Open MathsAll courses

Spectral intertwining rigidity

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

Let \(\varphi\) be a faithful normal state with trivial centralizer and \(b\) a continuous action of \(\mathbb R\) by \(\varphi\)-preserving automorphisms. Suppose that every bounded sequence \(w_n\) which approximately scales \(\varphi\) by \(e^s\) intertwines each \(a\in M\) with \(b_s(a)\), asymptotically. This lesson proves the rigidity theorem of [OpenAI-recovery]: then \(b\) is trivial. The bicentralizer flow of a self-bicentralizing factor satisfies exactly this hypothesis, so the theorem shows that such a flow is trivial. The proof combines the bounded recovery theorem of the previous lesson with a spectral argument: the modular generator \(D=\log\Delta\) and the generator \(Q\) of \(b\) commute, and the hypothesis forces every two points \((r,p),(q,\ell)\) of their joint spectral support to satisfy \(e^{i(r\ell+qp)}=1\). This leaves only supports that are countable or lie on a coordinate axis, and each of these is ruled out or forces \(b\) to be trivial.

We use Modular averaging and bounded recovery (its notation, (1.1), Lemmas 1.1, 1.2, 2.1, 2.3 and Theorem 3.1), the results (B3)–(B6) listed in Ultraproducts and the asymptotic centralizer, Stone's theorem, and the spectral calculus of self-adjoint operators.

1. State-preserving actions

Let \(M\), \(\varphi\), \(H\), \(\xi\), \(J\), \(S\), \(D\), \(U_t\), \(L\), \(R\) be as in the previous lesson.

Lemma 1.1. Every automorphism \(\alpha\) of \(M\) with \(\varphi\circ\alpha=\varphi\) commutes with \(\sigma^\varphi_t\) for every \(t\).

Proof. The map \(V(a\xi)=\alpha(a)\xi\) is isometric, since \(\|\alpha(a)\xi\|^2=\varphi(\alpha(a^*a))=\|a\xi\|^2\), with dense range \(M\xi\); so it extends to a unitary. On the core \(M\xi\) of \(S\), \(VSV^*(\alpha(a)\xi)=VSa\xi\), which is \(\alpha(a)^*\xi=S(\alpha(a)\xi)\), and \(V\) maps \(M\xi\) onto itself; so \(VSV^*=S\) on the closures. Hence \(V\Delta V^*=(VSV^*)^*(VSV^*)=S^*S=\Delta\), and \(V\Delta^{it}V^*=\Delta^{it}\). Applying this to \(\alpha(a)\xi\) gives \(\alpha(\sigma_t(a))\xi=V\Delta^{it}a\xi\), which is \(\Delta^{it}\alpha(a)\xi=\sigma_t(\alpha(a))\xi\), and \(\xi\) is separating. \(\square\)

Let \(b\) be an action of \(\mathbb R\) on \(M\) by \(\varphi\)-preserving automorphisms, continuous for the \(u\)-topology. Put \(V_s(a\xi)=b_s(a)\xi\). These are unitaries forming a group, and \(s\mapsto V_s\) is strongly continuous: \(\|V_sa\xi-a\xi\|^2\) equals \(2\varphi(a^*a)-2\operatorname{Re}\varphi(a^*b_s(a))\), and with \(\rho=\varphi(a^*\,\cdot\,)\), \(|\rho(b_s(a))-\rho(a)|\le\|\rho\circ b_s-\rho\|\|a\|\to0\); density of \(M\xi\) gives the rest. By Lemma 1.1, \(V_sU_t(a\xi)=b_s(\sigma_t(a))\xi\), which is \(\sigma_t(b_s(a))\xi=U_tV_s(a\xi)\), so the groups commute. By Stone's theorem \(V_s=e^{isQ}\) for a self-adjoint \(Q\). The spectral projections of \(D\) lie in the von Neumann algebra generated by the \(U_t\), and those of \(Q\) in the one generated by the \(V_s\); so all spectral projections \(1_A(D)\) and \(1_B(Q)\) commute.

2. From bounded recovery to interval estimates

From now on, \(M_\varphi=\mathbb C1\), and \(b\) satisfies:

(FP) for every \(s\in\mathbb R\), every bounded sequence \((w_n)\) in \(M\) and every \(a\in M\), \[ \begin{aligned} &\|\varphi w_n-e^sw_n\varphi\|\to0\quad\Longrightarrow\quad w_na\\ &\quad -b_s(a)w_n\to0\ \text{strongly}. \end{aligned} \]

Lemma 2.1 (estimate at a point). Let \(a,a^\flat\in M\) with \(a\xi=\xi a^\flat\), \(s\in\mathbb R\), and \(T_s=R(a^\flat)-L(b_s(a))\). If \(h_n\) are unit vectors with \(D\)-spectral support in \([s-d_n,s+d_n]\), \(d_n\to0\), then \(\|t\mapsto T_sU_th_n\|_{\mathrm{av}}\to0\).

Proof. Otherwise a subsequence has \(\liminf m_t\|T_sU_th_n\|^2>0\), and the bounded recovery theorem gives a bounded sequence \(v_j\in M\), with \(v_j\xi\) of \(D\)-spectral support shrinking to \(s\), and \(\|T_sv_j\xi\|\ge\eta>0\). By Lemma 1.1 of the previous lesson, \(\|\varphi v_j-e^sv_j\varphi\|\to0\), so (FP) gives \((v_ja-b_s(a)v_j)\xi\to0\). But left and right multiplications commute, so \(R(a^\flat)v_j\xi=v_j(\xi a^\flat)=v_ja\xi\), and \(T_sv_j\xi=(v_ja-b_s(a)v_j)\xi\). This is a contradiction. \(\square\)

Lemma 2.2 (estimate on an interval). Let \(a,a^\flat\in M\) with \(a\xi=\xi a^\flat\), \(I\subset\mathbb R\) a compact interval, \(c\in\mathbb C\), and \(h\in1_I(D)H\). Then \[ \begin{aligned} &\|t\mapsto(R(a^\flat)-cL(a))U_th\|_{\mathrm{av}}\\ &\quad \le\|h\|\sup_{s\in I}\|(b_s(a)-ca)\xi\| . \end{aligned} \]

Proof. Put \(S=R(a^\flat)-cL(a)\), \(C=P(S^*S)\) (Lemma 2.3 of the previous lesson) and \(B=\sup_{s\in I}\|(b_s(a)-ca)\xi\|\), which is at most \((1+|c|)\|a\|\). The left side squared is \(\langle Ch,h\rangle\). Suppose there is a unit vector \(h\in1_I(D)H\) with \(\langle Ch,h\rangle=B^2+\varepsilon\), \(\varepsilon>0\). For each \(n\) partition \(I\) into finitely many disjoint intervals \(J\) of length at most \(1/n\). Since \(C\) commutes with the spectral projections of \(D\), \(\langle Ch,h\rangle=\sum_J\langle C1_J(D)h,1_J(D)h\rangle\) and \(\sum_J\|1_J(D)h\|^2=1\); so for some \(J=J_n\) the unit vector \(h_n=1_{J_n}(D)h/\|1_{J_n}(D)h\|\) satisfies \(\langle Ch_n,h_n\rangle\ge B^2+\varepsilon\). Passing to a subsequence, the intervals \(J_n\) shrink to a point \(s\in I\), so \(h_n\) has spectral support in \([s-d_n,s+d_n]\) with \(d_n\to0\).

Put \(z=b_s(a)-ca\); then \(S=T_s+L(z)\). By Lemma 2.1, \(\|t\mapsto T_sU_th_n\|_{\mathrm{av}}\to0\), and by Lemma 2.3 of the previous lesson \[ \begin{aligned} &\|t\mapsto L(z)U_th_n\|_{\mathrm{av}}^2=\langle P(L(z^*z))h_n,h_n\rangle\\ &\quad =\varphi(z^*z)=\|z\xi\|^2\le B^2 . \end{aligned} \] So \(\limsup_n\langle Ch_n,h_n\rangle^{1/2}\le B\), a contradiction. \(\square\)

3. The joint spectral support

Let \(\mathcal S\) be the set of points \((r,p)\in\mathbb R^2\) such that \(1_I(D)1_J(Q)\ne0\) for all open intervals \(I\ni r\) and \(J\ni p\). Its complement is open, since a rectangle \(I\times J\) with \(1_I(D)1_J(Q)=0\) contains no point of \(\mathcal S\).

Lemma 3.1. Let \(K,L\) be compact intervals with \((K\times L)\cap\mathcal S=\emptyset\). Then \(1_K(D)1_L(Q)=0\).

Proof. Every point of \(K\times L\) has an open rectangle neighbourhood \(I\times J\) with \(1_I(D)1_J(Q)=0\). Finitely many cover \(K\times L\); let \(\delta>0\) be a Lebesgue number of this cover. Partition \(K\) and \(L\) into disjoint intervals \(A_\alpha\), \(B_\beta\) of length less than \(\delta/2\). Each \(A_\alpha\times B_\beta\) lies in one covering rectangle \(I\times J\), and since all these projections commute, \(1_{A_\alpha}(D)1_{B_\beta}(Q)\le1_I(D)1_J(Q)=0\). Summing, \(1_K(D)1_L(Q)=\sum_{\alpha,\beta}1_{A_\alpha}(D)1_{B_\beta}(Q)=0\). \(\square\)

Lemma 3.2.

  1. If \(\mathcal S\subset\mathbb R\times\{0\}\), then \(Q=0\).
  2. If \(\mathcal S\subset\{0\}\times\mathbb R\), then \(D=0\).
  3. If \(\mathcal S\) is countable, then \(D\) has pure point spectrum.

Proof. (1) For a compact interval \(L\) not containing \(0\) and \(K=[-N,N]\), Lemma 3.1 gives \(1_{[-N,N]}(D)1_L(Q)=0\); letting \(N\to\infty\), \(1_L(Q)=0\). Hence \(1_{\mathbb R\setminus\{0\}}(Q)=0\) and \(Q=0\). (2) is the same with the roles exchanged.

(3) Let \(\Lambda\) be the countable set of first coordinates of points of \(\mathcal S\). Fix \(N\), \(h\in H\), and the finite positive measure \(\mu(A)=\langle1_A(D)1_{[-N,N]}(Q)h,h\rangle\). If \(r\notin\Lambda\), the compact set \(\{r\}\times[-N,N]\) is disjoint from the closed set \(\mathcal S\), hence so is \([r-\delta,r+\delta]\times[-N,N]\) for some \(\delta>0\), and \(\mu([r-\delta,r+\delta])=0\) by Lemma 3.1. Countably many of these open neighbourhoods cover \(\mathbb R\setminus\Lambda\), so \(\mu(\mathbb R\setminus\Lambda)=0\). As \(h\) is arbitrary, \(1_{\mathbb R\setminus\Lambda}(D)1_{[-N,N]}(Q)=0\), and letting \(N\to\infty\), \(1_{\mathbb R\setminus\Lambda}(D)=0\). So \(H\) is the closed span of the eigenspaces \(1_{\{r\}}(D)H\), \(r\in\Lambda\). \(\square\)

Lemma 3.3 (joint spectral pairing). For all \((r,p),(q,\ell)\in\mathcal S\), \[ e^{i(r\ell+qp)}=1 . \]

Proof. Fix \(\delta>0\) and write \(I_v=[v-\delta,v+\delta]\). Localized elements. Choose smooth functions \(f,g\ge0\) supported in the interiors of \(I_r\), \(I_p\) and equal to \(1\) on smaller open intervals \(I'\ni r\), \(J'\ni p\). By the definition of \(\mathcal S\), the projection \(1_{I'}(D)1_{J'}(Q)\) is nonzero, and \(f(D)g(Q)\) restricts to the identity on its range; so \(f(D)g(Q)\ne0\), and by density there is \(y\in M\) with \(f(D)g(Q)y\xi\ne0\). Write \(g(u)=\int\hat g(s)e^{isu}\,ds\) with \(\hat g\in L^1\). The element \(y'=\int\hat g(s)b_s(y)\,ds\) of \(M\) satisfies \(y'\xi=g(Q)y\xi\) (Lemma 1.2 of the previous lesson and the remark after it), and \(x=y'_f\) satisfies \(x\xi=f(D)g(Q)y\xi\ne0\). So \(x\xi\in1_{I_r}(D)1_{I_p}(Q)H\). In the same way choose \(a\in M\) with \(0\ne a\xi\in1_{I_q}(D)1_{I_\ell}(Q)H\). Both vectors have compact \(D\)-spectral support, so there are \(x^\flat,a^\flat\in M\) with \(x\xi=\xi x^\flat\) and \(a\xi=\xi a^\flat\).

Two estimates. Put \(c=e^{ir\ell}\), \(d=e^{iqp}\), \(H_0=\|a\xi\|\|x\xi\|>0\), and \[ F(t)=\sigma_t(x)a\xi,\qquad G(t)=a\sigma_t(x)\xi . \] Since \(R(a^\flat)\sigma_t(x)\xi=\sigma_t(x)\xi a^\flat=\sigma_t(x)a\xi\), we have \(F(t)-cG(t)=(R(a^\flat)-cL(a))U_t(x\xi)\). Lemma 2.2 with \(h=x\xi\in1_{I_r}(D)H\), together with \((b_s(a)-ca)\xi=(e^{isQ}-c)a\xi\) and \(a\xi\in1_{I_\ell}(Q)H\), gives \[ \begin{aligned} \|F-cG\|_{\mathrm{av}}&\le\|x\xi\|\sup_{s\in I_r}\|(e^{isQ}-c)a\xi\|\\ &\le H_0\sup_{s\in I_r,\,u\in I_\ell}|e^{isu}-c| . \end{aligned} \] Exchanging the roles of \(a\) and \(x\), \[ \begin{aligned} &\|t\mapsto\sigma_t(a)x\xi-d\,x\sigma_t(a)\xi\|_{\mathrm{av}}\\ &\quad \le H_0\sup_{s\in I_q,\,u\in I_p}|e^{isu}-d| . \end{aligned} \] Now \(U_t\big(\sigma_{-t}(a)x\xi-d\,x\sigma_{-t}(a)\xi\big)\) equals \(G(t)-dF(t)\), \(U_t\) is unitary, and the mean is invariant under \(t\mapsto-t\); so \(\|G-dF\|_{\mathrm{av}}\) satisfies the same bound.

Normalization. By reflection invariance and Lemma 2.3 of the previous lesson, \[ \begin{aligned} \|F\|_{\mathrm{av}}^2&=m_t\langle L(\sigma_t(x^*x))a\xi,a\xi\rangle\\ &=\langle P(L(x^*x))a\xi,a\xi\rangle\\ &=\varphi(x^*x)\varphi(a^*a)=H_0^2 . \end{aligned} \] Since \((1-cd)F=(F-cG)+c(G-dF)\) and \(|c|=1\), \[ \begin{aligned} &|1-cd|\le\sup_{s\in I_r,\,u\in I_\ell}|e^{isu}-e^{ir\ell}|\\ &\quad +\sup_{s\in I_q,\,u\in I_p}|e^{isu}-e^{iqp}|\\ &\quad \le\delta(|r|+|\ell|+|q|+|p|)+2\delta^2 , \end{aligned} \] using \(|e^{iv}-e^{iw}|\le|v-w|\). Letting \(\delta\to0\) gives \(cd=1\). \(\square\)

Lemma 3.4 (modular eigenvalues; Marrakchi–Vaes [MV, Lemma 2.1]). If \(M_\varphi=\mathbb C1\), then \(D\) has no eigenvalue other than \(0\). If moreover \(D\) has pure point spectrum, then \(M=\mathbb C1\).

Proof. Suppose \(1_{\{r\}}(D)\ne0\), and choose \(x\in M\) with \(\eta=1_{\{r\}}(D)x\xi\ne0\). Let \(y_k=\int e^{-irt}\sigma_t(x)\,d\mu_k(t)\in M\), so \(\|y_k\|\le\|x\|\). By the mean ergodic theorem (Lemma 2.1 of the previous lesson) for the group \(e^{-irt}U_t\), whose fixed space is \(1_{\{r\}}(D)H\), \(y_k\xi\to\eta\). A \(\sigma\)-weak cluster point \(y\in M\) of \((y_k)\) satisfies \(y\xi=\eta\), so \(\sigma_t(y)\xi=U_t\eta=e^{irt}y\xi\), and \(\sigma_t(y)=e^{irt}y\) since \(\xi\) is separating. Then \(y^*y\) and \(yy^*\) are fixed by \(\sigma\), hence scalars; they equal \(\|y\|^2>0\). By (1.1) of the previous lesson, \[ \begin{aligned} &\varphi(yy^*)=\|\xi y\|^2=\|e^{D/2}y\xi\|^2\\ &\quad =e^r\|y\xi\|^2=e^r\varphi(y^*y), \end{aligned} \] so \(e^r=1\) and \(r=0\). If \(D\) has pure point spectrum, all of \(H\) is then its kernel, \(D=0\), \(\sigma\) is trivial, and \(M=M_\varphi=\mathbb C1\). \(\square\)

Theorem 3.5 (spectral intertwining rigidity). Let \(M\) be a von Neumann algebra with a faithful normal state \(\varphi\) such that \(M_\varphi=\mathbb C1\), and let \(b\) be a \(u\)-continuous action of \(\mathbb R\) on \(M\) by \(\varphi\)-preserving automorphisms satisfying (FP). Then \(b_s=\mathrm{id}\) for every \(s\).

Proof. By Lemma 3.3, the symmetric bilinear form \(\mathfrak b((r,p),(q,\ell))=r\ell+qp\) takes values in \(2\pi\mathbb Z\) on \(\mathcal S\times\mathcal S\). We show that \(\mathcal S\) is countable or contained in a coordinate axis. If \(\mathcal S\) contains two linearly independent vectors \(v_1,v_2\), the linear map \(z\mapsto(\mathfrak b(v_1,z),\mathfrak b(v_2,z))\) is invertible, because \(\mathfrak b\) is nondegenerate, and it maps \(\mathcal S\) into the countable set \((2\pi\mathbb Z)^2\); so \(\mathcal S\) is countable. Otherwise \(\mathcal S\) lies on a line through the origin. If that line is not a coordinate axis, its points are \(t(\alpha,\beta)\) with \(\alpha\beta\ne0\), and \(\mathfrak b(t(\alpha,\beta),t(\alpha,\beta))=2t^2\alpha\beta\in2\pi\mathbb Z\) leaves at most two values of \(t\) for each integer; so \(\mathcal S\) is countable.

If \(\mathcal S\) is countable, \(D\) has pure point spectrum by Lemma 3.2(3), and \(M=\mathbb C1\) by Lemma 3.4, so \(b\) is trivial. If \(\mathcal S\subset\{0\}\times\mathbb R\), then \(D=0\) by Lemma 3.2(2), and again \(M=M_\varphi=\mathbb C1\). If \(\mathcal S\subset\mathbb R\times\{0\}\), then \(Q=0\), so \(V_s=1\) and \(b_s(a)\xi=a\xi\) for all \(a\); as \(\xi\) is separating, \(b_s=\mathrm{id}\). \(\square\)

Corollary 3.6. Let \(M\) be a type III₁ factor with separable predual and \(\varphi\) a faithful normal state. Then the bicentralizer flow \(\beta^\varphi\) of The bicentralizer flow is trivial: \(\beta^\varphi_\lambda=\mathrm{id}\) on \(\mathrm B(M,\varphi)\) for every \(\lambda>0\).

Proof. Let \(\mathrm B=\mathrm B(M,\varphi)\), \(\varphi_{\mathrm B}=\varphi|_{\mathrm B}\) and \(b_s=\beta^\varphi_{e^{-s}}\). By Theorem 3.1 of The bicentralizer, \(\varphi_{\mathrm B}\) is a faithful normal state of \(\mathrm B\) with trivial centralizer. By Theorem 3.1(3), (4) of the flow lesson, \(b\) is a \(u\)-continuous action of \(\mathbb R\) on \(\mathrm B\) preserving \(\varphi_{\mathrm B}\). For bounded \((w_n)\) in \(\mathrm B\), \(\|\varphi_{\mathrm B}w_n-e^sw_n\varphi_{\mathrm B}\|\) equals \(e^s\|w_n\varphi_{\mathrm B}-e^{-s}\varphi_{\mathrm B}w_n\|\), so Corollary 3.2 of the flow lesson gives (FP) for \((\mathrm B,\varphi_{\mathrm B},b)\). Theorem 3.5 applies. \(\square\)

Corollary 3.6 does not yet show that \(\mathrm B(M,\varphi)\) is trivial: a trivial flow is compatible with a nontrivial algebra. What is missing is the ergodicity of the bicentralizer flow, proved in later lessons; an ergodic trivial flow can only act on \(\mathbb C1\).

4. Exercises

Exercise 4.1. Show that a finite-dimensional von Neumann algebra with a faithful state \(\varphi\) such that \(M_\varphi=\mathbb C1\) is \(\mathbb C\).

Exercise 4.2. Show that the bilinear form \(\mathfrak b((r,p),(q,\ell))=r\ell+qp\) on \(\mathbb R^2\) is nondegenerate, and find all \(t\) with \(\mathfrak b(t(1,1),t(1,1))\in2\pi\mathbb Z\).

Exercise 4.3. Show that Theorem 3.5 remains true if in (FP) "strongly" is replaced by "\(*\)-strongly", and that it is false if the hypothesis \(M_\varphi=\mathbb C1\) is dropped: take \(M=L^\infty(\mathbb T)\) with Haar measure and \(b_s\) the rotations.

Exercise 4.4. In the proof of Lemma 2.2, explain why the partition argument needs \(C=P(S^*S)\) to commute with the spectral projections of \(D\), and why the conclusion of Lemma 2.1 alone, without a rate uniform in \(s\), suffices.

5. Solutions

4.1. On a finite-dimensional Hilbert space \(D\) has pure point spectrum; apply Lemma 3.4.

4.2. \(\mathfrak b(v,z)=0\) for all \(z\) means, with \(v=(r,p)\), \(r\ell+qp=0\) for all \((q,\ell)\), so \(r=p=0\). Next, \(\mathfrak b(t(1,1),t(1,1))=2t^2\), so the condition is \(t^2\in\pi\mathbb Z\), that is \(t=\pm\sqrt{\pi k}\), \(k\ge0\).

4.3. The modified hypothesis has a stronger conclusion than (FP), so (FP) holds and the theorem applies. For \(M=L^\infty(\mathbb T)\) with the Haar state \(\varphi\), \(M\) is abelian, so \(\varphi\) is a trace, \(\Delta=1\), \(M_\varphi=M\ne\mathbb C1\), and \(\|\varphi w-e^sw\varphi\|=|1-e^s|\,\|w\varphi\|\). If \(s\ne0\) and \(\|\varphi w_n-e^sw_n\varphi\|\to0\), then \(\|w_n\|_\varphi^2\le\|w_n\|\,\|w_n\varphi\|\to0\), so \(w_na-b_s(a)w_n\to0\) strongly for every \(a\); for \(s=0\), \(b_0=\mathrm{id}\) and \(w_na-aw_n=0\) because \(M\) is abelian. So (FP) holds, the rotations are \(\varphi\)-preserving and \(u\)-continuous, and they are not trivial.

4.4. Commutation makes the quadratic form \(\langle Ch,h\rangle\) additive over the orthogonal spectral pieces \(1_J(D)h\), which lets one piece carry the excess \(\varepsilon\). The pieces concentrate at some point \(s\) of the compact interval, and Lemma 2.1 is applied at that single point, so no uniformity in \(s\) is needed.

References