras are determined by the separate moments, and conjugating one of two free algebras by a unitary of the other preserves freeness. Section 2 treats freely generating Haar tuples. Section 3 fixes a tuple \((A_1,\ldots,A_n,C)\) of this kind, takes a bounded logarithm \(S\) of \(C\), and shows that the conjugates \(A_gSA_g^*\), \(g\in\mathbb F_n\), are free. Section 4 proves a bound for the operator norm of a sum of free centred elements that does not depend on the number of summands: \(\|\sum_gk(g)A_gSA_g^*\|\le3\pi\|k\|_{\ell^2}\). We use: Haar unitaries, Definition 2.1 of [Haar unitaries, freeness and tracial ultraproducts](course:generators-of-ii1-factors/haar-unitaries-freeness-and-ultraproducts#2-haar-unitaries); group von Neumann algebras and the theorem that \(L(\Gamma)\) is a II₁ factor for an ICC group \(\Gamma\), Theorem 5.2 of [Classification of injective factors](course:injective-factors/classification-of-injective-factors#5-group-von-neumann-algebras); Kaplansky's density theorem, [Theorem 7.1 of its lesson](course:foundations-of-von-neumann-algebras/kaplansky-s-density-theorem-and-its-consequences#OA-FND-KD-07); the double commutant theorem, [Theorem 4.4 of its lesson](course:foundations-of-von-neumann-algebras/the-double-commutant-theorem#OA-FND-BI-06); and the Borel functional calculus of a normal operator, [Theorem 8.1 of the spectral theorem lesson](course:foundations-of-von-neumann-algebras/the-spectral-theorem-for-bounded-self-adjoint-operators#OA-FND-ST-08). A finite measure on the unit circle is determined by its Fourier coefficients. ## 1. Free independence Throughout, \((M,\tau)\) is a von Neumann algebra with a faithful normal tracial state. We write \(\|x\|\) for the operator norm, \(\|x\|_2=\tau(x^*x)^{1/2}\), and \(L^2(M)\) for the completion of \(M\) for \(\langle x,y\rangle=\tau(y^*x)\), with the vector \(\Omega\) given by \(1\in M\). Left multiplication \(x\mapsto L_x\) is a faithful normal representation of \(M\) on \(L^2(M)\), with \(L_x\Omega=x\Omega\) and \(\|x\Omega\|=\|x\|_2\). We have \(\|xy\|_2\le\|x\|\|y\|_2\) and \(\|xy\|_2\le\|x\|_2\|y\|\), so multiplying by unitaries on either side does not change \(\|\cdot\|_2\). An element \(x\) is *centred* if \(\tau(x)=0\), and \(x^\circ=x-\tau(x)1\). All subalgebras are unital. **Definition 1.1.** A family \((A_i)_{i\in I}\) of unital \(*\)-subalgebras of \(M\) is *free* if \[ \tau(x_1x_2\cdots x_r)=0 \] whenever \(r\ge1\), \(x_k\in A_{i_k}\), \(\tau(x_k)=0\), and \(i_k\ne i_{k+1}\) for \(1\le k0\), by the trace of a reduced product of centred elements of the \(A_{i_k}\), which is \(0\). (2) The \(*\)-algebra generated by \(\bigcup_{i\in J}A_i\) is spanned by products \(x_1\cdots x_r\) with \(x_k\in A_{i_k}\), \(i_k\in J\). Merging adjacent letters from the same algebra we may assume \(i_k\ne i_{k+1}\). Writing \(x_k=\tau(x_k)1+x_k^\circ\) and expanding, the product becomes the reduced product \(x_1^\circ\cdots x_r^\circ\) plus scalar multiples of products with fewer letters; induction on \(r\) proves the first statement. A nonempty reduced product has trace \(0\), so the centred elements of \(B_l\) are spanned by the nonempty reduced products of centred elements of the \(A_i\), \(i\in I_l\). Now let \(y_1\cdots y_r\) be a product of centred \(y_k\in B_{l_k}\) with \(l_k\ne l_{k+1}\). It is a linear combination of concatenations of nonempty reduced products, the \(k\)-th one in the algebras \(A_i\), \(i\in I_{l_k}\). At each junction the two adjacent letters lie in algebras \(A_i\) with indices from the disjoint sets \(I_{l_k}\) and \(I_{l_{k+1}}\), so each concatenation is reduced and has trace \(0\). (3) By induction on \(r\). Merging adjacent letters from the same algebra (the \(\pi_i\) are multiplicative) we may assume \(i_k\ne i_{k+1}\). For \(r=1\) the claim is \(\tau'\circ\pi_i=\tau\). Write \(x_k=\tau(x_k)1+x_k^\circ\) and expand both products. The term \(x_1^\circ\cdots x_r^\circ\) and its image \(\pi_{i_1}(x_1^\circ)\cdots\pi_{i_r}(x_r^\circ)\) are reduced, because \(\pi_i(x^\circ)=\pi_i(x)-\tau(x)1\) is centred; both have trace \(0\). Every other term is the product of scalars \(\tau(x_k)=\tau'(\pi_{i_k}(x_k))\) with the trace of a shorter product, to which the induction hypothesis applies. \(\square\) **Lemma 1.3** (conjugation by the other algebra). Let \(A\) and \(B\) be free unital \(*\)-subalgebras of \(M\) and \(U\in B\) a unitary. Then \(UAU^*\) and \(B\) are free. Moreover every product of elements \(Ua_kU^*\) (\(a_k\in A\)) and \(b_l\in B\), in any order, has the same trace as the product obtained by replacing each \(Ua_kU^*\) by \(a_k\). **Proof.** Let \(y_1\cdots y_r\) be a reduced product of centred elements of \(UAU^*\) and of \(B\). Then \(\tau(y_1\cdots y_r)=\tau(U^*y_1U\cdots U^*y_rU)\). A letter \(UaU^*\) becomes \(a\), centred in \(A\) because \(\tau(a)=\tau(UaU^*)\); a letter \(b\) becomes \(U^*bU\), centred in \(B\) because \(U\in B\). The new product is reduced, so its trace is \(0\). The second statement is Lemma 1.2(3) for the free pairs \((A,B)\) and \((UAU^*,B)\), with \(\pi_1=\operatorname{Ad}U|_A\) and \(\pi_2=\mathrm{id}_B\). \(\square\) **Lemma 1.4.** Let \(B\subseteq M\) be a unital \(*\)-subalgebra with \(W^*(B)=M\). Then \(B\Omega\) is dense in \(L^2(M)\), and a normal linear functional that vanishes on \(B\) vanishes on \(M\). **Proof.** By Theorem 4.4 of the double commutant lesson, \(M\) is the \(\sigma\)-weak closure of \(B\); a normal functional is \(\sigma\)-weakly continuous, which proves the second statement. If \(\eta\perp B\Omega\), the functional \(x\mapsto\langle x\Omega,\eta\rangle\) is normal and vanishes on \(B\), so \(\eta\perp M\Omega\), which is dense; hence \(\eta=0\). \(\square\) ## 2. Freely generating Haar tuples Let \(\mathbb F_d\) be the free group on \(x_1,\ldots,x_d\): words in the letters \(x_j^{\pm1}\), with concatenation as product, modulo inserting and deleting adjacent pairs \(yy^{-1}\) of inverse letters. A word is *reduced* if it contains no such pair; \(|g|\) denotes the length of the reduced word of \(g\). **Lemma 2.1** (normal form). Every element of \(\mathbb F_d\) is represented by exactly one reduced word. For every group \(G\) and elements \(g_1,\ldots,g_d\in G\) there is exactly one homomorphism \(\mathbb F_d\to G\) with \(x_j\mapsto g_j\). **Proof.** Deleting inverse pairs until none is left gives a reduced representative. For uniqueness, let \(W\) be the set of reduced words and, for a letter \(y\), let \(\sigma_y\colon W\to W\) send \(w\) to \(yw\) if \(w\) does not begin with \(y^{-1}\), and to \(w'\) if \(w=y^{-1}w'\). One checks in both cases that \(\sigma_{y^{-1}}\sigma_y=\mathrm{id}\), so \(\sigma_y\) is a bijection with inverse \(\sigma_{y^{-1}}\). The permutation \(\sigma_{y_1}\cdots\sigma_{y_r}\) attached to a word \(y_1\cdots y_r\) does not change when an inverse pair is inserted or deleted, and for a reduced word it sends the empty word to the word itself (induction from the right). So two reduced words representing the same element coincide. The same insertion argument shows that evaluating words at \(g_1,\ldots,g_d\) is well defined on \(\mathbb F_d\); it is a homomorphism, and any homomorphism with these values agrees with it on every word. \(\square\) For a countable group \(\Gamma\), \(L(\Gamma)\subseteq B(\ell^2(\Gamma))\) is generated by the left translations \(\lambda(g)\delta_h=\delta_{gh}\); its trace \(\tau(x)=\langle x\delta_e,\delta_e\rangle\) is faithful and normal, and \(\tau(\lambda(g))=0\) for \(g\ne e\). For a tuple \(U=(U_1,\ldots,U_d)\) of unitaries in \(M\) and \(g\in\mathbb F_d\), let \(U_g\) be the value of \(g\) at \(U\) given by Lemma 2.1 (so \(x_j^{-1}\) is evaluated as \(U_j^*\)); then \(U_gU_h=U_{gh}\) and \(U_g^*=U_{g^{-1}}\). The linear span of the \(U_g\) is the \(*\)-algebra generated by \(U\); we call it the *word algebra* of \(U\). **Lemma 2.2.** For \(d\ge2\), \(\mathbb F_d\) is ICC, so \(L(\mathbb F_d)\) is a factor of type II₁. **Proof.** Let \(g\ne e\), with reduced word beginning with the letter \(y\) and ending with the letter \(y'\). Choose one of the \(2d\ge4\) letters \(a\) with \(a\ne y^{-1}\) and \(a\ne y'\). Then the word \(a^kga^{-k}\) is reduced, of length \(|g|+2k\), so the conjugates \(a^kga^{-k}\), \(k\ge1\), are distinct. Theorem 5.2 of [Classification of injective factors](course:injective-factors/classification-of-injective-factors#5-group-von-neumann-algebras) finishes the proof. \(\square\) **Definition 2.3.** A *freely generating Haar \(d\)-tuple* in \(M\) is a tuple \(U=(U_1,\ldots,U_d)\) of unitaries with \(W^*(U)=M\) and \(\tau(U_g)=0\) for every \(g\in\mathbb F_d\setminus\{e\}\). The generators \(\lambda(x_1),\ldots,\lambda(x_d)\) form such a tuple in \(L(\mathbb F_d)\). **Lemma 2.4** (identification). Let \(U\) be a freely generating Haar \(d\)-tuple in \(M\). The vectors \(U_g\Omega\), \(g\in\mathbb F_d\), form an orthonormal basis of \(L^2(M)\); in particular the word algebra of \(U\) is dense in \(L^2(M)\). There is a unique normal unital \(*\)-isomorphism \(\Phi\colon L(\mathbb F_d)\to M\) with \(\Phi(\lambda(x_j))=U_j\), and it preserves the traces. **Proof.** The vectors are orthonormal: \(\langle U_g\Omega,U_h\Omega\rangle=\tau(U_h^*U_g)\), which is \(\tau(U_{h^{-1}g})=\delta_{g,h}\). Their span is the image of the word algebra, which is dense by Lemma 1.4. The unitary \(V\colon\ell^2(\mathbb F_d)\to L^2(M)\), \(V\delta_g=U_g\Omega\), satisfies \(V\lambda(x_j)\delta_g=U_jU_g\Omega=L_{U_j}V\delta_g\), so \(V\lambda(x_j)V^*=L_{U_j}\). Since \(L(\mathbb F_d)\) is generated by the unitaries \(\lambda(x_j)\), \(\operatorname{Ad}V\) maps it onto the von Neumann algebra generated by the \(L_{U_j}\), which is the image of \(W^*(U)=M\) under the faithful normal left representation. Composing with the inverse of that representation gives \(\Phi\). It preserves the traces because \(V\delta_e=\Omega\): \(\tau(\Phi(x))=\langle VxV^*\Omega,\Omega\rangle=\langle x\delta_e,\delta_e\rangle\). A normal \(*\)-homomorphism is determined by its values on generators, which gives uniqueness. \(\square\) **Corollary 2.5.** 1. If \(U,U'\) are freely generating Haar \(d\)-tuples in \(M\), there is a unique trace-preserving normal automorphism \(\alpha\) of \(M\) with \(\alpha(U_j)=U'_j\) for all \(j\). 2. If \(U\) is a freely generating Haar \(d\)-tuple and \(\theta\) is an automorphism of \(\mathbb F_d\), then \((U_{\theta(x_1)},\ldots,U_{\theta(x_d)})\) is a freely generating Haar \(d\)-tuple, so there is a trace-preserving normal automorphism \(\gamma\) of \(M\) with \(\gamma(U_j)=U_{\theta(x_j)}\). 3. If \(U\) is a freely generating Haar \(d\)-tuple, each \(U_j\) is a Haar unitary and \(W^*(U_1),\ldots,W^*(U_d)\) form a free family. 4. The image of a freely generating Haar tuple under a trace-preserving normal automorphism of \(M\) is a freely generating Haar tuple. **Proof.** (1) Compose the isomorphisms of Lemma 2.4. (2) The value of \(g\) at the new tuple is \(U_{\theta(g)}\), which has trace \(0\) for \(g\ne e\). The new tuple generates every \(U_{\theta(g)}\), in particular \(U_j=U_{\theta(\theta^{-1}(x_j))}\), so it generates \(M\). Apply (1). (3) \(\tau(U_j^k)=0\) for \(k\ne0\). The centred elements of the \(*\)-algebra generated by \(U_j\) are the Laurent polynomials \(\sum_{k\ne0}c_kU_j^k\). A reduced product of such polynomials in different variables expands into values of nonempty reduced words, of trace \(0\); Lemma 1.2(1) passes to the von Neumann algebras. (4) The values of words and the generated von Neumann algebra are carried along by the automorphism, and the traces are preserved. \(\square\) ## 3. A bounded logarithm and its conjugates From now on, \((A_1,\ldots,A_n,C)\) is a freely generating Haar \((n+1)\)-tuple in \(M\), \(n\ge1\), \(\Gamma=\mathbb F_n=\langle x_1,\ldots,x_n\rangle\), and \(A_g\) is the value of \(g\in\Gamma\) at \(A=(A_1,\ldots,A_n)\). By Corollary 2.5(3) and Lemma 1.2(1),(2), the von Neumann algebras \(W^*(A_1,\ldots,A_n)\) and \(W^*(C)\) are free. Let \[ S=\arg(C),\qquad S_g=A_gSA_g^*\quad(g\in\Gamma), \] with \(\arg\) the Borel function on the unit circle with values in \((-\pi,\pi]\), applied to the normal operator \(C\) by the Borel functional calculus. Then \(S\in W^*(C)\) is self-adjoint, \(\|S\|\le\pi\), and \(e^{iS}=C\), so \(W^*(S)=W^*(C)\). The distribution \(\mu(B)=\tau(1_B(C))\) of \(C\) is a probability measure on the circle with \(\int z^k\,d\mu=\tau(C^k)=0\) for \(k\ne0\), so \(\mu\) is the normalized arc length, and \(\tau(f(C))=\frac1{2\pi}\int_{-\pi}^{\pi}f(e^{i\theta})\,d\theta\) for bounded Borel \(f\). In particular \(\tau(S)=\frac1{2\pi}\int_{-\pi}^\pi\theta\,d\theta=0\). We also have \(A_gS=S_gA_g\) and \(A_gS_bA_g^*=S_{gb}\). **Lemma 3.1** (free conjugates). The von Neumann algebras \(A_gW^*(C)A_g^*\), \(g\in\Gamma\), form a free family. In particular \(W^*(S_g)=A_gW^*(C)A_g^*\) is free from the von Neumann algebra \(W^*(S_b:b\ne g)\). **Proof.** Consider a reduced product of centred elements \(A_{g_k}c_kA_{g_k}^*\), \(1\le k\le r\), with \(c_k\in W^*(C)\) centred and \(g_k\ne g_{k+1}\). Moving \(A_{g_1}\) to the end by the trace property, \[ \begin{aligned} &\tau\big(A_{g_1}c_1A_{g_1}^*\cdots A_{g_r}c_rA_{g_r}^*\big)\\ &\quad =\tau\big(c_1A_{g_1^{-1}g_2}c_2A_{g_2^{-1}g_3}\cdots c_rA_{g_r^{-1}g_1}\big). \end{aligned} \] The elements \(A_{g_k^{-1}g_{k+1}}\) are centred elements of \(W^*(A_1,\ldots,A_n)\). If \(g_r^{-1}g_1\ne e\), the last one is centred too; if \(g_r^{-1}g_1=e\), omit it. In both cases the product alternates between centred elements of the two free algebras \(W^*(C)\) and \(W^*(A_1,\ldots,A_n)\) and has trace \(0\) (for \(r=1\) it is \(\tau(c_1)=0\)). The second statement follows from Lemma 1.2(2),(1). \(\square\) Call the *label* of a product of the letters \(A_j,A_j^*,S\) the element of \(\Gamma\) obtained by multiplying its group letters in order and ignoring the letters \(S\). **Lemma 3.2** (labels). A product with label different from \(e\) has trace \(0\). A product with label \(e\) is a product of elements \(S_g\), or \(1\). **Proof.** Moving the group letters to the right with \(A_gS=S_gA_g\) turns a product with label \(h\) into a product of elements \(S_g\) followed by \(A_h\); this proves the second statement. For the first, we show by induction on the number \(m\) of letters \(S\) that a product with label \(h\ne e\) has trace \(0\). Merging adjacent group letters and adjacent letters \(S\), write the product as \(A_{g_0}S^{q_1}A_{g_1}S^{q_2}\cdots S^{q_p}A_{g_p}\) with \(q_l\ge1\), \(g_1,\ldots,g_{p-1}\ne e\) and \(g_0g_1\cdots g_p=h\). If \(p=0\), the product is \(A_h\), of trace \(0\). Otherwise write each \(S^{q_l}\) as \(\tau(S^{q_l})1+(S^{q_l})^\circ\) and expand. The term with all centred parts alternates between centred elements of \(W^*(C)\) and nonidentity group elements \(A_{g_l}\) (omit \(A_{g_0}\) or \(A_{g_p}\) if it is \(1\)), so it has trace \(0\) by freeness. Every other term is a scalar times a product with fewer letters \(S\) and the same label \(h\), of trace \(0\) by induction. \(\square\) ## 4. The norm of a free sum **Lemma 4.1.** Let \(B_1,\ldots,B_N\) be a free family of von Neumann subalgebras of \(M\) and \(z_j\in B_j\) centred and self-adjoint. Then \[ \begin{aligned} \Big\|\sum_jz_j\Big\|&\le2\Big(\sum_j\|z_j\|_2^2\Big)^{1/2}\\ &\quad +\max_j\|z_j\|. \end{aligned} \] **Proof.** Let \(B_0\) be the \(*\)-algebra generated by the \(B_j\), \(B=W^*(B_0)\), and \(H\) the closure of \(B_0\Omega\) in \(L^2(M)\). By Kaplansky's density theorem every element of \(B\) is the strong limit of a bounded net in \(B_0\), so \(H\) also contains \(B\Omega\), and \(H\) is invariant under \(L_B\). The representation \(x\mapsto L_x|_H\) of \(B\) is faithful (\(x\Omega=0\) forces \(x=0\)), hence isometric, so \(\|\sum_jz_j\|\) is the norm of \(\sum_jL_{z_j}\) on \(H\). A *word* is a vector \(y_1\cdots y_m\Omega\) with \(m\ge1\), \(y_k\in B_{i_k}\) centred and \(i_k\ne i_{k+1}\); its first index is \(i_1\). By Lemma 1.2(2), \(\Omega\) and the words span \(B_0\Omega\), which is dense in \(H\). For words \(\xi=y_1\cdots y_m\Omega\) and \(\xi'=y_1'\cdots y_{m'}'\Omega\) with index sequences \((i_k)\) and \((i_k')\), \[ \langle\xi,\xi'\rangle=\begin{cases}\prod_k\tau(y_k'^{\,*}y_k)&\text{if }(i_k)=(i_k'),\\0&\text{otherwise},\end{cases} \] and \(\langle\xi,\Omega\rangle=0\). Indeed, \(\langle\xi,\xi'\rangle=\tau(y_{m'}'^{\,*}\cdots y_1'^{\,*}y_1\cdots y_m)\). If \(i_1\ne i_1'\), the product is reduced and has trace \(0\). Otherwise write \(y_1'^{\,*}y_1=\tau(y_1'^{\,*}y_1)1+(y_1'^{\,*}y_1)^\circ\): the centred part gives a reduced product of trace \(0\), and the scalar part multiplies the same expression for the two shorter words. Repeating this, the trace is \(0\) unless both words are used up at the same step, and then it is the stated product. Similarly \(\langle\xi,\Omega\rangle=\tau(y_1\cdots y_m)=0\). Let \(P_j\) be the orthogonal projection onto the closed span of the words with first index \(j\). The subspaces \(P_jH\) are mutually orthogonal and \(H=\mathbb C\Omega\oplus\bigoplus_jP_jH\). If \(\xi\) is \(\Omega\) or a word with first index \(\ne j\), then \(z_j\xi\) is a word with first index \(j\), and by the inner product formula \(\langle z_j\xi,z_j\xi'\rangle=\|z_j\|_2^2\langle\xi,\xi'\rangle\) for any two such vectors. By linearity and continuity, \(L_{z_j}\) maps \((1-P_j)H\) into \(P_jH\) and multiplies norms there by \(\|z_j\|_2\). So \(T_j=P_jL_{z_j}(1-P_j)\) has \(\|T_j\|\le\|z_j\|_2\), and \((1-P_j)L_{z_j}(1-P_j)=0\). Since \(L_{z_j}\) is self-adjoint, \((1-P_j)L_{z_j}P_j=T_j^*\). Thus \[ \begin{gathered} L_{z_j}=T_j+T_j^*+D_j,\\ D_j=P_jL_{z_j}P_j,\quad\|D_j\|\le\|z_j\| . \end{gathered} \] The ranges of the \(T_j\) are mutually orthogonal, so \(\|\sum_jT_j\xi\|^2=\sum_j\|T_j\xi\|^2\), which is at most \(\sum_j\|z_j\|_2^2\|\xi\|^2\), and the same bound holds for \(\sum_jT_j^*=(\sum_jT_j)^*\). Each \(D_j\) maps \(P_jH\) into itself and vanishes on its orthogonal complement, so \(\|\sum_jD_j\|=\max_j\|D_j\|\). Adding the three bounds proves the lemma. \(\square\) For a finitely supported real function \(k\) on \(\Gamma\) put \[ s(k)=\sum_{g\in\Gamma}k(g)S_g . \] We write \(\|k\|_{\ell^2}\) for its norm in \(\ell^2(\Gamma)\), to distinguish it from the trace norm \(\|\cdot\|_2\) of operators. **Corollary 4.2.** \(\|s(k)\|\le K\|k\|_{\ell^2}\) with \(K=3\pi\). **Proof.** By Lemma 3.1 the algebras \(W^*(S_g)\), \(g\) in the support of \(k\), form a free family. Apply Lemma 4.1 to the centred self-adjoint elements \(z_g=k(g)S_g\), using \(\|S_g\|_2\le\|S_g\|\le\pi\) and \(|k(g)|\le\|k\|_{\ell^2}\). \(\square\) The point of Corollary 4.2 is that the bound does not grow with the support of \(k\); the trivial bound \(\pi\sum_g|k(g)|\) does. ## 5. Exercises **Exercise 5.1.** Show that if \(a\) and \(b\) are free centred elements, then \(\tau(abab)=0\) and \(\tau(aabb)=\tau(a^2)\tau(b^2)\). **Exercise 5.2.** Let \(U=(U_1,U_2)\) be a freely generating Haar pair. Show that \((U_1,U_2U_1)\) and \((U_1^*,U_2)\) are freely generating Haar pairs. **Exercise 5.3.** Show that Lemma 4.1 does not hold with the right side replaced by \((\sum_j\|z_j\|_2^2)^{1/2}\): take \(N=1\). **Exercise 5.4.** Compute \(\tau(S^2)\) for \(S=\arg(C)\), \(C\) a Haar unitary. ## 6. Solutions **5.1.** \(abab\) is a reduced product of centred letters, so its trace is \(0\). For \(aabb\), write \(a^2=\tau(a^2)1+(a^2)^\circ\) and \(b^2=\tau(b^2)1+(b^2)^\circ\). The product of the two centred parts is reduced, the two mixed terms are scalar multiples of centred elements, and what remains is \(\tau(a^2)\tau(b^2)\). **5.2.** Both are images of \(U\) under automorphisms of \(\mathbb F_2\): \(x_1\mapsto x_1\), \(x_2\mapsto x_2x_1\) has inverse \(x_1\mapsto x_1\), \(x_2\mapsto x_2x_1^{-1}\), and \(x_1\mapsto x_1^{-1}\), \(x_2\mapsto x_2\) is its own inverse. Apply Corollary 2.5(2). **5.3.** Let \(C\) be a Haar unitary and \(p,q\in W^*(C)\) the spectral projections of \(C\) for two disjoint arcs of length \(2\pi\varepsilon\), so \(\tau(p)=\tau(q)=\varepsilon\). Then \(z=p-q\) is centred and self-adjoint, \(\|z\|=1\) and \(\|z\|_2=\sqrt{2\varepsilon}\). **5.4.** \(\tau(S^2)=\frac1{2\pi}\int_{-\pi}^\pi\theta^2\,d\theta=\frac{\pi^2}3\). ## References - [OAI] OpenAI, *An isomorphism of the free group factors* (September 23, 2026), OpenAI Math Release preprint, Section 2. https://github.com/openai/math/blob/main/preprints/An-isomorphism-of-the-free-group-factors-September-23-2026/An-isomorphism-of-the-free-group-factors-September-23-2026.pdf - [Sh 2026] D. Shlyakhtenko, *On the isomorphism of semicircular algebras*, arXiv:2610.19590 (October 2026). https://arxiv.org/abs/2610.19590 - E. Ricard and Q. Xu, *Khintchine type inequalities for reduced free products and applications*, J. reine angew. Math. 599 (2006), 27–59; the decomposition into creation, annihilation and diagonal parts used in Lemma 4.1 comes from their Section 2. https://arxiv.org/abs/math/0505302