Expectations from peaking projections

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

The unitaries of this course are adjoined over abelian subalgebras generated by peaking sequences, and the construction needs trace-preserving conditional expectations onto these subalgebras. This lesson shows that a peaking sequence whose traces decrease strictly to zero generates a copy of the continuous functions on the one-point compactification of \(\mathbb N_0\), and that the trace and the peaked state together define the expectation. It then proves three facts used throughout the course: unique state extensions and a faithful expectation preserve simplicity; a faithful trace determines norms through moments; and simplicity and faithful traces survive closures of increasing unions. The notation is that of Pure states, excision and peaking projections.

1. Conditional expectations

Let \(B\subset C\) be a unital inclusion. A conditional expectation \(E\colon C\to B\) is a unital completely positive map with \(E(b)=b\) for \(b\in B\) and \[ \begin{gathered}E(b_1cb_2)=b_1E(c)b_2\\ (b_1,b_2\in B,\ c\in C).\end{gathered} \] It is faithful if \(E(c^*c)=0\) only for \(c=0\). If \((C,\tau)\) is tracial, \(E\) is trace-preserving when \(\tau\circ E=\tau\).

Lemma 1.1 ([OpenAI-NA, Lemma 2.8]). Let \((C,\tau)\) be a tracial algebra and \(B\subset C\) a unital C\(^*\)-subalgebra. There is at most one trace-preserving conditional expectation \(C\to B\), and such an expectation is faithful.

Proof. For a trace-preserving conditional expectation \(E\), bimodularity gives \[ \begin{gathered}\tau(b^*E(c))=\tau(E(b^*c))=\tau(b^*c)\\ (b\in B,\ c\in C).\end{gathered} \] If \(E'\) is another one, \(d=E(c)-E'(c)\in B\) satisfies \(\tau(b^*d)=0\) for every \(b\in B\). With \(b=d\) we get \(\tau(d^*d)=0\), so \(d=0\). Finally \(\tau(E(c^*c))=\tau(c^*c)>0\) for \(c\ne0\), so \(E\) is faithful. \(\square\)

2. The expectation onto a peaking algebra

Let \(\mathbb N_0\cup\{\infty\}\) be the one-point compactification of \(\mathbb N_0\). For a decreasing sequence of projections \((p_n)_{n\ge0}\) with \(p_0=1\) we write \(d_n=p_n-p_{n+1}\).

Proposition 2.1 ([OpenAI-NA, Lemma 2.9]). Let \((B,\tau)\) be a tracial algebra, and let \((p_n)_{n\ge0}\) be a decreasing sequence of projections of \(B\) that peaks to a state \(h\), with \(p_0=1\) and \[ \begin{gathered}1=\tau(p_0)>\tau(p_1)>\tau(p_2)>\cdots,\\\tau(p_n)\to0 .\end{gathered} \] Then \(D=C^*(p_n:n\ge0)\) is isomorphic to \(C(\mathbb N_0\cup\{\infty\})\) by an isomorphism carrying \(p_n\) to the indicator function of \(\{n,n+1,\ldots\}\cup\{\infty\}\). There is a trace-preserving conditional expectation \(E\colon B\to D\), given in these coordinates by \[ \begin{gathered}E(a)(n)=\frac{\tau(d_nad_n)}{\tau(d_n)}\\ (n\in\mathbb N_0),\\E(a)(\infty)=h(a).\end{gathered}\tag{2.1} \]

Proof. The algebra \(D\). The \(d_n\) are pairwise orthogonal projections, nonzero because \(\tau(d_n)=\tau(p_n)-\tau(p_{n+1})>0\), and \(p_n=1-\sum_{j<n}d_j\). For finitely many scalars, \(\|\sum_jc_jd_j\|=\max_j|c_j|\), so \((c_j)\mapsto\sum_jc_jd_j\) extends to an isometric \(*\)-isomorphism of \(c_0(\mathbb N_0)\) onto the closed linear span \(D_0\) of the \(d_j\). The unit is not in \(D_0\): for \(c\) in the span of \(d_0,\ldots,d_{n-1}\), \(p_nc=0\), so \(\|1-c\|\ge\|p_n(1-c)\|=\|p_n\|=1\) (here \(p_n\ne0\), as \(\tau(p_n)>0\)); by approximation \(\|1-c\|\ge1\) for every \(c\in D_0\). Hence \(D=\mathbb C1+D_0\) is the unitization of \(D_0\cong c_0(\mathbb N_0)\), which is \(C(\mathbb N_0\cup\{\infty\})\), with \(d_j\) corresponding to the indicator of \(\{j\}\) and \(p_n\) to the indicator of \(\{n,n+1,\ldots\}\cup\{\infty\}\). In these coordinates, an element \(f\in D\) satisfies \[ \begin{gathered}fd_n=d_nf=f(n)\,d_n\\ (n\in\mathbb N_0).\end{gathered}\tag{2.2} \]

The values. For \(z\in B\), the Cauchy–Schwarz inequality for \(\tau\) gives \[ |\tau(d_nzd_n)|\le\tau(d_n)^{1/2}\,\tau(d_nz^*zd_n)^{1/2}, \] and therefore \[ |\tau(d_nzd_n)|\le\|z\|\,\tau(d_n). \tag{2.3} \] So \(|E(a)(n)|\le\|a\|\). As \(d_n\le p_n\), \(d_n=d_np_n=p_nd_n\), and (2.3) with \(z=p_n(a-h(a)1)p_n\) gives \[ \begin{gathered}\Bigl|\frac{\tau(d_nad_n)}{\tau(d_n)}-h(a)\Bigr|\\\le\|p_n(a-h(a)1)p_n\|\longrightarrow0\end{gathered} \] by Lemma 3.3 of the first lesson. Hence the function \(E(a)\) of (2.1) is continuous on \(\mathbb N_0\cup\{\infty\}\), and \(E(a)\in D\).

Expectation. Each coordinate \(a\mapsto E(a)(n)\), and \(a\mapsto h(a)\), is a state, so \(E\) is unital and positive. Its range is abelian, so \(E\) is completely positive (Theorem 5.1(1) of Completely positive maps). For \(j\in\mathbb N_0\), \(E(d_j)(n)=\tau(d_nd_jd_n)/\tau(d_n)\) is \(1\) for \(n=j\) and \(0\) otherwise, and \(E(d_j)(\infty)=h(p_j)-h(p_{j+1})=0\). So \(E(d_j)=d_j\) and \(E(1)=1\); since \(E\) is bounded and these elements span a dense subspace of \(D\), \(E\) fixes \(D\).

For bimodularity, let \(f\in D\) and \(a\in B\). By (2.2), \(E(fa)(n)=f(n)E(a)(n)\) and \(E(af)(n)=E(a)(n)f(n)\). At \(\infty\): \(h(p_n)=1\) for every \(n\) gives \(h(d_j)=0\), so \(h(f)=f(\infty)\) for \(f\in D\), and \(h\) is multiplicative on \(D\). With \(g=f-f(\infty)1\), the Cauchy–Schwarz inequality gives \(|h(ga)|^2\le h(gg^*)h(a^*a)=0\) and \(|h(ag)|^2\le h(a a^*)h(g^*g)=0\). So \(h(fa)=f(\infty)h(a)\) and \(h(af)=h(a)f(\infty)\). Hence \(E(f_1af_2)=f_1E(a)f_2\).

Trace. For \(f\in D\), \(f=f(\infty)1+\sum_j(f(j)-f(\infty))d_j\) with norm convergence, and \(\sum_j\tau(d_j)=\lim_n(1-\tau(p_n))=1\); so \[ \tau(f)=\sum_jf(j)\,\tau(d_j). \] Therefore, using \(\tau(d_nad_n)=\tau(d_na)\), \[ \begin{aligned}\tau(E(a))&=\sum_n\tau(d_nad_n)\\&=\lim_N\tau\bigl((1-p_N)a\bigr)=\tau(a),\end{aligned} \] because \(|\tau(p_Na)|\le\tau(p_N)^{1/2}\tau(a^*p_Na)^{1/2}\), which is at most \(\tau(p_N)^{1/2}\|a\|\to0\). \(\square\)

By Lemma 1.1, \(E\) is the only trace-preserving conditional expectation onto \(D\), and it is faithful.

3. Simple extensions

A C\(^*\)-algebra is simple if its only closed two-sided ideals are \(\{0\}\) and itself.

Proposition 3.1 ([OpenAI-NA, Lemma 2.10]). Let \(B\subset C\) satisfy (U), let \(B\) be simple, and let \(E\colon C\to B\) be a faithful conditional expectation. Then \(C\) is simple.

Proof. Let \(J\) be a proper closed two-sided ideal of \(C\), and \(q\colon C\to C/J\) the quotient map. Its kernel meets \(B\) in an ideal of \(B\) not containing \(1\), so \(q|_B\) is injective, and hence isometric (Corollary 4.6 of the C\(^*\)-algebra lesson). Let \(h\) be a pure state of \(B\). The state \(h\circ(q|_B)^{-1}\) of \(q(B)\) extends to a state of \(C/J\), and composing with \(q\) gives a state of \(C\) that extends \(h\) and vanishes on \(J\). The state \(h\circ E\) also extends \(h\). By (U), the two coincide, so \(h(E(c))=0\) for every \(c\in J\). This holds for every pure state \(h\) of \(B\). The states of \(B\) form the weak\(^*\)-closed convex hull of the pure states (Proposition 8.4(4) of the GNS lesson), so every state vanishes on \(E(c)\). For \(c\in J\), \(c^*c\in J\) and \(E(c^*c)\ge0\); a positive element on which every state vanishes is zero (Proposition 7.1a there), so \(E(c^*c)=0\) and \(c=0\) by faithfulness. Hence \(J=0\). \(\square\)

4. Norms from trace moments, and limits

Lemma 4.1 (moments determine norms). Let \(t\) be a faithful state of a C\(^*\)-algebra \(C\). For every \(x\in C\), \[ \|x\|=\lim_{r\to\infty}t\bigl((x^*x)^r\bigr)^{1/(2r)} . \]

Proof. Put \(y=x^*x\ge0\) and identify \(C^*(1,y)\) with \(C(\sigma(y))\). The restriction of \(t\) is integration against a probability measure \(\mu\) on \(\sigma(y)\). Every nonempty open subset \(U\) of \(\sigma(y)\) has \(\mu(U)>0\): otherwise a nonzero continuous \(f\ge0\) supported in \(U\) gives \(t(f(y))=0\) with \(f(y)\ne0\), against faithfulness. Let \(M=\max\sigma(y)=\|y\|\). For \(0<\varepsilon<M\), the set \(U=\{\lambda>M-\varepsilon\}\) has \(\mu(U)>0\), so \[ \mu(U)^{1/r}(M-\varepsilon)\le\Bigl(\int\lambda^r\,d\mu\Bigr)^{1/r}\le M . \] Letting \(r\to\infty\) and then \(\varepsilon\to0\) gives \(t(y^r)^{1/r}\to M=\|x\|^2\). If \(M=0\), then \(x=0\) and both sides vanish. \(\square\)

Corollary 4.2. Let \(C_1,C_2\) be C\(^*\)-algebras with faithful states \(t_1,t_2\), generated by families \((g_i)_{i\in I}\) and \((g_i')_{i\in I}\). Suppose \(t_1(P(g))=t_2(P(g'))\) for every noncommutative \(*\)-polynomial \(P\). Then there is a \(*\)-isomorphism \(\Phi\colon C_1\to C_2\) with \(\Phi(g_i)=g_i'\) and \(t_2\circ\Phi=t_1\).

Proof. For a \(*\)-polynomial \(P\), the moments of \(P(g)^*P(g)\) are values of \(t_1\) on \(*\)-polynomials in \(g\), and they agree with the corresponding moments for \(g'\). By Lemma 4.1, \(\|P(g)\|=\|P(g')\|\). So \(P(g)\mapsto P(g')\) is a well-defined isometric \(*\)-homomorphism between dense \(*\)-subalgebras, and it extends to the required isomorphism. \(\square\)

Let \((C_\beta)_{\beta\in\Lambda}\) be a family of C\(^*\)-algebras indexed by a directed set, each a unital C\(^*\)-subalgebra of the next. The norms are compatible, and the completion of the union is a C\(^*\)-algebra \(C\), in which the union is dense.

Lemma 4.3 (limits). In this situation:

  1. If every \(C_\beta\) is simple, then \(C\) is simple.
  2. If \(\tau_\beta\) are faithful tracial states of the \(C_\beta\) with \(\tau_\gamma|_{C_\beta}=\tau_\beta\) for \(\beta\le\gamma\), their common extension \(\tau\) is a faithful tracial state of \(C\).

Proof. (1) Let \(J\) be a proper closed ideal of \(C\) and \(q\colon C\to C/J\) the quotient map. For each \(\beta\), \(J\cap C_\beta\) is an ideal of \(C_\beta\) not containing \(1\), hence zero, so \(q\) is injective and therefore isometric on \(C_\beta\). Thus \(q\) is isometric on the union, hence on \(C\), and \(J=0\).

(2) The functional defined on the union by the \(\tau_\beta\) has norm one, so it extends to a state \(\tau\) of \(C\), tracial by continuity. Let \(I=\{c\in C:\tau(c^*c)=0\}\). It is a closed left ideal (Lemma 3.3 of the GNS lesson), and a right ideal: \(\tau(b^*c^*cb)=\tau(cbb^*c^*)\le\|b\|^2\tau(cc^*)\), and \(\tau(cc^*)=\tau(c^*c)\). So \(I\) is a closed two-sided ideal, and \(I\cap C_\beta=0\) because \(\tau_\beta\) is faithful. As in (1), the quotient map \(C\to C/I\) is isometric on every \(C_\beta\), hence on \(C\), and \(I=0\). \(\square\)

5. Exercises

5.1. In Proposition 2.1, show that \(E(p_map_m)=p_mE(a)p_m\) for every \(m\), and compute \(E(p_m)\).

5.2. Let \(C=\mathbb C^2\) and \(t(x,y)=x\). Show that Lemma 4.1 fails for \(t\), and identify the hypothesis that fails.

5.3. Let \(B=M_2(\mathbb C)\), embedded in \(C=M_2(\mathbb C)\otimes M_2(\mathbb C)\) by \(b\mapsto b\otimes1\). Show that \(C\) is simple but (U) fails for \(B\subset C\).

5.4. Let \(C_n=M_{2^n}(\mathbb C)\), embedded in \(C_{n+1}\) by \(a\mapsto a\otimes1\). Show that the completion of the union is simple and carries a faithful tracial state.

5.5. In the setting of Proposition 2.1, show that \(\tau\) restricted to \(D\) corresponds to the probability measure \(\sum_j\tau(d_j)\delta_j\), with no mass at \(\infty\). Which hypothesis ensures this?

6. Solutions

5.1. Bimodularity with \(f_1=f_2=p_m\in D\) gives \(E(p_map_m)=p_mE(a)p_m\). As \(E\) fixes \(D\), \(E(p_m)=p_m\), the indicator of \(\{m,m+1,\ldots\}\cup\{\infty\}\).

5.2. For \(x=(0,1)\), \(t((x^*x)^r)=0\) for every \(r\), while \(\|x\|=1\). The state \(t\) is not faithful: \(t(x^*x)=0\) with \(x\ne0\).

5.3. \(C\cong M_4(\mathbb C)\) is simple. A pure state of \(B\) is a vector state \(\omega_\xi\), and \(\omega_\xi\otimes\varphi\) is a state extension for every state \(\varphi\) of \(M_2(\mathbb C)\); these differ for different \(\varphi\).

5.4. Each \(M_{2^n}(\mathbb C)\) is simple, and the normalized traces are compatible, since \(\operatorname{tr}(a\otimes1)=\operatorname{tr}(a)\), and faithful. Lemma 4.3 applies.

5.5. By the computation in the proof, \(\tau(f)=\sum_jf(j)\tau(d_j)\) and \(\sum_j\tau(d_j)=1\). This uses \(\tau(p_n)\to0\): the mass at \(\infty\) is \(\lim_n\tau(p_n)\).

References