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Stability of separably acting factors

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

This lesson assembles the previous ones into strong Kadison–Kastler stability for all factors on separable Hilbert spaces, of every type, with one modulus (Theorem 1.1). Finite factors were treated in Stability of finite factors. For infinite factors the steps are these. Tensor with the fixed factor \(D\) and prepare the pair; align the modular action; conjugate the two tensor products through their common crossed product; and descend twice, first undoing the preparation and then slicing off the auxiliary tensor factor. Each descent produces a completely positive map, which the completely positive criterion turns into a unitary near the identity [OpenAI-289, Section 8].

We use: Close operator algebras (Lemmas 1.4 and 1.5); Averaging and the completely positive criterion (Proposition 4.1); Irreducible hyperfinite subfactors (Lemma 1.1); Stability of finite factors (Theorem 2.1); Preparing infinite factors (Lemma 2.1, Proposition 2.2 and Proposition 5.2); A common modular action (Theorem 3.1); Common crossed products (Proposition 3.2); and slice maps of von Neumann tensor products (Theorem 9.2 of the tensor product lesson).

1. The theorem

Theorem 1.1 (strong stability of separably acting factors). There are \(t_{\rm fac}>0\) and a nondecreasing function \(\eta_{\rm fac}:(0,t_{\rm fac})\to(0,\infty)\) with \(\eta_{\rm fac}(t)\to0\) as \(t\to0\), with the following property. Let \(H\) be a separable Hilbert space and \(M_0,N_0\subseteq B(H)\) unital factors with \(d(M_0,N_0)<t<t_{\rm fac}\). Then there is a unitary \(u\in B(H)\) with \[ uM_0u^*=N_0,\qquad\|u-1\|\le\eta_{\rm fac}(t). \] The threshold and the function do not depend on the factors, their types or their representations.

Proof. The constants. Let \(c_{\rm mod},C_{\rm mod}\) be the constants of Theorem 3.1 of A common modular action, and \(\gamma_0,\zeta\) those of Proposition 3.2 of Common crossed products. For \(t>0\) put \(\gamma(t)=7\cdot10^7t\) and \[ \begin{aligned} r(t)&=3\cdot10^7t+C_{\rm mod}\gamma(t)\\ &\quad +\zeta\bigl(2C_{\rm mod}\gamma(t)\bigr). \end{aligned} \] Choose \(t_{\rm fac}\le t_{\rm fin}\) so small that, for \(t<t_{\rm fac}\), \(t<4\cdot10^{-9}\), \(\gamma(t)<c_{\rm mod}\), \(2C_{\rm mod}\gamma(t)<\gamma_0\) and \(2r(t)<10^{-3}\); and put \(\eta_{\rm fac}(t)=\max\{\eta_{\rm fin}(t),64r(t)\}\), which is nondecreasing and tends to \(0\).

Finite factors. If \(M_0\) is finite, Theorem 2.1 of the finite lesson gives the unitary. If \(N_0\) is finite, the same theorem applied to the pair \((N_0,M_0)\) gives a unitary \(u'\) with \(u'N_0u'^*=M_0\) and \(\|u'-1\|\le\eta_{\rm fin}(t)\); take \(u=u'^*\). By Lemma 1.1 of Irreducible hyperfinite subfactors, if one of the factors is finite, so is the other. So we may assume that both are infinite.

Preparation. Proposition 5.2 of Preparing infinite factors gives unitaries \(u_0\in B(H)\) and \(v\in B(H\otimes K)\) with \(\|u_0-1\|+\|v-1\|\le3\cdot10^7t\), such that \[ \begin{gathered} M=M_0\bar\otimes D,\\ N=v\bigl((u_0N_0u_0^*)\bar\otimes D\bigr)v^* \end{gathered} \] form a prepared pair, with \(d_{\rm cb}(M,N)\le6\cdot10^7t<\gamma(t)\).

Alignment. Theorem 3.1 of A common modular action gives a unitary \(q\) with \(\|q-1\|\le C_{\rm mod}\gamma(t)\) such that \(P=qMq^*\) is normalized by the modular group \(U_s=\Delta_N^{is}\) of the vector state on \(N\), and \(d_{\rm cb}(P,N)\le C_{\rm mod}\gamma(t)<2C_{\rm mod}\gamma(t)\). The algebra \(N\) is a type III\(_1\) factor with separable predual, by Proposition 2.2 of the preparation lesson, and \(C(N,U)\) is its continuous core for the vector state; by Lemma 2.1 of that lesson it is a factor of type II\(_\infty\). Nothing is claimed about the action of \(U\) on \(P\) beyond normalization.

The crossed products. Proposition 3.2 of Common crossed products gives a unitary \(a\) with \(aPa^*=N\) and \(\|a-1\|\le\zeta(2C_{\rm mod}\gamma(t))\).

Undoing the preparation. All these operations took place on \(H\otimes K\). Put \[ b=(u_0^*\otimes1)\,v^*aq . \] Then \(b(M_0\bar\otimes D)b^*\)\(=(u_0^*\otimes1)v^*Nv(u_0\otimes1)=N_0\bar\otimes D\), and by Lemma 1.5 of Close operator algebras, \(\|b-1\|\le r(t)\).

Slicing off the tensor factor. Fix a unit vector \(\xi\in K\), let \(\omega_\xi\) be its vector state on \(B(K)\), and put \[ \begin{gathered} F_0(x)=(\mathrm{id}\otimes\omega_\xi)\bigl(b(x\otimes1)b^*\bigr)\\ (x\in M_0). \end{gathered} \] With \(V_\xi\eta=\eta\otimes\xi\), the slice map is \(y\mapsto V_\xi^*yV_\xi\), which is UCP, and it maps \(N_0\bar\otimes D\) into \(N_0\) (Theorem 9.2 of the tensor product lesson). So \(F_0\) is a UCP map from \(M_0\) to \(N_0\). Since \(V_\xi^*(x\otimes1)V_\xi=x\), at every matrix level \[ \begin{aligned} &\|F_0-\mathrm{id}_{M_0}\|_{\rm cb}\\ &\quad \le\|b(\cdot\otimes1)b^*-(\cdot\otimes1)\|_{\rm cb}\\ &\quad \le2\|b-1\|\le2r(t). \end{aligned} \] The reverse ordinary gap from \(N_0\) to \(M_0\) is less than \(t\), and \(t+2r(t)<1\). Proposition 4.1 of the averaging lesson, with the inclusion of \(M_0\) as \(\rho\), gives a unitary \(u\in B(H)\) with \(uM_0u^*=N_0\) and \(\|u-1\|\le32\cdot2r(t)\le\eta_{\rm fac}(t)\). \(\square\)

2. Where the uniformity comes from

Every bound in the proof is a function of \(t\) alone. The finite case contributes the modulus \(\eta_{\rm fin}\), which is uniform because the common-position theorem, the boundary isomorphism and the uniform commutator estimate have absolute constants. The infinite case uses \(\eta_{\rm fin}\) once more, on a corner of a crossed product. Everything else consists of linear estimates with absolute constants: the preparation, the graph comparison, the infinitesimal estimates and the two applications of the completely positive criterion. For a single factor \(M_0\), one could hope for a tolerance \(\delta_{M_0}(\varepsilon)\) that works in every representation; Theorem 1.1 gives one tolerance for all factors at once, which is a stronger quantifier.

3. Exercises

Exercise 3.1. Show that the slice map \(y\mapsto V_\xi^*yV_\xi\) is UCP and satisfies \(V_\xi^*(x\otimes1)V_\xi=x\).

Exercise 3.2. Check that \(b(M_0\bar\otimes D)b^*=N_0\bar\otimes D\) for \(b=(u_0^*\otimes1)v^*aq\).

Exercise 3.3. Let \(M_0\) be a factor of type I\(_\infty\), that is \(M_0\cong B(L)\) with \(L\) infinite-dimensional. Which steps of the proof could be replaced by Christensen's theorem alone?

Exercise 3.4. Show that the conclusion of Theorem 1.1 implies \(d(M_0',N_0')\le2\eta_{\rm fac}(t)\) for factors with \(d(M_0,N_0)<t<t_{\rm fac}\).

4. Solutions

Solution 3.1. \(V_\xi\) is an isometry, so \(y\mapsto V_\xi^*yV_\xi\) is completely positive and unital. For \(\eta,\eta'\in H\), \(\langle V_\xi^*(x\otimes1)V_\xi\eta,\eta'\rangle=\langle x\eta,\eta'\rangle\langle\xi,\xi\rangle\).

Solution 3.2. \(aqMq^*a^*=aPa^*=N\), so \(v^*aqMq^*a^*v=(u_0N_0u_0^*)\bar\otimes D\), and conjugating by \(u_0^*\otimes1\) gives \(N_0\bar\otimes D\).

Solution 3.3. A type I factor is injective, so Christensen's theorem, applied to the mutual near inclusions of \(M_0\) and \(N_0\), already gives a unitary within \(150t\) of \(1\) conjugating \(M_0\) onto \(N_0\), for \(t<1/101\). The general proof is needed for infinite factors that are not injective.

Solution 3.4. \(N_0'=uM_0'u^*\), and Lemma 1.5 of Close operator algebras gives \(d(M_0',uM_0'u^*)\le2\|u-1\|\).

References