# The bridge

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

This lesson proves the theorem that drives the construction. Two inequivalent pure states with peaking sequences of matching traces are joined by a reduced HNN adjunction: in the new algebra the two classes merge, every pure state of the old algebra still has exactly one state extension, and no other classes merge. The adjunction is the one of [Reduced HNN adjunctions](reduced-hnn-adjunctions.md), formed over the abelian algebras generated by the two peaking sequences with the expectations of [Expectations from peaking projections](expectations-from-peaking-projections.md). The proof estimates norms of reduced words compressed by excision nets, using the tools of [Pure states, excision and peaking projections](pure-states-excision-and-peaking-projections.md) and the module estimates of [Reduced amalgamated free products of tracial algebras](reduced-amalgamated-free-products-of-tracial-algebras.md). It follows [OpenAI-NA, Section 4].

- The statement (Section 1).
- The two-by-two model (Section 2).
- Decay of compressed words (Section 3).
- Unique extensions and no other joins (Section 4).
- Back to the corner (Section 5).

## 1. The statement

**Theorem 1.1** (the bridge; [OpenAI-NA, Theorem 4.1]). Let \(A\) be a C\(^*\)-algebra with a faithful tracial state \(\tau\) and no finite-dimensional irreducible representation. For \(i=1,2\) let \((p^i_n)_{n\ge0}\) be a decreasing sequence of projections of \(A\) with \(p^i_0=1\) that peaks to a pure state \(f_i\). Suppose that \(f_1\not\sim f_2\) and
\[
\begin{gathered}
\tau(p^1_n)=\tau(p^2_n)=t_n,\\
1=t_0>t_1>t_2>\cdots,\qquad t_n\to0 .
\end{gathered}
\]
Let \(D_i=C^*(p^i_n:n\ge0)\), let \(E_i\colon A\to D_i\) be the trace-preserving expectations of Proposition 2.1 of the expectations lesson, and let \(\theta\colon D_1\to D_2\) be the isomorphism with \(\theta(p^1_n)=p^2_n\). Let \(H(A)\) be the reduced HNN adjunction of \(\theta\) over \(A\). Then:

1. every pure state \(h\) of \(A\) has exactly one state extension \(\tilde h\) to \(H(A)\), and \(\tilde h\) is pure;
2. for pure states \(h,k\) of \(A\),
\[
\begin{gathered}\tilde h\sim\tilde k\iff h\sim k\ \text{ or}\\\{[h],[k]\}=\{[f_1],[f_2]\};\end{gathered}
\]
3. \(H(A)\) has a faithful tracial state extending \(\tau\) and a faithful trace-preserving conditional expectation onto \(A\); if \(A\) is simple, so is \(H(A)\).

The data of the theorem are admissible. By Proposition 2.1 of the expectations lesson, \(D_i\) is a copy of \(C(\mathbb N_0\cup\{\infty\})\) in which \(p^i_n\) is the indicator of \(\{n,n+1,\ldots\}\cup\{\infty\}\), so \(\theta\) is a \(*\)-isomorphism. On \(D_i\) the trace is the measure \(\sum_j(t_j-t_{j+1})\delta_j\) (Exercise 5.5 there), so \(\theta\) preserves the trace. Part 3 follows from Theorem 1.1 of the adjunction lesson, except for simplicity, which follows from part 1 and Proposition 3.1 of the expectations lesson. The work is in parts 1 and 2.

## 2. The two-by-two model

We use the model of Theorem 1.1 of the adjunction lesson:
\[
\begin{gathered}F=M_2(A),\qquad G=M_2(D_1),\\D=D_1\oplus D_2,\qquad Z=F*_DG,\end{gathered}
\]
with \(E^F_D(a)=(E_1(a_{11}),E_2(a_{22}))\), the diagonal projections \(e_1,e_2\), \(s=e^G_{21}\), \(H(A)=e_1Ze_1\), \(Z=M_2(H(A))\) through the matrix units of \(F\), and \(v=e^F_{12}s\). Put
\[
\begin{gathered}P^i_n=p^i_ne_i\in D\subset F,\\\varphi_i(a)=f_i(a_{ii})\quad(a\in F),\end{gathered}
\]
where \(p^i_ne_i\) is the matrix with \(p^i_n\) in position \((i,i)\).

**Lemma 2.1.**

1. \((P^i_n)_n\) peaks to \(\varphi_i\) in \(F\); \(\varphi_1\) and \(\varphi_2\) are pure and inequivalent.
2. \(F\) has no finite-dimensional irreducible representation, and \(\tau_F(r_h)=0\) for every pure state \(h\) of \(F\), where \(\tau_F\) is the normalized trace of \(M_2(A)\).
3. For \(a\in F\), \(\varphi_i(a)\) is the value at \(\infty\) of the \(i\)-th component of \(E^F_D(a)\). In particular \(\varphi_1(a)=\varphi_2(a)=0\) when \(E^F_D(a)=0\).
4. For \(b\in\{s,s^*\}\) write \(b=e_{\ell(b)}be_{r(b)}\); so \((\ell(s),r(s))=(2,1)\) and \((\ell(s^*),r(s^*))=(1,2)\). Then
\[
\begin{gathered}P^{\ell(b)}_nb=bP^{r(b)}_n\\ (n\ge0).\end{gathered}\tag{2.1}
\]

*Proof.* (1) A state \(\psi\) of \(F\) with \(\psi(P^i_n)=1\) for every \(n\) has \(\psi(e_i)=1\), since \(P^i_0=e_i\). So \(\psi(x)=\psi(e_ixe_i)\) (Lemma 3.1 of the first lesson), and \(a\mapsto\psi(ae_i)\) is a state of \(A\) taking the value one on every \(p^i_n\); it is \(f_i\). Hence \(\psi=\varphi_i\), and \(\varphi_i\) is pure by Lemma 2.4 of the first lesson. Let \(\pi^{(2)}\) be the representation of \(M_2(A)\) on \(H_{f_i}\oplus H_{f_i}\) by matrices over \(\pi_{f_i}\). Its commutant consists of the scalar matrices over \(\pi_{f_i}(A)'=\mathbb C1\), so it is irreducible, and \(\varphi_i\) is the vector state of \(\xi_{f_i}\) placed in the \(i\)-th summand. If \(\varphi_1\sim\varphi_2\), a unitary intertwining the two matrix representations commutes with the action of \(e_1\), so it maps the first summand onto the first summand and intertwines the actions \(a\mapsto ae_1\) of \(A\) there; then \(\pi_{f_1}\cong\pi_{f_2}\), against \(f_1\not\sim f_2\).

(2) A finite-dimensional irreducible representation of \(F\), restricted along \(a\mapsto\operatorname{diag}(a,a)\), is a nonzero finite-dimensional representation of \(A\); it decomposes into irreducible ones, and \(A\) has none of finite dimension. The second claim is Lemma 4.2 of the first lesson.

(3) \(E_i(a_{ii})(\infty)=f_i(a_{ii})\) by (2.1) of the expectations lesson.

(4) In \(G\), \(P^1_n\) is \(p^1_ne^G_{11}\), and \(P^2_n=(0,p^2_n)\) is \(\theta^{-1}(p^2_n)e^G_{22}=p^1_ne^G_{22}\). So both sides of (2.1) are the matrix with \(p^1_n\) in the position of \(b\). \(\square\)

## 3. Decay of compressed words

By Lemma 2.3 of the adjunction lesson, \(Z\) is the closed span of \(F\) and the *tested words*
\[
\begin{gathered}W=a_0b_1a_1b_2\cdots b_ma_m,\\m\ge1,\quad b_t\in\{s,s^*\},\\a_t\in F,\end{gathered}\tag{3.1}
\]
with \(E^F_D(a_t)=0\) for \(0<t<m\).

Consider pure states \(h,k\) of \(F\) satisfying one of the two tests
\[
\begin{gathered}\text{(T1) } h=k;\\\text{(T2) } h\not\sim k\ \text{ and}\\\{[h],[k]\}\ne\{[\varphi_1],[\varphi_2]\}.\end{gathered}\tag{3.2}
\]
We also require: whenever one of the states lies in \([\varphi_i]\), it is \(\varphi_i\) itself. For \(h\) choose a net \(x=(x_\lambda)\): the peaking sequence \((P^i_n)_n\) if \(h=\varphi_i\), and otherwise a net of Theorem 3.2 of the first lesson. Choose \(y=(y_\nu)\) for \(k\) in the same way. In both cases the net consists of elements \(0\le x_\lambda\le1\) with \(h(x_\lambda)=1\), decreasing strongly to \(r_h\) in \(F^{**}\), and excision holds (Theorem 3.2 and Lemma 3.3 of the first lesson). Limits involving \(x\) and \(y\) are taken over the product of the two directed sets.

**Lemma 3.1** (two endpoints). Let \(W\) be a tested word, \(\ell=\ell(b_1)\) and \(r=r(b_m)\). If
\[
\begin{gathered}
\|E^F_D(e_\ell a_0^*x_\lambda^2a_0e_\ell)\|\to0,\\
\|E^F_D(e_ra_my_\nu^2a_m^*e_r)\|\to0,
\end{gathered}
\tag{3.3}
\]
then \(\|x_\lambda Wy_\nu\|\to0\).

*Proof.* Let \(w=b_1a_1\cdots a_{m-1}b_m\), a core in the sense of Section 4 of the free product lesson, and put \(d=x_\lambda a_0e_\ell\) and \(d'=e_ra_my_\nu\), so that \(x_\lambda Wy_\nu=dwd'\). By Lemma 4.1(2) there, \(w=P_Gw+(1-P_G)wP_G\), hence
\[
\begin{aligned}\|dwd'\|&\le\|dP_G\|\,\|w\|\,\|d'\|\\&\quad+\|d\|\,\|w\|\,\|P_Gd'\| .\end{aligned}
\]
By Lemma 4.1(3) there, \(\|dP_G\|^2\le\|E^F_D(d^*d)\|\), and \(\|P_Gd'\|^2=\|d'^*P_G\|^2\le\|E^F_D(d'd'^*)\|\). These are the quantities in (3.3), and \(\|d\|\le\|a_0\|\), \(\|d'\|\le\|a_m\|\). \(\square\)

**Lemma 3.2** (Dini). A decreasing net of continuous functions on a compact Hausdorff space that converges pointwise to \(0\) converges uniformly.

*Proof.* Given \(\varepsilon>0\), the open sets \(\{g_\lambda<\varepsilon\}\) cover the space. Choose finitely many of them and an index \(\lambda_0\) above their indices. For \(\lambda\ge\lambda_0\), monotonicity gives \(0\le g_\lambda<\varepsilon\) everywhere. \(\square\)

**Lemma 3.3** (when an endpoint decays). Let \(h\) be a pure state of \(F\) with its net \(x\), let \(a\in F\) and \(\ell\in\{1,2\}\). Then
\[
\|E^F_D(e_\ell a^*x_\lambda^2ae_\ell)\|\to0
\]
if \(h\not\sim\varphi_\ell\), and also if \(h=\varphi_\ell\) and \(\varphi_\ell(a)=0\). The same holds for \(\|E^F_D(e_\ell ay_\nu^2a^*e_\ell)\|\) with \(k\) and \(a^*\) in place of \(h\) and \(a\).

*Proof.* As \(x^2_\lambda\le x_\lambda\), it suffices to treat \(g_\lambda=E^F_D(e_\ell a^*x_\lambda ae_\ell)\). Only its \(\ell\)-th component is nonzero, and it is \(E_\ell((a^*x_\lambda a)_{\ell\ell})\in D_\ell\cong C(\mathbb N_0\cup\{\infty\})\). These are nonnegative continuous functions, decreasing in \(\lambda\). We compute their pointwise limits.

At \(n\in\mathbb N_0\), with \(Q=(p^\ell_n-p^\ell_{n+1})e_\ell\in F\),
\[
\begin{aligned}g_\lambda(n)&=\frac{\tau_F(Qa^*x_\lambda aQ)}{\tau_F(Q)}\\&\longrightarrow\frac{\tau_F(Qa^*r_haQ)}{\tau_F(Q)},\end{aligned}
\]
by normality of \(\tau_F\). By Lemma 1.2 of the first lesson and Lemma 2.1(2),
\[
\begin{aligned}&0\le\tau_F(Qa^*r_haQ)=\tau_F(r_haQa^*r_h)\\&\quad\le\|aQa^*\|\,\tau_F(r_h)=0 .\end{aligned}
\]
At \(\infty\), \(g_\lambda(\infty)=\varphi_\ell(a^*x_\lambda a)\to\hat\varphi_\ell(a^*r_ha)\). If \(h\not\sim\varphi_\ell\), the normal extension of \(\pi_{\varphi_\ell}\) annihilates \(r_h\) (first step of the proof of Lemma 4.1 of the first lesson), so the limit is \(0\). If \(h=\varphi_\ell\), then \(\bar\pi_h(r_h)\) is the projection onto \(\mathbb C\xi_h\): it fixes \(\xi_h\) (Lemma 3.1 of the first lesson), and \(\bar\pi_h(r_h)\pi_h(c)\xi_h=\bar\pi_h(r_hcr_h)\xi_h=h(c)\xi_h\) for \(c\in F\). So the limit is \(\|\bar\pi_h(r_h)\pi_h(a)\xi_h\|^2=|\varphi_\ell(a)|^2=0\). By Lemma 3.2 the convergence is uniform. The statement for \(y\) is the same with \(a^*\) in place of \(a\). \(\square\)

**Proposition 3.4** ([OpenAI-NA, (4.3)]). For every pair \((h,k)\) satisfying (3.2) and every tested word \(W\),
\[
\|x_\lambda Wy_\nu\|\longrightarrow0 .
\]

*Proof.* Call the left endpoint of \(W\) *matched* if \(h=\varphi_{\ell(b_1)}\), and the right endpoint matched if \(k=\varphi_{r(b_m)}\). At a matched endpoint write \(a_0=\varphi_\ell(a_0)1+a_0^0\), or \(a_m=\varphi_r(a_m)1+a_m^0\), and expand \(W\) into at most four terms. In a term in which each endpoint is unmatched, or carries the coefficient \(a_0^0\) or \(a_m^0\) of value zero, both conditions (3.3) hold by Lemma 3.3, and Lemma 3.1 gives decay. In the remaining terms an endpoint carries the scalar \(1\) and its net is a peaking sequence. Up to the scalar, move the peak through the adjacent letter by (2.1):
\[
\begin{gathered}P^{\ell(b_1)}_nb_1a_1\cdots=b_1\,P^{r(b_1)}_na_1\cdots,\\\cdots a_{m-1}b_mP^{r(b_m)}_t=\cdots a_{m-1}P^{\ell(b_m)}_tb_m .\end{gathered}
\]
The letters \(b_1\) and \(b_m\) are now bounded outside factors.

*Some core letter remains.* This happens for one transfer when \(m\ge2\), and for two transfers when \(m\ge3\). The compressed word between the outside factors is a tested word whose transferred endpoints carry the original internal coefficients \(a_1\) or \(a_{m-1}\), with peaking sequences for \(\varphi_{r(b_1)}\) or \(\varphi_{\ell(b_m)}\). By Lemma 2.1(3), these states vanish on \(a_1\) and \(a_{m-1}\), since these coefficients are centered. So each transferred endpoint satisfies its condition (3.3) by Lemma 3.3, whether or not its state matches the new adjacent letter, and an untransferred endpoint is unmatched or carries a coefficient of value zero, so it satisfies its condition by Lemma 3.3 as well. Lemma 3.1 gives decay.

*No core letter remains.* If \(m=1\), both endpoints cannot be matched: \(\ell(b_1)\ne r(b_1)\), and \((h,k)=(\varphi_{\ell(b_1)},\varphi_{r(b_1)})\) is excluded by (3.2). Suppose the left endpoint is matched, \(h=\varphi_i\) with \(i=\ell(b_1)\), and let \(j=r(b_1)\ne i\). The remaining expression is \(b_1P^j_na_1'y_\nu\), with \(a_1'\) the right coefficient of the term. Here \(k\not\sim\varphi_j\): under (T1), \(k=\varphi_i\); under (T2), \(k\sim\varphi_j\) would give the excluded pair \(\{[\varphi_i],[\varphi_j]\}\). By cross-excision (Lemma 4.1 of the first lesson), \(\|P^j_na_1'y_\nu\|\to0\). A matched right endpoint with \(m=1\) is symmetric. If \(m=2\) and both endpoints are matched, (3.2) forces \(h=k=\varphi_i\) with \(i=\ell(b_1)=r(b_2)\); since each letter interchanges the indices, \(r(b_1)=\ell(b_2)=j\ne i\). The remaining expression is \(b_1P^j_na_1P^j_tb_2\), and \(\varphi_j(a_1)=0\). With \(q=\min(n,t)\), Lemma 3.3 of the first lesson gives
\[
\|P^j_na_1P^j_t\|\le\|P^j_qa_1P^j_q\|\longrightarrow0 .
\]
This covers all cases. \(\square\)

## 4. Unique extensions and no other joins

**Proposition 4.1.**

1. Every pure state \(h\) of \(F\) has exactly one state extension to \(Z\), namely \(h\circ E_F\), and it is pure.
2. If \(h,k\) are pure states of \(F\) with \(h\not\sim k\) and \(\{[h],[k]\}\ne\{[\varphi_1],[\varphi_2]\}\), their extensions are inequivalent. If \(h\sim k\), their extensions are equivalent.

*Proof.* (1) First let \(h\) be \(\varphi_i\) or a pure state outside \([\varphi_1]\cup[\varphi_2]\), and apply Proposition 3.4 to the pair \((h,h)\). Let \(\psi\) be a state of \(Z\) extending \(h\). Since \(\psi(x_\lambda)=h(x_\lambda)=1\), Lemma 3.1 of the first lesson gives \(\psi(c)=\psi(x_\lambda cy_\nu)\) for every \(c\in Z\), with \(y=x\). For a tested word \(W\), \(|\psi(W)|\le\|x_\lambda Wy_\nu\|\to0\), so \(\psi(W)=0\). The state \(h\circ E_F\) also extends \(h\) and vanishes on tested words (Proposition 4.2 of the free product lesson). As \(F\) and the tested words span a dense subspace, \(\psi=h\circ E_F\). A pure state \(h\) in \([\varphi_i]\) is \(\varphi_i\circ\operatorname{Ad}(u^*)\) for a unitary \(u\in F\) (Lemma 2.2 of the first lesson); \(\operatorname{Ad}(u)\) is an automorphism of \(Z\) preserving \(F\), so \(\psi\mapsto\psi\circ\operatorname{Ad}(u^*)\) is a bijection between the extensions of \(\varphi_i\) and those of \(h\). The unique extensions are pure by Lemma 2.1 of the first lesson.

(2) By Lemma 2.2 of the first lesson, replacing \(h\) or \(k\) by an equivalent state replaces its extension by an equivalent one. So we may assume the pair satisfies (T2) with the convention on \([\varphi_i]\). Suppose the extensions \(\tilde h,\tilde k\) are equivalent, and realize them as vector states of unit vectors \(\xi,\eta\) in one irreducible representation \(\pi\) of \(Z\) (Lemma 1.1(2) of the first lesson). By Lemma 3.1 of the first lesson, \(\pi(x_\lambda)\xi=\xi\) and \(\pi(y_\nu)\eta=\eta\), so for \(c\in Z\)
\[
\langle\pi(c)\eta,\xi\rangle=\langle\pi(x_\lambda cy_\nu)\eta,\xi\rangle .
\]
For \(c\in F\) the right side tends to \(0\) by cross-excision in \(F\), and for a tested word by Proposition 3.4. So \(\langle\pi(c)\eta,\xi\rangle=0\) for every \(c\in Z\), and \(\xi\ne0\) is orthogonal to the dense subspace \(\pi(Z)\eta\), a contradiction. The last assertion is Lemma 2.2 of the first lesson. \(\square\)

## 5. Back to the corner

*Proof of Theorem 1.1.* For a state \(g\) of \(A\) let \(g^\sharp(a)=g(a_{11})\) on \(F\), and for a state \(\psi\) of \(H(A)=e_1Ze_1\) let \(\psi^\sharp(z)=\psi(e_1ze_1)\) on \(Z\). These are bijections onto the states taking the value one on \(e_1\) (Lemma 3.1 of the first lesson), and they commute with restriction from \(H(A)\) to \(A\). A state of \(Z\) extending \(g^\sharp\) takes the value one on \(e_1\). So the extensions of \(g\) to \(H(A)\) correspond to the extensions of \(g^\sharp\) to \(Z\), and Proposition 4.1(1) gives part 1.

As in the proof of Lemma 2.1(1), \(g^\sharp\) is the vector state of \(\xi_g\oplus0\) in the matrix representation of \(M_2(A)\) over \(\pi_g\), which is irreducible when \(g\) is pure; and an equivalence of two such matrix representations restricts to an equivalence of the corner representations. So \(g\mapsto g^\sharp\) preserves purity, and \(g\sim g'\) exactly when \(g^\sharp\sim g'^\sharp\). The same holds for \(H(A)\subset Z=M_2(H(A))\). Now \(f_1^\sharp=\varphi_1\), and \(f_2^\sharp=\varphi_2\circ\operatorname{Ad}(U^*)\) for the flip unitary \(U=e^F_{12}+e^F_{21}\), so \(f_2^\sharp\sim\varphi_2\). By Proposition 4.1(2), for pure \(h,k\) of \(A\) with \(h\not\sim k\) and \(\{[h],[k]\}\ne\{[f_1],[f_2]\}\), \(\tilde h\not\sim\tilde k\); and \(h\sim k\) gives \(\tilde h\sim\tilde k\).

It remains to join the designated pair. The state \(\tilde f_2\circ\operatorname{Ad}(v)\) of \(H(A)\) takes the value \(\tilde f_2(vp^1_nv^*)=\tilde f_2(p^2_n)=1\) on every \(p^1_n\), so its restriction to \(A\) is \(f_1\), and uniqueness gives
\[
\tilde f_1=\tilde f_2\circ\operatorname{Ad}(v).
\]
This is the vector state of \(\pi_{\tilde f_2}(v^*)\xi_{\tilde f_2}\) in the irreducible representation \(\pi_{\tilde f_2}\), so \(\tilde f_1\sim\tilde f_2\), and by Lemma 2.2 of the first lesson every \(\tilde h\) with \(h\sim f_1\) is equivalent to every \(\tilde k\) with \(k\sim f_2\). This proves part 2, and part 3 was shown in Section 1. \(\square\)

## 6. Exercises

**6.1.** Show that \(\tilde f_2(c)=\tilde f_1(v^*cv)\) for every \(c\in H(A)\).

**6.2.** In Lemma 2.1(4), verify (2.1) for \(b=s^*\) directly.

**6.3.** Show that \(\tilde h=h\circ E_A\) for every pure state \(h\) of \(A\), where \(E_A\) is the expectation of \(H(A)\) onto \(A\).

**6.4.** Let \(h\) be a pure state of \(A\). Show that \(\tilde h(v)=0\), and that \(\tilde h(vav^*)=h(\theta(E_1(a)))\) for \(a\in A\).

**6.5.** In the proof of Proposition 3.4 with \(m=1\), explain why the term with both endpoints carrying the scalar \(1\) never occurs.

## 7. Solutions

**6.1.** By Section 5, \(\tilde f_1=\tilde f_2\circ\operatorname{Ad}(v)\). Hence \(\tilde f_1(v^*cv)=\tilde f_2(vv^*cvv^*)=\tilde f_2(c)\).

**6.2.** \(s^*=e^G_{12}\), with \(\ell(s^*)=1\), \(r(s^*)=2\). In \(G\), \(P^1_n=p^1_ne^G_{11}\) and \(P^2_n=p^1_ne^G_{22}\). So \(P^1_ns^*=p^1_ne^G_{12}=s^*P^2_n\).

**6.3.** Both are states of \(H(A)\) extending \(h\), since \(E_A\) fixes \(A\); by uniqueness they agree.

**6.4.** By Exercise 6.3, \(\tilde h(v)=h(E_A(v))\), and \(v=1\cdot v\cdot1\) is a reduced word, so \(E_A(v)=0\). By Exercise 6.2 of the adjunction lesson, \(E_A(vav^*)=\theta(E_1(a))\), so \(\tilde h(vav^*)=h(\theta(E_1(a)))\).

**6.5.** Both endpoints carry the scalar only if both are matched: \(h=\varphi_{\ell(b_1)}\) and \(k=\varphi_{r(b_1)}\) with \(\ell(b_1)\ne r(b_1)\). Then \(h\ne k\) and \(\{[h],[k]\}=\{[\varphi_1],[\varphi_2]\}\), which (3.2) excludes.

## References

- [OpenAI-NA] OpenAI, *A counterexample to Naimark's problem in ZFC*, OpenAI Math Release preprint, 24 September 2026, Section 4. https://github.com/openai/math/tree/main/preprints/A-counterexample-to-Naimarks-problem-in-ZFC-September-24-2026