# Irreducible hyperfinite subfactors

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

This lesson begins the proof of strong stability for finite factors. It proves three facts about a finite factor \(M\). First, a von Neumann algebra on the same space at distance less than \(1/10\) from \(M\) is again a finite factor (Lemma 1.1). Second, Popa's theorem: if \(M\) is of type II\(_1\) with separable predual, it contains a hyperfinite II\(_1\) subfactor \(R\), generated by an increasing sequence of full matrix algebras, with trivial relative commutant \(R'\cap M=\mathbb C1\) (Lemma 2.1); Haar averaging over the matrix algebras then converges to the trace (Corollary 2.2). Third, a bounded operator on the trace Hilbert space that commutes with the left and right actions of \(R\) is controlled, up to the factor \(384\), by its coefficients between vectors of contractions of \(M\) (Lemma 3.1). The third fact is what turns closeness in operator norm into control of an operator on the trace Hilbert space in the following lessons [OpenAI-289, Section 5].

We use: [Close operator algebras](close-operator-algebras.md) (Lemma 1.4); on finite algebras, from [Traces on von Neumann algebras](course:OA-FOUND-REMAINDER/reader/supplements/traces-on-von-neumann-algebras-part-a-def-v-2-1-to-def-v-2-17#OA-FND-TA-11), Theorem 5.2 (the center-valued trace), Theorem 5.9 (finite traces are normal), [Corollary 5.4](course:OA-FOUND-REMAINDER/reader/supplements/traces-on-von-neumann-algebras-part-a-def-v-2-1-to-def-v-2-17#OA-FND-TA-12) (projections with equal trace are equivalent) and Lemma 5.1 (the \(\sigma\)-weakly closed convex hull of a unitary orbit meets the center); Lemma 4.3 of [Measurable operators for a trace](course:traces-and-noncommutative-integration/measurable-operators-for-a-trace-examples-convergence-and-the-commutant#4-minimal-projections-and-diffuse-algebras) (a projection of a diffuse algebra with finite trace has subprojections of every smaller trace); from [Integration for a trace](course:traces-and-noncommutative-integration/integration-for-a-trace-the-commutation-theorem-and-applications#9-conditional-expectations-that-preserve-a-trace), Theorem 9.1 (trace-preserving conditional expectations) and [Theorem 3.1](course:traces-and-noncommutative-integration/integration-for-a-trace-the-commutation-theorem-and-applications#3-the-commutation-theorem) (left and right multiplications on \(L^2(M,\tau)\) are mutual commutants); and the Haar probability measure of a compact group, [Theorem 8.3 of Haar measure on locally compact groups](course:harmonic-analysis-on-locally-compact-groups/reader/haar-measure-on-locally-compact-groups#OA-FND-HM-05).

## 1. Neighbours of finite factors

A finite factor \(M\) has a faithful normal tracial state \(\tau\): the center-valued trace of Theorem 5.2 takes values in \(\mathbb C1\), and it is normal by Theorem 5.9. A von Neumann algebra is finite when every isometry in it is unitary.

**Lemma 1.1** (finite neighbours). Let \(M\subseteq B(H)\) be a unital finite factor and \(N\subseteq B(H)\) a unital von Neumann algebra with \(d(M,N)<1/10\). Then \(N\) is a finite factor.

**Proof.** Choose \(t\) with \(d(M,N)<t<1/10\).

*Finiteness.* Suppose \(v\in N\) is an isometry with \(vv^*\ne1\). Choose \(x\in M_1\) with \(\|x-v\|<t\). Then \(\|x\xi\|\ge(1-t)\|\xi\|\), so \(1-t\le|x|\le1\) and the polar part \(w=x|x|^{-1}\) is an isometry of \(M\), hence a unitary because \(M\) is finite. Since \(w-x=w(1-|x|)\), \(\|w-x\|\le t\) and \(\|w-v\|<2t<1\). Then \(v=w\bigl(1-w^*(w-v)\bigr)\) is invertible, which contradicts \(vv^*\ne1\). So every isometry of \(N\) is unitary.

*Factoriality.* Let \(q\ne0\) be a projection in the center of \(N\), and choose \(a\in M_1\) with \(\|a-q\|<t\). For a unitary \(v\in M\) choose \(b\in N_1\) with \(\|v-b\|<t\). As \(b\) commutes with \(q\),
\[
\begin{aligned}
\|[v,q]\|&=\|[v-b,q]\|<2t,\\
\|vav^*-q\|&\le\|a-q\|+\|vqv^*-q\|\\
&<3t .
\end{aligned}
\]
The closed ball of radius \(3t\) around \(q\) is convex and \(\sigma\)-weakly closed, and it contains the unitary orbit of \(a\). By Lemma 5.1 of the traces lesson, the \(\sigma\)-weakly closed convex hull of that orbit contains a central element of \(M\), which is a scalar \(\lambda1\). Hence \(\|q-\lambda1\|\le3t<1/2\). A projection other than \(0\) and \(1\) has spectrum \(\{0,1\}\) and distance \(1/2\) from the scalars, so \(q=1\). The center of \(N\) is generated by its projections, so it is \(\mathbb C1\). \(\square\)

## 2. Popa's irreducible hyperfinite subfactor

Let \(M\) be a II\(_1\) factor with trace \(\tau\), \(\|x\|_2=\tau(x^*x)^{1/2}\), acting on \(L^2(M,\tau)\) with trace vector \(\Omega\). A *full matrix subalgebra* of \(M\) is a unital \(*\)-subalgebra \(F\) isomorphic to \(M_b(\mathbb C)\) for some \(b\). For such \(F\), Haar averaging over its unitary group,
\[
E_F(x)=\int_{\mathcal U(F)}uxu^*\,du ,
\]
is the trace-preserving conditional expectation of \(M\) onto the relative commutant \(F'\cap M\) of Theorem 9.1: \(E_F(x)\) commutes with \(F\) by invariance of the Haar measure, \(E_F\) fixes \(F'\cap M\), and \(\tau(E_F(x)y)=\tau(xy)\) for \(y\in F'\cap M\) because \(\tau(uxu^*y)=\tau(xu^*yu)=\tau(xy)\). In particular \(E_F\) is the orthogonal projection of \(L^2(M,\tau)\) onto the closure of \((F'\cap M)\Omega\), and it is contractive for \(\|\cdot\|_2\). On \(F\) itself, \(E_F\) is the normalized trace of \(F\), which agrees with \(\tau\).

The relative commutant of a full matrix subalgebra \(F\) with matrix units \((e_{ij})_{i,j\le b}\) is a II\(_1\) factor. Indeed \(x\mapsto xe_{11}\) is a \(*\)-isomorphism of \(F'\cap M\) onto the corner \(e_{11}Me_{11}\), with inverse \(y\mapsto\sum_ie_{i1}ye_{1i}\). The corner is a factor, since a central projection of \(e_{11}Me_{11}\) extends to the central projection \(\sum_ie_{i1}ze_{1i}\) of \(M\); it is finite; and it is infinite-dimensional, since \(M\) is the matrix algebra \(M_b(e_{11}Me_{11})\).

**Lemma 2.1** (Popa's theorem). Every II\(_1\) factor \(M\) with separable predual contains a hyperfinite II\(_1\) subfactor \(R=\bigl(\bigcup_jQ_j\bigr)''\), where \(Q_1\subseteq Q_2\subseteq\cdots\) are full matrix subalgebras of \(M\) whose sizes tend to infinity, with
\[
R'\cap M=\mathbb C1 .
\]

**Proof.** *One element.* Let \(P\) be a II\(_1\) factor with trace \(\tau\), \(y=y^*\in P\) and \(\varepsilon>0\). We find a full matrix subalgebra \(F\subseteq P\) with \(\|E_F(y)-\tau(y)1\|_2<\varepsilon\). By the spectral theorem choose \(y_0=\sum_{i=1}^l\lambda_ip_i\), with real \(\lambda_i\) and projections \(p_i\) summing to \(1\), and \(\|y-y_0\|<\varepsilon/4\). Let \(b\ge2\) be an integer. A II\(_1\) factor has no minimal projections, so by Lemma 4.3 of the measurable-operators lesson each \(p_i\) contains \(\lfloor b\tau(p_i)\rfloor\) mutually orthogonal projections of trace \(1/b\). The remaining projection \(r\) has trace
\[
\tau(r)=\frac{b-\sum_i\lfloor b\tau(p_i)\rfloor}{b}<\frac lb ,
\]
an integer multiple of \(1/b\), and it splits into projections of trace \(1/b\) in the same way. This gives \(b\) mutually orthogonal projections of trace \(1/b\) with sum \(1\). They are mutually equivalent by Corollary 5.4, so partial isometries between them provide matrix units of a full matrix subalgebra \(F\cong M_b(\mathbb C)\). Let \(y_1\in F\) be the diagonal element equal to \(\lambda_i\) on the pieces taken from \(p_i\) and to \(0\) on the pieces of \(r\). Then \(y_0-y_1=\sum_i\lambda_ir_i\), where \(r_i\le p_i\) are the leftover projections with \(\sum_ir_i=r\), so
\[
\|y-y_1\|_2\le\|y-y_0\|+\|y_0\|\sqrt{l/b}.
\]
Since \(E_F(y_1)=\tau(y_1)1\), \(E_F\) is \(\|\cdot\|_2\)-contractive and \(|\tau(z)|\le\|z\|_2\),
\[
\begin{aligned}
&\|E_F(y)-\tau(y)1\|_2\\
&\quad \le\|E_F(y-y_1)\|_2+|\tau(y_1-y)|\\
&\quad \le2\|y-y_1\|_2 ,
\end{aligned}
\]
which is less than \(\varepsilon\) once \(b\) is large.

*The sequence.* The predual of \(M\) is separable, so \(L^2(M,\tau)\) is separable; choose a sequence \((x_j)\) in the self-adjoint part of the unit ball of \(M\) whose vectors \(x_j\Omega\) are dense in the vectors of that set. Enumerate \(\mathbb N\times\mathbb N\) as a sequence of requirements \((j,k)\). Put \(Q_0=\mathbb C1\). Given a full matrix subalgebra \(Q_n\), its relative commutant \(P_n=Q_n'\cap M\) is a II\(_1\) factor. For the \(n\)-th requirement \((j,k)\), apply the first step in \(P_n\) to \(E_{P_n}(x_j)\), where \(E_{P_n}=E_{Q_n}\), with \(\varepsilon=1/k\); this gives a full matrix subalgebra \(F\subseteq P_n\). It commutes with \(Q_n\), so \(Q_{n+1}=(Q_n\cup F)''\) is a full matrix subalgebra, isomorphic to \(Q_n\otimes F\), of at least twice the size, and \(P_{n+1}=F'\cap P_n\). The expectations compose, \(E_{P_{n+1}}=E_{F}\circ E_{P_n}\) on \(M\), so
\[
\|E_{P_{n+1}}(x_j)-\tau(x_j)1\|_2<1/k ,
\]
and the bound persists for all later \(n\), because each later expectation is \(\|\cdot\|_2\)-contractive and fixes scalars. Hence \(\|E_{P_n}(x_j)-\tau(x_j)1\|_2\to0\) as \(n\to\infty\), for every \(j\).

*The relative commutant.* Put \(R=(\bigcup_nQ_n)''\). An element \(z\in R'\cap M\) lies in every \(P_n\), so, since \(E_{P_n}\) is the orthogonal projection onto the closure of \(P_n\Omega\),
\[
\begin{aligned}
\tau(z^*x_j)&=\langle E_{P_n}(x_j)\Omega,z\Omega\rangle\\
&\longrightarrow\tau(x_j)\tau(z^*).
\end{aligned}
\]
By density, \(\tau(z^*w)=\tau(w)\tau(z^*)\) for every \(w\in M\); with \(w=z-\tau(z)1\) this gives \(\|z-\tau(z)1\|_2^2=0\), so \(z=\tau(z)1\). Thus \(R'\cap M=\mathbb C1\). The center of \(R\) lies in \(R'\cap M\), so \(R\) is a factor; it is finite, since its isometries are isometries of \(M\); and it contains full matrix algebras of unbounded size, so it is of type II\(_1\). It is hyperfinite by construction. \(\square\)

**Corollary 2.2** (averages converge to the trace). In the situation of Lemma 2.1 let \(E_j=E_{Q_j}\). For every \(x\in M\),
\[
E_j(x)\longrightarrow\tau(x)1\qquad\text{in }\|\cdot\|_2 .
\]
More generally \(E_j\to\langle\cdot\,,\Omega\rangle\Omega\) strongly on \(L^2(M,\tau)\).

**Proof.** The \(E_j\) are decreasing orthogonal projections, so they converge strongly to a projection \(E_\infty\). For \(x\in M\) the operators \(E_j(x)\) are bounded by \(\|x\|\); every weak operator cluster point lies in \(Q_j'\cap M\) for every \(j\), hence in \(R'\cap M=\mathbb C1\), and its trace is \(\tau(x)\). So every cluster point is \(\tau(x)1\), and applying it to \(\Omega\) identifies \(E_\infty(x\Omega)=\tau(x)\Omega\). The vectors \(x\Omega\) are dense. \(\square\)

## 3. Testing a bimodular operator on contractions

Write \(\rho(y)\) for right multiplication, \(\rho(y)(x\Omega)=xy\Omega\); for \(y\in M\) it extends to a bounded operator with \(\|\rho(y)\|=\|y\|\), and it is unitary when \(y\) is unitary (Theorem 3.1 of the integration lesson).

**Lemma 3.1** (bimodular test). Let \(M\) be a II\(_1\) factor with separable predual and \(R=(\bigcup_jQ_j)''\) as in Lemma 2.1. If a bounded operator \(V\) on \(L^2(M,\tau)\) commutes with left multiplication by \(R\) and with \(\rho(R)\), then
\[
\|V\|\le384\sup_{a,b\in M_1}|\langle Va\Omega,b\Omega\rangle| .
\]

**Proof.** Let \(K\) be the supremum. Let \(a,b\in M\) with \(\|a\|_2,\|b\|_2\le1\); no bound on their operator norms is assumed.

*Spreading the elements.* By Corollary 2.2, for large \(j\) the averages \(E_j(aa^*)\) and \(E_j(bb^*)\) are within \(1/2\) of \(\tau(aa^*)1\) and \(\tau(bb^*)1\) in \(\|\cdot\|_2\). The Haar integral of a norm continuous function on the compact group \(\mathcal U(Q_j)\) is a norm limit of averages with rational weights, hence, after repetition, of averages with equal weights. So there are unitaries \(d_1,\dots,d_N\in R\) with
\[
\Bigl\|\tfrac1N\textstyle\sum_id_iaa^*d_i^*-\tau(aa^*)1\Bigr\|_2\le1,
\]
and the same for \(b\). In the same way choose unitaries \(e_1,\dots,e_{N'}\in R\) so that \(\frac1{N'}\sum_le_l^*a^*ae_l\) and \(\frac1{N'}\sum_le_l^*b^*be_l\) are within \(1\) of \(\tau(a^*a)1\) and \(\tau(b^*b)1\). Index the pairs \((i,l)\) by \(k=1,\dots,m\), with \(m=NN'\), and put \(a_k=d_iae_l\), \(b_k=d_ibe_l\). Then
\[
\begin{gathered}
L_a=\tfrac1m\textstyle\sum_ka_ka_k^*,\\
Q_a=\tfrac1m\textstyle\sum_ka_k^*a_k
\end{gathered}
\]
are the two averages above, so \(\|L_a\|_2\le2\) and \(\|Q_a\|_2\le2\), using \(\tau(aa^*)=\tau(a^*a)\le1\); the same holds for \(b\).

*Random signs.* Let \(\epsilon_1,\dots,\epsilon_m\) be independent fair signs and
\[
\begin{gathered}
A=m^{-1/2}\textstyle\sum_k\epsilon_ka_k,\\
B=m^{-1/2}\textstyle\sum_k\epsilon_kb_k .
\end{gathered}
\]
By orthogonality of the signs, \(\mathbb E\langle VA\Omega,B\Omega\rangle=\frac1m\sum_k\langle Va_k\Omega,b_k\Omega\rangle\). Since \(V\) commutes with left multiplication by the unitary \(d_i\) and with the unitary \(\rho(e_l)\), each term equals \(\langle Va\Omega,b\Omega\rangle\). Also \(\mathbb E\|A\|_2^2=\frac1m\sum_k\|a_k\|_2^2=\|a\|_2^2\le1\), and the same for \(B\).

*Fourth moments.* Put \(\|x\|_4^4=\tau((x^*x)^2)\). Expanding \(\tau((A^*A)^2)\) and keeping the sign patterns with nonzero expectation (the three pairings, with the all-equal indices counted once),
\[
\begin{aligned}
\mathbb E\|A\|_4^4&=\tau(Q_a^2)+\tau(L_a^2)\\
&\quad+\tfrac1{m^2}\textstyle\sum_{k,l}\tau((a_k^*a_l)^2)\\
&\quad-\tfrac2{m^2}\textstyle\sum_k\tau((a_k^*a_k)^2).
\end{aligned}
\]
By the Cauchy–Schwarz inequality for the trace, \(|\tau(X^2)|\le\tau(X^*X)\), so \(|\tau((a_k^*a_l)^2)|\le\tau(a_ka_k^*a_la_l^*)\), and the third term is at most \(\tau(L_a^2)\). Dropping the last, nonpositive, term,
\[
\mathbb E\|A\|_4^4\le\|Q_a\|_2^2+2\|L_a\|_2^2\le12,
\]
and likewise \(\mathbb E\|B\|_4^4\le12\).

*Clipping.* With the polar decomposition \(A=w|A|\), put \(A_s=w|A|1_{[0,s]}(|A|)\). Then \(\|A_s\|\le s\), \(\|A_s\|_2\le\|A\|_2\), and \(\|A-A_s\|_2^2=\tau(|A|^21_{(s,\infty)}(|A|))\)\(\le\|A\|_4^4/s^2\); so \(\mathbb E\|A-A_s\|_2^2\le12/s^2\). Define \(B_s\) in the same way. Writing \(\langle VA\Omega,B\Omega\rangle\) as \(\langle VA_s\Omega,B_s\Omega\rangle\)\(+\langle V(A-A_s)\Omega,B\Omega\rangle\)\(+\langle VA_s\Omega,(B-B_s)\Omega\rangle\), the first term is at most \(s^2K\) in modulus, and the Cauchy–Schwarz inequality in probability bounds the expectations of the other two by \(\|V\|\sqrt{12}/s\) each. Hence
\[
|\langle Va\Omega,b\Omega\rangle|\le s^2K+\frac{2\sqrt{12}}s\|V\| .
\]
With \(s=4\sqrt{12}\) this is \(192K+\|V\|/2\). The vectors \(a\Omega\) with \(a\in M\) and \(\|a\|_2\le1\) are dense in the unit ball of \(L^2(M,\tau)\), so \(\|V\|\le192K+\|V\|/2\), that is, \(\|V\|\le384K\). \(\square\)

The point of Lemma 3.1 is that the supremum involves only contractions of \(M\), on which closeness in operator norm gives information, while \(\|V\|\) is a norm on the whole Hilbert space, which contains vectors of unbounded elements.

## 4. Exercises

**Exercise 4.1.** Show that a projection \(q\ne0,1\) satisfies \(\|q-\lambda1\|\ge1/2\) for every \(\lambda\in\mathbb C\), with equality for \(\lambda=1/2\).

**Exercise 4.2.** Let \(F\cong M_b(\mathbb C)\) act on itself by conjugation. Show that \(\int_{\mathcal U(F)}uyu^*\,du=\operatorname{tr}_b(y)1\), where \(\operatorname{tr}_b\) is the normalized trace.

**Exercise 4.3.** Let \(Q\) and \(F\) be commuting full matrix subalgebras of \(M\), of sizes \(q\) and \(b\). Show that the algebra they generate is a full matrix subalgebra of size \(qb\).

**Exercise 4.4.** In the proof of Lemma 3.1, check that \(\frac1m\sum_ka_ka_k^*=\frac1N\sum_id_iaa^*d_i^*\).

**Exercise 4.5.** Let \(D_2\subseteq M_2(\mathbb C)\) be the diagonal matrices. Show that \(d(M_2(\mathbb C),D_2)=1\). So Lemma 1.1 needs a bound on the distance: a unital von Neumann algebra at distance \(1\) from a finite factor need not be a factor.

## 5. Solutions

**Solution 4.1.** The spectrum of \(q-\lambda1\) is \(\{-\lambda,1-\lambda\}\), and \(\max(|\lambda|,|1-\lambda|)\ge1/2\) since \(|\lambda|+|1-\lambda|\ge1\).

**Solution 4.2.** The average \(z\) commutes with every unitary of \(F\), hence with \(F\), so it is a scalar; the trace is invariant, so the scalar is \(\operatorname{tr}_b(y)\).

**Solution 4.3.** If \((f_{ij})\) and \((g_{kl})\) are matrix units of \(Q\) and \(F\), the products \(f_{ij}g_{kl}\) are matrix units indexed by pairs, with \(\sum_{i,k}f_{ii}g_{kk}=1\); they span the generated algebra.

**Solution 4.4.** \(a_ka_k^*=d_iae_le_l^*a^*d_i^*=d_iaa^*d_i^*\), and each \(i\) occurs \(N'\) times.

**Solution 4.5.** Since \(D_2\subseteq M_2(\mathbb C)\), \(\delta(D_2,M_2(\mathbb C))=0\). For the flip \(x=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\) and any diagonal \(y\), the matrix \(x-y\) has an off-diagonal entry \(1\), so \(\|x-y\|\ge1\). Hence \(d=1\), the largest possible value by Exercise 5.1 of [Close operator algebras](close-operator-algebras.md). The algebra \(D_2\) has the central projection \(\operatorname{diag}(1,0)\).

## References

- [OpenAI-289] OpenAI, Universal strong Kadison–Kastler stability, preprint, 23 September 2026, Section 5. https://github.com/openai/math/tree/main/preprints/Universal-strong-Kadison-Kastler-stability-September-23-2026