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Closed subiterations and countable determination

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

The construction iterates the bridge of The bridge transfinitely, adjoining one unitary at each active step and taking closures of unions at limits. This lesson proves that a pure state of the resulting algebra is already determined by its restriction to a separable subalgebra generated by countably many of the adjoined unitaries. The key step shows that any family of steps that contains the dependencies of each of its members can be kept, and all omitted steps performed afterwards, without losing unique extension. It follows [OpenAI-NA, Section 5]. Ordinals are von Neumann ordinals: each ordinal is the set of smaller ordinals.

1. Chains

A family \((B_\alpha)_{\alpha\le\lambda}\) of C\(^*\)-algebras indexed by the ordinals up to \(\lambda\), each a unital C\(^*\)-subalgebra of the next, is a continuous chain if \(B_\gamma\) is the closure of \(\bigcup_{\alpha<\gamma}B_\alpha\) for every nonzero limit ordinal \(\gamma\le\lambda\).

Lemma 1.1 ([OpenAI-NA, Lemma 5.1]). Let \((B_\alpha)_{\alpha\le\lambda}\) be a continuous chain whose successor inclusions \(B_\alpha\subset B_{\alpha+1}\) have property (U). Then every inclusion \(B_\alpha\subset B_\beta\), \(\alpha\le\beta\le\lambda\), has (U). The unique extensions of equivalent pure states are equivalent.

Proof. Fix \(\alpha\) and a pure state \(h\) of \(B_\alpha\); we show by induction on \(\beta\ge\alpha\) that \(h\) has exactly one extension to \(B_\beta\), and that it is pure. At a successor \(\beta+1\), an extension to \(B_{\beta+1}\) restricts to an extension to \(B_\beta\), which is the unique pure one by induction; by (U) for \(B_\beta\subset B_{\beta+1}\) the extension is unique, and pure by Lemma 2.1 of Pure states, excision and peaking projections. At a limit \(\gamma\), two extensions to \(B_\gamma\) agree on every \(B_\beta\) with \(\alpha\le\beta<\gamma\), hence on the dense union, hence everywhere; an extension exists, and it is pure by Lemma 2.1 there. The last assertion is Lemma 2.2 there. \(\square\)

Lemma 1.2 ([OpenAI-NA, Lemma 5.2]). Let \((B_\alpha)_{\alpha\le\lambda}\) be a continuous chain, \(\lambda\) a nonzero limit ordinal. Let \(h,k\) be pure states of \(B_\lambda\) whose restrictions to \(B_\alpha\) are pure and inequivalent for all \(\alpha\) in a tail \(\alpha_0\le\alpha<\lambda\). Then \(h\not\sim k\).

Proof. Otherwise \(h\) and \(k\) are vector states of unit vectors \(\xi,\eta\) in one irreducible representation \(\pi\) of \(B_\lambda\) (Lemma 1.1(2) of the first lesson). For \(\alpha\) in the tail, the closed subspaces \(M_\alpha=\overline{\pi(B_\alpha)\xi}\) and \(N_\alpha=\overline{\pi(B_\alpha)\eta}\) are invariant under \(\pi(B_\alpha)\), and by Theorem 5.5 of the GNS lesson they carry the GNS representations of \(h|_{B_\alpha}\) and \(k|_{B_\alpha}\), which are irreducible and inequivalent. The orthogonal projection onto \(N_\alpha\) commutes with \(\pi(B_\alpha)\), so its restriction \(T\colon M_\alpha\to N_\alpha\) intertwines the two representations. Then \(T^*T\) and \(TT^*\) commute with irreducible representations, so they are scalars (Corollary 2.3 of the GNS lesson); if \(T\ne0\), a multiple of \(T\) is a unitary intertwiner, against inequivalence. So \(T=0\), and \(\langle\pi(a)\xi,\eta\rangle=0\) for \(a\in B_\alpha\). By continuity of the chain this holds for all \(a\in B_\lambda\), against the cyclicity of \(\xi\). \(\square\)

Simplicity and faithful tracial states pass to closures of unions by Lemma 4.3 of Expectations from peaking projections.

2. The base algebra

Let \(A_0\) be the completion of the union of the matrix algebras \(M_{2^n}(\mathbb C)=M_2(\mathbb C)^{\otimes n}\), each embedded in the next by \(a\mapsto a\otimes1\), with the compatible normalized traces. Put \[ \begin{gathered}p_0=1,\\p_n=e_{11}^{\otimes n}\in M_{2^n}(\mathbb C)\subset A_0\quad(n\ge1).\end{gathered} \]

Lemma 2.1 ([OpenAI-NA, Lemma 5.3]). The algebra \(A_0\) is separable, simple and infinite-dimensional, its trace \(\tau\) is faithful, \(\tau(p_n)=2^{-n}\), and \((p_n)\) peaks to a pure state \(f_0\). Every C\(^*\)-algebra containing \(A_0\) as a unital subalgebra has no finite-dimensional irreducible representation.

Proof. Simplicity and faithfulness follow from Lemma 4.3 of the expectations lesson, since matrix algebras are simple and their normalized traces are faithful. Separability and infinite dimension are clear. The projection \(p_n\) has rank one in \(M_{2^n}(\mathbb C)\), so \(\tau(p_n)=2^{-n}\). A state \(\psi\) with \(\psi(p_n)=1\) satisfies \(\psi(x)=\psi(p_nxp_n)=\langle xe,e\rangle\) on \(M_{2^n}(\mathbb C)\), where \(e=e_1^{\otimes n}\), by Lemma 3.1 of the first lesson. These vector states are compatible, so they determine exactly one state on the dense union and on \(A_0\); this proves peaking, and \(f_0\) is pure by Lemma 2.4 of the first lesson. Finally, a finite-dimensional irreducible representation of an algebra containing \(A_0\) is unital (Lemma 1.1 of the first lesson) and restricts to a unital representation of \(A_0\), injective by simplicity; but \(A_0\) is infinite-dimensional. \(\square\)

3. Bridge iterations and their subiterations

Let \(\delta\) be an ordinal. A bridge iteration of length \(\delta\) is a continuous chain \((X_\beta)_{\beta\le\delta}\) with \(X_0=A_0\), with compatible faithful tracial states, in which every index \(j<\delta\) is of one of two kinds.

For this to be a valid recursion, the hypotheses of the bridge must hold at each active step. They do, by induction: each \(X_j\) contains \(A_0\), so it has no finite-dimensional irreducible representation (Lemma 2.1); its trace is faithful (the bridge at successors, Lemma 4.3 of the expectations lesson at limits); and \((p_n)\) peaks in \(X_j\). Indeed, by Lemma 1.1 every \(X_0\subset X_j\) has (U), and a state of \(X_j\) taking the value one on every \(p_n\) restricts to \(f_0\), hence is the unique extension of \(f_0\). In the same way, for an active \(j\), \((q^j_n)\) peaks in every \(X_\beta\) with \(\beta\ge j\), and by part 2 of the bridge theorem and Lemma 1.1, its state is equivalent to the state peaked by \((p_n)\) in every \(X_\beta\) with \(\beta>j\).

For \(T\subseteq\delta\) put \[ \begin{aligned}A[T]&=C^*\bigl(A_0,\{v_j:j\in T\text{ active}\}\bigr)\\&\subseteq X_\delta .\end{aligned} \] So \(A[\beta]=X_\beta\) for \(\beta\le\delta\), by induction on \(\beta\). Every element of \(X_\delta\) lies in \(A[T]\) for a countable \(T\): it is a limit of a sequence of \(*\)-polynomials, each involving finitely many stable unitaries. For each active \(j\) choose a countable set \(I_j\subseteq j\) with \(q^j_n\in A[I_j]\) for every \(n\); for inactive \(j\) put \(I_j=\varnothing\). A set \(T\subseteq\delta\) is dependency-closed if \[ j\in T\ \Longrightarrow\ I_j\subseteq T . \] Every countable set is contained in a countable dependency-closed set: adjoin the sets \(I_j\) of its members, repeat countably often, and take the union.

Lemma 3.1 (inherited iterations; [OpenAI-NA, Lemma 5.4]). Let \(T\subseteq\delta\) be dependency-closed. Then \(Y_\beta=A[T\cap\beta]\), \(\beta\le\delta\), is a bridge iteration with the same sequences \((q^k_n)\) at the active indices \(k\in T\) and inactive steps at the indices outside \(T\); for active \(k\in T\), \(Y_{k+1}\) is the bridge over \(Y_k\) with stable unitary \(v_k\). In particular \(A[T]\) is simple and has a faithful tracial state, and it is separable if \(T\) is countable.

Proof. Let \(k\in T\) be active. Then \(I_k\subseteq T\cap k\), so \((q^k_n)\) lies in \(Y_k\), and so does \((p_n)\). Both sequences peak in \(X_k\); by Lemma 2.5 of the first lesson, they peak in \(Y_k\subseteq X_k\), and each of the two peaked states of \(Y_k\) has a unique extension to \(X_k\). If these two states of \(Y_k\) were equivalent, Lemma 2.2 of the first lesson would make their extensions to \(X_k\) equivalent, against the choice of the active step \(k\). The algebra \(Y_k\) contains \(A_0\), and its trace is faithful. The trace-preserving expectations of \(X_k\) onto \(C^*(p_n)\) and \(C^*(q^k_n)\) restrict to trace-preserving expectations of \(Y_k\), which are therefore those of \(Y_k\) (Lemma 1.1 of the expectations lesson). By Corollary 3.1 of Reduced HNN adjunctions, \(C^*(Y_k,v_k)=Y_{k+1}\) is the adjunction over \(Y_k\) for these data. Continuity holds because a \(*\)-polynomial involves finitely many indices. Simplicity follows by induction: \(A_0\) is simple, the bridge preserves simplicity, and Lemma 4.3 of the expectations lesson handles limits. A countable set of unitaries and the separable algebra \(A_0\) generate a separable algebra. \(\square\)

4. Moving omitted steps to the end

Theorem 4.1 (closed subiterations; [OpenAI-NA, Theorem 5.5]). For every dependency-closed \(S\subseteq\delta\), the inclusion \(A[S]\subseteq X_\delta\) has property (U).

Proof. Fix an active \(j\notin S\). For \(j+1\le\beta\le\delta\) put \[ \begin{aligned}O_\beta&=A\bigl[j\cup(S\cap\beta)\bigr],\\ R_\beta&=A\bigl[(j+1)\cup(S\cap\beta)\bigr]\\&=C^*(O_\beta,v_j).\end{aligned} \] The index sets are dependency-closed. Since \(j\notin S\), \(O_{j+1}=X_j\) and \(R_{j+1}=X_{j+1}\). By Lemma 3.1, from \(\beta=j+1\) on, both families perform exactly the steps at the active indices \(k\in S\) with \(k>j\).

The omitted pair stays inequivalent. The sequences \((p_n)\) and \((q^j_n)\) peak in \(X_\delta\), hence in every \(O_\beta\) (Lemma 2.5 of the first lesson). We show by induction on \(\beta\) that their states on \(O_\beta\) are inequivalent. For \(\beta=j+1\) this is the choice of the active step \(j\). Let \(k\in S\), \(k>j\), be active, with the two states inequivalent on \(O_k\). The sequence \((q^k_n)\) also lies in \(O_k\) and peaks there, and the three states of \(O_k\) peaked by \((p_n)\), \((q^j_n)\) and \((q^k_n)\) have unique extensions to \(X_k\), by Lemma 2.5 of the first lesson, since the three sequences peak in \(X_k\). In \(X_k\) the \(q^j\)-state is equivalent to the \(p\)-state, because step \(j\) has been performed, and the \(q^k\)-state is not, because step \(k\) is active. If on \(O_k\) the \(q^k\)-state were equivalent to the \(p\)-state or to the \(q^j\)-state, the unique extensions would be equivalent (Lemma 2.2 of the first lesson), a contradiction. So the pair of classes of the \(p\)- and \(q^j\)-states on \(O_k\) differs from the pair joined by the \(k\)-bridge, and part 2 of the bridge theorem keeps them inequivalent on \(O_{k+1}\); their lifts are the states peaked on \(O_{k+1}\). At a limit, Lemma 1.2 applies.

The omitted adjunction commutes past the others. We show that for \(j+1\le\beta\le\delta\), \[ \begin{gathered}R_\beta\ \text{is the adjunction}\\\text{of the }j\text{-data over }O_\beta,\end{gathered}\tag{4.1} \] with stable unitary \(v_j\); by Proposition 2.2 of the adjunction lesson, this means that the trace of \(X_\delta\) vanishes on every reduced word in \(O_\beta\) and \(v_j\) for the expectations of \(O_\beta\) onto \(C^*(p_n)\) and \(C^*(q^j_n)\). For \(\beta=j+1\), (4.1) is the definition of step \(j\). At an active \(k\in S\) with \(k>j\), (4.1) for \(k\) and Lemma 3.1 give \[ \begin{aligned}R_{k+1}&=H_k(R_k)=H_k\bigl(H_j(O_k)\bigr)\\&\cong H_j\bigl(H_k(O_k)\bigr)=H_j(O_{k+1})\end{aligned} \] by Proposition 4.1 of the adjunction lesson. Its hypothesis on the expectations holds: the expectation of \(R_k\) onto \(O_k\), followed by an expectation of \(O_k\) onto a data algebra of step \(k\), is a trace-preserving expectation onto that algebra, so by Lemma 1.1 of the expectations lesson it is the expectation used by the step; the same holds with \(j\) and \(k\) interchanged. The isomorphism preserves the traces and fixes \(O_k\), \(v_j\) and \(v_k\), hence fixes \(O_{k+1}=C^*(O_k,v_k)\). In \(H_j(O_{k+1})\) the trace vanishes on reduced words in \(O_{k+1}\) and \(v_j\); so the trace of \(X_\delta\) vanishes on the same words in \(R_{k+1}\), which is (4.1) for \(k+1\). At a limit \(\gamma\), the data algebras and their expectations are compatible by Lemma 1.1 of the expectations lesson, and Lemma 5.1 of the adjunction lesson gives (4.1) for \(\gamma\).

Conclusion. At \(\beta=\delta\), \(O_\delta=A[S\cup j]\) and \(R_\delta=A[S\cup(j+1)]\). By the two claims and the bridge theorem, applied to \(O_\delta\) (which contains \(A_0\) and has a faithful trace), the inclusion \(A[S\cup j]\subseteq A[S\cup(j+1)]\) has (U). If \(j\in S\) or \(j\) is inactive, this inclusion is an equality. The family \((A[S\cup j])_{j\le\delta}\) is a continuous chain, because a \(*\)-polynomial involves finitely many indices outside \(S\). It starts at \(A[S]\) and ends at \(X_\delta\), and Lemma 1.1 proves the theorem. \(\square\)

5. Countable determination

Lemma 5.1 ([OpenAI-NA, Lemma 5.6]). For every pure state \(h\) of \(X_\delta\) there is a countable dependency-closed \(S\subseteq\delta\) such that \(h|_{A[S]}\) is pure and \(h\) is its only state extension to \(X_\delta\).

Proof. We construct countable dependency-closed sets \(S_0\subseteq S_1\subseteq\cdots\) with \(S_0=\varnothing\). Given \(S_n\), choose a countable dense subset \(C_n\) of the separable algebra \(A[S_n]\). For \(a\in C_n\) and an integer \(m\ge1\), Theorem 3.2 of the first lesson gives \(x_{a,m}\in X_\delta\) with \(0\le x_{a,m}\le1\), \(h(x_{a,m})=1\) and \[ \|x_{a,m}(a-h(a)1)x_{a,m}\|<1/m . \] Each \(x_{a,m}\) lies in \(A[T]\) for a countable \(T\). Let \(S_{n+1}\) be a countable dependency-closed set containing \(S_n\) and all these sets \(T\).

Put \(S=\bigcup_nS_n\), dependency-closed and countable, and \(B=A[S]\), which contains every \(x_{a,m}\). Suppose \(h|_B=th_1+(1-t)h_2\) with states \(h_i\) of \(B\) and \(0<t<1\). Then \(h_i(x_{a,m})=1\), since \(h(x_{a,m})=1\) and \(h_i(x_{a,m})\le1\). By Lemma 3.1 of the first lesson, \(h_i(a)=h_i(x_{a,m}ax_{a,m})\) and \(h_i(x_{a,m}^2)=1\), so \[ \begin{aligned}&|h_i(a)-h(a)|\\&\quad=\bigl|h_i\bigl(x_{a,m}(a-h(a)1)x_{a,m}\bigr)\bigr|<1/m .\end{aligned} \] Hence \(h_i=h\) on every \(C_n\), whose union is dense in \(B\), and \(h|_B\) is pure. By Theorem 4.1, \(h\) is its only extension. \(\square\)

A related countable restriction argument for algebras of free groups is [Akemann–Wassermann–Weaver 2010]. Here the separable algebra is chosen dependency-closed, so that Theorem 4.1 also gives uniqueness of the extension.

Corollary 5.2 ([OpenAI-NA, Corollary 5.7]). Let \(\mathfrak c=2^{\aleph_0}\). If \(|\delta|\le\mathfrak c\), then \(X_\delta\) has at most \(\mathfrak c\) pure states.

Proof. A countable subset of \(\delta\) is the range of a map \(\mathbb N\to\delta\) or is empty, so there are at most \(\mathfrak c^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=\mathfrak c\) of them, using \(|\mathbb N\times\mathbb N|=|\mathbb N|\). For each such \(S\), the separable algebra \(A[S]\) has at most \(\mathfrak c\) states, since a state is determined by its values on a countable dense subset and \(|\mathbb C^{\mathbb N}|=\mathfrak c\). By Lemma 5.1, assigning to a pure state \(h\) of \(X_\delta\) a set \(S\) as there and the restriction \(h|_{A[S]}\) is injective. So there are at most \(\mathfrak c\cdot\mathfrak c=\mathfrak c\) pure states. \(\square\)

6. Exercises

6.1. Show that a union of an increasing sequence of dependency-closed sets is dependency-closed, and that the union of any family of dependency-closed sets is dependency-closed.

6.2. Compute \(f_0\) on \(M_2(\mathbb C)\otimes M_2(\mathbb C)\subset A_0\): show that \(f_0(a\otimes b)=a_{11}b_{11}\).

6.3. Suppose every index of a bridge iteration is inactive. Show that \(A[T]=A_0\) for every \(T\), and that Theorem 4.1 is trivial in this case.

6.4. In the proof of Theorem 4.1, show that \(j\cup(S\cap\beta)\) is dependency-closed when \(S\) is.

6.5. Show that in a bridge iteration the state peaked by \((q^j_n)\) on \(X_\delta\) is equivalent to the state peaked by \((p_n)\), for every active \(j<\delta\).

7. Solutions

6.1. If \(j\) lies in the union, it lies in some member \(T\), and \(I_j\subseteq T\) is contained in the union. The argument does not use that the family is increasing.

6.2. \(p_2=e_{11}\otimes e_{11}\) is the projection onto \(e_1\otimes e_1\), and by the proof of Lemma 2.1, \(f_0\) is the vector state of \(e_1\otimes e_1\) on \(M_4(\mathbb C)\). So \(f_0(a\otimes b)=\langle ae_1,e_1\rangle\langle be_1,e_1\rangle=a_{11}b_{11}\).

6.3. No stable unitaries are adjoined, so \(A[T]=A_0=X_\delta\), and (U) for an equality is trivial.

6.4. Let \(k\in j\cup(S\cap\beta)\). If \(k<j\), then \(I_k\subseteq k\subseteq j\). If \(k\in S\cap\beta\), then \(I_k\subseteq S\) and \(I_k\subseteq k\subseteq\beta\).

6.5. In \(X_{j+1}\) the bridge joins the two classes (part 2 of the bridge theorem), and the states peaked in \(X_{j+1}\) are the unique extensions of those peaked in \(X_j\). By Lemma 1.1, the unique extensions to \(X_\delta\) of equivalent pure states are equivalent, and they are the states peaked in \(X_\delta\).

References