Rescaled Tomita graphs

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

For a prepared pair $M,N$ with common vector $\Omega$, each algebra has its Tomita operator $S_P$, the closure of $x\Omega\mapsto x^*\Omega$, and its modular operator $\Delta_P=S_P^*S_P$. This lesson shows that a small completely bounded distance between $M$ and $N$ forces the graphs of the rescaled operators $cS_P$ to be close, uniformly in $c>0$ (Proposition 2.1). Bounding a single element in operator norm would lose this uniformity as $c\to0$ or $c\to\infty$. Instead a graph vector is first flattened by random signs and Haar unitaries, then replicated into a row, where closeness at one matrix level supplies the estimate. A consequence is that the spectral projections of $\log\Delta_M$ and $\log\Delta_N$ for separated half-lines are almost orthogonal (Corollary 3.1). This is the comparison of the two modular actions on which the next lessons build [OpenAI-289, Section 6].

We use: Close operator algebras (Exercise 5.4 and Lemma 1.4); Preparing infinite factors (Definition 5.1 and the identity (5.1)); the Tomita–Takesaki theorem for a faithful normal state, Modular time from a faithful normal state; and the fixed-point criterion for centralizers.

1. Graphs and their symmetries

Fix a prepared pair $M,N\subseteq B(H)$ with vector $\Omega$, state $\varphi=\langle\cdot\,\Omega,\Omega\rangle$ and subfactor $R=(\bigcup_jR_j)''$. Let $\overline H$ be the conjugate Hilbert space, with vectors $\bar\xi$ and $\alpha\bar\xi=\overline{\bar\alpha\xi}$; an operator $T$ on $H$ gives the operator $\bar T\bar\xi=\overline{T\xi}$ on $\overline H$. For $P\in\{M,N\}$, $c>0$ and any $x\in B(H)$ put $g_c(x)=\bigl(x\Omega,\;c\,\overline{x^*\Omega}\bigr)\in H\oplus\overline H .$ This is complex linear in $x$. Let $E_c^P$ be the orthogonal projection onto the closure of $\{g_c(x):x\in P\}$, which is the graph of the closed linear operator $cS_P$, followed by $\xi\mapsto\bar\xi$, from $H$ to $\overline H$. Also $\|g_c(x)\|^2=\varphi(x^*x)+c^2\varphi(xx^*)$.

Lemma 1.1 (common symmetries). For unitaries $r,s\in R$ let $\rho_s=J_Ms^*J_M=J_Ns^*J_N$, right multiplication by $s$, and $U_{r,s}=r\rho_s\oplus\overline{s^*\rho_{r^*}} .$ Then $U_{r,s}$ is unitary, $g_c(rxs)=U_{r,s}g_c(x)$ for every $x\in B(H)$ commuting with $\rho_s$ and $\rho_{r^*}$, and both projections $E_c^M,E_c^N$ commute with every $U_{r,s}$. Moreover, for $x\in M$ or $x\in N$, the Haar averages $E_j(x)=\int_{\mathcal U(R_j)}uxu^*\,du$ converge weakly to $\varphi(x)1$.

Proof. The two expressions for $\rho_s$ agree by Definition 5.1(3) of the preparation lesson, and $\rho_s$ commutes with $M$ and $N$. For $x\in P$, $rxs\Omega=r\rho_sx\Omega$ and $(rxs)^*\Omega=s^*x^*r^*\Omega=s^*\rho_{r^*}x^*\Omega$, which gives the formula; the same computation applies to any $x$ commuting with $\rho_s$ and $\rho_{r^*}$. The adjoint of $U_{r,s}$ is $U_{r^*,s^*}$, so each graph closure is invariant under a self-adjoint group of unitaries and reduces it. For the averages: every weak operator cluster point of $E_j(x)$ commutes with all $R_j$, hence lies in $R'\cap P=\mathbb C1$, and $\varphi(E_j(x))=\varphi(x)$ because $R$ lies in the centralizer of $\varphi$. $\square$

2. Uniform comparison of the graphs

Proposition 2.1 (uniform graph comparison). Every prepared pair with $d_{\rm cb}(M,N)\le\gamma$ satisfies $\sup_{c>0}\|E_c^M-E_c^N\|\le5\cdot10^5\,\gamma .$

Proof. Put $T=(1-E_c^N)E_c^M$. By Lemma 1.1, $T$ commutes with every $U_{r,s}$. We first treat $c=\sqrt k$ with $k\in\mathbb N$, and we may suppose $T\ne0$.

A flattened element with the signal. Choose $x\in M$ with $\|g_c(x)\|=1$ and $\|Tg_c(x)\|\ge\frac12\|T\|$; this is possible since the vectors $g_c(x)$, $x\in M$, are dense in the range of $E_c^M$. Write $\alpha=\varphi(x^*x)$ and $\beta=\varphi(xx^*)$, so $\alpha+k\beta=1$. Choose independent Haar unitaries $u_i,v_i$ of $R_j$ and independent fair signs $\epsilon_i$, $i\le m$, and put $x_i=u_ixv_i,\qquad y=m^{-1/2}\textstyle\sum_{i=1}^m\epsilon_ix_i .$ By Lemma 1.1, $Tg_c(x_i)=U_{u_i,v_i}Tg_c(x)$, so all the vectors $w_i=Tg_c(x_i)$ have norm $s_0=\|Tg_c(x)\|$. For Hilbert space vectors, $\mathbb E_\epsilon\|\sum_i\epsilon_iw_i\|^2=\sum_i\|w_i\|^2$ and, expanding the fourth power over the three pairings of indices, $\mathbb E_\epsilon\|\sum_i\epsilon_iw_i\|^4\le3(\sum_i\|w_i\|^2)^2$. So $Z=\|Tg_c(y)\|^2$ has $\mathbb E_\epsilon Z=s_0^2$ and $\mathbb E_\epsilon Z^2\le3s_0^4$. The Cauchy–Schwarz inequality applied to $Z1_{\{Z\ge\mathbb EZ/2\}}$ gives $\mathbb P(Z\ge\mathbb EZ/2)\ge(\mathbb EZ)^2/(4\,\mathbb EZ^2)\ge1/12$. Hence, for every choice of the unitaries, with sign probability at least $1/12$, $\|Tg_c(y)\|\ge\frac{\|T\|}{2\sqrt2}. \tag{2.1}$

Fourth moments of the element. The unitaries of $R$ preserve $\varphi$, so $\mathbb E\varphi(y^*y)=\alpha$, $\mathbb E\varphi(yy^*)=\beta$ and $\mathbb E\|g_c(y)\|^2=1$. Expand $\varphi((y^*y)^2)$ and take the sign expectation. The terms with all four indices equal contribute $m^{-1}\varphi((x^*x)^2)$. For distinct $i,l$, use that elements of $R$ can be moved cyclically under $\varphi$, and that $v_iv_l^*$ and $u_i^*u_l$ are Haar distributed: $\begin{aligned} \mathbb E\varphi(x_i^*x_ix_l^*x_l)&=\varphi\bigl(x^*x\,E_j(x^*x)\bigr)\to\alpha^2,\\ \mathbb E\varphi(x_i^*x_lx_l^*x_i)&=\varphi\bigl(x^*E_j(xx^*)x\bigr)\to\alpha\beta, \end{aligned}$ as $j\to\infty$, by Lemma 1.1 applied to $\langle E_j(x^*x)\Omega,x^*x\Omega\rangle$ and $\langle E_j(xx^*)x\Omega,x\Omega\rangle$. The third pairing is bounded by the second: $|\varphi(a^2)|\le\varphi(aa^*)^{1/2}\varphi(a^*a)^{1/2}$ with $a=x_i^*x_l$, and the two factors have equal expectations after exchanging $i$ and $l$. The same computation for $yy^*$ exchanges $\alpha$ and $\beta$. Since $\alpha\le1$, $k\beta\le1$ and $k\ge1$, the weighted off-diagonal limit satisfies $\alpha^2+2\alpha\beta+k\beta^2+2k\alpha\beta\le6 .$ Choose $m$ with $m^{-1}(\varphi((x^*x)^2)+k\varphi((xx^*)^2))\le1$, and then $j$ so large that the off-diagonal contribution is at most $7$. Then $\begin{aligned} &\mathbb E\bigl[\|g_c(y)\|^2+\varphi((y^*y)^2)+k\varphi((yy^*)^2)\bigr]\\ &\quad \le10 .\quad(2.2) \end{aligned}$ By Markov's inequality the bracket exceeds $240$ with probability at most $1/24<1/12$. So some realization satisfies both (2.1) and the bound $240$ for the bracket.

Clipping. For $b>0$ let $f_b(t)=\min(1,b/t)$ and $f_b(0)=1$. Put $z=y\,f_{K_1}(|y|)$ with $K_1=4\sqrt{480}$. With the polar decomposition $y=w|y|$, $(y-z)^*(y-z)$$=|y|^2(1-f_{K_1}(|y|))^2\le|y|^4/K_1^2$ and $(y-z)(y-z)^*\le w|y|^4w^*/K_1^2$$=(yy^*)^2/K_1^2$, because $t(1-f_b(t))\le t^2/b$. Hence $\begin{aligned} &\|g_c(y-z)\|^2\\ &\quad\le K_1^{-2}\bigl[\varphi((y^*y)^2)+k\varphi((yy^*)^2)\bigr]\\ &\quad\le\frac{240}{K_1^2}, \end{aligned}$ and with $a_0=1/(4\sqrt2)$, since $\sqrt{240}/K_1=a_0$, $\begin{aligned} \|Tg_c(z)\|&\ge\Bigl(\frac1{2\sqrt2}-\frac{\sqrt{240}}{K_1}\Bigr)\|T\|\\ &=a_0\|T\|. \end{aligned}$ Also $\|z\|\le K_1$, $\alpha_z:=\varphi(z^*z)\le\varphi(y^*y)\le240$ and $\beta_z:=\varphi(zz^*)\le\varphi(yy^*)\le240/k$.

A row. A single element of norm $K_1$ would still cost $\sqrt k\,K_1\gamma$ on the second leg of the graph. To avoid this, let $r_1,\dots,r_k$ be independent Haar unitaries of a large $R_{j'}$ and form the row and its square $\begin{gathered} Y=[\,r_1z\ \cdots\ r_kz\,]\in M_{1,k}(M),\\ A=YY^*=\textstyle\sum_ir_izz^*r_i^* . \end{gathered}$ For rows of vectors use the normalized norm $\|(\eta_i)\|_{\rm av}^2=k^{-1}\sum_i\|\eta_i\|^2$. Since $Tg_c(r_iz)=U_{r_i,1}Tg_c(z)$, the row $(Tg_c(Y_i))_i$ has normalized norm at least $a_0\|T\|$ for every choice of the $r_i$. As before, as $j'\to\infty$, $\begin{aligned} &\mathbb E\varphi(A^2)\to k\varphi((zz^*)^2)+k(k-1)\beta_z^2,\\ &\mathbb E\,k^{-1}\textstyle\sum_i\varphi(Y_i^*AY_i)\\ &\qquad\to\varphi((z^*z)^2)+(k-1)\alpha_z\beta_z . \end{aligned}$ With $\varphi((zz^*)^2)\le K_1^2\beta_z$ and $\varphi((z^*z)^2)\le K_1^2\alpha_z$, both limits are at most $240K_1^2+240^2$. Choose $j'$ and a realization with $\begin{aligned} &\varphi(A^2)+k^{-1}\textstyle\sum_i\varphi(Y_i^*AY_i)\le C_1:\\ &\quad =3.81\cdot10^6 . \end{aligned}$

Clipping the row. Put $X=f_{K_2}(A^{1/2})Y$, so $\|X\|\le K_2$, and $Q=1-f_{K_2}(A^{1/2})$, so $Q^2\le A/K_2^2$ and $Y-X=QY$. Since $c^2=k$, $\begin{aligned} &k^{-1}\textstyle\sum_i\|g_c(Y_i-X_i)\|^2\\ &\quad =k^{-1}\sum_i\varphi(Y_i^*Q^2Y_i)\\ &\qquad +\varphi(QAQ)\le C_1/K_2^2 . \end{aligned}$ Choose $K_2=2\sqrt{C_1}/a_0$. Then $\begin{aligned} \|(Tg_c(X_i))_i\|_{\rm av}&\ge\Bigl(a_0-\frac{\sqrt{C_1}}{K_2}\Bigr)\|T\|\\ &=\frac{a_0}2\|T\| .\quad(2.3) \end{aligned}$

Comparison at one matrix level. Since $d_{\rm cb}(M,N)\le\gamma$, there is a row $Z\in M_{1,k}(N)$ with $\|X-Z\|\le K_2\gamma$, up to an arbitrarily small slack (Exercise 5.4 of Close operator algebras and compression to the first row). For any row $[e_i]$ of operators with norm at most $\eta$, $\begin{aligned} &k^{-1}\textstyle\sum_i\|g_c(e_i)\|^2\\ &\quad =k^{-1}\sum_i\varphi(e_i^*e_i)+\varphi\bigl(\sum_ie_ie_i^*\bigr)\\ &\quad \le2\eta^2 . \end{aligned}$ Since $g_c(X_i)$ lies in the range of $E_c^M$ and $g_c(Z_i)$ in that of $E_c^N$, $\begin{aligned} &\|(Tg_c(X_i))_i\|_{\rm av}\\ &\quad =\|((1-E_c^N)g_c(X_i-Z_i))_i\|_{\rm av}\\ &\quad \le\sqrt2K_2\gamma . \end{aligned}$ With (2.3), $\|(1-E_c^N)E_c^M\|\le(2\sqrt2K_2/a_0)\gamma$$=16K_2\gamma\le3.534\cdot10^5\gamma$. The same holds with $M$ and $N$ exchanged, and for orthogonal projections $\|E-F\|$ is the larger of $\|(1-F)E\|$ and $\|(1-E)F\|$.

All rescalings. The antiunitary flip $\Theta(\xi,\bar\eta)=(\eta,\bar\xi)$ of $H\oplus\overline H$ satisfies $\Theta g_c(x)=c\,g_{1/c}(x^*)$, so it carries the graph for $c$ onto the graph for $1/c$, for both algebras, and the bound holds for $c=1/\sqrt k$. For $c\ge1$ let $k=\lfloor c^2\rfloor$; then $r=c/\sqrt k\in[1,\sqrt2)$, and $1\oplus r$ maps the graphs for $\sqrt k$ onto those for $c$. An invertible operator $V$ multiplies the one-sided gap $\|(1-F)E\|$ between subspaces by at most $\|V\|\|V^{-1}\|$, here at most $\sqrt2$. For $0<c<1$ use $1/c$ and the flip. Hence $\sup_c\|E_c^M-E_c^N\|$$\le\sqrt2\cdot3.534\cdot10^5\gamma\le5\cdot10^5\gamma$. $\square$

3. Separated spectral cuts

Corollary 3.1 (logarithmic modular cuts). For a prepared pair with $d_{\rm cb}(M,N)\le\gamma$, put $L_P=\log\Delta_P$. For every $a\in\mathbb R$, $\begin{aligned} &\bigl\|1_{(-\infty,a-1/2]}(L_N)\,1_{[a+1/2,\infty)}(L_M)\bigr\|\\ &\quad \le2.1\cdot10^6\,\gamma , \end{aligned}$ and the same holds with $M$ and $N$ exchanged.

Proof. The projection onto the graph of a closed operator $A$ has upper left entry $(1+A^*A)^{-1}$. For the graph of $cS_P$ followed by conjugation, $A^*A=c^2\Delta_P$, so the upper left entry of $E_c^P$ is $(1+c^2\Delta_P)^{-1}=g(L_P+2\log c)$ with $g(s)=(1+e^s)^{-1}$. Taking $c=e^{-a/2}$, Proposition 2.1 gives $\|g(L_M-a)-g(L_N-a)\|\le5\cdot10^5\gamma .$ Let $p=1_{(-\infty,a-1/2]}(L_N)$, $q=1_{[a+1/2,\infty)}(L_M)$, and regard $X=pq$ as an operator from $qH$ to $pH$. Let $A_0$ and $B_0$ be the restrictions of $g(L_N-a)$ to $pH$ and of $g(L_M-a)$ to $qH$; since $g$ is decreasing, $A_0\ge g(-1/2)$ and $B_0\le g(1/2)$. As $p$ commutes with $g(L_N-a)$ and $q$ with $g(L_M-a)$, $\begin{aligned} &A_0X-XB_0\\ &\quad=p\bigl(g(L_N-a)-g(L_M-a)\bigr)q\\ &\quad=:D_0,\qquad\|D_0\|\le5\cdot10^5\gamma . \end{aligned}$ Differentiating $t\mapsto e^{-tA_0}Xe^{tB_0}$, which tends to $0$ since $\|e^{-tA_0}\|\|e^{tB_0}\|\le e^{-t(g(-1/2)-g(1/2))}$, gives $\begin{gathered} X=\int_0^\infty e^{-tA_0}D_0e^{tB_0}\,dt,\\ \|X\|\le\frac{\|D_0\|}{g(-1/2)-g(1/2)} . \end{gathered}$ Here $g(-1/2)-g(1/2)=\tanh(1/4)>0.24$. Exchanging the algebras gives the other estimate. $\square$

4. Exercises

Exercise 4.1. Let $w_1,\dots,w_m$ be vectors in a Hilbert space and $\epsilon_i$ independent fair signs. Prove $\mathbb E\|\sum_i\epsilon_iw_i\|^4\le3(\sum_i\|w_i\|^2)^2$.

Exercise 4.2. Prove the Paley–Zygmund bound used above: for a random variable $Z\ge0$ with $0<\mathbb EZ^2<\infty$, $\mathbb P(Z\ge\mathbb EZ/2)\ge(\mathbb EZ)^2/(4\,\mathbb EZ^2)$.

Exercise 4.3. Let $A$ be a closed densely defined operator from $H$ to $K$. Show that the orthogonal projection onto its graph has upper left entry $(1+A^*A)^{-1}$.

Exercise 4.4. For two orthogonal projections $E,F$, show that $\|E-F\|$ is the larger of $\|(1-F)E\|$ and $\|(1-E)F\|$.

Exercise 4.5. In the matrix example $M=M_2(\mathbb C)$ with the state of density $h=\operatorname{diag}(\frac23,\frac13)$, compute $\log\Delta$ on the matrix units and its spectrum.

5. Solutions

Solution 4.1. Expanding, $\mathbb E\|\sum\epsilon_iw_i\|^4$$=\sum_{i,j,k,l}\mathbb E[\epsilon_i\epsilon_j\epsilon_k\epsilon_l]\langle w_i,w_j\rangle\langle w_k,$$w_l\rangle$. Only the pairings $i=j,k=l$, $i=k,j=l$ and $i=l,j=k$ survive, with the all-equal terms counted once. Each pairing is at most $(\sum\|w_i\|^2)^2$ by $|\langle w_i,w_j\rangle|\le\|w_i\|\|w_j\|$.

Solution 4.2. $\mathbb EZ\le\mathbb E[Z1_{\{Z<\mathbb EZ/2\}}]$$+\mathbb E[Z1_{\{Z\ge\mathbb EZ/2\}}]\le\frac12\mathbb EZ$$+(\mathbb EZ^2)^{1/2}\mathbb P(Z\ge\mathbb EZ/2)^{1/2}$.

Solution 4.3. The graph is the range of the injective operator $\xi\mapsto(\xi,A\xi)$ on the domain. The vector $(\eta,0)$ decomposes as $((1+A^*A)^{-1}\eta,A(1+A^*A)^{-1}\eta)$ plus $(\eta-(1+A^*A)^{-1}\eta,-A(1+A^*A)^{-1}\eta)$, and the second vector is orthogonal to every $(\xi,A\xi)$ because $A^*A(1+A^*A)^{-1}\eta=\eta-(1+A^*A)^{-1}\eta$.

Solution 4.4. $E-F=E(1-F)-(1-E)F$, and the two terms map into orthogonal subspaces from orthogonal subspaces, so the norm of their sum is the maximum of their norms; and $\|E(1-F)\|=\|(1-F)E\|$.

Solution 4.5. $\Delta x=hxh^{-1}$, so $\Delta e_{12}=2e_{12}$, $\Delta e_{21}=\frac12e_{21}$ and $\Delta e_{ii}=e_{ii}$. The spectrum of $\log\Delta$ is $\{-\log2,0,\log2\}$.

References

[OpenAI-289] OpenAI, Universal strong Kadison–Kastler stability, preprint, 23 September 2026, Section 6. https://github.com/openai/math/tree/main/preprints/Universal-strong-Kadison-Kastler-stability-September-23-2026