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Reduced HNN adjunctions

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

Given two isomorphic subalgebras of a tracial algebra, one can adjoin a unitary that conjugates the first onto the second, in the freest possible way: every reduced word in the algebra and the new unitary has expectation zero. This is a reduced HNN extension, named after the corresponding construction for groups by Higman, Neumann and Neumann. This lesson constructs it as a corner of the reduced amalgamated free product of Reduced amalgamated free products of tracial algebras, following Ueda's matrix construction [Ueda 2008, Proposition 2.2 and Remark 2.3] in the form of [OpenAI-NA, Section 3]. It then proves the properties used later: uniqueness, restriction to a smaller base, interchange of two adjunctions, and passage to limits.

Setting. \((A,\tau)\) is a tracial algebra; \(D_1,D_2\subset A\) are unital C\(^*\)-subalgebras with trace-preserving conditional expectations \(E_i\colon A\to D_i\); and \(\theta\colon D_1\to D_2\) is a trace-preserving \(*\)-isomorphism. A word in \(A\) and a unitary \(v\) is a product \[ \begin{gathered}a_0v^{\epsilon_1}a_1v^{\epsilon_2}\cdots v^{\epsilon_m}a_m,\\a_t\in A,\ \epsilon_t=\pm1,\ m\ge1 .\end{gathered} \] It is reduced if every internal coefficient \(a_t\), \(0<t<m\), satisfies \[ \begin{gathered}E_1(a_t)=0\ \text{ if}\\ (\epsilon_t,\epsilon_{t+1})=(+1,-1),\\E_2(a_t)=0\ \text{ if}\\ (\epsilon_t,\epsilon_{t+1})=(-1,+1).\end{gathered}\tag{0.1} \] There is no condition when consecutive signs agree, and none on \(a_0\) and \(a_m\).

1. The reduced adjunction

Theorem 1.1 ([OpenAI-NA, Proposition 3.1]; [Ueda 2008]). There is a C\(^*\)-algebra \(H(A)\) containing \(A\) as a unital subalgebra, with a faithful tracial state \(\tau_H\) extending \(\tau\), generated by \(A\) and a unitary \(v\) with \[ vdv^*=\theta(d)\qquad(d\in D_1), \] and a faithful trace-preserving conditional expectation \(E_A\colon H(A)\to A\) that vanishes on every reduced word.

The unitary \(v\) is the stable unitary of the adjunction.

Proof. The free product. Put \[ \begin{gathered}F=M_2(A),\qquad G=M_2(D_1),\\D=D_1\oplus D_2 ,\end{gathered} \] with the normalized traces \(\tau_F(a)=\frac12(\tau(a_{11})+\tau(a_{22}))\) and \(\tau_G\) defined in the same way. Embed \(D\) in \(F\) by \((d_1,d_2)\mapsto\operatorname{diag}(d_1,d_2)\), and in \(G\) by \((d_1,d_2)\mapsto\operatorname{diag}(d_1,\theta^{-1}(d_2))\). Both traces restrict to \(\tau_D(d_1,d_2)=\frac12(\tau(d_1)+\tau(d_2))\), because \(\theta\) preserves the trace. The maps \[ \begin{gathered} E^F_D(a)=\bigl(E_1(a_{11}),E_2(a_{22})\bigr),\\ E^G_D(b)=\bigl(b_{11},\theta(b_{22})\bigr) \end{gathered} \] are trace-preserving conditional expectations onto the images of \(D\): the first is the diagonal compression followed by \(E_1\oplus E_2\), the second is the diagonal compression of \(M_2(D_1)\) written in the coordinates of \(D\). Let \(Z=F*_DG\), with trace \(\tau_Z\) and expectations \(E_D\) and \(E_F\) (Propositions 3.2 and 4.2 of the free product lesson).

The corner. The diagonal projections \(e_1=(1,0)\) and \(e_2=(0,1)\) of \(D\) are matrix units in both factors. Write \(e^F_{jk}\) and \(e^G_{jk}\) for the matrix units of \(F\) and \(G\), and put \[ \begin{gathered}s=e^G_{21},\qquad q=e^F_{12},\\v=qs,\qquad H(A)=e_1Ze_1 ,\end{gathered} \] identifying \(a\in A\) with \(ae^F_{11}\in e_1Fe_1\). Then \(v^*v=s^*e_2s=e^G_{11}=e_1\) and \(vv^*=qe_2q^*=e^F_{11}=e_1\), so \(v\) is a unitary of the corner. For \(d\in D_1\), \(de_1=(d,0)\in D\), and \(s(de_1)s^*\) is the matrix with \(d\) in position \((2,2)\) of \(G\), which is the element \((0,\theta(d))\) of \(D\), that is \(\theta(d)\) in position \((2,2)\) of \(F\). Hence \(vdv^*=q\,(0,\theta(d))\,q^*=\theta(d)\).

The matrix units of \(F\) identify \(Z\) with \(M_2(e_1Ze_1)\). In this identification \(M_2(C^*(A,v))\) contains \(F=M_2(A)\) and \(s=e^F_{21}v\). It also contains \(G\): for \(b\in D_1\), \(be^G_{11}=(b,0)\) and \(be^G_{22}=(0,\theta(b))\) lie in \(D\), while \(be^G_{21}=s\,(b,0)\) and \(be^G_{12}=s^*(0,\theta(b))\). So \(M_2(C^*(A,v))=Z\) and \(H(A)=C^*(A,v)\).

Trace and expectation. \(\tau_Z(e_1)=\frac12\), so \(\tau_H=2\tau_Z\) restricted to \(H(A)\) is a faithful tracial state, and \(\tau_H(a)=2\tau_F(ae^F_{11})=\tau(a)\) for \(a\in A\). As \(e_1\in F\), \(E_F(e_1ze_1)=e_1E_F(z)e_1\in e_1Fe_1=A\), so \(E_A=E_F|_{H(A)}\) is a faithful conditional expectation onto \(A\), and \(\tau\circ E_A=2\tau_F\circ E_F=\tau_H\).

Reduced words. Expand a word using \(v=qs\) and \(v^*=s^*q^*\). Between two consecutive letters \(s^{\pm1}\), the coefficient from \(F\) is \[ \begin{array}{c|c}(\epsilon_t,\epsilon_{t+1})&\text{coefficient}\\\hline(+,+)&a_tq\\ (+,-)&a_t\\ (-,+)&q^*a_tq\\ (-,-)&q^*a_t\end{array} \] The first and last are off-diagonal, so \(E^F_D\) vanishes on them. The second is \(a_t\) in position \((1,1)\), with \(E^F_D(a_t)=(E_1(a_t),0)\); the third is \(a_t\) in position \((2,2)\), with \(E^F_D=(0,E_2(a_t))\). By (0.1) all internal coefficients are centered. The letters \(s,s^*\) are centered for \(E^G_D\), being off-diagonal. So a reduced word is a product \(c_0wc_m\) with \(c_0,c_m\in F\) and \(w\) a core in the sense of Section 4 of the free product lesson, and \(E_F\) vanishes on it (Proposition 4.2 there). \(\square\)

2. Uniqueness, and a spanning set

Lemma 2.1. In any algebra generated by \(A\) and a unitary \(v\) with \(vdv^*=\theta(d)\) for \(d\in D_1\), every \(*\)-polynomial in \(A\) and \(v\) is the sum of an element of \(A\) and a linear combination of reduced words. The element of \(A\) and the words are obtained from the polynomial by operations that use only \(E_1,E_2,\theta\) and these relations.

Proof. A \(*\)-polynomial is a linear combination of elements of \(A\) and of words, since \(v^*=v^{-1}\). Let a word have an internal coefficient \(a_t\) at a turn \((+1,-1)\) with \(E_1(a_t)\ne0\). Write \(a_t=(a_t-E_1(a_t))+E_1(a_t)\). The first part gives a word in which this turn satisfies (0.1). The second gives a word with two fewer stable letters, since \(vE_1(a_t)v^*=\theta(E_1(a_t))\in A\). A turn \((-1,+1)\) is treated in the same way with \(v^*E_2(a_t)v=\theta^{-1}(E_2(a_t))\). Induction on the number of stable letters and on the number of non-reduced turns proves the claim. \(\square\)

Proposition 2.2 (uniqueness; [OpenAI-NA, Proposition 3.1]). Let \(C\) be a C\(^*\)-algebra with a faithful tracial state \(t\), generated by a copy of \(A\), on which \(t\) restricts to \(\tau\), and a unitary \(v'\) with \(v'dv'^*=\theta(d)\) for \(d\in D_1\). If \(t\) vanishes on every reduced word in \(A\) and \(v'\), there is a \(*\)-isomorphism \(\Phi\colon H(A)\to C\) fixing \(A\), with \(\Phi(v)=v'\) and \(t\circ\Phi=\tau_H\).

Proof. By Lemma 2.1, applied in both algebras with the same operations, a \(*\)-polynomial \(P\) satisfies \(P(A,v)=a_P+\sum\text{reduced words}\) and \(P(A,v')=a_P+\sum\text{same words in }v'\), with the same \(a_P\in A\). Hence \(\tau_H(P(A,v))=\tau(a_P)=t(P(A,v'))\). Corollary 4.2 of Expectations from peaking projections gives \(\Phi\). \(\square\)

So \(H(A)\) is determined, up to an isomorphism fixing \(A\) and \(v\), by the relation \(vdv^*=\theta(d)\), the trace on \(A\), and the vanishing of the trace on reduced words. We call it the reduced HNN adjunction of \(\theta\) over \(A\).

For the next lessons we record a spanning set of \(Z=F*_DG\) in the model of Theorem 1.1.

Lemma 2.3 ([OpenAI-NA, (3.4)]). The space \(F\) together with the products \[ \begin{gathered}a_0b_1a_1\cdots b_ma_m,\quad m\ge1,\\ b_t\in\{s,s^*\},\ a_t\in F,\end{gathered}\tag{2.1} \] with \(E^F_D(a_t)=0\) for \(0<t<m\), spans a dense subspace of \(Z\).

Proof. A centered element of \(G=M_2(D_1)\) is off-diagonal, hence of the form \(b'e^G_{21}+be^G_{12}=s\,(b',0)+s^*(0,\theta(b))\) with \(b,b'\in D_1\). In the products \(a_0wa_m\) of Proposition 4.2 of the free product lesson, replace each letter of the core \(w\) from \(G\) by this sum and absorb the elements of \(D\) into the neighbouring coefficients from \(F\). Internal coefficients stay centered, since \(E^F_D(\delta a)=\delta E^F_D(a)\) for \(\delta\in D\). By that proposition, the result spans a dense subspace. \(\square\)

3. Restriction

Corollary 3.1 ([OpenAI-NA, Corollary 3.3]). Let \(A'\subset A\) be a unital C\(^*\)-subalgebra containing \(D_1\) and \(D_2\). Then \(C^*(A',v)\subset H(A)\) is the reduced HNN adjunction \(H(A')\) formed with \(\tau|_{A'}\) and \(E_i|_{A'}\), and \(E_A\) restricts to the expectation \(H(A')\to A'\).

Proof. The restrictions \(E_i|_{A'}\) are trace-preserving conditional expectations onto \(D_i\subset A'\). The algebra \(C^*(A',v)\) with the restriction of \(\tau_H\) is faithfully tracial, satisfies the relation, and its trace vanishes on reduced words over \(A'\), which are reduced words over \(A\). Proposition 2.2 identifies it with \(H(A')\). By Lemma 2.1, a \(*\)-polynomial in \(A'\) and \(v\) is an element of \(A'\) plus reduced words over \(A'\), since \(E_i(A')\subseteq D_i\subseteq A'\) and \(\theta^{\pm1}\) maps \(D_1,D_2\) into \(A'\). So \(E_A\) maps it into \(A'\), and by continuity \(E_A(C^*(A',v))\subseteq A'\). The restriction is a trace-preserving conditional expectation onto \(A'\), unique by Lemma 1.1 of the expectations lesson. \(\square\)

4. Interchanging two adjunctions

Let \((O,\tau)\) be a tracial algebra with two sets of data \((D^t_1,D^t_2,E^t_1,E^t_2,\theta_t)\), \(t\in\{j,k\}\), each as in the setting. Write \(H_t(O)\) for the adjunction with stable unitary \(v_t\), and \(E^{(t)}_O\colon H_t(O)\to O\) for its expectation. To form the \(k\)-adjunction over \(H_j(O)\), use the expectations \(E^k_i\circ E^{(j)}_O\colon H_j(O)\to D^k_i\), which are trace-preserving conditional expectations because \(D^k_i\subset O\); similarly with \(j\) and \(k\) interchanged.

Proposition 4.1 ([OpenAI-NA, Proposition 3.4]). There is a trace-preserving \(*\)-isomorphism \[ H_k\bigl(H_j(O)\bigr)\cong H_j\bigl(H_k(O)\bigr) \] fixing \(O\), \(v_j\) and \(v_k\).

Proof. Let \(C=H_k(H_j(O))\). By Corollary 3.1, with \(A'=O\subset A=H_j(O)\), \(C\) contains \(H_k(O)=C^*(O,v_k)\), and the expectation \(C\to H_j(O)\) maps \(H_k(O)\) onto \(O\) as \(E^{(k)}_O\). Let \(E_O=E^{(j)}_O\circ E_{H_j(O)}\colon C\to O\). It restricts to \(E^{(j)}_O\) on \(H_j(O)\) and to \(E^{(k)}_O\) on \(H_k(O)\).

Freeness. Let \(x_1\cdots x_n\) be a product whose factors alternate between \(\ker E^{(j)}_O\subset H_j(O)\) and \(\ker E^{(k)}_O\subset H_k(O)\). We show \(E_O(x_1\cdots x_n)=0\). If \(n=1\), this is the hypothesis. Otherwise some factor lies in \(H_k(O)\). By Lemma 2.1, a polynomial in \(O\) and \(v_k\) approximating a centered element \(y\in H_k(O)\) is an element of \(O\) plus reduced words; subtracting its expectation, \(y\) is a norm limit of linear combinations of nonempty reduced \(k\)-words over \(O\). Replace each factor from \(H_k(O)\) by such a combination, and keep the factors from \(H_j(O)\). When two \(k\)-words are joined across a centered \(b\in H_j(O)\), the new internal coefficient is \(aba'\) with \(a,a'\in O\), and \[ E^k_i\bigl(E^{(j)}_O(aba')\bigr)=E^k_i\bigl(aE^{(j)}_O(b)a'\bigr)=0 . \] The other internal coefficients lie in \(O\) and keep their conditions, as \(E^k_i\circ E^{(j)}_O=E^k_i\) on \(O\). Initial and final coefficients from \(H_j(O)\) are allowed. So each resulting term is a nonempty reduced \(k\)-word over \(H_j(O)\), on which \(E_{H_j(O)}\), and hence \(E_O\), vanishes. Passing to the limit gives the claim.

Moments. Every \(*\)-polynomial in \(O,v_j,v_k\) is a linear combination of products \(y_1\cdots y_n\) of elements alternating between \(H_j(O)\) and \(H_k(O)\). Write \(y_t=E_O(y_t)+y_t^\circ\) and expand. The term with all factors centered has expectation zero, and every other term becomes, after absorbing the elements of \(O\) into neighbouring factors, a product with fewer alternating factors. So \(E_O(y_1\cdots y_n)\) is computed from the expectations \(E^{(j)}_O\) and \(E^{(k)}_O\) by a recursion symmetric in \(j\) and \(k\). The same holds in \(H_j(H_k(O))\). Hence the two algebras give the same value of the trace on every \(*\)-polynomial in \(O,v_j,v_k\), and Corollary 4.2 of the expectations lesson gives the isomorphism. \(\square\)

By Lemma 1.1 of the expectations lesson, restrictions and composites of these expectations agree whenever they have the same range and preserve the trace.

5. Limits

Lemma 5.1 ([OpenAI-NA, Lemma 3.5]). Let \((O_\beta)_{\beta\in\Lambda}\) be an increasing family of C\(^*\)-algebras indexed by a directed set, each a unital subalgebra of the next, with compatible faithful tracial states. Let \(D_1,D_2\) lie in every \(O_\beta\), with a fixed trace-preserving isomorphism \(\theta\colon D_1\to D_2\) and compatible trace-preserving conditional expectations \(E^\beta_i\colon O_\beta\to D_i\). Identify the adjunctions \(H(O_\beta)\) through Corollary 3.1, with one stable unitary \(v\), and let \(O\) and \(R\) be the completions of \(\bigcup_\beta O_\beta\) and \(\bigcup_\beta H(O_\beta)\). Then the expectations extend to \(E_i\colon O\to D_i\), and \(R=H(O)\) with its generators and trace.

Proof. The compatible maps \(E^\beta_i\) have norm one and extend continuously to \(O\), keeping complete positivity, bimodularity and trace preservation. The traces extend to faithful tracial states of \(O\) and \(R\) by Lemma 4.3 of the expectations lesson. The algebra \(R\) is generated by \(O\) and \(v\), and the relation \(vdv^*=\theta(d)\) holds. Let \(w=a_0v^{\epsilon_1}\cdots v^{\epsilon_m}a_m\) be a reduced word with coefficients in \(O\), and \(\varepsilon>0\). Approximate its finitely many coefficients by elements \(b_t\) of one \(O_\beta\) within \(\varepsilon\). At a turn requiring \(E_i(a_t)=0\), replace \(b_t\) by \(b_t-E^\beta_i(b_t)\), which lies in \(O_\beta\), is centered, and differs from \(a_t\) by at most \(2\varepsilon\), since \(\|E_i(b_t)\|=\|E_i(b_t-a_t)\|\le\varepsilon\). The new word is reduced over \(O_\beta\), so its trace is zero, and it is close to \(w\). Hence the trace of \(R\) vanishes on \(w\), and Proposition 2.2 identifies \(R\) with \(H(O)\). \(\square\)

6. Exercises

6.1. Let \(D_1=D_2=\mathbb C1\) and \(\theta=\mathrm{id}\). Show that \(\tau_H(v^n)=0\) for every integer \(n\ne0\).

6.2. Show that \(E_A(vav^*)=\theta(E_1(a))\) for \(a\in A\).

6.3. Show that \(\tau_H(vav^*b)=\tau\bigl(\theta(E_1(a))\,b\bigr)\) for \(a,b\in A\).

6.4. Show that \(v^*\theta(d)v=d\) for \(d\in D_1\), and that \(E_A(v^*av)=\theta^{-1}(E_2(a))\) for \(a\in A\).

6.5. In Lemma 2.3, write the reduced word \(v\,a\,v^*\), with \(E_1(a)=0\), in the form (2.1).

7. Solutions

6.1. For \(n\ge1\), \(v^n=1\cdot v\cdot1\cdots v\cdot1\) has equal consecutive signs, so it is a reduced word and \(E_A(v^n)=0\); likewise for \(n\le-1\). Hence \(\tau_H(v^n)=\tau(E_A(v^n))=0\).

6.2. Write \(a=E_1(a)+a^\circ\). Then \(vE_1(a)v^*=\theta(E_1(a))\in A\), and \(va^\circ v^*\) is a reduced word because its turn \((+1,-1)\) has a centered coefficient. So \(E_A(vav^*)=\theta(E_1(a))\).

6.3. By bimodularity and Exercise 6.2, \(\tau_H(vav^*b)=\tau(E_A(vav^*)b)\), which is \(\tau(\theta(E_1(a))b)\).

6.4. From \(v\,d\,v^*=\theta(d)\) we get \(d=v^*\theta(d)v\). Write \(a=E_2(a)+a^\circ\); then \(v^*E_2(a)v=\theta^{-1}(E_2(a))\), and \(v^*a^\circ v\) is reduced at its turn \((-1,+1)\).

6.5. With \(v=qs\) and \(v^*=s^*q^*\), \(vav^*=q\,s\,a\,s^*\,q^*\), which is (2.1) with \(m=2\), \(a_0=q\), \(a_1=a\) in position \((1,1)\), \(a_2=q^*\), \(b_1=s\) and \(b_2=s^*\); here \(E^F_D(a_1)=(E_1(a),0)=0\).

References