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Matrices with semicircular and circular entries

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

This lesson builds the elements from which the last lesson assembles \(L(\mathbb F_n)\) inside \(2\times2\) matrices. A semicircular matrix has free semicircular elements on the diagonal and a circular element and its adjoint off the diagonal. Theorem 2.1 shows, with the freeness criterion of Semicircular elements on the full Fock space, that such a matrix is semicircular for the normalized trace and free from all \(2\times2\) matrices over the algebra generated by the remaining semicircular elements. Section 3 records the properties of circular elements needed later: they generate copies of \(L(\mathbb F_2)\), \(c^*c\) has the limit law \(\nu\) of the third lesson, and \(c\) has trivial kernel, so its polar part is a unitary.

We use: Lemmas 3.1, 4.1, Theorem 3.2 and Theorem 4.2 of the previous lesson, with its Lemmas 1.1-1.3; Lemmas 1.2 and 1.3 of Free independence and Haar tuples; measures on a compact set are determined by their integrals of continuous functions, Corollary 1.2 of the spectral theorem lesson; the polar decomposition inside a von Neumann algebra and its uniqueness, Lemma 7.1 and Proposition 7.2 of the double commutant lesson; and the Weierstrass approximation theorem, Corollary 14.3 of the Stone–Weierstrass lesson.

1. Setting

Let \(e_1,\ldots,e_m\) be an orthonormal basis of a Hilbert space \(H\), \(s_j=s(e_j)\) on the full Fock space \(\mathcal F(H)\), and \(\mathcal S=W^*(s_1,\ldots,s_m)\) the semicircular algebra, with its faithful normal tracial state \(\varphi=\langle\,\cdot\,\Omega,\Omega\rangle\) (Theorem 4.2 of the previous lesson). For a set \(J\) of indices put \(\mathcal S_J=W^*(s_j:j\in J)\). The algebras \(\mathcal S_{J_1},\ldots,\mathcal S_{J_r}\) attached to disjoint sets of indices form a free family, by Theorem 4.2(3) there and Lemma 1.2(1)-(2) of the first lesson.

Let \(E_{pq}\) (\(p,q\in\{1,2\}\)) be the matrix units of \(M_2(\mathbb C)\). The algebra \(M_2(\mathcal S)\) of \(2\times2\) matrices over \(\mathcal S\) acts on \(\mathbb C^2\otimes\mathcal F(H)\), and \[ \begin{gathered} \psi(T)=\tfrac12\big(\varphi(T_{11})+\varphi(T_{22})\big)\\ (T=(T_{pq})\in M_2(\mathcal S)) \end{gathered} \] is a faithful normal tracial state on it. To write \(\psi\) as a vector state, let \(\varepsilon_1,\varepsilon_2\) be the standard basis of \(\mathbb C^2\), let operators \(T\) on \(\mathbb C^2\otimes\mathcal F(H)\) act on \(K=\mathbb C^2\otimes\mathcal F(H)\otimes\mathbb C^2\) as \(T\otimes1\), and put \(\xi=\frac1{\sqrt2}(\varepsilon_1\otimes\Omega\otimes\varepsilon_1+\varepsilon_2\otimes\Omega\otimes\varepsilon_2)\). Then \(\langle(T\otimes1)\xi,\xi\rangle=\psi(T)\).

For distinct indices \(j_1,j_2,j_3,j_4\) put \[ \begin{gathered} \gamma=\tfrac1{\sqrt2}\big(s_{j_3}+is_{j_4}\big),\\ X=\frac1{\sqrt2}\begin{pmatrix}s_{j_1}&\gamma\\ \gamma^*&s_{j_2}\end{pmatrix}\in M_2(\mathcal S). \end{gathered} \] We call \(\gamma\) the circular element of the pair \((j_3,j_4)\) and \(X\) the semicircular matrix of the quadruple \((j_1,j_2,j_3,j_4)\).

2. Freeness from matrices over the other variables

Theorem 2.1. Let \(X\) be the semicircular matrix of a quadruple \((j_1,j_2,j_3,j_4)\) and \(J\) a set of indices disjoint from it. Then \(X\) is semicircular with respect to \(\psi\), and \(W^*(X)\) and \(M_2(\mathcal S_J)\) are free in \((M_2(\mathcal S),\psi)\).

Proof. Put \(g_{11}=e_{j_1}\), \(g_{22}=e_{j_2}\), \(g_{12}=\frac1{\sqrt2}(e_{j_3}+ie_{j_4})\) and \(g_{21}=\frac1{\sqrt2}(e_{j_3}-ie_{j_4})\). These four vectors are orthonormal and orthogonal to \(K_0=\operatorname{span}\{e_j:j\in J\}\) (Exercise 5.4 of the previous lesson), and \(\gamma=\ell(g_{12})+\ell(g_{21})^*\). Let \[ L=\frac1{\sqrt2}\sum_{p,q=1}^2E_{pq}\otimes\ell(g_{pq})\otimes1 , \] an operator on \(K\). Its adjoint is \(\frac1{\sqrt2}\sum_{p,q}E_{qp}\otimes\ell(g_{pq})^*\otimes1\), and the \((p,q)\) entry of \(L+L^*\) is \(\frac1{\sqrt2}(\ell(g_{pq})+\ell(g_{qp})^*)\); so \(L+L^*=X\otimes1\). We check the conditions of Theorem 3.2 of the previous lesson for \(\mathcal D=M_2(\mathcal S_J)\otimes1\) and the vector \(\xi\).

(a) Let \(d=\sum_{k,l}E_{kl}\otimes b_{kl}\) with \(b_{kl}\in\mathcal S_J\subseteq\mathcal S(K_0)\). By Lemma 4.1(2) of the previous lesson, \(\ell(g_{pq})^*b_{kl}\ell(g_{p'q'})=\varphi(b_{kl})\langle g_{p'q'},g_{pq}\rangle1\), which vanishes unless \((p',q')=(p,q)\). Hence \[ \begin{aligned} &L^*(d\otimes1)L\\ &\quad =\frac12\sum_{p,q,k,l}\varphi(b_{kl})\,E_{qp}E_{kl}E_{pq}\otimes1\otimes1\\ &\quad =\frac12\sum_{p,q}\varphi(b_{pp})E_{qq}\otimes1\otimes1=\psi(d)\,1 . \end{aligned} \] (b) \(L^*\xi=0\), because \(\ell(g)^*\Omega=0\).

(c) \((d\otimes1)\xi=\frac1{\sqrt2}\sum_{k,p}\varepsilon_p\otimes b_{pk}\Omega\otimes\varepsilon_k\), and \(b_{pk}\Omega\in\mathcal F(K_0)\) is annihilated by every \(\ell(g_{pq})^*\) (Lemma 4.1(1) there); so \(L^*(d\otimes1)\xi=0\).

By Theorem 3.2 there, \(\psi(X^k)=\int t^k\,d\sigma\) for all \(k\), and the polynomials in \(X\) and \(M_2(\mathcal S_J)\) are free with respect to \(\psi\). Since \(L^*L=1\), \(\|X\|\le2\); the distribution \(\mu_X\) and \(\sigma\) are measures on \([-2,2]\) with the same moments, so they agree on continuous functions (Weierstrass) and are equal (Corollary 1.2 of the spectral theorem lesson). Freeness passes to \(W^*(X)\) by Lemma 1.2(1) of the first lesson. \(\square\)

Corollary 2.2. Let \(X_1,\ldots,X_r\) be the semicircular matrices of pairwise disjoint quadruples of indices, and \(J\) a set of indices disjoint from all of them. Then \(W^*(X_1),\ldots,W^*(X_r),M_2(\mathcal S_J)\) form a free family in \((M_2(\mathcal S),\psi)\).

Proof. By induction on \(r\); the case \(r=1\) is Theorem 2.1. Let \(J'\) be \(J\) together with the quadruples of \(X_1,\ldots,X_{r-1}\). By Theorem 2.1, \(W^*(X_r)\) is free from \(M_2(\mathcal S_{J'})\), which contains \(W^*(X_1),\ldots,W^*(X_{r-1})\) and \(M_2(\mathcal S_J)\). By the induction hypothesis these form a free family, and Lemma 3.1 of the previous lesson gives the claim. \(\square\)

3. Circular elements

Proposition 3.1. Let \(\gamma\) be the circular element of a pair \((j_3,j_4)\).

  1. \(W^*(\gamma)=W^*(s_{j_3},s_{j_4})\), and it is isomorphic to \(L(\mathbb F_2)\) by a trace-preserving normal isomorphism.
  2. \(\varphi(\gamma^k)=0\) for \(k\ge1\), and \(\gamma^*\gamma\) has the distribution \(\nu\) of the third lesson: \(\varphi(f(\gamma^*\gamma))=\Lambda(f)\) for continuous \(f\) on \([0,4]\).
  3. The kernel projection \(1_{\{0\}}(|\gamma|)\) of \(\gamma\) is \(0\), the polar decomposition is \(\gamma=v|\gamma|\) with a unitary \(v\in W^*(\gamma)\), and the distribution function of \(|\gamma|=(\gamma^*\gamma)^{1/2}\) is continuous and strictly increasing on \([0,2]\).

Proof. (1) \(s_{j_3}=\frac1{\sqrt2}(\gamma+\gamma^*)\) and \(s_{j_4}=\frac1{i\sqrt2}(\gamma-\gamma^*)\). The two semicircular elements are free, and Lemma 1.2 of the previous lesson gives Haar unitaries generating \(W^*(s_{j_3})\) and \(W^*(s_{j_4})\); apply Lemma 1.3 there.

(2) With \(f=\frac1{\sqrt2}(e_{j_3}+ie_{j_4})\) and \(g=\frac1{\sqrt2}(e_{j_3}-ie_{j_4})\), \(\gamma=\ell(f)+\ell(g)^*\). Expand \(\gamma^k\) into words in \(\ell(f)\) and \(\ell(g)^*\) and apply them to \(\Omega\). Since \(\ell(f)f^{\otimes j}=f^{\otimes(j+1)}\) and \(\ell(g)^*\) annihilates \(\Omega\) and every \(f^{\otimes j}\) (\(g\perp f\)), every word containing \(\ell(g)^*\) gives \(0\), and \(\ell(f)^k\Omega=f^{\otimes k}\perp\Omega\). So \(\varphi(\gamma^k)=0\) for \(k\ge1\).

For \(\gamma^*\gamma\), consider \(Y=\begin{pmatrix}0&\gamma\\ \gamma^*&0\end{pmatrix}\in M_2(\mathcal S)\) and \(L_Y=E_{12}\otimes\ell(f)\otimes1+E_{21}\otimes\ell(g)\otimes1\). Then \(L_Y+L_Y^*=Y\otimes1\), and \[ \begin{aligned} &L_Y^*L_Y=E_{22}\otimes\ell(f)^*\ell(f)\otimes1\\ &\quad +E_{11}\otimes\ell(g)^*\ell(g)\otimes1=1, \end{aligned} \] the cross terms vanishing because \(E_{21}E_{21}=E_{12}E_{12}=0\). Also \(L_Y^*\xi=0\). Theorem 3.2 of the previous lesson, with \(\mathcal D=\mathbb C1\), gives \(\psi(Y^k)=\int t^k\,d\sigma\). Now \(Y^{2r}=\operatorname{diag}((\gamma\gamma^*)^r,(\gamma^*\gamma)^r)\), and \(\varphi((\gamma\gamma^*)^r)=\varphi((\gamma^*\gamma)^r)\) because \(\varphi\) is tracial, so \[ \begin{aligned} &\varphi((\gamma^*\gamma)^r)=\psi(Y^{2r})\\ &\quad =\int t^{2r}\,d\sigma(t)=\Lambda(x^r) \end{aligned} \] by Lemma 2.1(5) of the previous lesson. The distribution of \(\gamma^*\gamma\) lives on \([0,\|\gamma\|^2]\subseteq[0,4]\), as does \(\nu\), and the two have the same moments; as in the proof of Theorem 2.1 they are equal.

(3) The kernel projection of \(\gamma\) is \(1_{\{0\}}(\gamma^*\gamma)\), whose trace is \(\nu(\{0\})=0\); it vanishes because \(\varphi\) is faithful. By Proposition 7.2 of the double commutant lesson, \(\gamma=v|\gamma|\) with a partial isometry \(v\in W^*(\gamma)\) whose initial projection \(v^*v\) is the complement of the kernel projection, that is \(1\). In a von Neumann algebra with a faithful tracial state an isometry is unitary: \(1-vv^*\) is a projection of trace \(1-\varphi(v^*v)=0\). Finally, the distribution of \(|\gamma|\) is the image of \(\nu\) under \(x\mapsto\sqrt x\) (Lemma 1.1 of the previous lesson), whose distribution function \(t\mapsto\nu([0,t^2])\) is continuous and strictly increasing on \([0,2]\) by Lemma 2.1(5) there. \(\square\)

The distribution of \(|\gamma|\) has density \(\frac1\pi\sqrt{4-t^2}\) on \([0,2]\), the quarter-circle law (Exercise 4.3).

4. Exercises

Exercise 4.1. For a semicircular matrix \(X\), compute \(\psi(XE_{11}XE_{11})\) directly and check that it equals \(\psi(X^2)\psi(E_{11})^2\), as freeness of \(X\) and \(E_{11}\) requires.

Exercise 4.2. Show that a circular element \(\gamma\) is not normal, by computing \(\gamma^*\gamma\,\Omega\) and \(\gamma\gamma^*\Omega\).

Exercise 4.3. Show that the distribution of \(|\gamma|\) has density \(\frac1\pi\sqrt{4-t^2}\) on \([0,2]\).

Exercise 4.4. Let \(J\) be disjoint from the quadruple of \(X\) and \(u\in\mathcal S_J\) a unitary. Show that \(UXU^*\), with \(U=\operatorname{diag}(1,u)\), has the same distribution as \(X\) jointly with \(M_2(\mathcal S_J)\), and that its \((1,2)\) entry is \(\frac1{\sqrt2}\gamma u^*\).

5. Solutions

4.1. \(XE_{11}XE_{11}\) has \((1,1)\) entry \(X_{11}X_{11}=\frac12s_{j_1}^2\) and second column \(0\), so \(\psi(XE_{11}XE_{11})=\frac12\cdot\frac12\varphi(s_{j_1}^2)=\frac14\). On the other hand \(\psi(X^2)=1\) (the second moment of \(\sigma\)) and \(\psi(E_{11})=\frac12\). For free \(a,b\) with \(\psi(a)=0\), freeness gives \(\psi(abab)=\psi(a^2)\psi(b)^2\): expand \(b=\psi(b)+b^\circ\) and use Exercise 5.1 of the first lesson.

4.2. With \(\gamma=\ell(f)+\ell(g)^*\): \(\gamma\Omega=f\) and \(\gamma^*\gamma\Omega=\gamma^*f=g\otimes f+\Omega\), while \(\gamma^*\Omega=g\) and \(\gamma\gamma^*\Omega=f\otimes g+\Omega\). These differ.

4.3. For continuous \(f\), \(\varphi(f(|\gamma|))=\int f(\sqrt x)\,d\nu(x)\), which is \(\frac4\pi\int_0^{\pi/2}f(2\sin\vartheta)\cos^2\vartheta\,d\vartheta\). Substitute \(t=2\sin\vartheta\): \(dt=2\cos\vartheta\,d\vartheta\) and \(\cos\vartheta=\frac12\sqrt{4-t^2}\), so the integrand becomes \(\frac1\pi f(t)\sqrt{4-t^2}\,dt\).

4.4. \(W^*(X)\) and \(M_2(\mathcal S_J)\) are free (Theorem 2.1) and \(U\in M_2(\mathcal S_J)\) is unitary, so Lemma 1.3 of the first lesson applies. Since \(U\) is diagonal, \((UXU^*)_{12}=U_{11}X_{12}(U^*)_{22}=\frac1{\sqrt2}\gamma u^*\).

References