A full cone in a tensor power

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

Let \(D\) be a nonzero unital C*-algebra without characters. This lesson finds, in a finite maximal tensor power \(P\) of \(D\), a full element \(w\) with \(w^2=0\) (Lemma 2.4), and turns it into a unital homomorphism from the matrix cone \[ C=\{f\in C([0,1],M_2):f(0)\in\mathbb C1\} \] into \(P\) whose first matrix corner is full (Proposition 3.1). The last lesson, From the cone to the dimension-drop algebra, turns this into a homomorphism from \(I(2,3)\).

The lesson follows [OpenAI-KRC, Sections 4.3 and 4.4]. Kirchberg and Rørdam combine finitely many elements by orthogonal full tensor labels [KR, Proposition 6.1]; here the labels of Labels that survive representations do this through linear independence. All tensor products are maximal, \(D^{\otimes k}\) is the maximal tensor power, and \(\iota\in C\) is the function \(\iota(t)=t\), times \(1\) or times a matrix unit as indicated.

1. Full elements

An element \(x\) of a unital C*-algebra \(P\) is full if the closed two-sided ideal generated by \(x\) is \(P\).

Lemma 1.1. For \(x\in P\) the following are equivalent:

  1. \(x\) is full;
  2. there are \(h_1,\dots,h_N,l_1,\dots,l_N\in P\) with \(\bigl\|1-\sum_ih_ixl_i\bigr\|<1\);
  3. \(\pi(x)\ne0\) for every irreducible representation \(\pi\) of \(P\).

Proof. (1)\(\Rightarrow\)(2): the elements \(\sum_ih_ixl_i\) form a dense subset of the closed ideal, which is \(P\). (2)\(\Rightarrow\)(3): \(y=\sum_ih_ixl_i\) is invertible, and an irreducible \(\pi\) is unital, so \(\pi(y)\pi(y^{-1})=1\ne0\) and \(\pi(x)\ne0\). (3)\(\Rightarrow\)(1): if the closed ideal \(I\) generated by \(x\) is proper, \(P/I\) is a nonzero C*-algebra (Order, local units and quotients of C*-algebras, Theorem 15.1). It has an irreducible representation, because norms are attained in irreducible representations (Building representations from positive functionals, Theorem 8.5). Composed with the quotient map, it is an irreducible representation of \(P\) that vanishes at \(x\). \(\blacksquare\)

2. A full square-zero element

Fix the space \(V\subseteq D_{\mathrm{sa}}\) of Labels that survive representations, Lemma 3.1, with a real basis \(v_1,\dots,v_m\), and labels \(x_1,\dots,x_m\in D^{\otimes n}\) from Corollary 5.1 there, for \(r=m\). In \[ P_0=D\otimes D^{\otimes n}\otimes D^{\otimes n}=D^{\otimes(2n+1)} \] put \[ \begin{gathered} s=\sum_{i=1}^mv_i\otimes x_i\otimes1,\\ t=\sum_{i=1}^mv_i\otimes1\otimes x_i . \end{gathered} \] Both are self-adjoint.

Lemma 2.1. \(\pi([s,t])\ne0\) for every irreducible representation \(\pi\) of \(P_0\).

Proof. Since the second and third blocks commute, \[ [s,t]=\sum_{i,j=1}^m[v_i,v_j]\otimes x_i\otimes x_j . \] Let \(\pi_0\) be the restriction of \(\pi\) to the first leg, \(X_i=\pi(1\otimes x_i\otimes1)\) and \(Y_j=\pi(1\otimes1\otimes x_j)\). The von Neumann algebras \(N_0\), \(N_L\), \(N_R\) generated by the images of the three blocks are pairwise commuting factors (Lemma 2.1 of the previous lesson). The \(X_i\in N_L\) are linearly independent, and so are the \(Y_j\in N_R\) (Corollary 5.1 there). By Lemma 1.1(2) there, the products \(X_iY_j\) are linearly independent; they lie in \(N_0'\). If \[ \pi([s,t])=\sum_{i,j}\pi_0([v_i,v_j])\,X_iY_j=0, \] Lemma 1.1(1) there gives \(\pi_0([v_i,v_j])=0\) for all \(i,j\). Then \(\pi_0(V)\) is commutative, while \(\pi_0\) is a unital representation on a nonzero space, which contradicts Lemma 3.1 there. \(\blacksquare\)

Lemma 2.2. The self-adjoint element \(z=[s,[s,t]]\) has full positive and negative parts \(z_+\) and \(z_-\).

Proof. Let \(\pi\) be an irreducible representation on \(H\), and \(S=\pi(s)\), \(T=\pi(t)\). The exponential series shows that \(\lambda\mapsto e^{i\lambda S}Te^{-i\lambda S}\) is differentiable in norm, with derivative \(e^{i\lambda S}\,i[S,T]\,e^{-i\lambda S}\); here \(i[S,T]\) is self-adjoint. For \(\xi\in H\), the real function \[ f_\xi(\lambda)=\bigl\langle e^{i\lambda S}Te^{-i\lambda S}\xi,\xi\bigr\rangle\qquad(\lambda\in\mathbb R) \] is bounded by \(\|T\|\|\xi\|^2\) and twice differentiable, with \[ \begin{gathered} f_\xi'(0)=\langle i[S,T]\xi,\xi\rangle,\\ f_\xi''(\lambda)=-\bigl\langle\pi(z)\,e^{-i\lambda S}\xi,\,e^{-i\lambda S}\xi\bigr\rangle . \end{gathered} \] Suppose \(\pi(z)\ge0\). Then \(f_\xi''\le0\), so \(f_\xi'\) is nonincreasing. A bounded function on \(\mathbb R\) with nonincreasing derivative has derivative \(0\): if \(f_\xi'(\lambda_0)>0\), the mean value theorem gives \(f_\xi(\lambda)\le f_\xi(\lambda_0)-f_\xi'(\lambda_0)(\lambda_0-\lambda)\) for \(\lambda<\lambda_0\), which tends to \(-\infty\); a negative value is excluded in the same way as \(\lambda\to+\infty\). So \(\langle i[S,T]\xi,\xi\rangle=0\) for every \(\xi\), and the self-adjoint operator \(i[S,T]\) is \(0\), contradicting Lemma 2.1. If \(\pi(z)\le0\), the same argument applies to \(-f_\xi\). So \(\pi(z)\) is neither positive nor negative, that is, \(\pi(z_+)=\pi(z)_+\ne0\) and \(\pi(z_-)=\pi(z)_-\ne0\). By Lemma 1.1, \(z_+\) and \(z_-\) are full. \(\blacksquare\)

Lemma 2.3 (bridges). There are \(d_1,\dots,d_r\in P_0\) such that the elements \(c_i=z_-d_iz_+\) satisfy \(c_ic_j=0\) for all \(i,j\), and \(\bigl\|1-\sum_ih_ic_il_i\bigr\|<1\) for some \(h_i,l_i\in P_0\).

Proof. Since \(z_\pm\) is full (Lemma 2.2), the elements \(\sum_kh_kz_\pm l_k\) are dense in \(P_0\); choose \(y_\pm=\sum_kh^\pm_kz_\pm l^\pm_k\) with \(\|1-y_\pm\|<1/4\). Then \(\|y_-\|\le5/4\) and \[ \begin{aligned} &\|1-y_-y_+\|\\ &\quad\le\|1-y_-\|+\|y_-\|\,\|1-y_+\|\\ &\quad<\tfrac14+\tfrac5{16}<1 . \end{aligned} \] Expanding \(y_-y_+\) gives a sum of terms \(h^-_k\bigl(z_-(l^-_kh^+_{k'})z_+\bigr)l^+_{k'}\), of the required form with \(d_i=l^-_kh^+_{k'}\). Finally \(c_ic_j=z_-d_i(z_+z_-)d_jz_+=0\), since \(z_+z_-=0\). \(\blacksquare\)

Lemma 2.4 (full square-zero element; [OpenAI-KRC, Lemma 4.6]). Some finite maximal tensor power \(P\) of \(D\) contains a full element \(w\) with \(w^2=0\) and \(\|w\|=1\).

Proof. Take labels \(x_1',\dots,x_r'\in D^{\otimes n'}\) from Corollary 5.1 of the previous lesson, for the number \(r\) of bridges in Lemma 2.3. Put \[ \begin{gathered} P=P_0\otimes D^{\otimes n'}=D^{\otimes(2n+1+n')},\\ w_0=\sum_{i=1}^rc_i\otimes x_i' . \end{gathered} \] Then \(w_0^2=\sum_{i,j}c_ic_j\otimes x_i'x_j'=0\). Let \(\rho\) be an irreducible representation of \(P\), and \(\rho_0\) its restriction to \(P_0\otimes1\), a unital representation of \(P_0\). The von Neumann algebra \(N\) generated by \(\rho(P_0\otimes1)\) is a factor, and the operators \(B_i=\rho(1\otimes x_i')\in N'\) are linearly independent (Lemma 2.1 and Corollary 5.1 of the previous lesson). If \(\rho(w_0)=\sum_i\rho_0(c_i)B_i=0\), Lemma 1.1(1) there gives \(\rho_0(c_i)=0\) for all \(i\). Then \(\rho_0\bigl(\sum_ih_ic_il_i\bigr)=0\), although this element is invertible and \(\rho_0\) is unital. So \(\rho(w_0)\ne0\), and \(w_0\) is full by Lemma 1.1; in particular \(w_0\ne0\). Put \(w=w_0/\|w_0\|\). \(\blacksquare\)

The exponents \(n\) and \(n'\), hence the exponent of \(P\), depend on \(D\) through the space \(V\) and the number of bridges. Two full orthogonal positive elements in a finite maximal tensor power were found earlier in [OpenAI-ND, Theorem 1.4].

3. The cone homomorphism

\(C\) is the unitization of \(C_0((0,1],M_2)\): every \(F\in C\) is \(F(0)\) plus a function vanishing at \(0\).

Proposition 3.1 ([OpenAI-KRC, Proposition 4.7]). For every nonzero unital C*-algebra \(D\) without characters there are \(m\ge1\) and a unital \(*\)-homomorphism \(\eta\colon C\to P=D^{\otimes m}\) such that \(\eta(\iota e_{11})\) is full in \(P\).

Proof. Take \(P\) and \(w\) from Lemma 2.4, and the polar decomposition \(w=u|w|\) in \(P^{**}\), as in Cuntz comparison and order-zero maps: \(u^*u=e_1=s(|w|)\), \(uu^*=e_2=s(|w^*|)\), \(|w^*|=u|w|u^*\), and \(u=ue_1=e_2u\). Since \(w^2=0\), \(|w|^2|w^*|^2=w^*(ww)w^*=0\). For positive \(a,b\) with \(ab=0\), \(a(a+\tau)^{-1}b=0\) for \(\tau>0\), so \(s(a)b=0\), and then \(s(a)s(b)=0\). As \(s(|w|)=s(|w|^2)\) and \(s(|w^*|)=s(|w^*|^2)\), \(e_1e_2=0\). Hence \[ \begin{gathered} E_{11}=e_1,\quad E_{22}=e_2,\\ E_{21}=u,\quad E_{12}=u^* \end{gathered} \] are matrix units in \(P^{**}\); for instance \(E_{21}E_{21}=ue_1e_2u=0\).

Put \(h=|w|+|w^*|\in P\). Its two summands are positive with orthogonal supports \(e_1\) and \(e_2\), so \(\|h\|=\|w\|=1\). Using \(|w|u=|w|e_1e_2u=0\), \(|w^*|u=u|w|\) and their adjoints, \[ \begin{gathered} hE_{11}=E_{11}h=|w|,\\ hE_{22}=E_{22}h=|w^*|,\\ hE_{21}=E_{21}h=w,\\ hE_{12}=E_{12}h=w^* . \end{gathered} \] So \(h\) commutes with the matrix units, and so does every continuous function of \(h\). The formula \[ \varphi\Bigl(\sum_{i,j}f_{ij}e_{ij}\Bigr)=\sum_{i,j}f_{ij}(h)E_{ij} \] defines a \(*\)-homomorphism \(\varphi\colon C_0((0,1],M_2)\to P^{**}\), by the matrix-unit relations. Its values lie in \(P\): each \(f\in C([0,1])\) with \(f(0)=0\) is a uniform limit of polynomials without constant term, and \(h^kE_{ij}=h^{k-1}(hE_{ij})\in P\) for \(k\ge1\). Its unitization \[ \begin{gathered} \eta(F)=\varphi\bigl(F-F(0)\bigr)+c_F1_P,\\ F(0)=c_F1, \end{gathered} \] is a unital \(*\)-homomorphism \(C\to P\), with \(\eta(\iota e_{21})=hE_{21}=w\) and \(\eta(\iota e_{11})=hE_{11}=|w|\).

Finally, \(w\) and \(|w|\) generate the same closed ideal. The ideal of \(w\) contains \(w^*w\) and its square root \(|w|\). Conversely, with \(g_\varepsilon(t)=\varepsilon/(t+\varepsilon)\), \[ \begin{aligned} &\bigl\|w-w|w|(|w|+\varepsilon)^{-1}\bigr\|^2\\ &\quad=\bigl\|g_\varepsilon(|w|)\,|w|^2g_\varepsilon(|w|)\bigr\|\le\varepsilon^2, \end{aligned} \] so \(w\) lies in the closed ideal of \(|w|\). As \(w\) is full, \(\eta(\iota e_{11})=|w|\) is full. \(\blacksquare\)

The partial isometry \(u\) need not lie in \(P\), and \(e_1+e_2\) need not be \(1\); neither is needed.

4. Exercises

4.1. Show that in \(P=M_2\) the element \(e_{21}\) is full and square-zero, and compute the homomorphism \(\eta\) of Proposition 3.1 for \(w=e_{21}\).

4.2. Show that a nonzero element of a simple unital C*-algebra is full.

4.3. Let \(z\) be a self-adjoint element of \(M_n\) of the form \([s,[s,t]]\) with \(s,t\) self-adjoint. Show that the trace of \(z\) is \(0\), and deduce directly that \(z\) is neither positive nor negative unless \(z=0\).

4.4. Why does the proof of Lemma 2.4 need labels, rather than the plain sum \(\sum_ic_i\)?

4.5. Show that \(\|h\|=\max\{\|a\|,\|b\|\}\) for positive \(a,b\) with \(ab=0\) and \(h=a+b\).

5. Solutions

4.1. \(e_{21}^2=0\), and \(M_2\) is simple, so \(e_{21}\) is full. Here \(|w|=e_{11}\), \(|w^*|=e_{22}\), \(u=e_{21}\), \(h=1\), and \(\eta(F)=F(1)\): evaluation at \(1\).

4.2. The closed ideal generated by a nonzero element is a nonzero closed ideal, hence the whole algebra.

4.3. \(\operatorname{tr}[s,y]=0\) for every \(y\), with \(y=[s,t]\). A positive matrix with trace \(0\) has all eigenvalues \(0\), hence is \(0\); similarly for a negative one.

4.4. The sum \(\sum_ic_i\) also squares to zero, but it may vanish in some irreducible representation even though some \(c_i\) does not: the images of different \(c_i\) can cancel. With labels, Lemma 1.1(1) of the previous lesson rules out such a cancellation.

4.5. \(a\) and \(b\) commute and generate a commutative C*-algebra \(C_0(\Omega)\), in which they are functions with disjoint supports; the norm of \(a+b\) is the maximum of the two suprema.

References