# Cuntz comparison and order-zero maps *Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).* This lesson collects the tools for the Rørdam–Winter criterion of the next lesson. Cuntz comparison compares positive elements up to conjugation and limits; only a few elementary rules are needed, and no cancellation (Lemma 1.2). A comparison can be made exact after cutting the smaller element slightly (Lemma 2.1). Families of contractions with orthogonal "Gram" products define completely positive maps on matrix algebras that preserve orthogonality, with a supporting homomorphism into the bidual (Lemma 3.1). Two such maps on \(M_3\) and \(M_2\) with commuting ranges and complementary units give a unital homomorphism from the dimension-drop algebra \(I(2,3)\) (Lemma 4.1). - Cuntz comparison (Definition 1.1, Lemma 1.2). - Exact cuts (Lemma 2.1). - Gram maps and their supporting homomorphisms (Lemma 3.1). - Commuting Gram maps and \(I(2,3)\) (Lemma 4.1). The lesson follows [OpenAI-KRC, Sections 2.1 and 3]. Throughout, \(B\) is a unital C\*-algebra. *The bidual.* \(B^{**}\) denotes the universal enveloping von Neumann algebra of \(B\): the von Neumann algebra generated by \(B\) in its universal representation, in which \(B\) is identified with a C\*-subalgebra ([The universal enveloping von Neumann algebra of a C\*-algebra and W\*-algebras](course:foundations-of-von-neumann-algebras/the-universal-enveloping-von-neumann-algebra-of-a-c-star-algebra-and-w-star-algebras), Theorem 3.3). A positive \(T\in B^{**}\) has the *support projection* \(s(T)\in B^{**}\) onto the closure of its range, the strong limit of \(T(T+\tau)^{-1}\) as \(\tau\downarrow0\). Every \(x\in B^{**}\) has a *polar decomposition* \(x=u|x|\) with a partial isometry \(u\in B^{**}\), \(u^*u=s(|x|)\), \(|x^*|=u|x|u^*\) and \(uu^*=s(|x^*|)\) ([The double commutant theorem](course:foundations-of-von-neumann-algebras/the-double-commutant-theorem), Proposition 7.2 and Corollary 7.3). Support projections and polar decompositions need not lie in \(B\); membership in \(B\) is checked whenever an element of \(B\) is constructed from them. For \(a\in B_+\) and \(\varepsilon\ge0\), \((a-\varepsilon)_+\) is the functional calculus of \(t\mapsto\max\{t-\varepsilon,0\}\). For \(b\in B_+\), \(\operatorname{Her}(b)\) is the norm closure of \(b^{1/2}Bb^{1/2}\). ## 1. Cuntz comparison **Definition 1.1.** For positive matrices \(a\in M_n(B)\) and \(b\in M_m(B)\), write \(a\precsim b\) if there are \(m\times n\) matrices \(x_j\) over \(B\) with \(x_j^*bx_j\to a\) in norm. Write \([a]\le[b]\) for \(a\precsim b\), \([a]=[b]\) if both hold, \([a]+[b]=[a\oplus b]\), and \(k[a]\) for the class of the direct sum of \(k\) copies of \(a\). **Lemma 1.2** (rules). Let \(a,b,c\) be positive matrices over \(B\). 1. If \(a\precsim b\) and \(b\precsim c\), then \(a\precsim c\); if \(a\precsim b\) and \(a'\precsim b'\), then \(a\oplus a'\precsim b\oplus b'\). 2. \([v^*v]=[vv^*]\) for every rectangular matrix \(v\) over \(B\). 3. If \(a_1,\dots,a_l\in M_n(B)\) are positive and pairwise orthogonal, then \([\sum_ia_i]=\sum_i[a_i]\). 4. A \(*\)-homomorphism \(\varphi\colon B\to C\), applied entrywise, preserves \(\precsim\). For a unital C\*-algebra \(C\) and \(e\in C_+\), \(a\precsim b\) over \(B\) implies \(a\otimes e\precsim b\otimes e\) over \(B\otimes_{\max}C\), where \(M_n(B)\otimes C\) is identified with \(M_n(B\otimes_{\max}C)\); likewise with \(e\) as left factor. 5. \([a]\le[1_B]\) for \(a\in B_+\). *Proof.* (1) If \(x_j^*bx_j\to a\) and \(y_k^*cy_k\to b\), then \(\|x_j^*y_k^*cy_kx_j-x_j^*bx_j\|\) is at most \(\|x_j\|^2\|y_k^*cy_k-b\|\); choose \(k\) depending on \(j\). Block diagonal witnesses give the second claim. (2) Put \(b=v^*v\) and \(x_\gamma=v(b+\gamma)^{-1/2}\) for \(\gamma>0\). Then \(x_\gamma^*(vv^*)x_\gamma=b^2(b+\gamma)^{-1}\), which differs from \(b\) by \(\gamma b(b+\gamma)^{-1}\), of norm at most \(\gamma\). So \(v^*v\precsim vv^*\), and the same with \(v^*\) in place of \(v\) gives the converse. (3) For positive orthogonal \(a_i,a_j\), \((a_i^{1/2}a_j)^*(a_i^{1/2}a_j)=a_ja_ia_j=0\), so \(a_i^{1/2}a_j=0\), and in the same way \(a_i^{1/2}a_j^{1/2}=0\). The row \(v=(a_1^{1/2},\dots,a_l^{1/2})\) has \(vv^*=\sum_ia_i\) and \(v^*v=\operatorname{diag}(a_1,\dots,a_l)\); apply (2). (4) Apply \(\varphi\) to the witnesses, or tensor them with \(1_C\): \((x_j\otimes1)^*(b\otimes e)(x_j\otimes1)=x_j^*bx_j\otimes e\), which differs from \(a\otimes e\) by at most \(\|x_j^*bx_j-a\|\,\|e\|\) ([Central sequences and the character criterion](central-sequences-and-the-character-criterion.md), Lemma 3.1(7)). (5) The witness \(a^{1/2}\) gives \(a^{1/2}1_Ba^{1/2}=a\). \(\blacksquare\) By (1), a comparison \(a\precsim b\) of two elements of \(B\), obtained through matrices of any size, has witnesses in \(B\) itself. No cancellation of the Cuntz order is used anywhere. ## 2. Exact cuts **Lemma 2.1** (exact cut; [OpenAI-KRC, Lemma 3.1]). If \(a,b\in B_+\) and \(a\precsim b\), then for every \(\delta>0\) there is \(z\in B\) with \[ z^*z=(a-\delta)_+,\qquad zz^*\in\operatorname{Her}(b). \] *Proof.* Choose \(x\in B\) and \(0<\eta<\delta\) with \(\|a-x^*bx\|<\eta\), so \(x^*bx\ge a-\eta\). Put \(e=(a-\delta)_+\) and \(y=b^{1/2}xe^{1/2}\). Then \(T=y^*y=e^{1/2}x^*bxe^{1/2}\) satisfies \[ \begin{aligned} (\delta-\eta)e&\le e^{1/2}(a-\eta)e^{1/2}\\ &\le T\le\|x^*bx\|\,e ; \end{aligned} \] the first inequality is the functional calculus of \(a\), since \((t-\eta)(t-\delta)_+\ge(\delta-\eta)(t-\delta)_+\). So \(T\) and \(e\) have the same kernel, hence the same support. Let \(y=uT^{1/2}\) be the polar decomposition in \(B^{**}\) and \(z=ue^{1/2}\); then \(z^*z=e^{1/2}s(T)e^{1/2}=e\). To see \(z\in B\), put \(z_\tau=y(T+\tau)^{-1/2}e^{1/2}\in b^{1/2}B\) for \(\tau>0\). With \(f_\tau(t)=1-\sqrt{t/(t+\tau)}\), we have \(z-z_\tau=uf_\tau(T)e^{1/2}\), and since \(e\le T/(\delta-\eta)\), \[ \begin{aligned} \|z-z_\tau\|^2&\le\|f_\tau(T)ef_\tau(T)\|\\ &\le\frac{\sup_{t\ge0}tf_\tau(t)^2}{\delta-\eta}\le\frac{\tau}{\delta-\eta}, \end{aligned} \] because \(f_\tau(t)\le1-t/(t+\tau)=\tau/(t+\tau)\). No commutation of \(e\) with \(T\) is used. So \(z\in B\), and \(zz^*\) is the norm limit of \(z_\tau z_\tau^*\in b^{1/2}Bb^{1/2}\). \(\blacksquare\) ## 3. Gram maps and their supporting homomorphisms A linear map \(\alpha\colon M_q\to B\) is *of order zero* if \(\alpha(x)\alpha(y)=0\) for positive \(x,y\) with \(xy=0\). **Lemma 3.1** (Gram maps; [OpenAI-KRC, Section 3]). Let \(x_1,\dots,x_q\in B\) be contractions with \[ x_ix_j^*=\delta_{ij}h\qquad(1\le i,j\le q) \] for a positive contraction \(h\), and define \(\alpha\colon M_q\to B\) by \(\alpha(e_{ij})=x_i^*x_j\). Then: 1. \(\alpha\) is completely positive and contractive, and \(H=\alpha(1_q)=\sum_ix_i^*x_i\) is a positive contraction; 2. there is a \(*\)-homomorphism \(\bar\alpha\colon M_q\to B^{**}\) with \(\alpha(x)=H\bar\alpha(x)=\bar\alpha(x)H\) for all \(x\) and \(\bar\alpha(1_q)=s(H)\); 3. \(\alpha(x)\alpha(y)=H\alpha(xy)\) for all \(x,y\in M_q\); in particular \(\alpha\) is of order zero and \(H\) commutes with the range of \(\alpha\). *Proof.* (1) With the column \(R=(x_1,\dots,x_q)^{\mathrm T}\), \(\alpha(X)=R^*(X\otimes1)R\), which is completely positive, of norm at most \(\|R\|^2=\|R^*R\|=\|H\|\). The \(x_i^*x_i\) are positive contractions and pairwise orthogonal, since \(x_i^*x_ix_j^*x_j=x_i^*(x_ix_j^*)x_j=0\) for \(i\ne j\); so \(\|H\|\le1\). (2) Let \(x_i^*=v_i|x_i^*|\) be the polar decompositions in \(B^{**}\). Here \(|x_i^*|=(x_ix_i^*)^{1/2}=h^{1/2}=:k\), so \(v_i^*v_i=s(k)=s(h)=:s\) and \(x_i=kv_i^*\). For \(i\ne j\), \(kv_i^*v_jk=x_ix_j^*=0\); as the range of \(k\) is dense in the range of \(s\) and \(v_i=v_is\), this gives \(v_i^*v_j=0\). So the \(v_i\) are partial isometries with the common initial projection \(s\) and mutually orthogonal final projections, and \(\bar\alpha(e_{ij})=v_iv_j^*\) defines matrix units, that is, a \(*\)-homomorphism \(M_q\to B^{**}\). Now \(\alpha(e_{ij})=v_ihv_j^*\) and \(H=\sum_mv_mhv_m^*\), and \(v_m^*v_i=\delta_{mi}s\), \(hs=sh=h\) give \[ H\bar\alpha(e_{ij})=v_ihv_j^*=\bar\alpha(e_{ij})H . \] The summands \(v_ihv_i^*\) of \(H\) are positive with support \(v_iv_i^*\), and these are orthogonal; so \(s(H)=\sum_iv_iv_i^*=\bar\alpha(1_q)\). (3) \(\alpha(x)\alpha(y)=H\bar\alpha(x)\bar\alpha(y)H\), which is \(H\bar\alpha(xy)H=H\alpha(xy)\), and with \(y=1\), \(\alpha(x)H=H\alpha(x)\). For positive orthogonal \(x,y\), \(\alpha(x)\alpha(y)=H\alpha(xy)=0\). \(\blacksquare\) We call \(\alpha\) the *Gram map* of the family \((x_i)\) and \(\bar\alpha\) its *supporting homomorphism*. Its matrix units are also the strong limits of \(\alpha(e_{ij})(H+\tau)^{-1}\) as \(\tau\downarrow0\), since \(\alpha(e_{ij})(H+\tau)^{-1}=H(H+\tau)^{-1}\bar\alpha(e_{ij})\). ## 4. Commuting Gram maps and \(I(2,3)\) Recall \[ \begin{aligned} I(2,3)=\{f\in{}&C([0,1],M_2\otimes M_3):\\ &f(0)\in M_2\otimes1,\\ &f(1)\in1\otimes M_3\}. \end{aligned} \] **Lemma 4.1** ([OpenAI-KRC, Lemma 3.2], after [RW, Proposition 2.5]). Let \(\alpha\colon M_3\to B\) and \(\beta\colon M_2\to B\) be linear maps preserving adjoints, with positive \(\alpha(1_3)\) and \(\beta(1_2)\), with commuting ranges, and with \[ \psi(x)\psi(y)=\psi(1)\psi(xy)\qquad(\psi=\alpha,\beta). \] If \(\alpha(1_3)+\beta(1_2)=1_B\), there is a unital \(*\)-homomorphism \(I(2,3)\to B\) sending \(\alpha_0(y)\) to \(\alpha(y)\) and \(\beta_0(x)\) to \(\beta(x)\), where \[ \begin{gathered} \alpha_0(y)(t)=t(1_2\otimes y),\\ \beta_0(x)(t)=(1-t)(x\otimes1_3). \end{gathered} \] *Proof.* *Generators of \(I(2,3)\).* The C\*-algebra \(G\subseteq I(2,3)\) generated by the ranges of \(\alpha_0\) and \(\beta_0\) contains \(\alpha_0(1)+\beta_0(1)=1\) and \(\alpha_0(1)=t1\), hence all scalar functions \(g(t)1\). It contains \(\alpha_0(y)\beta_0(x)=t(1-t)(x\otimes y)\), hence, by the Stone–Weierstrass theorem, all functions \(g(t)X\) with \(g\in C_0((0,1))\) and \(X\in M_6\), and so all of \(C_0((0,1),M_6)\). For \(F\in I(2,3)\) with \(F(0)=X_0\otimes1\) and \(F(1)=1\otimes Y_1\), the difference \(F-\beta_0(X_0)-\alpha_0(Y_1)\) vanishes at both endpoints. So \(G=I(2,3)\). *Irreducible representations of the target.* Let \(E\) be the C\*-algebra generated by \(1_B\) and the ranges of \(\alpha\) and \(\beta\). The identity with \(y=1\) shows that \(\alpha(1)\) commutes with the range of \(\alpha\); it commutes with the range of \(\beta\) by hypothesis, so it is central in \(E\). In an irreducible representation \(\pi\) of \(E\), \(\pi(\alpha(1))=r\) is a scalar, with \(0\le r\le1\) since \(\beta(1)=1-\alpha(1)\ge0\). If \(0