Hausdorffifized algebraic $K_1$ group and invariants for $C^*$-algebras with the ideal property

TL;DR

This paper introduces a refined invariant for $C^*$-algebras with the ideal property, incorporating Hausdorffified algebraic $K_1$-groups, to distinguish non-isomorphic algebras that share existing invariants.

Contribution

It constructs new examples of $A ext{T}$ algebras with the ideal property that are indistinguishable by known invariants but differ in the Hausdorffified algebraic $K_1$-groups, enhancing classification tools.

Findings

Constructed two non-isomorphic $A ext{T}$ algebras with identical existing invariants.

Demonstrated the Hausdorffified algebraic $K_1$-groups can distinguish these algebras.

Proved the augmented invariant is complete for certain classes of $C^*$-algebras with the ideal property.

Abstract

A $C^*$-algebra $A$ is said to have the ideal property if each closed two-sided ideal of $A$ is generated by the projections inside the ideal, as a closed two sided ideal. $C^*$-algebras with the ideal property are generalization and unification of real rank zero $C^*$-algebras and unital simple $C^*$-algebras. It is long to be expected that an invariant (see [Stev] and [Ji-Jiang], [Jiang-Wang] and [Jiang1]) , we call it $Inv^0(A)$ (see the introduction), consisting of scaled ordered total $K$-group $(\underline{K}(A), \underline{K}(A)^{+},\Sigma A)_{\Lambda}$ (used in the real rank zero case), the tracial state space $T(pAp)$ of cutting down algebra $pAp$ as part of Elliott invariant of $pAp$ (for each $[p]\in\Sigma A$) with a certain compatibility, is the complete invariant for certain well behaved class of $C^*$-algebras with the ideal property (e.g., $AH$ algebras with no dimension growth). In this paper, we will construct two non isomorphic $A\mathbb{T}$ algebras $A$ and $B$ with the ideal property such that $Inv^0(A)\cong Inv^0(B)$. The invariant to differentiate the two algebras is the Hausdorffifized algebraic $K_1$-groups $U(pAp)/\overline{DU(pAp)}$ (for each $[p]\in\Sigma A$) with a certain compatibility condition. It will be proved in [GJL] that, adding this new ingredients, the invariant will become the complete invariant for $AH$ algebras (of no dimension growth) with the ideal property.

Equations

$$A_1\stackrel{{\varphi }_{1,2}}{\to }{A}_2\stackrel{{\varphi }_{2,3}}{\to }{A}_3⟶\cdots ⟶A_n⟶\cdots$$

$${\varphi }_{n,m}={\varphi }_{m-1,m}\circ {\varphi }_{m-2,m-1}\circ \cdots \circ {\varphi }_{n,n+1},$$

$$A_1\stackrel{{\varphi }_{1,2}}{\to }{A}_2\stackrel{{\varphi }_{2,3}}{\to }{A}_3⟶\cdots ⟶\cdots ,$$

$$(\ast )K_0(A{)}^{+}\bigcap (-K_0(A{)}^{+})=\{0\}\text{mbox}and{K}_0(A{)}^{+}-K_0(A{)}^{+}=K_0(A).$$

$$\Sigma A=\{[p]\in K_0(A{)}^{+},pisaprojectioninA\}.$$

$$\Gamma :KK(A,B)\to Ho{m}_{\wedge }(\underline{K}(A),\underline{K}(B)).$$

$$\alpha :(\underline{K}(A),\underline{K}(A{)}^{+},\Sigma A)\to (\underline{K}(B),\underline{K}(B{)}^{+},\Sigma B),$$

$${\alpha }_k^i:K_i(A,\mathbb{Z}/k\mathbb{Z})⟶K_i(B,\mathbb{Z}/k\mathbb{Z}),i=0,1,k=0,2,3,\cdots$$

$$AffT\varphi :AffTA⟶AffT(\varphi ({\text{1}}_A)B\varphi ({\text{1}}_A))$$

$$\big{(}(AffT\phi)(l)\big{)}(\tau)=\tau(\phi(x))~{}~{}\mbox{for any}~{}~{}\tau\in T(\phi(\textbf{1}_{A})B\phi(\textbf{1}_{A})),$$

$$\big{(}(\phi_{T})(l)\big{)}(\tau)=\tau(\phi(x))~{}~{}\mbox{for any}~{}~{}\tau\in T(B),$$

$${\xi }^{p,q}:AffT(pAp)⟶AffT(qBq)$$

$$\begin{array}{ccc}AffT(p_{1}A{p}_1)& \stackrel{{\xi }^{p_1,q_1}}{\to }& AffT(q_{1}B{q}_1)\\ {ı}_T\downarrow & & {ı}_T\downarrow \\ AffT(p_{2}A{p}_2)& \stackrel{{\xi }^{p_2,q_2}}{\to }& AffT(q_{2}B{q}_2),\end{array}(2.A)$$

$$(\underline{K}(A),\underline{K}(A{)}^{+},\Sigma A,\{AffT(pAp){\}}_{[p]\in \Sigma A})$$

$$\alpha :(\underline{K}(A),\underline{K}(A{)}^{+},\Sigma A)⟶(\underline{K}(B),\underline{K}(B{)}^{+},\Sigma B)$$

$${\xi }^{p,q}:AffT(pAp)⟶AffT(qBq)$$

$$\begin{array}{ccc}{K}_0(pAp)& \stackrel{\rho }{\to }& AffT(pAp)\\ \alpha \downarrow & & {\xi }^{p,q}\downarrow \\ K_0(qBq)& \stackrel{\rho }{\to }& AffT(qBq).\end{array}(2.B)$$

$${\xi }^j(f)(\tau )=\sum _{\{i:\tau (p_i^j)\ne 0\}}\tau (p_i^j){\xi }_i^j(f)(\frac{\tau {|}_{p_i^j{B}^j{p}_i^j}}{\tau (p_i^j)})\text{mbox}forf\in AffTA\text{mbox}and\tau \in T{B}^j.$$

$$(AffT\varphi {)}^j(f)(\tau )=\sum _{\{i:\tau (p_i^j)\ne 0\}}\tau (p_i^j)AffT{\varphi }_i^j(f)(\frac{\tau {|}_{p_i^j{B}^j{p}_i^j}}{\tau (p_i^j)}),$$

$$\Vert AffT\varphi (f)-\xi (f)\Vert <\epsilon .$$

$$D_A(u,v)=inf\{\Vert u{v}^{\ast }-c\Vert :c\in \overline{DU(A)}\},$$

$$d^{\prime }(f,g)=inf\{\Vert f-g-h\Vert ,h\in \overline{\rho K_0(A)}\}.$$

$$d_{A}(f,g)=\left\{\begin{lgathered}2,\;\;\;\mbox{if}\;d^{\prime}(f,g)\geq\frac{1}{2}\\ |e^{2\pi id^{\prime}(f,g)}-1|,\;\;\;\mbox{if}\;d^{\prime}(f,g)<\frac{1}{2}~{}~{}~{}~{}.\end{lgathered}2,\;\;\;\mbox{if}\;d^{\prime}(f,g)\geq\frac{1}{2}\\ |e^{2\pi id^{\prime}(f,g)}-1|,\;\;\;\mbox{if}\;d^{\prime}(f,g)<\frac{1}{2}~{}~{}~{}~{}.\right.$$

$$u=\left(\begin{array}[]{cccc}e^{2\pi i\frac{k-1}{k}t}&&&\\ &e^{2\pi i(\frac{-t}{k})}&&\\ &&\ddots&\\ &&&e^{2\pi i(\frac{-t}{k})}\\ \end{array}\right)\in I_{k}.$$

$$U_0(A)\cap SU(A)=\{e^{2\pi i(\frac{j}{k})},j=0,1,\cdots ,k-1\}\cdot \overline{DU(A)}.$$

$$U_0(A)\cap SU(A)=\{e^{2\pi i(\frac{j}{kl})},j=0,1,\cdots ,kl-1\}\cdot \overline{DU(A)}.$$

$$\Delta (\xi )(\tau )=\frac{1}{2\pi i}{\int }_0^1\tau (\frac{d\xi }{dt}\cdot {\xi }^{\ast })dt.$$

$$\Delta ({\xi }_1)-\Delta ({\xi }_2)=\Delta ({\xi }_1\cdot {\xi }_2^{\ast })\subset {\Delta }^{\circ }({\pi }_1(U_0(A))).$$

$$\frac{d}{dt}(diag(u(t),1_{n-m})=diag(\frac{d}{dt}(u(t)),0_{n-m}),$$

$${\overline{\Delta }}_n{|}_{U_0(M_m(A))}={\overline{\Delta }}_m.$$

Full text

The case $A=M_{l}(I_{k})$ is similar. ∎

2.22.   Let $path(U(A))$ denote the set of piecewise smooth paths $\xi:[0,1]\rightarrow U(A)$. Recall that de la Harp-Skandalis determinant (see [dH-S]) $\Delta:path(U(A))\rightarrow AffTA$ is defined by

[TABLE]

It is proved in [dH-S](see also [Thm4]) that $\Delta$ induces a map $\Delta^{\circ}:\pi_{1}(U_{0}(A))\rightarrow AffTA.$ For any two paths $\xi_{1},\xi_{2}$ starting at $\xi_{1}(0)=\xi_{2}(0)=1\in A$ and ending at the same unitary $u=\xi_{1}(1)=\xi_{2}(1)$, we have that

[TABLE]

Consequently $\Delta$ induces a map

$\overline{\Delta}:U_{0}(A)\rightarrow AffTA/\Delta^{\circ}(\pi_{1}(U_{0}(A))).$ (See [Thm4, section 3].)

Passing to matrix over $A$, we have a map $\overline{\Delta}_{n}:U_{0}(M_{n}(A))\rightarrow AffTA/\Delta^{\circ}_{n}(\pi_{1}(U_{0}(M_{n}(A)))).$

If $1\leq m<n$, then $path(U(M_{m}(A)))$ (and $U_{0}(M_{m}(A))$ ) can be embedded into $path(U(M_{n}(A)))$ (and $U_{0}(M_{n}(A))$ ) by sending $u(t)$ to $diag(u(t),1_{n-m})$. From the above definition, and the formula

[TABLE]

one gets

[TABLE]

Recall that the Bott isomorphism $b:K_{0}(A)\rightarrow K_{1}(SA)$ is given by the following: for any $x\in K_{0}(A)$ represented by a projection $p\in M_{n}(A)$, we have

[TABLE]

If $\xi(t)=e^{2\pi it}p+(\textbf{1}_{n}-p)$, then

[TABLE]

Since Bott map is an isomorphism, it follows that each loop in $\pi_{1}(U_{0}(A))$ is homotopic to a product of loops of the above form $\xi(t)$. Consequently $\Delta^{\circ}(\pi_{1}(U_{0}(M_{n}(A))))\subset\rho_{A}K_{0}(A)$. Hence $\overline{\Delta}_{n}$ can be regarded as a map

[TABLE]

Proposition 2.23.   For $A\in\mathcal{HD}$ or $A\in A\mathcal{HD}$, $\overline{DU_{0}(A)}=\overline{DU(A)}$.

[TABLE]

implies that $\overline{DU(A)}\subset\overline{DU_{0}(M_{3}(A))}.$ Therefore by Lemma 3.1 of [Thm4], $\overline{DU(A)}\subset ker\overline{\Delta}_{3}.$ If $x\in U_{0}(A)\cap\overline{DU(A)}$, then $\overline{\Delta}_{1}$ is defined at $x$. By calculation in 2.22, $\overline{\Delta}_{3}|_{U_{0}(A)}=\overline{\Delta}_{1}$. Hence we have $\overline{\Delta}_{1}(x)=0.$ And therefore $x\in\overline{DU_{0}(A)}=ker\overline{\Delta}_{1},$ by Lemma 3.1 of [Thm4]. Note that if $A\in\mathcal{HD}$ or $A\mathcal{HD}$, then $\overline{DU(A)}\subset U_{0}(A)$.

∎

(It is not known to the authors whether it is always true that $\overline{DU_{0}(A)}=\overline{DU(A)}.$)

2.24.   There is a natural map $\alpha:\pi_{1}(U(A))\longrightarrow K_{0}(A)$, or more generally, $\alpha_{n}:\pi_{1}(U(M_{n}(A))\longrightarrow K_{0}(A))$ for any $n\in\mathbb{N}.$ We need the following notation. For a unital $C^{*}$-algebra $A$, let $\mathcal{P}_{n}K_{0}(A)$ (see [GLX]) be the subgroup of $K_{0}(A)$ generated by the formal difference of projections $p,q\in M_{n}(A)$ (instead of $M_{\infty}(A)$). Then

[TABLE]

In particular, if $\rho:K_{0}(A)\longrightarrow AffTA$ satisfies $\rho(\mathcal{P}_{n}K_{0}(A))=\rho K_{0}(A)$, then by Theorem 3.2 of [Thm4],

[TABLE]

Note that for all $A\in\mathcal{HD}$, we have $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ (see below). Consequently,

[TABLE]

If $A$ does not contain building blocks of form $PM_{n}(C(T_{II,k}))P$, then such $A$ is the special case of [Thm2], and the above fact is observed in [Thm2] (for circle algebras in [NT] earlier)—in this special case, we ever have $\mathcal{P}_{1}K_{0}(A)=K_{0}(A)$ (as used in [NT] and [Thm2] in the form of surjectivity of $\alpha:\pi_{1}(U(A))\longrightarrow K_{0}(A)$). For $A=PM_{n}(C(T_{II,k}))P$, we do not have the surjectivity of $\alpha:\pi_{1}(U(A))\longrightarrow K_{0}(A)$ any more. But $K_{0}(A)=\mathbb{Z}\oplus\mathbb{Z}/k\mathbb{Z}$ and image$(\alpha)=\mathcal{P}_{1}K_{0}(A)$ contains at least one element which corresponds to a rank one projection (any bundle over $T_{II,k}$ has a subbundle of rank 1)—that is,

[TABLE]

consisting all constant functions from $T_{II,k}$ to $\frac{1}{rank(P)}\mathbb{Z}$.

As in [NT, Lemma 3.1] and [Thm 2, Lemma 6.4], the map $\overline{\Delta}:~{}U_{0}(A)\rightarrow AffTA/\overline{\rho_{A}(K_{0}(A))}$ (in 2.22) has Ker$\overline{\Delta}=\overline{DU(A)}$ and the following lemma holds.

Lemma 2.25.   If a unital $C^{*}$-algebra $A$ satisfies $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ and $\overline{DU_{0}(A)}=\overline{DU(A)}$ (see 2.24 and 2.23), in particular, if $A\in\mathcal{HD}$ or $A\in A\mathcal{HD}$, then the following hold:

(1)   There is a split exact sequence

[TABLE]

(2)   $\lambda_{A}$ is an isometry with respect to the metrics $d_{A}$ and $D_{A}$.

2.26.   Recall from $\S 3$ of [Thm4], the de la Harpe—Skandalis determinant (see [dH-S]) can be used to define

[TABLE]

With the condition of Lemma 2.25 above, this map is an isometry with respect to the metrics $d_{A}$ and $D_{A}$. In fact, the inverse of this map is $\lambda_{A}$ in Lemma 2.25.

It follows from the definition of $\overline{\Delta}$ (see $\S 3$ of [Thm4]) that

[TABLE]

where $[p]\in K_{0}(A)$ is the element represented by projection $p\in A$.

It is convenient to introduce the extended commutator group $DU^{+}(A)$, which is generated by

$DU(A)\subset U(A)$ and the set $\{e^{2\pi itp}=e^{2\pi it}p+(\textbf{1}-p)\in U(A)~{}|~{}~{}t\in\mathbb{R},p\in A~{}\mbox{ is a projection}\}.$ Let $\widetilde{DU(A)}$ denote the closure of $DU^{+}(A)$. That is, $\widetilde{DU(A)}=\overline{DU^{+}(A)}$.

Let us use $\widetilde{\rho K_{0}(A)}$ to denote the real vector space spanned by $\overline{\rho K_{0}(A)}$. That is,

[TABLE]

Suppose that $\overline{\rho K_{0}(A)}=\overline{\rho(\mathcal{P}_{1}K_{0}(A))}$. It follows from (2.c), the image of $\widetilde{DU(A)}/\overline{DU(A)}$ under the map $\overline{\Delta}$ is exactly $\widetilde{\rho K_{0}(A)}/\overline{\rho K_{0}(A)}$. Therefore $\lambda_{A}$ takes $\widetilde{\rho K_{0}(A)}/\overline{\rho K_{0}(A)}$ to $\widetilde{DU(A)}/\overline{DU(A)}$. Hence $\overline{\Delta}:U_{0}(A)/\overline{DU(A)}\longrightarrow AffTA/\overline{\rho K_{0}(A)}$ also induces a quotient map (still denoted by $\overline{\Delta}$)

[TABLE]

which is an isometry using the quotient metrics of $d_{A}$ and $D_{A}$. The inverse of this quotient map $\overline{\Delta}$ gives rise to the isometry

[TABLE]

which is an isometry with respect to the quotient metrics $\widetilde{d}_{A}$ and $\overline{D_{A}}$ as described below.

For any $u,v\in U(A)/\widetilde{DU(A)},$

[TABLE]

Let $\widetilde{d^{\prime}}$ denote the quotient metric on $AffTA/\widetilde{\rho K_{0}(A)}$ of $AffTA$, that is,

[TABLE]

Define $\widetilde{d}_{A}$ by

[TABLE]

The following result is a consequence of Lemma 2.25.

Lemma 2.27.   If a unital $C^{*}$-algebra $A$ satisfies $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ and $\overline{DU_{0}(A)}=\overline{DU(A)}$ (see 2.24 and 2.23), in particular, if $A\in\mathcal{HD}$ or $A\in A\mathcal{HD}$, then we have

(1)   There is a split exact sequence

[TABLE]

(2)   $\widetilde{\lambda}_{A}$ is an isometry with respect to $\widetilde{d}_{A}$ and $\overline{D_{A}}$.

Note that $\widetilde{d}_{A}$ on $AffTA/\widetilde{\rho K_{0}(A)}$ is the quotient metric induced by $d_{A}$ on $AffTA/\overline{\rho K_{0}(A)}$ and $\overline{D_{A}}$ on $U(A)/\widetilde{DU(A)}$ is the quotient metric induced by $D_{A}$ on $U(A)/\overline{DU(A)}$. Hence $\widetilde{\lambda}_{A}$ is an isometry, since so is $\lambda_{A}$.

∎

2.28.   Instead of $\widetilde{DU(A)}$, we will need the group

[TABLE]

For $A\in\mathcal{HD}$, say $A=PM_{l}(C(X))P$ ($X=[0,1],S^{1}$ or $T_{II,k}$) or $A=M_{l}(I_{k})$,   $\widetilde{SU(A)}$ is the set of all unitaries $u\in P(M_{l}C(X))P$ or $u\in M_{l}(I_{k})$ such that the determinant function

[TABLE]

is a constant function. Comparing with the set $SU(A)$ in [EGL1] or 2.17 above (which only defines for $\mathcal{HD}$ blocks), where the function will be constant 1, here we allow the function to be arbitrary constant in $\mathbb{T}$. Hence for a basic building block $A=PM_{n}(C(X))P\in\cal{HD}$ or $A=M_{l}(I_{k})$,

[TABLE]

The notations $\widetilde{\rho K_{0}(A)}$, $\widetilde{DU(A)}$ and $\widetilde{SU(A)}$ reflect that they are constructed from $\rho K_{0}(A)$, $DU(A)$ and $SU(A)$, respectively. To make the notation simpler, from now on, we will use $\widetilde{\rho K_{0}}(A)$ to denote $\widetilde{\rho K_{0}(A)}=\widetilde{\rho_{A}(K_{0}(A))}$, $\widetilde{DU}(A)$ to denote $\widetilde{DU(A)}$, and $\widetilde{SU}(A)$ to denote $\widetilde{SU(A)}$.

Lemma 2.29.   Let $\alpha,\beta:K_{1}(A)\longrightarrow U(A)/\widetilde{DU}(A)$ be two splittings of $\pi_{A}$ in Lemma 2.27. Then

[TABLE]

and $\alpha(tor\;K_{1}(A))\subset\widetilde{SU}(A)/\widetilde{DU}(A)$. Furthermore, $\alpha$ identifies $tor(K_{1}(A))$ with $\widetilde{SU}(A)/\widetilde{DU}(A)$.

[TABLE]

By the exactness of the sequence, there is an element $f\in AffTA/\widetilde{\rho K_{0}}(A)$ such that

[TABLE]

Since $k\alpha(z)-k\beta(z)=\alpha(kz)-\beta(kz)=0,$ we have $\widetilde{\lambda}_{A}(kf)=0$. By the injectivity of $\widetilde{\lambda}_{A}$, $kf=0$. Note that $AffTA/\widetilde{\rho K_{0}}(A)$ is an $\mathbb{R}$-vector space, $f=0$. Furthermore, $k\alpha(z)=0$ in $U(A)/\widetilde{DU}(A)$ implies that

[TABLE]

Hence we get $\alpha(tor\;K_{1}(A))\subset\widetilde{SU}(A)$. If $u\in\widetilde{SU}(A)/\widetilde{DU}(A)$ then $\alpha(\pi_{A}(u))=u$.

∎

2.30.   Let $U_{tor}(A)$ denote the set of unitaries $u\in A$ such that $[u]\in tor\;K_{1}(A)$. For any $C^{*}$ algebra $A$ we have $\widetilde{SU}(A)\subset U_{tor}(A).$ If we further assume $\overline{DU_{0}(A)}=\overline{DU(A)}$, then

[TABLE]

Evidently, we have $U_{0}(A)/\widetilde{DU}(A)\cong U_{tor}(A)/\widetilde{SU}(A).$ The metric $\overline{D_{A}}$ on $U(A)/\widetilde{DU}(A)$ induces a metric $\widetilde{D}_{A}$ on $U(A)/\widetilde{SU}(A)$. And the above identification $U_{0}(A)/\widetilde{DU}(A)$ with $U_{tor}(A)/\widetilde{SU}(A)$ is an isometry with respect to $\overline{D}_{A}$ and $\widetilde{D}_{A}$. Hence $\widetilde{\lambda}_{A}$ in 2.26 can be regarded as a map (still denoted by $\widetilde{\lambda}_{A}$):

[TABLE]

Similar to Lemma 2.27, we have

Lemma 2.31.   If a unital $C^{*}$-algebra $A$ satisfies $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ and $\overline{DU_{0}(A)}=\overline{DU(A)}$ (see 2.24 and 2.23), in particular, if $A\in\mathcal{HD}$ or $A\in A\mathcal{HD}$, then the following hold:

(1)   There is a split exact sequence

[TABLE]

(2)   $\widetilde{\lambda}_{A}$ is an isometry with respect to the metrics $\widetilde{d}_{A}$ and $\widetilde{D}_{A}$.

2.32.   For each pair of projections $p_{1},p_{2}\in A$ with $p_{1}=up_{2}u^{*},$

[TABLE]

Also, since in any unital $C^{*}$-algebra $A$ and unitaries $u,v\in U(A)$, $v$ and $uvu^{*}$ represent a same element in $U(A)/\widetilde{SU}(A)$, and the above identification does not depend on the choice of $u$ to implement $p_{1}=up_{2}u^{*}$. That is for any $[p]\in\Sigma A$, the group $U(pAp)/\widetilde{SU}(pAp)$ is well defined, which does not depend on choice of $p\in[p]$. We will include this group (with metric) as part of our invariant. If $[p]\leq[q]$, then we can choose $p,q$ such that $p\leq q$. In this case, there is a natural inclusion map $\imath:pAp\longrightarrow qAq$ which induces

[TABLE]

where $\imath_{*}$ is defined by

[TABLE]

A unital homomorphism $\phi:A\to B$ induces a contractive group homomorphism

[TABLE]

If $\phi$ is not unital, then the map $\phi^{\natural}:U(A)/\widetilde{SU}(A)\longrightarrow U(\phi(\textbf{1}_{A})B\phi(\textbf{1}_{A}))/\widetilde{SU}(\phi(\textbf{1}_{A})B\phi(\textbf{1}_{A}))$ is induced by the corresponding unital homomorphism. In this case, $\phi$ also induces the map $\imath_{*}\circ\phi^{\natural}:U(A)/\widetilde{SU}(A)\longrightarrow U(B)/\widetilde{SU}(B)$, which is denoted by $\phi_{*}$ to avoid confusion. If $\phi$ is unital, then $\phi^{\natural}=\phi_{*}$. If $\phi$ is not unital, then $\phi^{\natural}$ and $\phi_{*}$ have different codomains. That is, $\phi^{\natural}$ has codomain $U(\phi(\textbf{1}_{A})B\phi(\textbf{1}_{A}))/\widetilde{SU}(\phi(\textbf{1}_{A})B\phi(\textbf{1}_{A}))$, but $\phi_{*}$ has codomain $U(B)/\widetilde{SU}(B)$. (See some further explanation with an example in the last paragraph of 3.7 below.)

Since $U(A)/\widetilde{SU}(A)$ is an Abelian group, we will call the unit $[{\bf 1}]\in U(A)/\widetilde{SU}(A)$ the zero element. If $\phi:A\to B$ satisfies $\phi(U(A))\subset\widetilde{SU}(\phi(\textbf{1}_{A})B\phi(\textbf{1}_{A}))$, then $\phi^{\natural}=0$. In particular, if the image of $\phi$ is of finite dimensional, then $\phi^{\natural}=0$.

2.33.   In this paper and [GJL], we will denote

[TABLE]

by $Inv(A)$. By a map from $Inv(A)$ to $Inv(B)$, we mean

[TABLE]

as in 2.7, and for each pair $([p],[\overline{p}])\in\Sigma A\times\Sigma B$ with $\alpha([p])=[\overline{p}]$, there exist an associate unital positive (continuous) linear map

[TABLE]

and an associate contractive group homomorphism

[TABLE]

satisfying the following compatibility conditions. (Note that $\chi^{p,\overline{p}}$ is continuous, as it is a contractive group homomorphism from a metric group to another metric group.)

(a)   If $p<q$, then the diagrams

[TABLE]

and

[TABLE]

commutes, where the vertical maps are induced by inclusions.

(b)   The following diagram commutes

[TABLE]

and therefore $\xi^{p,\overline{p}}$ induces a map (still denoted by $\xi^{p,\overline{p}}$):

[TABLE]

(The commutativity of $(III)$ follows from the commutativity of $(I)$, by 1.20 of [Ji-Jiang]. So this is not an extra requirement.)

(c)   The following diagrams

[TABLE]

and

[TABLE]

commute, where $\alpha_{1}$ is induced by $\alpha$.

We will denote the map from $Inv(A)$ to $Inv(B)$ by

[TABLE]

[TABLE]

Completely similar to [NT, Lemma 3.2] and [Thm2, Lemma 6.5], we have the following propositions.

Proposition 2.34.   Let unital $C^{*}$-algebra $A$ ($B$, resp.) satisfy $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ ($\rho(\mathcal{P}_{1}K_{0}(B))=\rho K_{0}(B)$, resp.) and $\overline{DU_{0}(A)}=\overline{DU(A)}$ ($\overline{DU_{0}(B)}=\overline{DU(B)}$, resp.). In particular, let $A,B\in\mathcal{HD}$ or $A\mathcal{HD}$ be unital $C^{*}$-algebras. Assume that

[TABLE]

are group homomorphisms such that $\psi_{0}$ is a contraction with respect to $d_{A}$ and $d_{B}$. Then there is a group homomorphism

[TABLE]

which is a contraction with respect to $D_{A}$ and $D_{B}$ such that the diagram

[TABLE]

commutes. If $\psi_{0}$ is an isometric isomorphism and $\psi_{1}$ is an isomorphism, then $\psi$ is an isometric isomorphism.

Proposition 2.35.   Let unital $C^{*}$-algebra $A$ ($B$, resp.) satisfy $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ ($\rho(\mathcal{P}_{1}K_{0}(B))=\rho K_{0}(B)$, resp.) and $\overline{DU_{0}(A)}=\overline{DU(A)}$ ($\overline{DU_{0}(B)}=\overline{DU(B)}$, resp.). In particular, let $A,B\in\mathcal{HD}$ or $A\mathcal{HD}$ be unital $C^{*}$-algebras. Assume that

[TABLE]

are group homomorphisms such that $\psi_{0}$ is a contraction with respect to $\widetilde{d}_{A}$ and $\widetilde{d}_{B}$. Then there is a group homomorphism

[TABLE]

which is a contraction with respect to $\widetilde{D}_{A}$ and $\widetilde{D}_{B}$ such that the diagram

[TABLE]

commutes. If $\psi_{0}$ is an isometric isomorphism and $\psi_{1}$ is an isomorphism, then $\psi$ is an isometric isomorphism.

Remark 2.36.  As in Proposition 2.35 (or Proposition 2.34), for each fixed pair $p\in A$, $\overline{p}\in B$ with

[TABLE]

if we have an isometric isomorphism between $AffT(pAp)/\widetilde{\rho K_{0}}(pAp)$ and $AffT(\overline{p}B\overline{p})/\widetilde{\rho K_{0}}(\overline{p}B\overline{p})$ (or between $AffT(pAp)/\overline{\rho K_{0}(pAp)}$ and $AffT(\overline{p}B\overline{p})/\overline{\rho K_{0}(\overline{p}B\overline{p})}$) and isomorphism between $K_{1}(pAp)$ and $K_{1}(pBp)$, then we have an isometric isomorphism between $U(pAp)/\widetilde{SU}(pAp)$ and $U(\overline{p}B\overline{p})/\widetilde{SU}(\overline{p}B\overline{p})$ (or $U(pAp)/\overline{DU(pAp)}$ and $U(\overline{p}B\overline{p})/\overline{DU(\overline{p}B\overline{p})}$) making both diagrams  $(IV)$ and $(V)$ commute. This is the reason $U(A)/\overline{DU(A)}$ is not included in the Elliott invariant in the classification of simple $C^{*}$-algebras. For our setting, even though for each pair of projections $(p,\bar{p})$ with $\alpha([p])=[\bar{p}]$, we can find an isometric isomorphism between $U(pAp)/\widetilde{SU}(pAp)$ and $U(\overline{p}B\overline{p})/\widetilde{SU}(\overline{p}B\overline{p})$, provided that the other parts of invariants $Inv^{0}(A)$ and $Inv^{0}(B)$ are isomorphic, we still can not make such system of isometric isomorphisms compatible—that is, can not make the diagram $II$ commutes for $p<q$. We will present two non isomorphic $C^{*}$-algebras $A$ and $B$ in our class such that $Inv^{0}(A)\cong Inv^{0}(B)$, in next section, where $Inv^{0}(B)$ is defined in 2.10. Hence it is essential to include $\{U(pAp)/\widetilde{SU}(pAp)\}_{p\in\Sigma}$ with the compatibility as part of $Inv(A)$.

2.37.  Replacing $U(pAp)/\widetilde{SU}(pAp)$, one can also use $U(pAp)/\overline{DU(pAp)}$ as the part of the invariant. That is, one can define $Inv^{\prime}(A)$ as

[TABLE]

with corresponding compatibility condition—one needs to change diagrams $(IV)$ and $(V)$ to the corresponding ones. It is not difficult to see that $Inv^{\prime}(A)\cong Inv^{\prime}(B)$ implies $Inv(A)\cong Inv(B)$. We choose the formulation of $Inv(A)$, since it is much more convenient for the proof of the main theorem in [GJL] and it is formally a weaker requirement than the one to require the isomorphism between $Inv^{\prime}(A)$ and $Inv^{\prime}(B)$, and the theorem is formally stronger. (Let us point out that, in the construction of the example (and its proof) in section 3 of this article, $Inv^{\prime}(A)$ is as convenient as $Inv(A)$, and therefore if only for the sake of example in section 3 of this paper, it is not necessary to introduce $\widetilde{SU}(A)$.)

Furthermore, it is straight forward to check the following proposition:

Proposition 2.38. Let unital $C^{*}$-algebra $A$ ($B$, resp.) satisfy $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ ($\rho(\mathcal{P}_{1}K_{0}(B))=\rho K_{0}(B)$, resp.) and $\overline{DU_{0}(A)}=\overline{DU(A)}$ ($\overline{DU_{0}(B)}=\overline{DU(B)}$, resp.). In particular, let $A,B\in\mathcal{HD}$ or $A\mathcal{HD}$ be unital $C^{*}$-algebras. Suppose that $K_{1}(A)=tor(K_{1}(A))$ and $K_{1}(B)=tor(K_{1}(B))$. It follows that $Inv^{0}(A)\cong Inv^{0}(B)$ implies that $Inv(A)\cong Inv(B)$.

$\xi^{p,\overline{p}}:~{}~{}AffT(pAp)/\widetilde{\rho K_{0}}(pAp)\longrightarrow AffT(\overline{p}B\overline{p})/\widetilde{\rho K_{0}}(\overline{p}B\overline{p})$

induces a unique isomorphism

$\chi^{p,\overline{p}}:U(pAp)/\widetilde{SU}(pAp)\longrightarrow U(\overline{p}B\overline{p})/\widetilde{SU}(\overline{p}B\overline{p})$

(Note that by the split exact sequence in Lemma 2.31, we have $AffT(pAp)/\widetilde{\rho K_{0}}(pAp)\cong U(pAp)/\widetilde{SU}(pAp)$).

∎

The following calculations and notations will be used in [GJL].

2.39.  In general, for $A=\oplus A^{i}$, $\widetilde{SU}(A)=\oplus_{i}\widetilde{SU}(A^{i}).$ For $A=PM_{l}(C(X))P\in\mathcal{HD}$, $\widetilde{SU}(A)=\widetilde{DU}(A).$ For $A=M_{l}(I_{k})$, $\widetilde{SU}(A)=\widetilde{DU}(A)\oplus K_{1}(A).$ For both cases, $U(A)/\widetilde{SU}(A)$ can be identified with

$C_{1}(X,S^{1}):=C(X,S^{1})/\{constant\;functions\}$ or with $C_{1}([0,1],S^{1})=C([0,1],S^{1})/\{constant\;functions\},$ for $A=M_{l}(I_{k}).$

Furthermore, $C_{1}(X,S^{1})$ can be identified as the set of continuous functions from $X$ to $S^{1}$ such that $f(x_{0})=1$ for certain fixed base point $x_{0}\in X$. For $X=[0,1]$, we choose 0 to be the base point. For $X=S^{1}$, we choose $1\in S^{1}$ to be the base point.

2.40.  Let $A=\oplus^{n}_{i=1}A^{i}\in\mathcal{HD}$, $B=\oplus^{m}_{j=1}B^{j}\in\mathcal{HD}$. In this subsection we will discuss some consequences of the compatibility of the maps between $AffT$ spaces. Let

[TABLE]

be projections satisfying $\alpha([p])=[\overline{p}]$ and $\alpha([q])=[\overline{q}]$. Suppose that two unital positive linear maps $\xi_{1}:AffTpAp\longrightarrow AffT\overline{p}B\overline{p}$ and $\xi_{2}:AffTqAq\longrightarrow AffT\overline{q}B\overline{q}$ are compatible with $\alpha$ (see diagram (2.$B$) in 2.11) and compatible with each other (see diagram (2.$A$) in 2.9). Since the (not necessarily unital) maps $AffTpAp\longrightarrow AffTqAq$ and $AffT\overline{p}B\overline{p}\longrightarrow AffT\overline{q}B\overline{q}$ induced by inclusions, are injective, we know that the map $\xi_{1}$ is completely decided by $\xi_{2}$. Let

[TABLE]

be the corresponding component of the map $\xi_{2}$ (or $\xi_{1}$). If $p^{i}\neq 0$ and $\overline{p}^{j}\neq 0$, then $\xi_{1}^{i,j}$ is given by the following formula, for any $f\in AffTp^{i}A^{i}p^{i}=C_{\mathbb{R}}(SpA^{i})(\cong AffTq^{i}Aq^{i})$,

[TABLE]

In particular, if $q=\textbf{1}_{A}$ with $\overline{q}=\alpha_{0}[\textbf{1}_{A}]$, and $\xi_{2}=\xi:AffTA\longrightarrow Aff\alpha_{0}[\textbf{1}_{A}]B\alpha_{0}[\textbf{1}_{A}]$ (note that since $AffTQBQ$ only depends on the unitary equivalence class of $Q$, it is convenient to denote it as $AffT[Q]B[Q]$), then we will denote $\xi_{1}$ by $\xi|_{([p],\alpha[p])}$. Even for the general case, we can also write $\xi_{1}=\xi_{2}|_{([p],\alpha[p])}$, when $p<q$ as above.

2.41.  As in 2.40, let $A=\oplus_{i=1}^{n}A^{i}$, $B=\oplus_{j=1}^{m}B^{j}$ and $p<q\in A$, $\overline{p}<\overline{q}\in B$, with $\alpha_{0}[p]=[\overline{p}]$ and $\alpha_{0}[q]=[\overline{q}]$. If

[TABLE]

is compatible with

[TABLE]

then $\gamma_{1}$ is completely determined by $\gamma_{2}$ (since both maps

[TABLE]

are injective). Therefore we can denote $\gamma_{1}$ by $\gamma_{2}|_{([p],\alpha[p])}$.

2.42.  Let us point out that, in 2.40 and 2.41, if $A\in A\mathcal{HD}$ and $B\in A\mathcal{HD}$, $\xi_{1}$ is not completely determined by $\xi_{2}$ and $\gamma_{1}$ is not completely determined by $\gamma_{2}$.

§3. The counter example

3.1.  In this section, we will present an example of $A\mathbb{T}$ algebras to prove that $Inv^{\prime}(A)$ or $Inv(A)$ is not completely determined by $Inv^{0}(A)$. That is, the Hausdorffifized algebraic $K_{1}$ group $\{U(pAp)/\overline{DU(pAp)}\}_{p\in proj(A)}$ or $\{U(pAp)/\widetilde{SU}(pAp)\}_{p\in proj(A)}$ with the corresponding compatibilities are indispensable as a part of the invariant for $Inv^{\prime}(A)$ or $Inv(A)$. This is one of the essential differences between the simple $C^{*}$-algebras and the $C^{*}$-algebras with the ideal property. In fact, for all the unital $C^{*}$-algebras $A$ satisfy a reasonable condition (e.g., $\rho(\mathcal{P}_{1}K_{0}(A))=\rho K_{0}(A)$ and $\overline{DU_{0}(A)}=\overline{DU(A)}$), we have

[TABLE]

[TABLE]

i.e., the metric groups $U(pAp)/\overline{DU(pAp)}$ and $U(pAp)/\widetilde{SU}(pAp)$ themselves are completely determined by $AffTpAp$ and $K_{1}(pAp)$, which are included in othe parts of the invariants i.e., there are decided by $Inv^{0}(A)$, but the compatibilities make the difference. The point is that the above isomorphisms are not natural and therefor the isomorphisms corresponding to the cutting down algebras $pAp$ and $qAq$ ($p<q$) may not be chosen to be compatible.

As pointed out in 2.37, $Inv^{\prime}(A)\cong Inv^{\prime}(B)$ implies $Inv(A)\cong Inv(B)$. For the $C^{*}$ algebras $A$ and $B$ constructed in this paper, we only need to prove $Inv^{0}(A)\cong Inv^{0}(B)$ but $Inv(A)\not\cong Inv(B)$. Consequently, $Inv^{\prime}(A)\not\cong Inv^{\prime}(B)$.

3.2.  Let $p_{1}=2,~{}~{}~{}p_{2}=3,~{}~{}~{}p_{3}=5,~{}~{}~{}p_{4}=7,~{}~{}~{}p_{5}=11,\cdots,~{}~{}~{}p_{n}$ be the n-th prime number, let $1<k_{1}<k_{2}<k_{3}<\cdots$ be a sequence of positive integers. Let

[TABLE]

[TABLE]

[TABLE]

[TABLE]

In general, let

$A_{n}=B_{n}=\bigoplus\limits_{i=1}\limits^{n-1}M_{p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{i}^{k_{i}}p_{i}^{k_{i+1}}\cdots p_{i}^{k_{n-1}}}(C[0,1])\oplus M_{p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{n-1}^{k_{n-1}}}(C(S^{1}))$

$=\bigoplus\limits_{i=1}\limits^{n-1}M_{\prod\limits_{j=1}\limits^{i}p_{j}^{k_{j}}\cdot\prod\limits_{j=i+1}\limits^{n-1}p_{i}^{k_{j}}}(C[0,1])\oplus M_{\prod\limits_{i=1}\limits^{n-1}p_{i}^{k_{i}}}(C(S^{1})).$

For $1\leq i\leq n-1$, let $[n,i]=\prod\limits_{j=1}\limits^{i}p_{j}^{k_{j}}\cdot\prod\limits_{j=i+1}\limits^{n-1}p_{i}^{k_{j}}$ and $[n,n]=[n,n-1]$. Then

[TABLE]

(Note that last two blocks have same size $[n,n]=[n,n-1]$.)

Note that $[n+1,i]=[n,i]\cdot p_{i}^{k_{n}}$ for all $i\in\{1,2,\cdots,n-1\}$ and $[n+1,n+1]=[n+1,n]=[n,n]\cdot p_{n}^{k_{n}}$.

3.3.  Let $\{t_{n}\}_{n=1}^{\infty}$ be a dense subset of $[0,1]$ and $\{z_{n}\}_{n=1}^{\infty}$ be a dense subset of $S^{1}$.

In this subsection, we will define the connecting homomorphisms

[TABLE]

For $i\leq n-1$, define $\phi_{n,n+1}^{i,i}=\psi_{n,n+1}^{i,i}:~{}~{}M_{[n,i]}(C[0,1])\longrightarrow M_{[n+1,i]}(C[0,1])(=M_{[n,i]\cdot p_{i}^{k_{n}}}(C[0,1]))$ by

[TABLE]

Define $\phi_{n,n+1}^{n,n+1}=\psi_{n,n+1}^{n,n+1}:~{}~{}M_{[n,n]}(C(S^{1}))\longrightarrow M_{[n+1,n+1]}(C(S^{1}))=M_{[n,n]\cdot p_{n}^{k_{n}}}(C(S^{1}))$ by

[TABLE]

But $\phi_{n,n+1}^{n,n}$ and $\psi_{n,n+1}^{n,n}$ are defined differently—this is the only non-equal component of $\phi_{n,n+1}$ and $\psi_{n,n+1}$.

Let $l=p_{n}^{k_{n}}-1$, then

[TABLE]

[TABLE]

for any $f\in M_{[n,n]}(C(S^{1}))$, where $l_{n}=4^{n}\cdot[n+1,n]\in\mathbb{N}$.

Let all other parts $\phi_{n,n+1}^{i,j},\psi_{n,n+1}^{i,j}$ of $\phi_{n,n+1},\psi_{n,n+1}$ (except $i=j\leq n$ or $i=n,j=n+1$, defined above) be zero.

Note that all $\phi_{n,n+1}^{i,j},\psi_{n,n+1}^{i,j}$ are either injective or zero.

Let $A=lim(A_{n},\phi_{n,m}),B=lim(B_{n},\psi_{n,m})$. Then it follows from the density of the sets $\{t_{n}\}_{n=1}^{\infty}$ and $\{z_{n}\}_{n=1}^{\infty}$ that both $A$ and $B$ have the ideal property (see the characterization theorem for $AH$ algebras with the ideal property [Pa]).

Proposition 3.4.   There is an isomorphism between $Inv^{0}(A)$ and $Inv^{0}(B)$ (see 2.10)—that is, there is an isomorphism

[TABLE]

which is compatible with Bockstein operations, and for pairs $(p,q)$ with $p\in\Sigma A,q\in\Sigma B$ and $\alpha([p])=[q]$, there are associated unital positive linear maps

[TABLE]

which are compatible in the sense of 2.9 (see diagram (2.A) in 2.9).

[TABLE]

Note that $\phi_{n,n+1}^{i,i}=\psi_{n,n+1}^{i,i}$ for $i\leq n-1$, $\phi_{n,n+1}^{n,n+1}=\psi_{n,n+1}^{n,n+1}$, and

[TABLE]

(see the definition of $\phi_{n,n+1}$ and $\psi_{n,n+1}$). Therefore,

[TABLE]

induce the approximately intertwining diagram

[TABLE]

in the sense of Elliott [Ell1]. Therefore, there is a unital positive isomorphism

[TABLE]

Also, for any projection $[P]\in K_{0}(A)$, there is a projection $P_{n}\in A_{n}=B_{n}$ (for $n$ large enough) with $P_{n}^{i}=diag(1,\cdots,1,0,\cdots,0)\in M_{[n,i]}(C(X_{n,i}))$, where $X_{n,i}=[0,1]$ for $i\leq n-1$, and $X_{n,n}=S^{1}$, such that $\phi_{n,\infty}([P_{n}])=[P]\in K_{0}(A)$. Note that for any constant functions $f\in A_{n}^{i}=B_{n}^{i}$ (e.g., $P_{n}^{i}$ above) and for any $j$, $\phi^{i,j}_{n,n+1}(f)$ and $\psi^{i,j}_{n,n+1}(f)$ are still constant functions and $\phi^{i,j}_{n,n+1}(f)=\psi^{i,j}_{n,n+1}(f)$. That is, we have

[TABLE]

[TABLE]

Let $P_{\infty}=\phi_{n,\infty}(P_{n})$ and $Q_{\infty}=\psi_{n,\infty}(P_{n})$. Then the identity maps $\{\eta_{m}\}_{m>n}$ also induce the following approximate intertwining diagram:

[TABLE]

and hence induce a positive linear isomorphism

[TABLE]

(Note that $[P_{\infty}]=[P],\;[Q_{\infty}]=\alpha[P]\;in\;K_{0}(A)\;and\;K_{0}(B)$, respectively.) Evidently those maps are compatible since, they are induced by the same sequence of homomorphisms $\{\eta_{n}\}$ and $\{\eta_{n}^{-1}\}$.

∎

The following Definition 3.5 and Proposition 3.6 are inspired by [Ell3].

Definition 3.5.   Let $C\!=\!\lim(C_{n},\phi_{n,m})$ be an $A\mathcal{HD}$ inductive limit. We say the system $(C_{n},\phi_{n,m})$ has the uniformly varied determinant if for any $C^{i}_{n}=M_{[n,i]}(C(S^{1}))$ (that is, $C^{i}_{n}$ has spectrum $S^{1}$) and $C^{j}_{n+1}$ and $f\in C_{n}^{i}$ defined by

[TABLE]

we have that det($\phi^{i,j}_{n,n+1}(f)(x))=$ constant for $x\in Sp(C^{j}_{n+1})\neq S^{1}$ or det($\phi^{i,j}_{n,n+1}(f)(z))=\lambda z^{k}$ ($\lambda\in\mathbb{C}$) for $z\in Sp(C^{j}_{n+1})=S^{1}$, where $j$ satisfy $\phi^{i,j}_{n,n+1}\not=0$ and the determinant is taken inside $\phi^{i,j}_{n,n+1}(1_{C^{i}_{n}})C^{j}_{n+1}\phi^{i,j}_{n,n+1}(1_{C^{i}_{n}})$.

Proposition 3.6.   If the inductive limit system $C=(C_{n},\phi_{n,m})$ has the uniformly varied determinant, then for any elements $[p]\in\sum C$, there are a splitting maps

[TABLE]

of the exact sequences

[TABLE]

(that is, $\pi_{pCp}\circ S_{pCp}=id\;on\;K_{1}(pCp)/tor\;K_{1}(pCp)$) such that the system of maps $\{S_{pCp}\}_{[p]\in\sum C}$ are compatible in the following sense: if $p<q$, then the following diagram commutes

[TABLE]

where the vertical maps are induced by the inclusions $pCp\longrightarrow qCq$.

Write $p_{n}C_{n}p_{n}=D=\oplus D^{i}$. Let $I=\{i~{}|~{}~{}Sp(D^{i})=S^{1}\}$. For $i\in I$, $D^{i}$ can be identified with $M_{l_{i}}(C(S^{1}))$. Let $u_{i}\in D^{i}$ be defined by

[TABLE]

which represents the standard generator of $K_{1}(D^{i}).$ Then $x_{n}$ can be represented by

[TABLE]

Define $S(x)=[\phi_{n,\infty}(u)]\in U(pCp)/\widetilde{SU}(pCp)$. Note that all unitaries with constant determinants are in $\widetilde{SU}$, and that the inductive system has the uniformly varied determinant, it is routine to verify that $S(x)$ is well defined and the system $\{S_{pCp}\}_{[p]\in\sum C}$ makes the diagram (3.6.A) commute.

∎

3.7.  Let ${\cal A}$ be a unital $C^{*}$-algebra. Then $AffT{\cal A}$ is a real Banach space with quotient space $AffT{\cal A}/\widetilde{\rho K_{0}}({\cal A})$. Let us use $\|\cdot\|^{\sim}$ to denote the quotient norm. Note that $\widetilde{\lambda}_{\cal A}$ identifies $U_{tor}({\cal A})/\widetilde{SU}({\cal A})$ with $AffT{\cal A}/\widetilde{\rho K_{0}}({\cal A})$. In this way, $U_{tor}({\cal A})/\widetilde{SU}({\cal A})$ is regarded as a real Banach space, whose norm is also denoted by $\|\cdot\|^{\sim}$. In general, we have

[TABLE]

but the identification is not canonical. Even though $U({\cal A})/\widetilde{SU}({\cal A})$ is not a Banach space, it is an Abelian group: for $[u],[v]\in U({\cal A})/\widetilde{SU}({\cal A})$, define $[u]-[v]=[uv^{*}]$.

The norm $\|\cdot\|^{\sim}$ is related to the metrices $\widetilde{d}_{\cal A}$ (on $AffT{\cal A}/\widetilde{\rho K_{0}}({\cal A})$; see 2.26) and $\widetilde{D}_{\cal A}$ (on $U_{tor}({\cal A})/\widetilde{SU}({\cal A})$; see 2.30) as below. Let $\varepsilon<1$. For any $f,g\in AffT{\cal A}/\widetilde{\rho K_{0}}({\cal A})$,

[TABLE]

And for any $[u],[v]\in U({\cal A})/\widetilde{SU}({\cal A})$ with $[u]-[v]=[uv^{*}]\in U_{tor}({\cal A})/\widetilde{SU}({\cal A})$,

[TABLE]

For ${\cal A}=PM_{l}(C(X))P\in\mathcal{HD}$ or ${\cal A}=M_{l}(I_{k})$ (at this case we also denote $[0,1]$ by $X$), there are canonical identification (see 2.39)

[TABLE]

Choose a base point $x_{0}\in X$. Let $C_{x_{0}}(X,\mathbb{R})$ be the set of functions $f\in C(X,\mathbb{R})$ with $f(x_{0})=0$. Then $C(X,\mathbb{R})/\{constant\;functions\}\cong C_{x_{0}}(X,\mathbb{R})$. For $[f]\in AffT{\cal A}/\widetilde{\rho K_{0}}({\cal A})$ (or $[f]\in U_{tor}({\cal A})/\widetilde{SU}({\cal A})$) identified with a function $f\in C_{x_{0}}(X,\mathbb{R})$, we have

[TABLE]

(rather than $sup_{x\in X}\{|f(x)|\}$).

In the above case, if $p\in{\cal A}$ is a non zero projection, then $U_{tor}(p{\cal A}p)/\widetilde{SU}(p{\cal A}p)\cong AffT(p{\cal A}p)/\widetilde{\rho K_{0}}(p{\cal A}p)$ is also identified with $C_{x_{0}}(X,\mathbb{R})$. Consider the inclusion map $\imath:pAp\to A$. Then the map $\imath_{*}$ as map from $U_{tor}(p{\cal A}p)/\widetilde{SU}(p{\cal A}p)\cong AffT(p{\cal A}p)/\widetilde{\rho K_{0}}(p{\cal A}p)$ to $U_{tor}({\cal A})/\widetilde{SU}({\cal A})$ can be described as below: if

$u\in U_{tor}(p{\cal A}p)/\widetilde{SU}(p{\cal A}p)\cong AffT(p{\cal A}p)/\widetilde{\rho K_{0}}(p{\cal A}p)$ is identified with $f\in C_{x_{0}}(X,\mathbb{R})$, then

$\imath_{*}(u)\in U_{tor}({\cal A})/\widetilde{SU}({\cal A})$ is identified with $\frac{rank(p)}{rank(\textbf{1}_{calA})}f$. But $\imath^{\natural}$ is the identity map from $U_{tor}(p{\cal A}p)/\widetilde{SU}(p{\cal A}p)\cong AffT(p{\cal A}p)/\widetilde{\rho K_{0}}(p{\cal A}p)$ to itself (not to $U_{tor}({\cal A})/\widetilde{SU}({\cal A})$).

3.8.   It is easy to see that $K_{1}(A)=K_{1}(B)=\mathbb{Z}$.

In the definition of $A_{n}=\oplus_{i=1}^{n}A_{n}^{i}$, only one block $A_{n}^{n}=M_{[n,n]}(C(S^{1}))$ has spectrum $S^{1}$, and only two partial maps $\phi^{n,j}_{n,n+1}$ for $~{}j=n,~{}j=n+1$ (of $\phi_{n,n+1}$ from $A_{n}^{n}$) are nonzero. Let $f\in A_{n}^{n}$ be defined as in Definition 3.5. Then det$(\phi^{n,n+1}_{n,n+1}(f)(z))=z$ and det$(\phi^{n,n}_{n,n+1}(f)(t))=e^{2\pi it}e^{-2\pi it}e^{2\pi i\frac{1}{l}}e^{2\pi i\frac{2}{l}}\cdots e^{2\pi i\frac{l-1}{l}}=\pm 1$ (see 3.3). So the inductive limit system $(A_{n},\phi_{n,m})$ has the uniformly varied determinant, and therefore the limit algebra $A$ has compatible splitting maps $S_{p}:K_{1}(pAp)\rightarrow U(pAp)/\widetilde{SU}(pAp)$.

We will prove that $B=lim(B_{n},\psi_{n,m})$ does not have such compatible system of splitting maps $\{K_{1}(pBp)\longrightarrow U(pBp)/\widetilde{SU}(pBp)\}_{[p]\in\sum B}$.

Before proving the above fact, let us describe the $K_{0}$-group of $A$ and $B$. Let

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

where $p_{1}=2,p_{2}=3,\cdots,p_{i},\cdots$ and $k_{1},k_{2},\cdots,k_{i}\cdots$ are defined in 3.2. Then

[TABLE]

Furthermore, their positive cones consist of the elements whose coordinates are non-negative, and their order units are $[\textbf{1}_{A}]=[\textbf{1}_{B}]=(1,1,\cdots,1,\cdots)\in\prod\limits_{n=1}\limits^{\infty}G_{n}$. Let

[TABLE]

be a scaled ordered isomorphism. Then \alpha_{0}\big{(}(1,1,\cdots,1,\cdots)\big{)}=(1,1,\cdots,1,\cdots). Note that an element $x\in{\tilde{G}}$ is divisible by power $p_{1}^{n}$ (for any $n$) of the first prime number $p_{1}=2$ if and only if $x=(t,0,0,\cdots,0,\cdots)\in G_{1}\subset{\tilde{G}}$. Hence \alpha_{0}\big{(}(1,0,0,\cdots,0,\cdots)\big{)}=(t,0,0,\cdots,0,\cdots) for some $t\in G_{1}$ with $t>0$. Hence

[TABLE]

Since $\alpha_{0}$ preserves the positive cone, we have $1-t\geq 0$ which implies $t\leq 1$. On the other hand, $(\alpha_{0})^{-1}$ takes $(1,0,0,\cdots,0,\cdots)$ to $(1/t,0,0,\cdots,0,\cdots)$. But $(\alpha_{0})^{-1}$ also preserves the positive cone. Symmetrically, we get $t\geq 1$. That is, \alpha_{0}\big{(}(1,0,0,\cdots,0,\cdots)\big{)}=(1,0,0,\cdots,0,\cdots). Similarly, using the fact that $G_{k}$ is the subgroup of all elements in ${\tilde{G}}$ which can be divisible by any power of $p_{k}$—the $k^{th}$ prime number, we can prove that

[TABLE]

That is, $\alpha_{0}$ is the identity on ${\tilde{G}}$.

Note that $Sp(A)=Sp(B)$ is the one point compactification of $\{1,2,3\cdots\}$—or, in other words, $\{1,2,3\cdots,\infty\}$. If we let $I_{n}$ (or $J_{n}$) be the primitive ideal $A$ (or $B$) corresponding to $n$ (including $n=\infty$), then

[TABLE]

Note also that if $m^{\prime}>m>n\in\mathbb{N}$, then $\phi_{m,m^{\prime}}(A_{m}^{n})\subset A_{m^{\prime}}^{n}$ and $\psi_{m,m^{\prime}}(B_{m}^{n})\subset B_{m^{\prime}}^{n}$. Hence

$A/I_{n}=\lim_{n<m\to\infty}(A_{m}^{n},\phi_{m,m^{\prime}}|_{A_{m}^{n}})$ (and $B/J_{n}=\lim_{n<m\to\infty}(B_{m}^{n},\psi_{m,m^{\prime}}|_{B_{m}^{n}})$ resp.) are ideals of $A$ (and $B$ resp.). But $A/I_{\infty}$ (or $B/J_{\infty}$) is not an ideal of $A$ (or $B$).

Let $\alpha:(\underline{K}(A),\underline{K}(A)^{+},\Sigma A)\longrightarrow(\underline{K}(B),\underline{K}(B)^{+},\Sigma B)$ be an isomorphism. By 3.8 the induced map $\alpha_{0}$ on $K_{0}$ group is identity, when both $K_{0}(A)$ and $K_{0}(B)$ are identified with ${\tilde{G}}$ as scaled ordered groups. That is, $\alpha_{0}$ is the same as the $\alpha_{0}$ induced by the shape equivalence in the proof of Proposition 3.4. In particular, if there is an isomorphism $\wedge:A\longrightarrow B$, then for all $i\leq n-1$, $\wedge_{*}[(\phi_{n,\infty}(\textbf{1}_{A_{n}^{i}}))]=[\psi_{n,\infty}(\textbf{1}_{B_{n}^{i}})]$. This implies $\wedge(\phi_{n,\infty}(\textbf{1}_{A_{n}^{i}}))=\psi_{n,\infty}(\textbf{1}_{B_{n}^{i}})$, since $\psi_{n,\infty}(\textbf{1}_{B_{n}^{i}})=\textbf{1}_{B/I_{i}}$, which is in the center of $B$ (any element in the center of the $C^{*}$-algebra can only unitary equivalent to itself). Hence it is also true that $\wedge(\phi_{n,\infty}(\textbf{1}_{A_{n}^{i}}))=\psi_{n,\infty}(\textbf{1}_{B_{n}^{i}})$ for $i=n$.

3.9.  Let $P_{1}=1_{B}=\psi_{1,\infty}(\textbf{1}_{B_{1}}),~{}~{}P_{2}=\psi_{2,\infty}(\textbf{1}_{B_{2}^{2}}),~{}~{}P_{3}=\psi_{3,\infty}(\textbf{1}_{B_{3}^{3}}),\cdots,~{}~{}P_{n}=\psi_{n,\infty}(\textbf{1}_{B_{n}^{n}}),\cdots.$ Then

$P_{1}>P_{2}>\cdots>P_{n}\cdots.$ We will prove that there are no splitings

[TABLE]

which are compatible for all pairs of projections $P_{n}>P_{m}$ (see diagram (3.6.A)), in the next subsection. Before doing so, we need some preparations.

Set $Q_{1}=P_{1}-P_{2},~{}~{}Q_{2}=P_{2}-P_{3},\cdots,~{}~{}Q_{n}=P_{n}-P_{n+1}.$ Then for each $n$, we have the inductive limit

[TABLE]

(note that for $m>n$, $\psi_{m,m+1}^{n,j}=0$ if $j\neq n$), which is the quotient algebra corresponding to the primitive ideal of $n\in Sp(B)=\{1,2,3\cdots,\infty\}$. Note that $Q_{n}BQ_{n}$ is a simple $AI$ algebra. The inductive limit of the $C^{*}$-algebras

[TABLE]

induces the inductive limit of the ordered Banach spaces

[TABLE]

whose connecting maps $\xi_{m,m+1}:C_{\mathbb{R}}([0,1])\longrightarrow C_{\mathbb{R}}([0,1])$ (for $m>n$) satisfy that

[TABLE]

Hence we have the following approximate intertwining diagram

[TABLE]

Consequently, $AffTQ_{n}BQ_{n}\cong C_{\mathbb{R}}[0,1]$, and the maps

[TABLE]

(under the identification) satisfy

[TABLE]

Therefore $\|\xi_{m,\infty}(f)\|\geq\frac{3}{4}\|f\|.$

Note that $\widetilde{\rho K_{0}}(Q_{n}BQ_{n})=\mathbb{R}=\widetilde{\rho K_{0}}(B_{m}^{n})$ consists of constant functions on $[0,1]$. Let $h\in C_{\mathbb{R}}[0,1]=AffT(B_{m}^{n}).$ Considering the element $\xi_{m,\infty}(h)$ as in $AffT(Q_{n}BQ_{n})/\widetilde{\rho K_{0}}(Q_{n}BQ_{n})$, we have

[TABLE]

where $\|\cdot\|^{\sim}$ is defined in 3.7.

3.10.  We now prove that no compatible splittings

[TABLE]

exists. Suppose such splittings exist. Then consider the generator $x\in K_{1}(B)=\mathbb{Z}.$

Note that $x\in K_{1}(P_{n}BP_{n})\cong K_{1}(B)$, for all $P_{n}$. Note also that the diagram

[TABLE]

commutes $(P_{1}BP_{1}=B)$. The composition

[TABLE]

is the zero map. (Note that $Q_{i}BQ_{i}$ is an ideal of $B$ and is also the quotient $B/J_{i}$.) Consequently, we have

[TABLE]

where $\pi_{n}:B\rightarrow Q_{n}BQ_{n}$ is the quotient map. Let $S_{1}(x)$ be represented by a unitary $u\in U(B)$. Then there are an $n$ (large enough) and $[u_{n}]\in U(B_{n})/\widetilde{SU}(B_{n})$, represented by unitary $u_{n}\in B_{n}$, such that $\psi_{n,\infty}^{\natural}([u_{n}])-S_{1}(x)\in U_{tor}(B_{n})/\widetilde{SU}(B_{n})$ and

[TABLE]

Note that

[TABLE]

is the identify map from $\mathbb{Z}$ to $\mathbb{Z}$. Let $g\in M_{[n,n]}(C(S^{1}))=B_{n}^{n}$ be defined by

[TABLE]

Then $[g^{-1}u_{n}]=0$ in $K_{1}(B_{n})$. By the exactness of the sequence

[TABLE]

there is an $h\in\bigoplus\limits_{i=1}\limits^{n}C_{\mathbb{R}}[0,1]\oplus C_{\mathbb{R}}(S^{1})=AffTB_{n}$ such that

[TABLE]

Let $\|h\|=M$. Choose $m>n$ such that $4^{m-1}>8M+8$.

Consider

[TABLE]

which is the composition

[TABLE]

Let $g^{\prime}=\psi_{n,m}^{n,m-1}(g)$. We know that

[TABLE]

where the $*^{\prime}s$ represent constant functions on $[0,1]$, and therefore

[TABLE]

with $h^{\prime}(t)=\frac{l_{m-1}}{[m,m-1]}\cdot t\cdot 1_{[m,m-1]}.$ When we identify $U(B_{m}^{m-1})/\widetilde{SU}(B_{m}^{m-1})$ with

[TABLE]

$g^{\prime}$ is identified with $\widetilde{h}\in C_{\mathbb{R}}[0,1]$ with

[TABLE]

Since $\frac{l_{m-1}}{[m,m-1]}\geq 8M+8,$ we have

[TABLE]

(see 3.7). On the other hand,

[TABLE]

where $[h]\in AffTB_{n}/\widetilde{\rho K_{0}}(B_{n})$ is the element defined by $h$, and

[TABLE]

is the map defined in 2.30 (also see 2.26). Consequently,

[TABLE]

with

[TABLE]

since $\|h\|\leq M.$ Therefore,

[TABLE]

satisfies

[TABLE]

where $\pi_{m-1}:B\longrightarrow Q_{m-1}BQ_{m-1}$ is the quotient map. On the other hand,

[TABLE]

as calculated in ($*$). Recall that

[TABLE]

We get

[TABLE]

which is a contradiction. This contradiction proves that such system of splittings do not exist. Hence $Inv(A)\ncong Inv(B)$ and $A\ncong B$.

3.11. One can easily verify that

[TABLE]

[TABLE]

Since $\overline{\rho K_{0}(A)}(=\overline{\rho K_{0}(B)})$ is already a vector space, we have $\widetilde{\rho K_{0}}(A)=\overline{\rho K_{0}(A)}$ and $\widetilde{\rho K_{0}}(B)=\overline{\rho K_{0}(B)}$. Therefore

[TABLE]

On the other hand, $U_{tor}(A)=U_{0}(A)$. Hence $\widetilde{SU}(A)=\overline{DU(A)}$. Furthermore the map ${\lambda}_{A}:AffTA/\overline{\rho K_{0}(A)}\longrightarrow U(A)/\overline{DU(A)}$ can be identified with the map $\widetilde{\lambda}_{A}:AffTA/\widetilde{\rho K_{0}}(A)\longrightarrow U(A)/\widetilde{SU(A)}.$ That is $Inv^{\prime}(A)=Inv(A)$. Similarly, $Inv(B)=Inv^{\prime}(B)$.

3.12. A routine calculation shows (we omit the details) that for any finite subset $F\subset A_{n}$, and $\varepsilon>0$, there is an $m>n$ and two finite dimensional unital sub $C^{*}$-algebras $C,D\subset A_{m}$ with non abelian central projection such that

[TABLE]

Consequently, both $C^{*}$algebras $A$ and $B$ are approximately divisible in the sense of Definition 1.2 of [BKR]. By Theorem 2.3 of [TW], both $A$ and $B$ are ${\cal Z}$-stable. That is, $A\otimes{\cal Z}\cong A$ and $B\otimes{\cal Z}\cong B$, where ${\cal Z}$ is the Jiang-Su algebra (see [JS]). Furthermore, by using [Ti] (see [Cow-Ell-I] also), one can prove that $Cu(A)\cong Cu(B)$ and $Cu(A\otimes C(S^{1}))\cong Cu(B\otimes C(S^{1}))$.