Polar decomposition of a circular element

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

A circular element $\gamma$ has a polar decomposition $\gamma=v|\gamma|$ with a unitary $v$ (Matrices with semicircular and circular entries, Proposition 3.1). This lesson proves Voiculescu's theorem that $v$ is a Haar unitary and that $v$ and $|\gamma|$ are free (Theorem 3.1). The proof compares $\gamma$ with $\gamma w^*$, where $w$ is a Haar unitary free from $\gamma$: the two have the same $*$-distribution, because the semicircular matrix containing $\gamma$ is free from $2\times2$ matrices over the algebra of $w$; and the polar parts of $\gamma w^*$ are visibly a free Haar unitary times a conjugate of $|\gamma|$.

Section 1 shows that equal distributions give isomorphic von Neumann algebras and transports polar decompositions. Section 2 collects three facts about free Haar unitaries. Section 3 proves the theorem and the form in which the last lesson uses it.

We use: Lemmas 1.2, 1.3, 1.4 and 2.4 and Corollary 2.5 of Free independence and Haar tuples; Lemma 3.1 and Theorem 4.2 of Semicircular elements on the full Fock space; Theorem 2.1 and Proposition 3.1 of the previous lesson; and the uniqueness of the polar decomposition, Lemma 7.1 of the double commutant lesson.

1. Equal distributions

Let $(M_1,\tau_1)$ and $(M_2,\tau_2)$ be von Neumann algebras with faithful normal tracial states. Elements $x\in M_1$ and $y\in M_2$ have the same $*$-distribution if $\tau_1(P(x,x^*))=\tau_2(P(y,y^*))$ for every noncommutative polynomial $P$ in two variables.

Lemma 1.1. If $x$ and $y$ have the same $*$-distribution, there is a unique normal $*$-isomorphism $\pi\colon W^*(x)\to W^*(y)$ with $\pi(x)=y$, and $\tau_2\circ\pi=\tau_1$. If moreover $x=v|x|$ with a unitary $v$ and the kernel projection of $x$ is $0$, then $y=\pi(v)\,|y|$ is the polar decomposition of $y$, with $\pi(v)$ unitary and $\pi(|x|)=|y|$.

Proof. Let $\mathcal K_1$ be the closure of $W^*(x)\Omega_1$ in $L^2(M_1)$, and similarly $\mathcal K_2$. Left multiplication represents $W^*(x)$ faithfully and normally on $\mathcal K_1$ ($\Omega_1$ is separating, $\tau_1$ being faithful). The map $P(x,x^*)\Omega_1\mapsto P(y,y^*)\Omega_2$ is well defined and isometric, since $\|P(x,x^*)\Omega_1\|^2=\tau_1(P^*P(x,x^*))$ is determined by the $*$-distribution. Its domain and range are dense in $\mathcal K_1$ and $\mathcal K_2$ (Lemma 1.4 of the first lesson, applied in $W^*(x)$ and $W^*(y)$), so it extends to a unitary $V\colon\mathcal K_1\to\mathcal K_2$ with $V\Omega_1=\Omega_2$ and $VL_xV^*=L_y$. Conjugation by $V$ maps the von Neumann algebra generated by $L_x$ onto the one generated by $L_y$; composing with the left representations gives $\pi$. It preserves the traces because $V\Omega_1=\Omega_2$. It is unique because a normal $*$-homomorphism is determined by its value on a generator.

Since $|x|=(x^*x)^{1/2}$ is a norm limit of polynomials in $x^*x$, $\pi(|x|)=|y|$; and $\pi(v)$ is unitary with $y=\pi(v)\pi(|x|)$. The kernel projection $1_{\{0\}}(|x|)$ is $0$, and $\pi$ preserves traces of polynomials in $|x|$, hence the distribution of $|x|$; so the kernel projection of $|y|$ has trace $0$ and vanishes. By Lemma 7.1 of the double commutant lesson, $y=\pi(v)|y|$ is the polar decomposition. $\square$

2. Free Haar unitaries

In this section $(M,\tau)$ is a von Neumann algebra with a faithful normal tracial state.

Lemma 2.1 (conjugation by a free Haar unitary). Let $v\in M$ be a Haar unitary and $\mathcal A\subseteq M$ a von Neumann subalgebra such that $W^*(v)$ and $\mathcal A$ are free. For von Neumann subalgebras $P,Q\subseteq\mathcal A$, the algebras $P$ and $vQv^*$ are free.

Proof. The centred elements of $vQv^*$ are the $vbv^*$ with $b\in Q$ centred. A reduced product of centred elements of $P$ and $vQv^*$, written out, reads $\cdots a_1\,v\,b_1\,v^*\,a_2\,v\,b_2\,v^*\cdots$ with centred $a_k\in P\subseteq\mathcal A$ and $b_k\in Q\subseteq\mathcal A$. It alternates between centred elements of $\mathcal A$ and the centred elements $v$, $v^*$ of $W^*(v)$, so its trace is $0$. $\square$

Lemma 2.2 (a free Haar unitary absorbs a unitary). Let $w\in M$ be a Haar unitary, $\mathcal A\subseteq M$ a von Neumann subalgebra free from $W^*(w)$, and $v\in\mathcal A$ a unitary. Then $wv$ is a Haar unitary and $W^*(wv)$ is free from $\mathcal A$.

Proof. Consider a product whose factors alternate between powers $(wv)^k$, $k\ne0$, and centred elements $a_j\in\mathcal A$, beginning and ending with either kind and containing at least one power; for $k<0$, $(wv)^k=(v^*w^*)^{|k|}$. Multiply out: the product becomes an alternating product of letters $w^{\pm1}$ and elements of $\mathcal A$ (each maximal block of consecutive elements of $\mathcal A$ being multiplied into one element). Between two letters of the same sign the block is $v$, $v^*$, $va_j$ or $a_jv^*$, and between letters of opposite signs it is $va_jv^*$ (after $w$, before $w^*$) or $a_j$ (after $w^*$, before $w$), which are centred. Write every block that is not centred as its trace plus its centred part and expand. Only blocks between letters of the same sign, and a first or last block, can fail to be centred. Replacing blocks between letters of the same sign by scalars merges runs of letters of one sign into powers $w^{\pm j}$, $j\ge1$, which are centred; replacing a first or last block by a scalar just removes it. So every term is a product alternating between nonzero powers of $w$ and centred elements of $\mathcal A$, with at least one power of $w$, and has trace $0$ by freeness. The product consisting of a single power gives $\tau((wv)^k)=0$ for $k\ne0$. Since the centred elements of the $*$-algebra generated by $wv$ are the combinations of powers $(wv)^k$, $k\ne0$, the $*$-algebras are free, and so are the von Neumann algebras (Lemma 1.2(1) of the first lesson). $\square$

Lemma 2.3 (changing a free Haar basis). Let $v_1,\ldots,v_r\in M$ be Haar unitaries with $W^*(v_1),\ldots,W^*(v_r)$ free. Then $v_1$, $v_2v_1^*,\ldots,v_rv_1^*$ are Haar unitaries, their von Neumann algebras are free, and together they generate $W^*(v_1,\ldots,v_r)$.

Proof. As in Lemma 1.3 of Semicircular elements on the full Fock space, $(v_1,\ldots,v_r)$ is a freely generating Haar tuple of $W^*(v_1,\ldots,v_r)$. The endomorphism of $\mathbb F_r$ that fixes $x_1$ and sends $x_l$ to $x_lx_1^{-1}$ for $l\ge2$ is an automorphism (its inverse sends $x_l$ to $x_lx_1$). Corollary 2.5(2)-(3) of the first lesson gives the claim. $\square$

3. The theorem

Theorem 3.1 (Voiculescu). Let $\gamma$ be the circular element of a pair of indices in the semicircular algebra $\mathcal S$ of a finite orthonormal basis, and $\gamma=v|\gamma|$ its polar decomposition. Then $v$ is a Haar unitary, and $W^*(v)$ and $W^*(|\gamma|)$ are free. The same holds for every element, in any von Neumann algebra with a faithful normal tracial state, with the same $*$-distribution as $\gamma$.

Proof. The circular elements of all pairs of indices, in the semicircular algebras of all bases, have the same $*$-distribution: by Theorem 4.2 of the lesson on the Fock space, their two semicircular parts are free and semicircular, and Lemma 1.2(3) of the first lesson determines the mixed moments. By Lemma 1.1, the statement is invariant under passage to an element with the same $*$-distribution: $\pi$ carries the polar parts and preserves all traces. So we may take the basis $e_1,\ldots,e_5$, $\gamma$ the circular element of the pair $(3,4)$, and the semicircular matrix $X$ of the quadruple $(1,2,3,4)$, whose $(1,2)$ entry is $\frac1{\sqrt2}\gamma$.

Let $J=\{5\}$ and $w=e^{2\pi iF_\sigma(s_5)}\in\mathcal S_J$, a Haar unitary (Lemma 1.2 of the lesson on the Fock space). By Theorem 2.1 of the previous lesson, $W^*(X)$ and $M_2(\mathcal S_J)$ are free, and $U=\operatorname{diag}(1,w)\in M_2(\mathcal S_J)$ is unitary. By Lemma 1.3 of the first lesson, $UXU^*$ has the same distribution as $X$ jointly with $M_2(\mathcal S_J)$, in particular jointly with the matrix units. The $(1,2)$ entry of $UXU^*$ is $\frac1{\sqrt2}\gamma w^*$ (Exercise 4.4 of the previous lesson). For $T\in M_2(\mathcal S)$, the matrix $Z_1=E_{11}TE_{21}$ has $(1,1)$ entry $T_{12}$ and all other entries $0$, and $Z_*=E_{12}T^*E_{11}$ has $(1,1)$ entry $(T^*)_{21}=(T_{12})^*$ and all other entries $0$. Hence, for $\epsilon_1,\ldots,\epsilon_k\in\{1,*\}$, $\varphi\big((T_{12})^{\epsilon_1}\cdots(T_{12})^{\epsilon_k}\big)=2\,\psi\big(Z_{\epsilon_1}\cdots Z_{\epsilon_k}\big),$ a trace of a product of $T$, $T^*$ and matrix units. Applied to $T=X$ and $T=UXU^*$, this shows that every $*$-moment of $\gamma w^*$ equals the corresponding $*$-moment of $\gamma$: they have the same $*$-distribution.

By Proposition 3.1 of the previous lesson, $v$ is unitary and the kernel projection of $\gamma$ is $0$. Lemma 1.1 gives a trace-preserving isomorphism $\pi\colon W^*(\gamma)\to W^*(\gamma w^*)$ with $\pi(\gamma)=\gamma w^*$, $\pi(|\gamma|)=|\gamma w^*|$, and $\pi(v)$ the polar part of $\gamma w^*$. Now $\gamma w^*=v|\gamma|w^*=(vw^*)\,(w|\gamma|w^*),$ with $vw^*$ unitary and $w|\gamma|w^*\ge0$ with zero kernel projection; by Lemma 7.1 of the double commutant lesson this is the polar decomposition, so $\pi(v)=vw^*$ and $\pi(|\gamma|)=w|\gamma|w^*$. For every noncommutative polynomial $P$, using that $\pi$ and conjugation by $w^*$ preserve the trace, $\begin{aligned} &\varphi(P(v,v^*,|\gamma|))\\ &\quad =\varphi\big(P(vw^*,wv^*,w|\gamma|w^*)\big)\\ &\quad =\varphi\big(P(w^*v,v^*w,|\gamma|)\big). \end{aligned}$ So $(v,|\gamma|)$ has the same joint distribution as $(w^*v,|\gamma|)$.

The algebras $\mathcal S_{\{3,4\}}=W^*(\gamma)$, which contains $v$ and $|\gamma|$, and $\mathcal S_{\{5\}}\ni w$ are free (Section 1 of the previous lesson). By Lemma 2.2, with the Haar unitary $w^*$ and $\mathcal A=W^*(\gamma)$, the unitary $w^*v$ is a Haar unitary and $W^*(w^*v)$ is free from $W^*(\gamma)\ni|\gamma|$. Being a Haar unitary, and freeness of the $*$-algebras generated by two elements, are statements about their joint distribution. So $v$ is a Haar unitary and the $*$-algebras generated by $v$ and by $|\gamma|$ are free; by Lemma 1.2(1) of the first lesson, so are $W^*(v)$ and $W^*(|\gamma|)$. $\square$

The last lesson uses the theorem in the following form.

Corollary 3.2. Let $X_1,\ldots,X_r$ be the semicircular matrices of pairwise disjoint quadruples $(j_1^l,j_2^l,j_3^l,j_4^l)$ of indices, $\gamma_l$ their circular entries with polar decompositions $\gamma_l=v_l|\gamma_l|$, and $J$ a set of indices disjoint from the quadruples. Then the von Neumann algebras $\mathcal S_J$, $W^*(s_{j_1^l})$, $W^*(s_{j_2^l})$, $W^*(v_l)$ and $W^*(|\gamma_l|)$ for $1\le l\le r$ form a free family, and $v_1,\ldots,v_r$ are Haar unitaries.

Proof. The algebras $\mathcal S_J$, $\mathcal S_{\{j_1^l\}}$, $\mathcal S_{\{j_2^l\}}$ and $\mathcal S_{\{j_3^l,j_4^l\}}=W^*(\gamma_l)$ are attached to disjoint sets of indices and form a free family. By Theorem 3.1, inside each $W^*(\gamma_l)$ the algebras $W^*(v_l)$ and $W^*(|\gamma_l|)$ are free. Lemma 3.1 of the lesson on the Fock space (associativity) gives the claim. $\square$

4. Exercises

Exercise 4.1. Let $w$ be a Haar unitary free from a von Neumann subalgebra $\mathcal A$. Show that $\mathcal A$ and $w\mathcal Aw^*$ are free, and that $W^*(w)$ and $w\mathcal Aw^*$ are free.

Exercise 4.2. Let $v$ be a Haar unitary and $q\ge0$ free from $v$. Compute $\varphi((vq)^2)$ and $\varphi((vq)(vq)^*)$, and compare with $\varphi(\gamma^2)=0$ and $\varphi(\gamma\gamma^*)=1$ for a circular $\gamma$.

Exercise 4.3. In Lemma 2.2, show that $W^*(wv)$ need not be free from $W^*(w)$.

5. Solutions

4.1. The first statement is Lemma 2.1 with $P=Q=\mathcal A$. For the second, the inner automorphism $\operatorname{Ad}w$ of $M$ preserves the trace, fixes $W^*(w)$ and maps $\mathcal A$ onto $w\mathcal Aw^*$; it carries the free pair $W^*(w)$, $\mathcal A$ to the pair $W^*(w)$, $w\mathcal Aw^*$, and freeness is preserved because reduced products of centred elements are mapped to reduced products of centred elements.

4.2. $\varphi((vq)^2)=\varphi(vqvq)$: write $q=\varphi(q)+q^\circ$; every term is a product alternating between centred powers of $v$ (here $v$, $v^2$) and centred elements of $W^*(q)$, so $\varphi((vq)^2)=0$. And $\varphi(vq\,qv^*)=\varphi(q^2)$. For $q=|\gamma|$, $\varphi(q^2)=\varphi(\gamma^*\gamma)=1$, consistent with Theorem 3.1.

4.3. Take $\mathcal A=\mathbb C1$ and $v=1$: then $wv=w$ is not free from itself, since $\tau(w\,w^*)=1\ne0$.

References

[OAI] OpenAI, An isomorphism of the free group factors (September 23, 2026), OpenAI Math Release preprint. 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
[Sp] R. Speicher, Lecture notes on "Free probability theory", arXiv 1908.08125, the remark on the polar decomposition of circular elements. https://arxiv.org/abs/1908.08125