Haar unitaries, freeness and tracial ultraproducts

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This course proves that every factor of type II₁ with separable predual is generated, as a von Neumann algebra, by a single operator. The proof works with unitaries that behave like free group generators relative to a given algebra, and it constructs them in a tracial ultraproduct. This first lesson collects the tools: norm inequalities in a tracial algebra, Haar unitaries and the decay of their correlations, a working criterion for freeness of a unitary from a C*-algebra with an absorption lemma, and the facts about tracial ultraproducts that the later lessons use.

We use: Theorem 3.2 of Ultraproducts and the asymptotic centralizer, which shows that a tracial ultraproduct is a von Neumann algebra with a faithful normal tracial state, and Lemma 1.4 of that lesson (Section 1), completeness of the unit ball for the trace norm; Lemma 55.1 of Finite Fourier bases produce one small quantized corner, a maximal abelian subalgebra of a II₁ factor with projections of every trace; the polar decomposition inside a von Neumann algebra, The double commutant theorem, Section 7; the comparison theorem for projections, Projections and types of von Neumann algebras, Section 5; Kaplansky's density theorem, Section 7 of its lesson; and the trace-preserving conditional expectation, Theorem 9.1 of Integration for a trace.

1. Tracial von Neumann algebras

A tracial von Neumann algebra \((M,\tau)\) is a von Neumann algebra \(M\) with a faithful normal tracial state \(\tau\). Put \[ \begin{gathered} \|x\|_2=\tau(x^*x)^{1/2},\\ \|x\|_1=\tau(|x|),\qquad \langle a,b\rangle=\tau(b^*a), \end{gathered} \] and let \(L^2(M)\) be the completion of \(M\) for \(\|\cdot\|_2\); \(M\) acts on it by left multiplication, and \(\|x\|\) denotes the operator norm. A II₁ factor is a factor with a faithful normal tracial state and no nonzero abelian projection. For a subset \(S\subseteq M\), \(W^*(S)\) is the smallest von Neumann subalgebra of \(M\) containing \(S\) and \(1\); \(C^*(S)\) is the smallest unital C*-subalgebra containing \(S\).

Lemma 1.1. Let \(a,b,c\in M\).

  1. \(|\tau(ab)|\le\|a\|_2\|b\|_2\) and \(\|abc\|_2\le\|a\|\,\|b\|_2\,\|c\|\).
  2. \(|\tau(ab)|\le\|a\|\,\|b\|_1\), and \(\|b\|_1=\max\{|\tau(db)|:\|d\|\le1\}\). Consequently \(\|abc\|_1\le\|a\|\,\|b\|_1\,\|c\|\).
  3. \(\|b\|_1\le\|b\|_2\). If \(b=bq\) for a projection \(q\), then \(\|b\|_1\le\tau(q)^{1/2}\|b\|_2\).

Proof. (1) The first inequality is the Cauchy–Schwarz inequality for the positive form \((a,b)\mapsto\tau(b^*a)\), applied to \(a\) and \(b^*\). For the second, \(a^*a\le\|a\|^21\) gives \(\|ab\|_2^2=\tau(b^*a^*ab)\le\|a\|^2\tau(b^*b)\), and \(\|bc\|_2=\|c^*b^*\|_2\le\|c\|\,\|b^*\|_2=\|c\|\,\|b\|_2\) because \(\tau(bb^*)=\tau(b^*b)\).

(2) Write \(b=w|b|\) with \(w\in M\) a partial isometry and \(w^*w|b|=|b|\). Traciality gives \(\tau(ab)=\tau(|b|^{1/2}aw|b|^{1/2})\). The functional \(x\mapsto\tau(|b|^{1/2}x|b|^{1/2})\) is positive with norm \(\tau(|b|)\), so \(|\tau(ab)|\le\|aw\|\,\tau(|b|)\le\|a\|\,\|b\|_1\). The choice \(d=w^*\) gives \(\tau(w^*b)=\tau(|b|)\), so the maximum is attained. Finally \(\|abc\|_1=\max_{\|d\|\le1}|\tau((cda)b)|\), which is at most \(\|a\|\,\|c\|\,\|b\|_1\).

(3) By (1), \(\tau(|b|)=\tau(|b|\cdot1)\le\||b|\|_2\|1\|_2=\|b\|_2\). If \(b=bq\), then \(b^*b=qb^*bq\); \(|b|\) is a norm limit of polynomials in \(b^*b\) without constant term, so \(|b|=|b|q\), and \(\tau(|b|)=\tau(|b|q)\), which is at most \(\|b\|_2\|q\|_2=\tau(q)^{1/2}\|b\|_2\). \(\square\)

By (1), products and adjoints are continuous for \(\|\cdot\|_2\) on bounded sets: if \(\sup_i\|x_i\|\) and \(\sup_i\|y_i\|\) are finite, \(x_i\to x\) and \(y_i\to y\) in \(2\)-norm, then \[ \begin{aligned} &\|x_iy_i-xy\|_2\le\|x_i\|\,\|y_i-y\|_2\\ &\quad +\|x_i-x\|_2\,\|y\|,\\ &\|x_i^*-x^*\|_2=\|x_i-x\|_2 . \end{aligned} \] Hence a product of finitely many bounded letters depends continuously on the letters, and so does its trace. A \(2\)-norm limit of a bounded net of unitaries (projections) is a unitary (projection). The closed unit ball of \(M\) is complete for \(\|\cdot\|_2\) (Lemma 1.4 of the ultraproduct lesson, where \(\|x\|^\sharp=\sqrt2\|x\|_2\) for a trace).

For a von Neumann subalgebra \(N\subseteq M\) with \(1\in N\), the trace-preserving conditional expectation \(E_N\colon M\to N\) is the unique map with \(\tau(E_N(x)y)=\tau(xy)\) for \(x\in M\), \(y\in N\) (Theorem 9.1 of the integration lesson; a finite trace is semifinite on every subalgebra). The identity says that \(x-E_N(x)\) is orthogonal to \(N\) in \(L^2(M)\), so \(E_N\) is the restriction to \(M\) of the orthogonal projection \(e_N\) of \(L^2(M)\) onto the closure \(L^2(N)\) of \(N\).

2. Haar unitaries

Definition 2.1. Let \(q\in M\) be a nonzero projection. A unitary of the corner \(qMq\) is an \(h\in qMq\) with \(h^*h=hh^*=q\); its powers are \(h^0=q\), \(h^{j+1}=h^jh\) and \(h^{-j}=(h^*)^j\) for \(j\ge1\), so that \(h^ih^j=h^{i+j}\) for all \(i,j\in\mathbb Z\). It is a Haar unitary of \(qMq\) if \(\tau(h^j)=0\) for every \(j\ne0\). A Haar unitary of \(M\) is one of the corner \(1M1\).

Lemma 2.2. Let \(A\subseteq M\) be an abelian von Neumann subalgebra and \(e\in A\) a nonzero projection such that, for every projection \(f\le e\) of \(A\) and every \(0\le c\le\tau(f)\), there is a projection \(f'\le f\) in \(A\) with \(\tau(f')=c\). Then \(Ae\) contains a Haar unitary of \(eMe\).

Proof. Put \(s=\tau(e)\). Choose projections \(p_{m,k}\in A\), \(m\ge0\), \(0\le k<2^m\), with \(p_{0,0}=e\), \(p_{m,k}=p_{m+1,2k}+p_{m+1,2k+1}\) and \(\tau(p_{m,k})=2^{-m}s\): given \(p_{m,k}\), take \(p_{m+1,2k}\le p_{m,k}\) of trace \(2^{-m-1}s\) and let \(p_{m+1,2k+1}\) be the rest. Put \[ h_m=\sum_{k=0}^{2^m-1}k2^{-m}p_{m,k}. \] Then \(h_{m+1}-h_m=\sum_k2^{-m-1}p_{m+1,2k+1}\), so \(0\le h_{m+1}-h_m\le2^{-m-1}e\), and \(h_m\) converges in norm to a self-adjoint \(x\in Ae\) with \(0\le x\le e\). For a continuous \(g\) on \([0,1]\) with \(g(0)=0\), \(g(h_m)=\sum_kg(k2^{-m})p_{m,k}\) because the \(p_{m,k}\) are orthogonal projections, so \[ \begin{aligned} &\tau(g(x))=\lim_m\tau(g(h_m))\\ &\quad =\lim_m\sum_{k=0}^{2^m-1}g(k2^{-m})2^{-m}s\\ &\quad =s\int_0^1g(r)\,dr . \end{aligned} \] Let \(h=\exp(2\pi ix)e\), the functional calculus of \(r\mapsto e^{2\pi ir}\) in the corner \(eMe\) (with unit \(e\)). Then \(h\) is a unitary of \(eMe\), and for \(j\ne0\), \(h^j-e\) is \(g_j(x)\) with \(g_j(r)=e^{2\pi ijr}-1\), so \(\tau(h^j)=\tau(e)+s\int_0^1(e^{2\pi ijr}-1)\,dr\), which is \(s-s=0\). \(\square\)

Lemma 2.3. In a factor \(Q\) with a faithful tracial state \(\tau\), two projections \(e,f\) with \(\tau(e)=\tau(f)\) are equivalent: there is \(v\in Q\) with \(v^*v=e\), \(vv^*=f\).

Proof. By the comparison theorem, one is equivalent to a subprojection of the other, say \(e\sim f'\le f\). Then \(\tau(f')=\tau(e)=\tau(f)\) because \(\tau(v^*v)=\tau(vv^*)\), so \(\tau(f-f')=0\) and \(f'=f\) by faithfulness. \(\square\)

Corollary 2.4. Let \(Q\) be a II₁ factor with trace \(\tau\) and \(q\in Q\) a nonzero projection. Then \(qQq\) has a Haar unitary, and \(Q\) has projections of every trace in \([0,1]\).

Proof. Lemma 55.1 of the quantization lesson gives a maximal abelian subalgebra \(A\subseteq Q\) in which every projection \(f\) has subprojections of every trace in \([0,\tau(f)]\); in particular \(A\) has projections of every trace in \([0,1]\). Take \(e\in A\) with \(\tau(e)=\tau(q)\) and \(v\in Q\) with \(v^*v=e\), \(vv^*=q\) (Lemma 2.3). Lemma 2.2 gives a Haar unitary \(h'\) of \(eQe\) in \(Ae\). Then \(h=vh'v^*\) satisfies \(h^*h=hh^*=q\), \(h^j=vh'^jv^*\) and \(\tau(h^j)=\tau(h'^jv^*v)=\tau(h'^j)=0\) for \(j\ne0\). \(\square\)

Lemma 2.5 (decay of correlations). Let \(h\) be a Haar unitary of \(qMq\) and \(b\in M\). Then \[ \sum_{j\in\mathbb Z}|\tau(h^jb)|^2\le\tau(q)\|b\|_2^2 ; \] in particular \(\tau(h^jb)\to0\) as \(|j|\to\infty\).

Proof. For \(i,j\in\mathbb Z\), \(\langle h^j,h^i\rangle=\tau(h^{-i}h^j)=\tau(h^{j-i})\), which is \(\tau(q)\) for \(i=j\) and \(0\) otherwise. So the vectors \(\tau(q)^{-1/2}h^j\) are orthonormal in \(L^2(M)\), and \(\tau(h^jb)=\tau(bh^j)=\langle h^j,b^*\rangle\). Bessel's inequality gives the claim. \(\square\)

3. Freeness of a unitary from a C*-algebra

Definition 3.1. Let \(D\subseteq M\) be a unital C*-subalgebra and \(t\in M\) a unitary. Call \(t\) free from \(D\) if \(\tau(c_1c_2\cdots c_r)=0\) whenever \(r\ge1\), each \(c_i\) has trace zero and lies in \(C^*(t)\) or in \(D\), and consecutive letters \(c_i,c_{i+1}\) come from different ones of these two algebras.

Elements of trace zero are called centred. If \(t\) is a Haar unitary, every power \(t^n\), \(n\ne0\), is a centred element of \(C^*(t)\). The next lemma replaces the outer letters of an alternating word by arbitrary elements of \(D\); this is the form used throughout the course.

Lemma 3.2. Let \(t\) be a Haar unitary of \(M\) and \(D\subseteq M\) a unital C*-subalgebra. Then \(t\) is free from \(D\) if and only if \[ \tau(d_0t^{n_1}d_1t^{n_2}\cdots t^{n_j}d_j)=0 \tag{3.1} \] for every \(j\ge1\), all \(n_1,\dots,n_j\in\mathbb Z\setminus\{0\}\) and all \(d_0,\dots,d_j\in D\) with \(\tau(d_i)=0\) for \(1\le i\le j-1\). The outer letters \(d_0,d_j\) are arbitrary.

Proof. Suppose \(t\) is free from \(D\). Write \(d_0=\tau(d_0)1+d_0^\circ\) and \(d_j=\tau(d_j)1+d_j^\circ\) and expand (3.1) into four words. In each of them the letters \(t^{n_i}\) are centred elements of \(C^*(t)\), they are separated by the centred letters \(d_1,\dots,d_{j-1}\) of \(D\), and the outer letters are either absent or the centred \(d_0^\circ,d_j^\circ\). So each word is alternating and centred, and has trace \(0\).

Conversely assume (3.1). The Laurent polynomials in \(t\) are norm dense in \(C^*(t)\). If \(c\in C^*(t)\) is centred and Laurent polynomials \(L_k\) converge to \(c\) in norm, then \(L_k-\tau(L_k)1\to c\), and \(L_k-\tau(L_k)1\) is a combination of powers \(t^n\) with \(n\ne0\), since \(\tau(t^n)=0\) for \(n\ne0\). The trace of a word depends multilinearly and norm continuously on its letters, so in an alternating centred word it suffices to treat letters \(c_i=t^{n_i}\) from \(C^*(t)\). Such a word has the form (3.1), with \(d_0=1\) if it begins with a power of \(t\), \(d_j=1\) if it ends with one, and centred interior letters from \(D\). Its trace is \(0\). A word without letters from \(C^*(t)\) is a single centred element of \(D\), and has trace \(0\) as well. \(\square\)

Lemma 3.3 (absorption). Let \(t\) be a Haar unitary free from \(D\), and let \(a,b\) be unitaries of \(D\). Then \(t^*\) and \(atb\) are Haar unitaries free from \(D\).

Proof. For \(t^*\), (3.1) for \(t^*\) is (3.1) for \(t\) with the exponents negated. Put \(v=atb\) and consider a word \(d_0v^{n_1}d_1\cdots v^{n_j}d_j\) as in (3.1). For \(n>0\) and \(n<0\) respectively, \[ \begin{gathered} v^n=at(ba)t(ba)\cdots(ba)tb,\\ v^{n}=b^*t^*(a^*b^*)t^*\cdots(a^*b^*)t^*a^*, \end{gathered} \] with \(|n|\) letters \(t^{\pm1}\). Substituting, the word becomes \(e_0t^{\sigma_1}e_1t^{\sigma_2}\cdots t^{\sigma_N}e_N\) with \(\sigma_r=\pm1\) and \(e_r\in D\). Inside one power all letters have the same sign. A separator between letters of opposite signs therefore sits at a junction of \(v^{n_i}\) and \(v^{n_{i+1}}\), and equals \(bd_ib^*\) (from \(t\) to \(t^*\)) or \(a^*d_ia\) (from \(t^*\) to \(t\)); it is centred, since \(\tau(bd_ib^*)=\tau(d_i)=0\). Write every separator between letters of equal sign as \(\tau(e_r)1+e_r^\circ\) and expand. In each resulting word, the letters joined by scalar separators combine into a power \(t^m\) with \(m\ne0\), because they have equal signs, and successive powers are separated by centred elements of \(D\). The outer letters \(e_0,e_N\) lie in \(D\). By (3.1) for \(t\), every term has trace \(0\). Thus (3.1) holds for \(v\). With \(j=1\) and \(d_0=d_1=1\) it says \(\tau(v^n)=0\) for \(n\ne0\), so \(v\) is a Haar unitary, and Lemma 3.2 shows that \(v\) is free from \(D\). \(\square\)

4. Tracial ultraproducts

Fix a free ultrafilter \(\omega\) on \(\mathbb N\) and tracial von Neumann algebras \((M_n,\tau_n)\). The ultraproduct \(\mathbf M=\prod_\omega M_n\) is the quotient of the C*-algebra of bounded sequences \((x_n)\), \(x_n\in M_n\), by the ideal of sequences with \(\lim_\omega\|x_n\|_2=0\), with trace \(\tau((x_n))=\lim_\omega\tau_n(x_n)\). By Theorem 3.2 of the ultraproduct lesson, \((\mathbf M,\tau)\) is a tracial von Neumann algebra. A bounded sequence mapping to \(x\) is a representing sequence of \(x\). Because the quotient map is a \(*\)-homomorphism, a word in elements of \(\mathbf M\) is represented by the same word in representing sequences, and \(\tau(x)=\lim_\omega\tau_n(x_n)\) for every representing sequence. In particular a sequence of projections (unitaries, partial isometries \(w_n\) with \(w_n^*w_n=w_nw_n^*\)) represents a projection (a unitary, a partial isometry \(w\) with \(w^*w=ww^*\)). When all \(M_n\) equal \(M\), we write \(M^\omega\) and regard \(M\subseteq M^\omega\) through constant sequences.

Lemma 4.1. Let \(N_n\subseteq M_n\) be von Neumann subalgebras with \(1\in N_n\). The classes in \(\mathbf M\) of bounded sequences \((y_n)\) with \(y_n\in N_n\) form a von Neumann subalgebra of \(\mathbf M\), identified with \(\prod_\omega N_n\).

Proof. A bounded sequence in \(\prod N_n\) has the same \(2\)-norms in \(N_n\) and in \(M_n\), so the map \(\prod_\omega N_n\to\mathbf M\) is a well-defined, injective, trace-preserving unital \(*\)-homomorphism. Its image \(\mathcal N\) is a unital \(*\)-subalgebra. An injective \(*\)-homomorphism of C*-algebras is isometric, so the unit ball of \(\mathcal N\) is the image of the unit ball of \(\prod_\omega N_n\), which is complete for \(\|\cdot\|_2\) by Theorem 3.2 and Lemma 1.4 of the ultraproduct lesson. Let \(\mathbf M\) act on \(L^2(\mathbf M)\) and let \(x\) be in the unit ball of \(\mathcal N''\). By Kaplansky's density theorem there is a net \(x_i\) in the unit ball of \(\mathcal N\) converging strongly to \(x\); applied to the vector \(1\), this says \(\|x_i-x\|_2\to0\). So \((x_i)\) is Cauchy for \(\|\cdot\|_2\), converges in the unit ball of \(\mathcal N\), and its limit is \(x\). Hence \(\mathcal N=\mathcal N''\). \(\square\)

Lemma 4.2 (unitaries lift). If every \(N_n\) is a factor, every unitary \(U\in\prod_\omega N_n\) has a representing sequence of unitaries \(U_n\in N_n\).

Proof. Let \((b_n)\) be a bounded representing sequence with \(b_n\in N_n\), and \(b_n=w_n|b_n|\) its polar decompositions in \(N_n\). Since \(\tau_n(w_n^*w_n)=\tau_n(w_nw_n^*)\), Lemma 2.3 gives \(y_n\in N_n\) with \(y_n^*y_n=1-w_n^*w_n\) and \(y_ny_n^*=1-w_nw_n^*\). Then \(U_n=w_n+y_n\) is a unitary (the cross terms vanish because \(w_n^*(1-w_nw_n^*)=0\) and \(w_n(1-w_n^*w_n)=0\)), and \(U_n|b_n|=b_n\) because \(w_n^*w_n\) is the support of \(|b_n|\). Hence \(b_n-U_n=U_n(|b_n|-1)\). For \(r\ge0\), \(|\sqrt r-1|\le|r-1|\), so functional calculus gives \[ \begin{aligned} &\|b_n-U_n\|_2=\||b_n|-1\|_2\\ &\quad \le\|b_n^*b_n-1\|_2 . \end{aligned} \] The right side tends to \(\|U^*U-1\|_2=0\) along \(\omega\), so \((U_n)\) represents \(U\). \(\square\)

5. Exercises

Exercise 5.1. Show that a unitary \(u\) of \(M\) is a Haar unitary if and only if \(\tau(g(u))=\int_0^1g(e^{2\pi ir})\,dr\) for every continuous function \(g\) on the unit circle.

Exercise 5.2. Show that a finite-dimensional tracial von Neumann algebra has no Haar unitary.

Exercise 5.3. Let \(h\) be a Haar unitary of \(M\). Show that the class of the sequence \((h^n)_{n\ge1}\) in \(M^\omega\) is a unitary orthogonal to \(M\) in \(L^2(M^\omega)\). Conclude that \(M^\omega\ne M\) when \(M\) has a Haar unitary.

Exercise 5.4. Let \(t\) be a Haar unitary free from \(D\) and \(k\ne0\). Show that \(t^k\) is a Haar unitary free from \(D\).

Exercise 5.5. Show that every projection of \(\prod_\omega N_n\) has a representing sequence of projections \(p_n\in N_n\). (Hint: \(|\chi_{[1/2,\infty)}(r)-r|\le2|r^2-r|\) for real \(r\).)

6. Solutions

5.1. The identity holds for \(g(z)=z^j\) exactly when \(\tau(u^j)=0\) for \(j\ne0\) (for \(j=0\) both sides are \(1\)). Both sides are linear and continuous in \(g\) for the supremum norm, and trigonometric polynomials are dense in the continuous functions on the circle.

5.2. If \(M\) is finite-dimensional, \(u\) has finitely many spectral projections \(p_1,\dots,p_r\) with eigenvalues \(\lambda_i\), and \(\tau(g(u))=\sum_ig(\lambda_i)\tau(p_i)\). Choose a continuous \(g\ge0\) vanishing at every \(\lambda_i\) and positive elsewhere: the left side of Exercise 5.1 is \(0\), the right side is positive. So \(u\) is not Haar.

5.3. The class \(x\) of \((h^n)\) is a unitary, and for \(y\in M\), \(\langle x,y\rangle=\lim_\omega\tau(y^*h^n)=0\) by Lemma 2.5. If \(M^\omega=M\), then \(x\in M\) would be orthogonal to itself, while \(\|x\|_2=1\).

5.4. \(\tau((t^k)^n)=\tau(t^{kn})=0\) for \(n\ne0\), and (3.1) for \(t^k\) with exponents \(n_i\) is (3.1) for \(t\) with exponents \(kn_i\ne0\). Lemma 3.2 applies.

5.5. Take a self-adjoint representing sequence \((y_n)\) of the projection \(p\) (replace \(y_n\) by \((y_n+y_n^*)/2\)), and \(p_n=\chi_{[1/2,\infty)}(y_n)\). The scalar inequality and functional calculus give \(\|p_n-y_n\|_2\le2\|y_n^2-y_n\|_2\), which tends to \(\|p^2-p\|_2=0\) along \(\omega\). (For \(r\ge\frac12\) the inequality reads \(|1-r|\le2r|r-1|\); for \(r<\frac12\) it reads \(|r|\le2|r|\,|r-1|\).)

References