# 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](from-the-cone-to-the-dimension-drop-algebra.md), turns this into a homomorphism from \(I(2,3)\).

- Full elements (Lemma 1.1).
- A full square-zero element (Lemmas 2.1 to 2.4).
- The cone homomorphism (Proposition 3.1).

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](labels-that-survive-representations.md) 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](course:foundations-of-von-neumann-algebras/c-star-algebras-continuous-functional-calculus-automatic-continuity-positive-cones), Theorem 15.1). It has an irreducible representation, because norms are attained in irreducible representations ([Building representations from positive functionals](course:foundations-of-von-neumann-algebras/representations-and-positive-functionals-the-gns-construction-and-the-gelfand-naimark), 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](labels-that-survive-representations.md), 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](cuntz-comparison-and-order-zero-maps.md): \(u^*u=e_1=s(|w|)\), \(uu^*=e_2=s(|w^*|)\), \(|w^*|=u|w|u^*\), 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

- [OpenAI-KRC] OpenAI, *The Kirchberg–Rørdam character criterion*, OpenAI Math Release preprint, 25 September 2026, Sections 4.3 and 4.4. https://github.com/openai/math/tree/main/preprints/The-Kirchberg-Rordam-character-criterion-September-25-2026
- [OpenAI-ND] OpenAI, *Nuclear dimension and Jiang–Su stability without elementary subquotients*, OpenAI Math Release preprint, 23 September 2026, Theorem 1.4. https://github.com/openai/math/tree/main/preprints/Nuclear-dimension-and-Jiang-Su-stability-without-elementary-subquotients-September-23-2026
- [KR] E. Kirchberg and M. Rørdam, *When central sequence C\*-algebras have characters*, arXiv:1409.1395v2, Proposition 6.1. https://arxiv.org/abs/1409.1395