On the magic square C*-algebra of size 4

0. Introduction

Let $n=1,2,\ldots$ . The magic square C*-algebra $A(n)$ of size $n$ is the underlying C*-algebra of the quantum group $A_{s}(n)$ defined by Wang in [W] as a free analogue of the symmetric group $\mathfrak{S}_{n}$ . In [BC, Proposition 1.1], it is claimed that for $n=1,2,3$ , $A(n)$ is isomorphic to $\mathbb{C}^{n!}$ , and hence commutative and finite dimensional. We give the proof of this fact in Proposition 2.1. In [BM, Proposition 1.2] it is proved that for $n\geq 4$ $A(n)$ is non-commutative and infinite dimensional. We see that for $n\geq 5$ $A(n)$ is not exact (Proposition 2.5). Something interesting happens for $A(4)$ (see [BM, BB, BC]). In [BM], Banica and Moroianu constructed a $*$ -homomorphism from $A(4)$ to $M_{4}(C(SU(2)))$ by using the Pauli matrices, and showed that it is faithful in some weak sense. In [BC], Banica and Collins showed that the $*$ -homomorphism above is in fact faithful by using integration techniques. We reprove this fact in Corollary 7.9. Our method uses a twisted crossed product. The following is the first main result.

Theorem A (Theorem 3.6). The twisted crossed product $A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ is isomorphic to $M_{4}(C(\mathbb{R}P^{3}))$ .

The notation in this theorem is explained in Section 3. From this theorem, we see that the magic square C*-algebra $A(4)$ of size $4$ is isomorphic to a $C^{*}$ -subalgebra of the homogeneous $C^{*}$ -algebra $M_{4}(C(\mathbb{R}P^{3}))$ . The next theorem, which is the second main result, expresses this $C^{*}$ -subalgebra as a fixed point algebra of $M_{4}(C(\mathbb{R}P^{3}))$ .

Theorem B (Theorem 8.2). The fixed point algebra $M_{4}(C(\mathbb{R}P^{3}))^{\beta}$ of the action $\beta$ is isomorphic to $A(4)$ .

See Section 8 for the definition of the action $\beta$ . Since $\beta$ is concrete, we can analyze $M_{4}(C(\mathbb{R}P^{3}))^{\beta}$ very explicitly. In particular, we can compute the K-groups of $M_{4}(C(\mathbb{R}P^{3}))^{\beta}$ explicitly. As a corollary we get the following which is the third main result.

Theorem C (Theorem 15.16). We have $K_{0}(A(4))\cong\mathbb{Z}^{10}$ and $K_{1}(A(4))\cong\mathbb{Z}$ . More specifically, $K_{0}(A(4))$ is generated by $\{[p_{i,j}]_{0}\}_{i,j=1}^{4}$ , and $K_{1}(A(4))$ is generated by $[u]_{1}$ .

The positive cone $K_{0}(A(4))_{+}$ of $K_{0}(A(4))$ is generated by $\{[p_{i,j}]_{0}\}_{i,j=1}^{4}$ as a monoid.

Note that $\{p_{i,j}\}_{i,j=1}^{4}$ is the generating set of $A(4)$ consisting of projections, and $u$ is the defining unitary (see Definition 15.15). We should remark that the computation $K_{0}(A(4))\cong\mathbb{Z}^{10}$ and $K_{1}(A(4))\cong\mathbb{Z}$ and that $K_{0}(A(4))$ is generated by $\{[p_{i,j}]_{0}\}_{i,j=1}^{4}$ were already obtained by Voigt in [V] by using Baum-Connes conjecture for quantum groups. In fact, Voigt got the corresponding results for $A(n)$ with $n\geq 4$ . Theorem C gives totally different proofs for the results by Voigt in [V] by analyzing the structure of $A(4)$ directly which seems not to be applied to $A(n)$ for $n>4$ . That $K_{1}(A(4))$ is generated by $[u]_{1}$ was not obtained in [V], and is a new result. Combining this result with the computation that $K_{1}(A(n))\cong\mathbb{Z}$ for $n\geq 4$ in [V] and the easy fact that the surjection $A(n)\to A(4)$ in Corollary 2.4 for $n\geq 4$ sends the defining unitary to the direct sum of the defining unitary and the units, we obtain that $K_{1}(A(n))\cong\mathbb{Z}$ is generated by the $K_{1}$ class of the defining unitary for $n\geq 4$ . We would like to thank Christian Voigt for the discussion about this observation.

This paper is organized as follows. In Section 1, we define magic square C*-algebras $A(n)$ and their abelianizations $A^{\text{ab}}(n)$ . In Section 2, we investigate $A(n)$ for $n\neq 4$ . From Section 3, we study $A(4)$ . In Section 3, we introduce the twisted crossed product $A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ , and state Theorem A. We give the proof of Theorem A from Section 4 to Section 7. In Section 8, we state and prove Theorem B. From Section 9 to Section 15, we prove Theorem C.

Acknowledgments. The first author thank Junko Muramatsu for helping the research in the beginning of the research. The authors are grateful to Makoto Yamashita for calling attention to [V], and to Christian Voigt for the discussion on the results in [V]. The first author was supported by JSPS KAKENHI Grant Number JP18K03345.

1. Definitions of and basic facts on magic square C*-algebras

Definition 1.1.

Let $n=1,2,\ldots$ . The magic square C*-algebra of size $n$ is the universal unital C*-algebra $A(n)$ generated by $n\times n$ projections $\{p_{i,j}\}_{i,j=1}^{n}$ satisfying

\displaystyle\sum_{i=1}^{n}p_{i,j}=1

\displaystyle\quad(j=1,2,\ldots,n),

\displaystyle\sum_{j=1}^{n}p_{i,j}=1

\displaystyle\quad(i=1,2,\ldots,n).

Remark 1.2.

The magic square C*-algebra $A(n)$ is the underlying C*-algebra of the quantum group $A_{s}(n)$ defined by Wang in [W] as a free analogue of the symmetric group $\mathfrak{S}_{n}$ .

We fix a positive integer $n$ . Let $\mathfrak{S}_{n}$ be the symmetric group of degree $n$ whose element is considered to be a bijection on the set $\{1,2,\ldots,n\}$ .

Definition 1.3.

By the universality of $A(n)$ , there exists an action $\alpha\colon\mathfrak{S}_{n}\times\mathfrak{S}_{n}\curvearrowright A(n)$ defined by

\alpha_{(\sigma,\mu)}(p_{i,j})=p_{\sigma(i),\mu(j)}

for $(\sigma,\mu)\in\mathfrak{S}_{n}\times\mathfrak{S}_{n}$ and $i,j=1,2,\ldots,n$ .

Definition 1.4.

Let $A^{\text{ab}}(n)$ be the universal unital C*-algebra generated by $n\times n$ projections $\{p_{i,j}\}_{i,j=1}^{n}$ satisfying the relations in Definition 1.1 and

p_{i,j}p_{k,l}=p_{k,l}p_{i,j}\qquad(i,j,k,l=1,2,\ldots,n).

The following lemma follows immediately from the definitions.

Lemma 1.5.

The C*-algebra $A^{\text{ab}}(n)$ is the abelianization of $A(n)$ . More specifically, there exists a natural surjection $A(n)\twoheadrightarrow A^{\text{ab}}(n)$ sending each projection $p_{i,j}$ to $p_{i,j}$ , and every $*$ -homomorphism from $A(n)$ to an abelian C*-algebra factors through this surjection.

Proposition 1.6.

The abelian C*-algebra $A^{\text{ab}}(n)$ is isomorphic to the C*-algebra $C(\mathfrak{S}_{n})$ of continuous functions on the discrete set $\mathfrak{S}_{n}$ .

Proof.

For each $\sigma\in\mathfrak{S}_{n}$ , we define a character $\chi_{\sigma}$ of $A^{\text{ab}}(n)$ by

\chi_{\sigma}(p_{i,j})=\begin{cases}1&(i=\sigma(j))\\ 0&(i\neq\sigma(j)).\end{cases}

Note that such a character $\chi_{\sigma}$ uniquely exists by the universality of $A^{\text{ab}}(n)$ . It is easy to see that any character of $A^{\text{ab}}(n)$ is in the form of $\chi_{\sigma}$ for some $\sigma\in\mathfrak{S}_{n}$ . This shows that $A^{\text{ab}}(n)$ is isomorphic to $C(\mathfrak{S}_{n})$ by the Gelfand theorem. ∎

We can compute minimal projections of $A^{\text{ab}}(n)$ as follows.

Proposition 1.7.

For $\sigma\in\mathfrak{S}_{n}$ , we set

p_{\sigma}\coloneqq p_{\sigma(1),1}p_{\sigma(2),2}\cdots p_{\sigma(n),n}\in A^{\text{ab}}(n).

Then $\{p_{\sigma}\}_{\sigma\in\mathfrak{S}_{n}}$ is the set of minimal projections of $A^{\text{ab}}(n)$ .

Proof.

Since $A^{\text{ab}}(n)$ is commutative, $p_{\sigma}$ is a projection for every $\sigma\in\mathfrak{S}_{n}$ . For $\sigma\in\mathfrak{S}_{n}$ , let $\chi_{\sigma}$ be the character defined in the proof of Proposition 1.6. Then we have

\chi_{\sigma^{\prime}}(p_{\sigma})=\begin{cases}1&(\sigma^{\prime}=\sigma)\\ 0&(\sigma^{\prime}\neq\sigma)\end{cases}

for $\sigma,\sigma^{\prime}\in\mathfrak{S}_{n}$ . This shows that $\{p_{\sigma}\}_{\sigma\in\mathfrak{S}_{n}}$ is the set of minimal projections of $A^{\text{ab}}(n)$ . ∎

For each $\sigma\in\mathfrak{S}_{n}$ , we can define a character $\chi_{\sigma}$ of $A(n)$ by the same formula as in the proof of Proposition 1.6 (or to be the composition of the character $\chi_{\sigma}$ in the proof of Proposition 1.6 and the natural surjection $A(n)\twoheadrightarrow A^{\text{ab}}(n)$ ). With these characters we have the following as a corollary of of Proposition 1.6 (It is easy to show it directly).

Corollary 1.8.

The set of all characters of the magic square C*-algebra $A(n)$ is $\{\chi_{\sigma}\mid\sigma\in\mathfrak{S}_{n}\}$ whose cardinality is $n!$ .

2. General results on magic square C*-algebras

Proposition 2.1.

For $n=1,2,3$ , $A(n)$ is commutative. Hence the surjection $A(n)\twoheadrightarrow A^{\text{ab}}(n)$ is an isomorphism for $n=1,2,3$ .

Proof.

For $n=1$ and $n=2$ , it is easy to see $A(1)\cong\mathbb{C}$ and $A(2)\cong\mathbb{C}^{2}$ . To show that $A(3)$ is commutative, it suffices to show $p_{1,1}$ commutes with $p_{2,2}$ . In fact if $p_{1,1}$ commutes with $p_{2,2}$ , we can see that $p_{1,1}$ commutes with $p_{2,3}$ , $p_{3,2}$ and $p_{3,3}$ using the action $\alpha$ defined in Definition 1.3. Then $p_{1,1}$ commutes with every generators because $p_{1,1}$ is orthogonal to and hence commutes with $p_{1,2}$ , $p_{1,3}$ , $p_{2,1}$ and $p_{3,1}$ . Using the action $\alpha$ again, we see that every generators commutes with every generators.

Now we are going to show that $p_{1,1}$ commutes with $p_{2,2}$ . We have

	$\displaystyle p_{1,1}p_{2,2}=(1-p_{1,2}-p_{1,3})p_{2,2}$	$\displaystyle=p_{2,2}-p_{1,3}p_{2,2}$
		$\displaystyle=p_{2,2}-(1-p_{2,3}-p_{3,3})p_{2,2}=p_{3,3}p_{2,2}.$

By symmetry, we have $p_{2,2}p_{3,3}=p_{1,1}p_{3,3}$ and $p_{3,3}p_{1,1}=p_{2,2}p_{1,1}$ . Hence we get

\displaystyle p_{1,1}p_{2,2}=p_{3,3}p_{2,2}=(p_{2,2}p_{3,3})^{*}=(p_{1,1}p_{3,3})^{*}=p_{3,3}p_{1,1}=p_{2,2}p_{1,1}.

This completes the proof. ∎

Proposition 2.2.

Let $n_{1},n_{2},\ldots,n_{k}$ be positive integers, and set $n=\sum_{j=1}^{k}n_{j}$ . There exists a surjection from $A(n)$ to the unital free product $\mathop{\scalebox{1.3}{\raisebox{-0.86108pt}{$\ast$}}}_{j=1}^{k}A(n_{j})$ .

Proof.

The desired surjection is obtained by sending the generators $\{p_{i,j}\}_{i,j=1}^{n_{1}}$ of $A(n)$ to the generators of $A(n_{1})\subset\mathop{\scalebox{1.3}{\raisebox{-0.86108pt}{$\ast$}}}_{j=1}^{k}A(n_{j})$ , the generators $\{p_{i,j}\}_{i,j=n_{1}+1}^{n_{1}+n_{2}}$ of $A(n)$ to the generators of $A(n_{2})\subset\mathop{\scalebox{1.3}{\raisebox{-0.86108pt}{$\ast$}}}_{j=1}^{k}A(n_{j})$ and so on, and by sending the other generators of $A(n)$ to $0$ . ∎

Corollary 2.3.

Let $n$ be a positive integer. There exists a surjection from $A(n+1)$ to $A(n)$ .

Proof.

This follows from Proposition 2.2 because $A(n)*A(1)\cong A(n)*\mathbb{C}\cong A(n)$ . ∎

Corollary 2.4.

Let $n,m$ be positive integers with $n\geq m$ . There exists a surjection from $A(n)$ to $A(m)$ .

Proof.

This follows from Corollary 2.3. ∎

Proposition 2.5.

For $n\geq 5$ , $A(n)$ is not exact.

Proof.

Note that an image of an exact C*-algebra is exact (see [BO, Corollary 9.4.3]). By Corollary 2.4, it suffices to show that $A(5)$ is not exact. By Proposition 2.2, there exists a surjection from $A(5)$ to $A(2)*A(3)\cong\mathbb{C}^{2}*\mathbb{C}^{6}$ which is not exact (see [BO, Proposition 3.7.11]). This completes the proof. ∎

The $C^{*}$ -algebra $A(4)$ is not commutative, but is exact, in fact is subhomogeneous (Corollary 7.9). From the next section, we investigate the structure of $A(4)$ .

3. Twisted crossed product

We denote elements $\sigma\in\mathfrak{S}_{4}$ by $(\sigma(1)\sigma(2)\sigma(3)\sigma(4))$ . We define the Klein (four) group $K$ by

K\coloneqq\{t_{1},t_{2},t_{3},t_{4}\}\subset\mathfrak{S}_{4}

where $t_{1}$ is the identity $(1234)$ of $\mathfrak{S}_{4}$ , $t_{2}=(2143)$ , $t_{3}=(3412)$ and $t_{4}=(4321)$ . The group $K$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ .

We choose the indices so that we have $t_{i}t_{j}=t_{t_{i}(j)}$ for $i,j=1,2,3,4$ . Note that we have $t_{i}(j)=t_{j}(i)$ for $i,j=1,2,3,4$ .

Definition 3.1.

Define unitaries $c_{1},c_{2},c_{3},c_{4}$ in $M_{2}(\mathbb{C})$ by

c_{1}\coloneqq\begin{pmatrix}1&0\\ 0&1\end{pmatrix},\quad c_{2}\coloneqq\begin{pmatrix}\sqrt{-1}&0\\ 0&-\sqrt{-1}\end{pmatrix},\quad c_{3}\coloneqq\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\quad c_{4}\coloneqq\begin{pmatrix}0&\sqrt{-1}\\ \sqrt{-1}&0\end{pmatrix}.

The unitaries $c_{1},c_{2},c_{3},c_{4}$ are called the Pauli matrices.

Definition 3.2.

Put $\omega=(1342)\in\mathfrak{S}_{4}$ . Define the map $\varepsilon\colon\{1,2,3,4\}^{2}\to\{1,-1\}$ by

\varepsilon(i,j)\coloneqq\begin{cases}1&\text{($i=1$ or $j=1$ or $\omega(i)=j$)}\\ -1&\text{(otherwise)},\end{cases}

for each $i,j=1,2,3,4$ .

$i\diagdown j$	$1$	$2$	$3$	$4$
$1$	1	1	1	1
$2$	1	$-1$	1	$-1$
$3$	1	$-1$	$-1$	1
$4$	1	1	$-1$	$-1$

Table 3.1. Values of

\varepsilon(i,j)

We have the following calculation which can be proved straightforwardly.

Lemma 3.3.

For $i,j=1,2,3,4$ , we have $c_{i}c_{j}=\varepsilon(i,j)c_{t_{i}(j)}$ .

From this lemma and the computation $t_{i}t_{j}=t_{t_{i}(j)}$ , we have the following lemma which means that $K^{2}\ni(t_{i},t_{j})\mapsto\varepsilon(i,j)\in\{1,-1\}$ becomes a cocycle of $K$ .

Lemma 3.4.

For $i,j,k=1,2,3,4$ , we have $\varepsilon(i,j)\varepsilon(t_{i}(j),k)=\varepsilon(i,t_{j}(k))\varepsilon(j,k)$ .

Proof.

Compute $c_{i}c_{j}c_{k}$ in the two ways, namely $(c_{i}c_{j})c_{k}$ and $c_{i}(c_{j}c_{k})$ . ∎

Hence the following definition makes sense. Let us denote by the same symbol $\alpha$ the restriction of the action $\alpha\colon\mathfrak{S}_{4}\times\mathfrak{S}_{4}\curvearrowright A(4)$ to $K\times K\subset\mathfrak{S}_{4}\times\mathfrak{S}_{4}$ .

Definition 3.5.

Let $A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ be the twisted crossed product of the action $\alpha$ and the cocycle

(K\times K)^{2}\ni((t_{i},t_{j}),(t_{k},t_{l}))\mapsto\varepsilon(i,k)\varepsilon(j,l)\in\{1,-1\}.

By definition, $A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ is the universal $C^{*}$ -algebra generated by the unital subalgebra $A(4)$ and unitaries $\{u_{i,j}\}_{i,j=1}^{4}$ such that

\displaystyle u_{i,j}xu_{i,j}^{*}=\alpha_{(t_{i},t_{j})}(x)\qquad\text{for all $i,j$ and all $x\in A(4)$}

and

\displaystyle u_{i,j}u_{k,l}=\varepsilon(i,k)\varepsilon(j,l)u_{t_{i}(k),t_{j}(l)}\quad\text{for all $i,j,k,l$.}

We denote by $\mathcal{R}_{\text{u}}$ the latter relation. The former relation is equivalent to the relation

\displaystyle u_{i,j}p_{k,l}=p_{t_{i}(k),t_{j}(l)}u_{i,j}\qquad\text{for all $i,j,k,l$}

which is denoted by $\mathcal{R}_{\text{up}}$ .

Recall that $A(4)$ is the universal unital $C^{*}$ -algebra generated by the set $\{p_{i,j}\}_{i,j=1}^{4}$ of projections satisfying the following relation denoted by $\mathcal{R}_{\text{p}}$

\displaystyle\sum_{i=1}^{4}p_{i,j}=1

\displaystyle\quad(j=1,2,3,4),

\displaystyle\sum_{j=1}^{4}p_{i,j}=1

\displaystyle\quad(i=1,2,3,4).

The following is the first main theorem.

Theorem 3.6.

The twisted crossed product $A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ is isomorphic to $M_{4}(C(\mathbb{R}P^{3}))$ .

We finish the proof of this theorem in the end of Section 7.

To prove this theorem, we start with finite presentation of the $C^{*}$ -algebra $C(\mathbb{R}P^{3})$ in the next section.

4. Real projective space $\mathbb{R}P^{3}$

Definition 4.1.

We set an equivalence relation $\sim$ on the manifold

S^{3}\coloneqq\Big{\{}a=(a_{1},a_{2},a_{3},a_{4})\in\mathbb{R}^{4}\ \Big{|}\ \sum_{i=1}^{4}a_{i}^{2}=1\Big{\}}

so that $a\sim b$ if and only if $a=b$ or $a=-b$ . The quotient space $S^{3}/{\sim}$ is the real projective space $\mathbb{R}P^{3}$ of dimension 3. The equivalence class of $(a_{1},a_{2},a_{3},a_{4})\in S^{3}$ is denoted as $[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}$ .

Definition 4.2.

For $i,j=1,2,3,4$ , we define a continuous function $f_{i,j}$ on $\mathbb{R}P^{3}$ by $f_{i,j}([a_{1},a_{2},a_{3},a_{4}])=a_{i}a_{j}$ for $[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}$ .

Note that $f_{i,j}$ is a well-defined continuous function.

Lemma 4.3.

The functions $\{f_{i,j}\}_{i,j=1}^{4}$ satisfy the following relation

	$\displaystyle f_{i,j}=f_{i,j}^{*}=f_{j,i}\quad\text{for all $i,j$,}$
	$\displaystyle f_{i,j}f_{k,l}=f_{i,k}f_{j,l}\quad\text{for all $i,j,k,l$,}$
	$\displaystyle\sum_{i=1}^{4}f_{i,i}=1.$

Proof.

This follows from easy computation. ∎

Definition 4.4.

We denote by $\mathcal{R}_{\text{f}}$ the relation in Lemma 4.3.

Proposition 4.5.

The $C^{*}$ -algebra $C(\mathbb{R}P^{3})$ is the universal unital $C^{*}$ -algebra generated by elements $\{f_{i,j}\}_{i,j=1}^{4}$ satisfying $\mathcal{R}_{\textrm{f}}$ .

Proof.

Let $A$ be the universal unital $C^{*}$ -algebra generated by elements $\{f_{i,j}\}_{i,j=1}^{4}$ satisfying $\mathcal{R}_{\text{f}}$ . For $i,j,k,l=1,2,3,4$ , we have

\displaystyle f_{i,j}f_{k,l}=f_{i,k}f_{j,l}=f_{k,i}f_{l,j}=f_{k,l}f_{i,j}.

Hence $A$ is commutative. Thus there exists a compact set $X$ such that $A\cong C(X)$ .

By Lemma 4.3, we have a unital $*$ -homomorphism $A\to C(\mathbb{R}P^{3})$ . This induces a continuous map $\varphi\colon\mathbb{R}P^{3}\to X$ . It suffices to show that this continuous map is homeomorphic.

We first show that $\varphi$ is injective. Take $[a_{1},a_{2},a_{3},a_{4}]$ and $[b_{1},b_{2},b_{3},b_{4}]\in\mathbb{R}P^{3}$ with $\varphi([a_{1},a_{2},a_{3},a_{4}])=\varphi([b_{1},b_{2},b_{3},b_{4}])$ . Then, for $i,j=1,2,3,4$ , we have $a_{i}a_{j}=b_{i}b_{j}$ . Since $\sum_{i=1}^{4}a_{i}^{2}=1$ , there exists $i_{0}$ such that $a_{i_{0}}\neq 0$ . Set $\sigma=b_{i_{0}}/a_{i_{0}}\in\mathbb{R}$ . Since $a_{i}a_{i_{0}}=b_{i}b_{i_{0}}$ , we have $a_{i}=\sigma b_{i}$ for $i=1,2,3,4$ . Since $\sum_{i=1}^{4}a_{i}^{2}=\sum_{i=1}^{4}b_{i}^{2}=1$ , we get $\sigma=\pm 1$ . Hence $[a_{1},a_{2},a_{3},a_{4}]=[b_{1},b_{2},b_{3},b_{4}]$ . This shows that $\varphi$ is injective.

Next we show that $\varphi$ is surjective. Take a unital character $\chi\colon A\to\mathbb{C}$ of $A$ . To show that $\varphi$ is surjective, it suffices to find $[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}$ such that $\chi(f_{i,j})=a_{i}a_{j}$ for all $i,j=1,2,3,4$ . Since $\sum_{i=1}^{4}\chi(f_{i,i})=\chi\big{(}\sum_{i=1}^{4}f_{i,i}\big{)}=1$ , there exists $i_{0}$ such that $\chi(f_{i_{0},i_{0}})\neq 0$ . Since

\displaystyle f_{i_{0},i_{0}}=f_{i_{0},i_{0}}\sum_{i=1}^{4}f_{i,i}=\sum_{i=1}^{4}f_{i_{0},i_{0}}f_{i,i}=\sum_{i=1}^{4}f_{i_{0},i}f_{i_{0},i}=\sum_{i=1}^{4}f_{i_{0},i}f_{i_{0},i}^{*}.

we have $\chi(f_{i_{0},i_{0}})>0$ . Put $a_{i}\coloneqq\frac{\chi(f_{i_{0},i})}{\sqrt{\chi(f_{i_{0},i_{0}})}}$ . We have

\displaystyle\sum_{i=1}^{4}a_{i}^{2}

\displaystyle=\sum_{i=1}^{4}\frac{\chi(f_{i_{0},i})^{2}}{\chi(f_{i_{0},i_{0}})}=\sum_{i=1}^{4}\frac{\chi(f_{i_{0},i_{0}})\chi(f_{i,i})}{\chi(f_{i_{0},i_{0}})}=\sum_{i=1}^{4}\chi(f_{i,i})=1.

We also have

\displaystyle\chi(f_{i,j})

\displaystyle=\frac{\chi(f_{i_{0},i})\chi(f_{i_{0},j})}{\chi(f_{i_{0},i_{0}})}=a_{i}a_{j},

for $i,j=1,2,3,4$ . This shows that $\varphi$ is surjective.

Since $\mathbb{R}P^{3}$ is compact and $X$ is Hausdorff, $\varphi\colon\mathbb{R}P^{3}\to X$ is a homeomorphism. Thus we have shown that $A$ is isomorphic to $C(\mathbb{R}P^{3})$ . ∎

Let $\{e_{i,j}\}_{i,j=1}^{4}$ be the matrix unit of $M_{4}(\mathbb{C})$ . Then $\{e_{i,j}\}_{i,j=1}^{4}$ satisfies the following relation denoted by $\mathcal{R}_{\text{e}}$ ;

	$\displaystyle e_{i,j}=e_{j,i}^{*}\quad\text{for all $i,j$,}$
	$\displaystyle e_{i,j}e_{k,l}=\delta_{j,k}e_{i,l}\quad\text{for all $i,j,k,l$,}$
	$\displaystyle\sum_{i=1}^{4}e_{i,i}=1,$

here $\delta_{j,k}$ is the Kronecker delta. It is well-known, and easy to see, that $M_{4}(\mathbb{C})$ is the universal unital C*-algebra generated by $\{e_{i,j}\}_{i,j=1}^{4}$ satisfying $\mathcal{R}_{\text{e}}$ .

The $C^{*}$ -algebra $M_{4}(C(\mathbb{R}P^{3}))=C(\mathbb{R}P^{3},M_{4}(\mathbb{C}))=C(\mathbb{R}P^{3})\otimes M_{4}(\mathbb{C})$ is the universal unital $C^{*}$ -algebra generated by $\{f_{i,j}\}_{i,j=1}^{4}$ and $\{e_{i,j}\}_{i,j=1}^{4}$ satisfying $\mathcal{R}_{\text{f}}$ , $\mathcal{R}_{\text{e}}$ and the following relation denoted by $\mathcal{R}_{\text{fe}}$ ;

\displaystyle f_{i,j}e_{k,l}=e_{k,l}f_{i,j}\quad\text{for all $i,j,k,l$.}

5. Unitaries

Definition 5.1.

For $i,j=1,2,3,4$ , we define a unitary $U_{i,j}\in M_{4}(\mathbb{C})\subset M_{4}(C(\mathbb{R}P^{3}))$ by

U_{i,j}\coloneqq\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)e_{t_{i}(k),t_{j}(k)}

From a direct calculation, we have

	$\displaystyle U_{1,1}=\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix},$	$\displaystyle U_{1,2}=\begin{pmatrix}0&1&0&0\\ -1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0\end{pmatrix},$
	$\displaystyle U_{1,3}=\begin{pmatrix}0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0\end{pmatrix},$	$\displaystyle U_{1,4}=\begin{pmatrix}0&0&0&1\\ 0&0&-1&0\\ 0&1&0&0\\ -1&0&0&0\end{pmatrix},$
	$\displaystyle U_{2,1}=\begin{pmatrix}0&-1&0&0\\ 1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0\end{pmatrix},$	$\displaystyle U_{2,2}=\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix},$
	$\displaystyle U_{2,3}=\begin{pmatrix}0&0&0&-1\\ 0&0&1&0\\ 0&1&0&0\\ -1&0&0&0\end{pmatrix},$	$\displaystyle U_{2,4}=\begin{pmatrix}0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\end{pmatrix},$
	$\displaystyle U_{3,1}=\begin{pmatrix}0&0&-1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&-1&0&0\end{pmatrix},$	$\displaystyle U_{3,2}=\begin{pmatrix}0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\end{pmatrix},$
	$\displaystyle U_{3,3}=\begin{pmatrix}1&0&0&0\\ 0&-1&0&0\\ 0&0&1&0\\ 0&0&0&-1\end{pmatrix},$	$\displaystyle U_{3,4}=\begin{pmatrix}0&-1&0&0\\ -1&0&0&0\\ 0&0&0&1\\ 0&0&1&0\end{pmatrix},$
	$\displaystyle U_{4,1}=\begin{pmatrix}0&0&0&-1\\ 0&0&-1&0\\ 0&1&0&0\\ 1&0&0&0\end{pmatrix},$	$\displaystyle U_{4,2}=\begin{pmatrix}0&0&-1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&1&0&0\end{pmatrix},$
	$\displaystyle U_{4,3}=\begin{pmatrix}0&1&0&0\\ 1&0&0&0\\ 0&0&0&1\\ 0&0&1&0\end{pmatrix},$	$\displaystyle U_{4,4}=\begin{pmatrix}1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&1\end{pmatrix}.$

We have the following. We denote the transpose matrix of a matrix $M$ by $M^{\mathrm{T}}$ .

Proposition 5.2.

For $(a_{1},a_{2},a_{3},a_{4})\in\mathbb{C}^{4}$ ,

(b_{1},b_{2},b_{3},b_{4})^{\mathrm{T}}\coloneqq U_{i,j}(a_{1},a_{2},a_{3},a_{4})^{\mathrm{T}},

satisfies $\sum_{k=1}^{4}b_{k}c_{k}=c_{i}\left(\sum_{k=1}^{4}a_{k}c_{k}\right)c_{j}^{*}$ .

Proof.

For $i,j,k=1,2,3,4$ , we have

\displaystyle c_{i}c_{t_{j}(k)}

\displaystyle=\varepsilon(i,t_{j}(k))c_{t_{i}(t_{j}(k))}

\displaystyle c_{t_{i}(k)}c_{j}

\displaystyle=\varepsilon(t_{i}(k),j)c_{t_{j}(t_{i}(k))}.

Hence $c_{i}c_{t_{j}(k)}c_{j}^{*}=\varepsilon(i,t_{j}(k))\varepsilon(t_{i}(k),j)^{-1}c_{t_{i}(k)}$ . Since $\varepsilon(i,t_{j}(k))\varepsilon(k,j)=\varepsilon(i,k)\varepsilon(t_{i}(k),j)$ , we have

\displaystyle\varepsilon(i,t_{j}(k))\varepsilon(t_{i}(k),j)^{-1}=\varepsilon(i,k)\varepsilon(k,j)^{-1}=\varepsilon(i,k)\varepsilon(k,j)

This shows that $U_{i,j}=\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)e_{t_{i}(k),t_{j}(k)}$ satisfies the desired property. ∎

Proposition 5.3.

For $i,j,k,l=1,2,3,4$ , we have

U_{i,j}U_{k,l}=\varepsilon(i,k)\varepsilon(j,l)U_{t_{i}(k),t_{j}(l)}.

Proof.

We have

	$\displaystyle U_{i,j}U_{k,l}$	$\displaystyle=\Big{(}\sum_{m=1}^{4}\varepsilon(i,m)\varepsilon(m,j)e_{t_{i}(m),t_{j}(m)}\Big{)}\Big{(}\sum_{n=1}^{4}\varepsilon(k,n)\varepsilon(n,l)e_{t_{k}(n),t_{l}(n)}\Big{)}$
		$\displaystyle=\Big{(}\sum_{m=1}^{4}\varepsilon(i,t_{k}(m))\varepsilon(t_{k}(m),j)e_{t_{i}(t_{k}(m)),t_{j}(t_{k}(m))}\Big{)}$
		$\displaystyle\phantom{=\Big{(}\sum_{m=1}^{4}\varepsilon(i,t_{k}(m))\varepsilon(t_{k}(m),j)}\Big{(}\sum_{n=1}^{4}\varepsilon(k,t_{j}(n))\varepsilon(t_{j}(n),l)e_{t_{k}(t_{j}(n)),t_{l}(t_{j}(n))}\Big{)}$
		$\displaystyle=\sum_{m=1}^{4}\varepsilon(i,t_{k}(m))\varepsilon(t_{k}(m),j)\varepsilon(k,t_{j}(m))\varepsilon(t_{j}(m),l)e_{t_{i}(t_{k}(m)),t_{l}(t_{j}(m))}$

Since we have

	$\displaystyle\varepsilon(i,t_{k}(m))\varepsilon(k,m)$	$\displaystyle=\varepsilon(i,k)\varepsilon(t_{i}(k),m),$	$\displaystyle\varepsilon(k,t_{j}(m))\varepsilon(m,j)$	$\displaystyle=\varepsilon(k,m)\varepsilon(t_{k}(m),j)$
	$\displaystyle\varepsilon(m,j)\varepsilon(t_{j}(m),l)$	$\displaystyle=\varepsilon(m,t_{j}(l))\varepsilon(j,l),$

we get

\varepsilon(i,t_{k}(m))\varepsilon(t_{k}(m),j)\varepsilon(k,t_{j}(m))\varepsilon(t_{j}(m),l)=\varepsilon(i,k)\varepsilon(j,l)\varepsilon(t_{i}(k),m)\varepsilon(m,t_{j}(l)).

Hence we obtain

	$\displaystyle U_{i,j}U_{k,l}$	$\displaystyle=\sum_{m=1}^{4}\varepsilon(i,k)\varepsilon(j,l)\varepsilon(t_{i}(k),m)\varepsilon(m,t_{j}(l))e_{t_{i}(t_{k}(m)),t_{j}(t_{l}(m))}$
		$\displaystyle=\varepsilon(i,k)\varepsilon(j,l)U_{t_{i}(k),t_{j}(l)}.$

∎

One can prove this proposition using Proposition 5.2.

6. Projections

Definition 6.1.

We define $P_{1,1}\coloneqq\sum_{i,j=1}^{4}f_{i,j}e_{i,j}\in M_{4}(C(\mathbb{R}P^{3}))$ . For $i,j=1,2,3,4$ , we define $P_{i,j}\in M_{4}(C(\mathbb{R}P^{3}))$ by

P_{i,j}=U_{i,j}P_{1,1}U_{i,j}^{*}.

Note that $U_{1,1}=1$ .

Proposition 6.2.

For each $i,j=1,2,3,4$ , $P_{i,j}$ is a projection.

Proof.

It suffices to show that $P_{1,1}$ is a projection. We have

\displaystyle P_{1,1}^{*}=\sum_{i,j=1}^{4}f_{i,j}^{*}e_{i,j}^{*}=\sum_{i,j=1}^{4}f_{j,i}e_{j,i}=P_{1,1},

and

	$\displaystyle P_{1,1}^{2}$	$\displaystyle=\sum_{i,j=1}^{4}f_{i,j}e_{i,j}\sum_{k,l=1}^{4}f_{k,l}e_{k,l}=\sum_{i,j,k,l=1}^{4}f_{i,j}e_{i,j}f_{k,l}e_{k,l}$
		$\displaystyle=\sum_{i,j,l=1}^{4}f_{i,j}f_{j,l}e_{i,l}=\sum_{i,j,l=1}^{4}f_{i,l}f_{j,j}e_{i,l}=\sum_{i,l=1}^{4}f_{i,l}e_{i,l}=P_{1,1}.$

Hence $P_{1,1}$ is a projection. ∎

Proposition 6.3.

The set $\{P_{i,j}\}_{i,j=1}^{4}$ of projections and the set $\{U_{i,j}\}_{i,j=1}^{4}$ of unitaries satisfy $\mathcal{R}_{\text{up}}$ .

Proof.

This follows from the computation

	$\displaystyle U_{i,j}P_{k,l}U_{i,j}^{*}$	$\displaystyle=U_{i,j}U_{k,l}P_{1,1}U_{k,l}^{}U_{i,j}^{}$
		$\displaystyle=(\varepsilon(i,k)\varepsilon(j,l))^{2}U_{t_{i}(k),t_{j}(l)}P_{1,1}U_{t_{i}(k),t_{j}(l)}^{*}=P_{t_{i}(k),t_{j}(l)}$

using Proposition 5.3. ∎

Proposition 6.4.

The set $\{P_{i,j}\}_{i,j=1}^{4}$ of projections satisfies $\mathcal{R}_{\text{p}}$ .

Proof.

From Proposition 6.3, it suffices to show

\displaystyle P_{1,1}+P_{1,2}+P_{1,3}+P_{1,4}

\displaystyle=1,

\displaystyle P_{1,1}+P_{2,1}+P_{3,1}+P_{4,1}

\displaystyle=1.

This follows from the following direct computations

	$\displaystyle P_{1,1}=\begin{pmatrix}f_{1,1}&f_{1,2}&f_{1,3}&f_{1,4}\\ f_{2,1}&f_{2,2}&f_{2,3}&f_{2,4}\\ f_{3,1}&f_{3,2}&f_{3,3}&f_{3,4}\\ f_{4,1}&f_{4,2}&f_{4,3}&f_{4,4}\end{pmatrix},$
	$\displaystyle P_{1,2}=\begin{pmatrix}f_{2,2}&-f_{2,1}&-f_{2,4}&f_{2,3}\\ -f_{1,2}&f_{1,1}&f_{1,4}&-f_{1,3}\\ -f_{4,2}&f_{4,1}&f_{4,4}&-f_{4,3}\\ f_{3,2}&-f_{3,1}&-f_{3,4}&f_{3,3}\end{pmatrix},$	$\displaystyle P_{2,1}=\begin{pmatrix}f_{2,2}&-f_{2,1}&f_{2,4}&-f_{2,3}\\ -f_{1,2}&f_{1,1}&-f_{1,4}&f_{1,3}\\ f_{4,2}&-f_{4,1}&f_{4,4}&-f_{4,3}\\ -f_{3,2}&f_{3,1}&-f_{3,4}&f_{3,3}\end{pmatrix},$
	$\displaystyle P_{1,3}=\begin{pmatrix}f_{3,3}&f_{3,4}&-f_{3,1}&-f_{3,2}\\ f_{4,3}&f_{4,4}&-f_{4,1}&-f_{4,2}\\ -f_{1,3}&-f_{1,4}&f_{1,1}&f_{1,2}\\ -f_{2,3}&-f_{2,4}&f_{2,1}&f_{2,2}\end{pmatrix},$	$\displaystyle P_{3,1}=\begin{pmatrix}f_{3,3}&-f_{3,4}&-f_{3,1}&f_{3,2}\\ -f_{4,3}&f_{4,4}&f_{4,1}&-f_{4,2}\\ -f_{1,3}&f_{1,4}&f_{1,1}&-f_{1,2}\\ f_{2,3}&-f_{2,4}&-f_{2,1}&f_{2,2}\end{pmatrix},$
	$\displaystyle P_{1,4}=\begin{pmatrix}f_{4,4}&-f_{4,3}&f_{4,2}&-f_{4,1}\\ -f_{3,4}&f_{3,3}&-f_{3,2}&f_{3,1}\\ f_{2,4}&-f_{2,3}&f_{2,2}&-f_{2,1}\\ -f_{1,4}&f_{1,3}&-f_{1,2}&f_{1,1}\end{pmatrix},$	$\displaystyle P_{4,1}=\begin{pmatrix}f_{4,4}&f_{4,3}&-f_{4,2}&-f_{4,1}\\ f_{3,4}&f_{3,3}&-f_{3,2}&-f_{3,1}\\ -f_{2,4}&-f_{2,3}&f_{2,2}&f_{2,1}\\ -f_{1,4}&-f_{1,3}&f_{1,2}&f_{1,1}\end{pmatrix}.$

∎

By Proposition 5.3, Proposition 6.2, Proposition 6.3 and Proposition 6.4, we have a $*$ -homomorphism $\varPhi\colon A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)\to M_{4}(C(\mathbb{R}P^{3}))$ sending $p_{i,j}$ to $P_{i,j}$ and $u_{i,j}$ to $U_{i,j}$ . In the next section, we construct the inverse map of $\varPhi$ .

7. The inverse map

Definition 7.1.

For $i,j=1,2,3,4$ , we set

E_{i,j}\coloneqq\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)u_{t_{i}(k),t_{j}(k)}\in A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)

Definition 7.2.

For $i,j=1,2,3,4$ , we set

F_{i,j}\coloneqq\sum_{k=1}^{4}E_{k,i}p_{1,1}E_{j,k}\in A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K).

Lemma 7.3.

For $i,j=1,2,3,4$ , we have $u_{i,1}E_{1,1}u_{1,j}=E_{i,j}$ . For $i=1,2,3,4$ , we have $u_{i,i}E_{1,1}=E_{1,1}u_{i,i}=E_{1,1}$ . We also have $E_{1,1}^{2}=E_{1,1}$ .

Proof.

We have $E_{1,1}=\frac{1}{4}\sum_{k=1}^{4}u_{k,k}$ . For $i,j=1,2,3,4$ , we have

\displaystyle u_{i,1}E_{1,1}u_{1,j}

\displaystyle=\frac{1}{4}\sum_{k=1}^{4}u_{i,1}u_{k,k}u_{1,j}=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)u_{t_{i}(k),t_{j}(k)}=E_{i,j}.

For $i=1,2,3,4$ , we have

\displaystyle u_{i,i}E_{1,1}

\displaystyle=\frac{1}{4}\sum_{k=1}^{4}u_{i,i}u_{k,k}=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)^{2}u_{t_{i}(k),t_{i}(k)}=\frac{1}{4}\sum_{k=1}^{4}u_{k,k}=E_{1,1}.

Similarly, we get $E_{1,1}u_{i,i}=E_{1,1}$ . Finally, we have $E_{1,1}^{2}=\frac{1}{4}\sum_{k=1}^{4}u_{k,k}E_{1,1}=E_{1,1}$ . ∎

Proposition 7.4.

The set $\{E_{i,j}\}_{i,j=1}^{4}$ satisfies $\mathcal{R}_{\text{e}}$ .

Proof.

We have $E_{1,1}=\frac{1}{4}\sum_{k=1}^{4}u_{k,k}$ . We also have

	$\displaystyle E_{2,2}$	$\displaystyle=\frac{1}{4}(u_{1,1}+u_{2,2}-u_{3,3}-u_{4,4})$
	$\displaystyle E_{3,3}$	$\displaystyle=\frac{1}{4}(u_{1,1}-u_{2,2}+u_{3,3}-u_{4,4})$
	$\displaystyle E_{4,4}$	$\displaystyle=\frac{1}{4}(u_{1,1}-u_{2,2}-u_{3,3}+u_{4,4}).$

Hence $\sum_{i=1}^{4}E_{i,i}=u_{1,1}=1$ .

It is easy to see $E_{1,1}^{*}=E_{1,1}$ . For $i=1,2,3,4$ , we have

E_{1,1}u_{i,1}^{*}=E_{1,1}u_{i,i}u_{i,1}^{*}=E_{1,1}u_{1,i}u_{i,1}u_{i,1}^{*}=E_{1,1}u_{1,i}

and $u_{1,i}^{*}E_{1,1}=u_{i,1}E_{1,1}$ similarly. Hence by Lemma 7.3, we obtain

\displaystyle E_{i,j}^{*}

\displaystyle=(u_{i,1}E_{1,1}u_{1,j})^{*}=u_{1,j}^{*}E_{1,1}u_{i,1}^{*}=u_{j,1}E_{1,1}u_{1,i}=E_{j,i}

for $i,j=1,2,3,4$ .

By Lemma 7.3, we obtain

	$\displaystyle E_{i,j}E_{j,k}=u_{i,1}E_{1,1}u_{1,j}u_{j,1}E_{1,1}u_{1,k}$	$\displaystyle=u_{i,1}E_{1,1}u_{j,j}E_{1,1}u_{1,k}$
		$\displaystyle=u_{i,1}E_{1,1}^{2}u_{1,k}=u_{i,1}E_{1,1}u_{1,k}=E_{i,k}$

for $i,j,k=1,2,3,4$ . The proof ends if we show $E_{i,j}E_{k,l}=0$ for $i,j,k,l=1,2,3,4$ with $j\neq k$ . It suffices to show $E_{1,1}u_{1,j}u_{k,1}E_{1,1}=0$ for $j,k=1,2,3,4$ with $j\neq k$ . Since $u_{1,j}u_{k,1}=u_{k,j}=\varepsilon(k,t_{k}(j))u_{k,k}u_{1,t_{k}(j)}$ , it suffices to show $E_{1,1}u_{1,j}E_{1,1}=0$ for $j=2,3,4$ . For $j=2$ , we get

	$\displaystyle 4E_{1,1}u_{1,2}E_{1,1}$	$\displaystyle=\sum_{k=1}^{4}u_{k,k}u_{1,2}E_{1,1}$
		$\displaystyle=u_{1,2}E_{1,1}+u_{1,2}u_{2,2}E_{1,1}-u_{1,2}u_{3,3}E_{1,1}-u_{1,2}u_{4,4}E_{1,1}$
		$\displaystyle=0$

By similar computations, we get $E_{1,1}u_{1,3}E_{1,1}=E_{1,1}u_{1,4}E_{1,1}=0$ . This completes the proof. ∎

Proposition 7.5.

The set $\{F_{i,j}\}_{i,j=1}^{4}$ satisfy $\mathcal{R}_{\text{f}}$ .

Proof.

For $i,j=1,2,3,4$ , we have

	$\displaystyle F_{i,j}^{}=\Big{(}\sum_{k=1}^{4}E_{k,i}p_{1,1}E_{j,k}\Big{)}^{}$	$\displaystyle=\sum_{k=1}^{4}E_{j,k}^{}p_{1,1}^{}E_{k,i}^{*}$
		$\displaystyle=\sum_{k=1}^{4}E_{k,j}p_{1,1}E_{i,k}=F_{j,i}.$

Next, we show $F_{i,j}=F_{j,i}$ for $i,j=1,2,3,4$ . We are going to prove $F_{2,4}=F_{4,2}$ . The other 5 cases can be proved similarly. To show that $F_{2,4}=F_{4,2}$ , it suffices to show $E_{1,2}p_{1,1}E_{4,1}=E_{1,4}p_{1,1}E_{2,1}$ because it implies $E_{k,2}p_{1,1}E_{4,k}=E_{k,4}p_{1,1}E_{2,k}$ for $k=1,2,3,4$ by multiplying $E_{k,1}$ from left and $E_{1,k}$ from right. By Lemma 7.3, we have

	$\displaystyle 4E_{1,2}p_{1,1}E_{4,1}$	$\displaystyle=(u_{1,2}-u_{2,1}-u_{3,4}+u_{4,3})p_{1,1}u_{4,1}E_{1,1}$
		$\displaystyle=(p_{1,2}u_{1,2}-p_{2,1}u_{2,1}-p_{3,4}u_{3,4}+p_{4,3}u_{4,3})u_{4,1}E_{1,1}$
		$\displaystyle=(p_{1,2}u_{4,2}+p_{2,1}u_{3,1}-p_{3,4}u_{2,4}-p_{4,3}u_{1,3})E_{1,1}$
		$\displaystyle=(p_{1,2}u_{1,3}u_{4,4}-p_{2,1}u_{1,3}u_{3,3}+p_{3,4}u_{1,3}u_{2,2}-p_{4,3}u_{1,3})E_{1,1}$
		$\displaystyle=(p_{1,2}-p_{2,1}+p_{3,4}-p_{4,3})u_{1,3}E_{1,1}$
	$\displaystyle 4E_{1,4}p_{1,1}E_{2,1}$	$\displaystyle=(u_{1,4}-u_{2,3}+u_{3,2}-u_{4,1})p_{1,1}u_{2,1}E_{1,1}$
		$\displaystyle=(p_{1,4}u_{1,4}-p_{2,3}u_{2,3}+p_{3,2}u_{3,2}-p_{4,1}u_{4,1})u_{2,1}E_{1,1}$
		$\displaystyle=(p_{1,4}u_{2,4}+p_{2,3}u_{1,3}-p_{3,2}u_{4,2}-p_{4,1}u_{3,1})E_{1,1}$
		$\displaystyle=(-p_{1,4}u_{1,3}u_{2,2}+p_{2,3}u_{1,3}-p_{3,2}u_{1,3}u_{4,4}+p_{4,1}u_{1,3}u_{3,3})E_{1,1}$
		$\displaystyle=(-p_{1,4}+p_{2,3}-p_{3,2}+p_{4,1})u_{1,3}E_{1,1}.$

Since

	$\displaystyle p_{1,1}$	$\displaystyle+p_{1,2}+p_{1,3}+p_{1,4}+p_{3,1}+p_{3,2}+p_{3,3}+p_{3,4}$
		$\displaystyle=2=p_{1,1}+p_{2,1}+p_{3,1}+p_{4,1}+p_{1,3}+p_{2,3}+p_{3,3}+p_{4,3},$

we have

p_{1,2}-p_{2,1}+p_{3,4}-p_{4,3}=-p_{1,4}+p_{2,3}-p_{3,2}+p_{4,1}.

Therefore, we obtain $E_{1,2}p_{1,1}E_{4,1}=E_{1,4}p_{1,1}E_{2,1}$ . Thus we have proved $F_{2,4}=F_{4,2}$ .

Next we show $F_{i,j}F_{k,l}=F_{i,k}F_{j,l}$ for $i,j,k,l=1,2,3,4$ , To show this, it suffices to show $p_{1,1}E_{j,k}p_{1,1}=p_{1,1}E_{k,j}p_{1,1}$ for $j,k=1,2,3,4$ . We are going to prove $p_{1,1}E_{3,4}p_{1,1}=p_{1,1}E_{4,3}p_{1,1}$ . The other 5 cases can be proved similarly. This follows from the following computation

	$\displaystyle 4p_{1,1}E_{3,4}p_{1,1}$	$\displaystyle=p_{1,1}(u_{3,4}+u_{4,3}-u_{1,2}-u_{2,1})p_{1,1}$
		$\displaystyle=p_{1,1}(u_{3,4}+u_{4,3})p_{1,1}-p_{1,1}p_{1,2}u_{1,2}-p_{1,1}p_{2,1}u_{2,1}$
		$\displaystyle=p_{1,1}(u_{3,4}+u_{4,3})p_{1,1},$
	$\displaystyle 4p_{1,1}E_{4,3}p_{1,1}$	$\displaystyle=p_{1,1}(u_{4,3}+u_{3,4}+u_{2,1}+u_{1,2})p_{1,1}$
		$\displaystyle=p_{1,1}(u_{3,4}+u_{4,3})p_{1,1}+p_{1,1}p_{2,1}u_{2,1}+p_{1,1}p_{1,2}u_{1,2}$
		$\displaystyle=p_{1,1}(u_{3,4}+u_{4,3})p_{1,1}.$

Finally we show $\sum_{i=1}^{4}F_{i,i}=1$ . For $i=1,2,3,4$ , we have

	$\displaystyle F_{i,i}$	$\displaystyle=\sum_{k=1}^{4}E_{k,i}p_{1,1}E_{i,k}=\sum_{k=1}^{4}u_{k,1}E_{1,1}u_{1,i}p_{1,1}u_{i,1}E_{1,1}u_{1,k}$
		$\displaystyle=\sum_{k=1}^{4}u_{k,1}E_{1,1}p_{1,i}u_{1,i}u_{i,1}E_{1,1}u_{1,k}=\sum_{k=1}^{4}u_{k,1}E_{1,1}p_{1,i}u_{i,i}E_{1,1}u_{1,k}$
		$\displaystyle=\sum_{k=1}^{4}u_{k,1}E_{1,1}p_{1,i}E_{1,1}u_{1,k}.$

Hence we obtain

	$\displaystyle\sum_{i=1}^{4}F_{i,i}$	$\displaystyle=\sum_{i=1}^{4}\sum_{k=1}^{4}u_{k,1}E_{1,1}p_{1,i}E_{1,1}u_{1,k}$
		$\displaystyle=\sum_{k=1}^{4}u_{k,1}E_{1,1}^{2}u_{1,k}=\sum_{k=1}^{4}u_{k,1}E_{1,1}u_{1,k}=\sum_{k=1}^{4}E_{k,k}=1$

by Lemma 7.3 and Proposition 7.4. We are done. ∎

Proposition 7.6.

The sets $\{E_{i,j}\}_{i,j=1}^{4}$ and $\{F_{i,j}\}_{i,j=1}^{4}$ satisfy $\mathcal{R}_{\text{fe}}$ .

Proof.

For $i,j,k,l=1,2,3,4$ , we have $E_{i,j}F_{k,l}=F_{k,l}E_{i,j}$ because

	$\displaystyle E_{i,j}F_{k,l}$	$\displaystyle=E_{i,j}\sum_{m=1}^{4}E_{m,k}p_{1,1}E_{l,m}=E_{i,k}p_{1,1}E_{l,j},$
	$\displaystyle F_{k,l}E_{i,j}$	$\displaystyle=\sum_{m=1}^{4}E_{m,k}p_{1,1}E_{l,m}E_{i,j}=E_{i,k}p_{1,1}E_{l,j}$

by Proposition 7.4. ∎

By Proposition 7.4, Proposition 7.5 and Proposition 7.6, we have a $*$ -homomorphism $\varPsi\colon M_{4}(C(\mathbb{R}P^{3}))\to A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ sending $f_{i,j}$ to $F_{i,j}$ and $e_{i,j}$ to $E_{i,j}$ .

We are going to see that this map $\varPsi$ is the inverse of $\varPhi$ . We first show $\varPsi\circ\varPhi=\operatorname{id}_{A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)}$ .

Proposition 7.7.

For $x\in A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ , we have $\varPsi(\varPhi(x))=x$ .

Proof.

For $i,j=1,2,3,4$ , we have

	$\displaystyle\varPsi(\varPhi(u_{i,j}))$	$\displaystyle=\varPsi(U_{i,j})=\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)\varPsi(e_{t_{i}(k),t_{j}(k)})$
		$\displaystyle=\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)E_{t_{i}(k),t_{j}(k)}$
		$\displaystyle=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)\sum_{m=1}^{4}\varepsilon(t_{i}(k),m)\varepsilon(m,t_{j}(k))u_{t_{i}(t_{k}(m)),t_{j}(t_{k}(m))}$
		$\displaystyle=\frac{1}{4}\sum_{k=1}^{4}\sum_{l=1}^{4}\varepsilon(i,k)\varepsilon(k,j)\varepsilon(t_{i}(k),t_{k}(l))\varepsilon(t_{k}(l),t_{j}(k))u_{t_{i}(l),t_{j}(l)}.$

Since we have

	$\displaystyle\frac{1}{4}\sum_{k=1}^{4}$	$\displaystyle\varepsilon(i,k)\varepsilon(k,j)\varepsilon(t_{i}(k),t_{k}(l))\varepsilon(t_{k}(l),t_{j}(k))$
		$\displaystyle=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(t_{i}(k),t_{k}(l))\varepsilon(t_{k}(l),t_{j}(k))\varepsilon(k,j)$
		$\displaystyle=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,l)\varepsilon(k,t_{k}(l))\varepsilon(t_{k}(l),k)\varepsilon(l,j)=\delta_{l,1},$

we obtain $\varPsi(\varPhi(u_{i,j}))=u_{i,j}$ . By the computation in the proof of Proposition 7.6, we have

\displaystyle\varPsi(P_{1,1})=\varPsi\Big{(}\sum_{i,j=1}^{4}f_{i,j}e_{i,j}\Big{)}=\sum_{i,j=1}^{4}F_{i,j}E_{i,j}=\sum_{i,j=1}^{4}E_{i,i}p_{1,1}E_{j,j}=p_{1,1}.

For $i,j=1,2,3,4$ , we have

\displaystyle\varPsi(\varPhi(p_{i,j}))

\displaystyle=\varPsi(P_{i,j})=\varPsi(U_{i,j})\varPsi(P_{1,1})\varPsi(U_{i,j})^{*}=u_{i,j}p_{1,1}u_{i,j}^{*}=p_{i,j}.

These show that $\varPsi(\varPhi(x))=x$ for all $x\in A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ . ∎

Proposition 7.8.

For $x\in M_{4}(C(\mathbb{R}P^{3}))$ , we have $\varPhi(\varPsi(x))=x$ .

Proof.

For $i,j=1,2,3,4$ , we have

	$\displaystyle\varPhi(\varPsi(e_{i,j}))$	$\displaystyle=\varPhi(E_{i,j})=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)\varPhi(u_{t_{i}(k),t_{j}(k)})$
		$\displaystyle=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)U_{t_{i}(k),t_{j}(k)}$
		$\displaystyle=\frac{1}{4}\sum_{k=1}^{4}\varepsilon(i,k)\varepsilon(k,j)\sum_{m=1}^{4}\varepsilon(t_{i}(k),m)\varepsilon(m,t_{j}(k))e_{t_{i}(t_{k}(m)),t_{j}(t_{k}(m))}$
		$\displaystyle=\frac{1}{4}\sum_{k=1}^{4}\sum_{l=1}^{4}\varepsilon(i,k)\varepsilon(k,j)\varepsilon(t_{i}(k),t_{k}(l))\varepsilon(t_{k}(l),t_{j}(k))e_{t_{i}(l),t_{j}(l)}$
		$\displaystyle=e_{i,j}$

as in the proof of Proposition 7.7. For $i,j=1,2,3,4$ , we have

	$\displaystyle\varPhi(\varPsi(f_{i,j}))$	$\displaystyle=\varPhi(F_{i,j})=\sum_{k=1}^{4}\varPhi(E_{k,i})\varPhi(p_{1,1})\varPhi(E_{j,k})$
		$\displaystyle=\sum_{k=1}^{4}e_{k,i}P_{1,1}e_{j,k}$
		$\displaystyle=\sum_{k=1}^{4}e_{k,i}\Big{(}\sum_{l,m=1}^{4}f_{l,m}e_{l,m}\Big{)}e_{j,k}$
		$\displaystyle=\sum_{k=1}^{4}f_{i,j}e_{k,k}=f_{i,j}.$

These show that $\varPhi(\varPsi(x))=x$ for all $x\in M_{4}(C(\mathbb{R}P^{3}))$ . ∎

By these two propositions, we get Theorem 3.6. As its corollary, we have the following.

Corollary 7.9 (cf. [BC, Theorem 4.1]).

There is an injective $*$ -homomorphism $A(4)\to M_{4}(C(\mathbb{R}P^{3}))$ .

Proof.

This follows from Theorem 3.6 because the $*$ -homomorphism $A(4)\to A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ is injective. ∎

One can see that the injective $*$ -homomorphism constructed in this corollary is nothing but the Pauli representation constructed in [BM] and considered in [BC]. Note that Banica and Collins remarked after [BC, Definition 2.1] that the target of the Pauli representation is replaced by $M_{4}(C(SO_{3}))$ instead of $M_{4}(C(SU_{2}))$ . Here $SO_{3}$ is a homeomorphic to $\mathbb{R}P^{3}$ whereas $SU_{2}$ is a homeomorphic to $S^{3}$ .

8. Action

One can see that the dual group of $K\times K$ is isomorphic to $K\times K$ using the product of the cocycle $\varepsilon$ (see below).

$i\diagdown j$	$1$	$2$	$3$	$4$
$1$	1	1	1	1
$2$	1	1	$-1$	$-1$
$3$	1	$-1$	1	$-1$
$4$	1	$-1$	$-1$	1

Table 8.1. Values of

\varepsilon(i,j)\varepsilon(j,i)

Let $\widehat{\alpha}\colon K\times K\curvearrowright A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)$ be the dual action of $\alpha$ . Namely $\widehat{\alpha}$ is determined by the following equation for all $i,j,k,l$

\displaystyle\widehat{\alpha}_{i,j}(p_{k,l})

\displaystyle=p_{k,l},

\displaystyle\widehat{\alpha}_{i,j}(u_{k,l})

\displaystyle=\varepsilon(i,k)\varepsilon(k,i)\varepsilon(j,l)\varepsilon(l,j)u_{k,l},

where we write $\widehat{\alpha}_{(t_{i},t_{j})}$ as $\widehat{\alpha}_{i,j}$ .

For $i,j=1,2,3,4$ , define $\sigma_{i,j}\colon\mathbb{R}P^{3}\to\mathbb{R}P^{3}$ by $\sigma_{i,j}([a_{1},a_{2},a_{3},a_{4}])=[b_{1},b_{2},b_{3},b_{4}]$ for $[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}$ where $(b_{1},b_{2},b_{3},b_{4})\in S^{3}$ is determined by

(b_{1},b_{2},b_{3},b_{4})^{\mathrm{T}}=U_{i,j}(a_{1},a_{2},a_{3},a_{4})^{\mathrm{T}},

in other words $\sum_{k=1}^{4}b_{k}c_{k}=c_{i}\left(\sum_{k=1}^{4}a_{k}c_{k}\right)c_{j}^{*}$ by Proposition 5.2. Let $\beta\colon K\times K\curvearrowright M_{4}(C(\mathbb{R}P^{3}))$ be the action determined by $\beta_{i,j}(F)=\operatorname{Ad}U_{i,j}\circ F\circ\sigma_{i,j}$ for $F\in M_{4}(C(\mathbb{R}P^{3}))=C(\mathbb{R}P^{3},M_{4}(\mathbb{C}))$ where we write $\beta_{(t_{i},t_{j})}$ as $\beta_{i,j}$ .

Proposition 8.1.

The $*$ -homomorphism $\varPhi\colon A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)\to M_{4}(C(\mathbb{R}P^{3}))$ is equivariant with respect to $\widehat{\alpha}$ and $\beta$ .

Proof.

For $i,j=1,2,3,4$ , we have $P_{1,1}\circ\sigma_{i,j}=\operatorname{Ad}U_{i,j}\circ P_{1,1}$ . In fact for $[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}$ , on one hand we have

\displaystyle(P_{1,1}\circ\sigma_{i,j})([a_{1},a_{2},a_{3},a_{4}])

\displaystyle=(b_{1},b_{2},b_{3},b_{4})^{\mathrm{T}}(b_{1},b_{2},b_{3},b_{4}),

where

(b_{1},b_{2},b_{3},b_{4})^{\mathrm{T}}=U_{i,j}(a_{1},a_{2},a_{3},a_{4})^{\mathrm{T}},

and on the other hand we have

\displaystyle(\operatorname{Ad}U_{i,j}\circ P_{1,1})([a_{1},a_{2},a_{3},a_{4}])

\displaystyle=U_{i,j}(a_{1},a_{2},a_{3},a_{4})^{\mathrm{T}}(a_{1},a_{2},a_{3},a_{4})U_{i,j}^{*}

here note $U_{i,j}^{*}=U_{i,j}^{\mathrm{T}}$ because the entries of $U_{i,j}$ are $-1$ , $0$ or $1$ . For $i,j,k,l=1,2,3,4$ , we have

	$\displaystyle\beta_{i,j}(P_{k,l})$	$\displaystyle=\operatorname{Ad}U_{i,j}\circ(\operatorname{Ad}U_{k,l}\circ P_{1,1})\circ\sigma_{i,j}$
		$\displaystyle=\operatorname{Ad}U_{i,j}\circ\operatorname{Ad}U_{k,l}\circ\operatorname{Ad}U_{i,j}\circ P_{1,1}$
		$\displaystyle=\operatorname{Ad}(U_{i,j}U_{k,l}U_{i,j})\circ P_{1,1}$
		$\displaystyle=\operatorname{Ad}U_{k,l}\circ P_{1,1}=P_{k,l}.$

For $i,j,k,l=1,2,3,4$ , we also have

	$\displaystyle\beta_{i,j}(U_{k,l})$	$\displaystyle=\operatorname{Ad}U_{i,j}\circ U_{k,l}\circ\sigma_{i,j}$
		$\displaystyle=U_{i,j}U_{k,l}U_{i,j}^{*}$
		$\displaystyle=\varepsilon(i,k)\varepsilon(j,l)U_{t_{i}(k),t_{j}(l)}U_{i,j}^{*}$
		$\displaystyle=\varepsilon(i,k)\varepsilon(j,l)\varepsilon(k,i)^{-1}\varepsilon(l,j)^{-1}U_{k,l}U_{i,j}U_{i,j}^{*}$
		$\displaystyle=\varepsilon(i,k)\varepsilon(j,l)\varepsilon(k,i)\varepsilon(l,j)U_{k,l}$

here note that $U_{k,l}\in M_{4}(C(\mathbb{R}P^{3}))$ is a constant function. These complete the proof. ∎

The following is the second main theorem.

Theorem 8.2.

The fixed point algebra $M_{4}(C(\mathbb{R}P^{3}))^{\beta}$ of the action $\beta$ is isomorphic to $A(4)$ .

Proof.

This follows from Theorem 3.6 and Proposition 8.1 because the fixed point algebra $\big{(}A(4)\rtimes_{\alpha}^{\text{tw}}(K\times K)\big{)}^{\widehat{\alpha}}$ of $\widehat{\alpha}$ is $A(4)$ . ∎

9. Quotient Space $\mathbb{R}P^{3}/(K\times K)$

Definition 9.1.

We set $A\coloneqq M_{4}(C(\mathbb{R}P^{3}))^{\beta}$ .

By Theorem 8.2, the $C^{*}$ -algebra $A(4)$ is isomorphic to $A$ . From this section, we compute the structure of $A$ and its K-groups.

In this section, we study the quotient Space $\mathbb{R}P^{3}/(K\times K)$ of $\mathbb{R}P^{3}$ by the action $\sigma$ of $K\times K$ . In [O], it is proved that this quotient space $\mathbb{R}P^{3}/(K\times K)$ is homeomorphic to $S^{3}$ .

Definition 9.2.

We denote by $X$ the quotient space $\mathbb{R}P^{3}/(K\times K)$ of the action $\sigma$ of $K\times K$ . We denote by $\pi\colon\mathbb{R}P^{3}\to X$ the quotient map.

We use the following lemma later.

Lemma 9.3.

For $i,j=2,3,4$ and $[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}$ with $\sigma_{i,j}([a_{1},a_{2},a_{3},a_{4}])=[a_{1},a_{2},a_{3},a_{4}]$ , we have $P_{k,l}([a_{1},a_{2},a_{3},a_{4}])=P_{t_{i}(k),t_{j}(l)}([a_{1},a_{2},a_{3},a_{4}])$ for $k,l=1,2,3,4$ .

Proof.

This follows from

	$\displaystyle P_{k,l}([a_{1},a_{2},a_{3},a_{4}])$	$\displaystyle=\beta_{i,j}(P_{k,l})([a_{1},a_{2},a_{3},a_{4}])$
		$\displaystyle=\operatorname{Ad}U_{i,j}\big{(}P_{k,l}(\sigma_{i,j}([a_{1},a_{2},a_{3},a_{4}]))\big{)}$
		$\displaystyle=\operatorname{Ad}U_{i,j}\big{(}P_{k,l}([a_{1},a_{2},a_{3},a_{4}])\big{)},$
	$\displaystyle P_{t_{i}(k),t_{j}(l)}([a_{1},a_{2},a_{3},a_{4}])$	$\displaystyle=\big{(}\operatorname{Ad}U_{i,j}(P_{k,l})\big{)}([a_{1},a_{2},a_{3},a_{4}])$
		$\displaystyle=\operatorname{Ad}U_{i,j}\big{(}P_{k,l}([a_{1},a_{2},a_{3},a_{4}])\big{)}.\qed$

Definition 9.4.

For each $i,j=2,3,4$ , define

\widetilde{F}_{i,j}\coloneqq\{[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}\mid\sigma_{i,j}([a_{1},a_{2},a_{3},a_{4}])=[a_{1},a_{2},a_{3},a_{4}]\}\subset\mathbb{R}P^{3}

to be the set of fixed points of $\sigma_{i,j}$ , and define $F_{i,j}\subset X$ to be the image $\pi(\tilde{F}_{i,j})$ .

We have $\tilde{F}_{i,j}=\pi^{-1}(F_{i,j})$ . The following two propositions can be proved by direct computation using the computation of $U_{i,j}$ after Definition 5.1

Proposition 9.5.

For each $i=2,3,4$ , $\sigma_{1,i}$ and $\sigma_{i,1}$ have no fixed points.

Proposition 9.6.

For each $i,j=2,3,4$ , $\widetilde{F}_{i,j}$ is homeomorphic to a disjoint union of two circles. More precisely, we have

	$\displaystyle\widetilde{F}_{2,2}=\{[a,b,0,0],[0,0,a,b]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ a^{2}+b^{2}=1\}$
	$\displaystyle\widetilde{F}_{2,3}=\{[a,b,-b,a],[a,b,b,-a]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ 2(a^{2}+b^{2})=1\}$
	$\displaystyle\widetilde{F}_{2,4}=\{[a,b,a,b],[a,b,-a,-b]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ 2(a^{2}+b^{2})=1\}$
	$\displaystyle\widetilde{F}_{3,2}=\{[a,b,b,a],[a,b,-b,-a]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ 2(a^{2}+b^{2})=1\}$
	$\displaystyle\widetilde{F}_{3,3}=\{[a,0,b,0],[0,a,0,b]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ a^{2}+b^{2}=1\}$
	$\displaystyle\widetilde{F}_{3,4}=\{[a,a,b,-b],[a,-a,b,b]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ 2(a^{2}+b^{2})=1\}$
	$\displaystyle\widetilde{F}_{4,2}=\{[a,b,a,-b],[a,b,-a,b]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ 2(a^{2}+b^{2})=1\}$
	$\displaystyle\widetilde{F}_{4,3}=\{[a,a,b,b],[a,-a,b,-b]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ 2(a^{2}+b^{2})=1\}$
	$\displaystyle\widetilde{F}_{4,4}=\{[a,0,0,b],[0,a,b,0]\in\mathbb{R}P^{3}\mid a,b\in\mathbb{R},\ a^{2}+b^{2}=1\}$

Definition 9.7.

We set $\widetilde{F}\coloneqq\bigcup_{i,j=2}^{4}\widetilde{F}_{i,j}$ and $F\coloneqq\bigcup_{i,j=2}^{4}F_{i,j}$ . We also set $\widetilde{O}\coloneqq\mathbb{R}P^{3}\setminus\widetilde{F}$ and $O\coloneqq X\setminus F$ .

We have $\widetilde{F}=\pi^{-1}(F)$ and hence $\widetilde{O}=\pi^{-1}(O)$ . Note that $\widetilde{O}$ is the set of points $[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}$ such that $\sigma_{i,j}([a_{1},a_{2},a_{3},a_{4}])\neq[a_{1},a_{2},a_{3},a_{4}]$ for all $i,j=1,2,3,4$ other than $(i,j)=(1,1)$ . Note also that $\widetilde{F}$ and $F$ are closed, and hence $\widetilde{O}$ and $O$ are open.

Definition 9.8.

For each $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ , define $\widetilde{F}_{(i_{2}i_{3}i_{4})}\subset\mathbb{R}P^{3}$ by

\widetilde{F}_{(i_{2}i_{3}i_{4})}\coloneqq\widetilde{F}_{i_{2},2}\cap\widetilde{F}_{i_{3},3}\cap\widetilde{F}_{i_{4},4},

and define $F_{(i_{2}i_{3}i_{4})}\subset X$ to be the image $\pi(\widetilde{F}_{(i_{2}i_{3}i_{4})})$ .

Proposition 9.9.

For each $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ , we have

\widetilde{F}_{(i_{2}i_{3}i_{4})}=\widetilde{F}_{i_{2},2}\cap\widetilde{F}_{i_{3},3}=\widetilde{F}_{i_{2},2}\cap\widetilde{F}_{i_{4},4}=\widetilde{F}_{i_{3},3}\cap\widetilde{F}_{i_{4},4}.

We also have

	$\displaystyle\widetilde{F}_{(234)}=\big{\{}[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]\big{\}},$
	$\displaystyle\widetilde{F}_{(342)}=\left\{\Big{[}\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\Big{]},\Big{[}\frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\Big{]},\Big{[}\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2}\Big{]},\Big{[}\frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2}\Big{]}\right\},$
	$\displaystyle\widetilde{F}_{(423)}=\left\{\Big{[}-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\Big{]},\Big{[}\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2}\Big{]},\Big{[}\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{2}\Big{]},\Big{[}\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2}\Big{]}\right\},$
	$\displaystyle\widetilde{F}_{(243)}=\left\{\Big{[}\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0\Big{]},\Big{[}\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,0\Big{]},\Big{[}0,0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big{]},\Big{[}0,0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\Big{]}\right\},$
	$\displaystyle\widetilde{F}_{(432)}=\left\{\Big{[}\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},0\Big{]},\Big{[}\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}},0\Big{]},\Big{[}0,\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\Big{]},\Big{[}0,\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}\Big{]}\right\},$
	$\displaystyle\widetilde{F}_{(324)}=\left\{\Big{[}\frac{1}{\sqrt{2}},0,0,\frac{1}{\sqrt{2}}\Big{]},\Big{[}\frac{1}{\sqrt{2}},0,0,-\frac{1}{\sqrt{2}}\Big{]},\Big{[}0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big{]},\Big{[}0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\Big{]}\right\}.$

Proof.

This follows from Proposition 9.6. ∎

Proposition 9.10.

For each $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ , $F_{(i_{2}i_{3}i_{4})}$ consists of one point.

Proof.

This follows from Proposition 9.9. ∎

Definition 9.11.

For each $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ , we set $x_{(i_{2}i_{3}i_{4})}\in X$ by $F_{(i_{2}i_{3}i_{4})}=\{x_{(i_{2}i_{3}i_{4})}\}$ .

Proposition 9.12.

For each $i,j=2,3,4$ , $F_{i,j}$ is homeomorphic to a closed interval whose endpoints are $x_{(i_{2}i_{3}i_{4})}$ with $i_{j}=i$ ,

Proof.

This follows from Proposition 9.6. See also Figure 13.2 and the remark around it. ∎

Note that $F\subset X$ is the complete bipartite graph between $\{x_{(234)},x_{(342)},x_{(423)}\}$ and $\{x_{(243)},x_{(432)},x_{(324)},\}$ . See Figure 13.2.

Definition 9.13.

For $i,j=2,3,4$ , we define

F_{i,j}^{\circ}\coloneqq F_{i,j}\setminus\{x_{(i_{2}i_{3}i_{4})}\mid i_{j}=i\},

and define

F^{\circ}\coloneqq\bigcup_{i_{j}=2}^{4}F_{i,j}^{\circ},\qquad F^{\text{\textbullet}}\coloneqq\{x_{(234)},x_{(243)},x_{(324)},x_{(342)},x_{(423)},x_{(432)}\}.

Definition 9.14.

We set $\widetilde{F}_{i,j}^{\circ}\coloneqq\pi^{-1}(F_{i,j}^{\circ})$ for $i,j=2,3,4$ , $\widetilde{F}^{\circ}\coloneqq\pi^{-1}(F^{\circ})$ and $\widetilde{F}^{\text{\textbullet}}\coloneqq\pi^{-1}(F^{\text{\textbullet}})$ .

10. Exact sequences

For a locally compact subset $Y$ of $\mathbb{R}P^{3}$ which is invariant under the action $\sigma$ , the action $\beta\colon K\times K\curvearrowright M_{4}(C(\mathbb{R}P^{3}))$ induces the action $K\times K\curvearrowright M_{4}(C_{0}(Y))$ which is also denoted by $\beta$ . We use the following lemma many times.

Lemma 10.1.

Let $Y$ be a locally compact subset of $\mathbb{R}P^{3}$ which is invariant under the action $\sigma$ . Let $Z$ be a closed subset of $Y$ which is invariant under the action $\sigma$ . Then we have a a short exact sequence

Proof.

It suffices to show that $M_{4}(C_{0}(Y))^{\beta}\to M_{4}(C_{0}(Z))^{\beta}$ is surjective. The other assertions are easy to see.

Take $f\in M_{4}(C_{0}(Z))^{\beta}$ . Since $M_{4}(C_{0}(Y))\to M_{4}(C_{0}(Z))$ is surjective, there exists $g\in M_{4}(C_{0}(Y))$ with $g|_{Z}=f$ . Set $g_{0}\in M_{4}(C_{0}(Y))$ by

g_{0}\coloneqq\frac{1}{16}\sum_{i,j=1}^{4}\beta_{i,j}(g).

Then $g_{0}\in M_{4}(C_{0}(Y))^{\beta}$ and $g_{0}|_{Z}=f$ . This completes the proof. ∎

We also use the following lemma many times.

Lemma 10.2.

Let $Y$ be a locally compact subset of $\mathbb{R}P^{3}$ which is invariant under the action $\sigma$ . Let $Z$ be a closed subset of $Y$ such that $Y=\bigcup_{i,j=1}^{4}\sigma_{i,j}(Z)$ and that $\sigma_{i,j}(Z)\cap Z=\emptyset$ for $i,j=1,2,3,4$ with $(i,j)\neq(1,1)$ . Then we have $M_{4}(C_{0}(Y))^{\beta}\cong M_{4}(C_{0}(Z))$ .

Proof.

The restriction map $M_{4}(C_{0}(Y))^{\beta}\to M_{4}(C_{0}(Z))$ is an isomorphism because its inverse is given by

M_{4}(C_{0}(Z))\ni f\mapsto\sum_{i,j=1}^{4}\beta_{i,j}(f)\in M_{4}(C_{0}(Y))^{\beta}.\qed

Under the situation of the lemma above, $\pi\colon Z\to\pi(Z)=\pi(Y)$ is a homeomorphism. Hence we have $M_{4}(C_{0}(Y))^{\beta}\cong M_{4}(C_{0}(Z))\cong M_{4}\big{(}C_{0}(\pi(Z))\big{)}=M_{4}\big{(}C_{0}(\pi(Y))\big{)}$ .

The following lemma generalize Lemma 10.2.

Lemma 10.3.

Let $G$ be a subgroup of $K\times K$ . Let $Y$ be a locally compact subset of $\mathbb{R}P^{3}$ which is invariant under the action $\sigma$ . Suppose that each point of $Y$ is fixed by $\sigma_{i,j}$ for all $(t_{i},t_{j})\in G$ . Let $Z$ be a closed subset of $Y$ such that $Y=\bigcup_{i,j=1}^{4}\sigma_{i,j}(Z)$ and that $\sigma_{i,j}(Z)\cap Z=\emptyset$ for $i,j=1,2,3,4$ with $(t_{i},t_{j})\not\in G$ . Then we have $M_{4}(C_{0}(Y))^{\beta}\cong C_{0}(Z,D)$ where

D\coloneqq\{T\in M_{4}(\mathbb{C})\mid\text{$\operatorname{Ad}U_{i,j}(T)=T$ for all $(t_{i},t_{j})\in G$}\}.

Proof.

We have a restriction map $M_{4}(C_{0}(Y))^{\beta}\to C_{0}(Z,D)$ which is an isomorphism because its inverse is given by

C_{0}(Z,D)\ni f\mapsto\sum_{(i,j)\in I}\beta_{i,j}(f)\in M_{4}(C_{0}(Y))^{\beta},

where an index set $I$ was chosen so that $\{(t_{i},t_{j})\in K\times K\mid(i,j)\in I\}$ becomes a complete representative of the quotient $(K\times K)/G$ . ∎

Under the situation of the lemma above, $\pi\colon Z\to\pi(Z)=\pi(Y)$ is a homeomorphism. Hence we have $M_{4}(C_{0}(Y))^{\beta}\cong C_{0}(Z,D)\cong C_{0}(\pi(Z),D)=C_{0}(\pi(Y),D)$ .

Definition 10.4.

We set $I\coloneqq M_{4}(C_{0}(\widetilde{O}))^{\beta}$ and $B\coloneqq M_{4}(C(\widetilde{F}))^{\beta}$ .

By Lemma 10.1 we get the short exact sequence

0\longrightarrow I\longrightarrow A\longrightarrow B\longrightarrow 0,

From this sequence, we get a six-term exact sequence

From next section, we compute $K_{i}(B)$ , $K_{i}(I)$ and $\delta_{i}$ for $i=0,1$ . Consult [RLL] for basics of K-theory.

11. The Structure of the Quotient $B$

Definition 11.1.

For $i,j=2,3,4$ , let $D_{i,j}$ be the fixed algebra of $\operatorname{Ad}U_{i,j}$ on $M_{4}(\mathbb{C})$ .

From the direct computation, we have the following.

Proposition 11.2.

For each $i,j=2,3,4$ , $D_{i,j}$ is isomorphic to $M_{2}(\mathbb{C})\oplus M_{2}(\mathbb{C})$ . More precisely, we have

	$\displaystyle D_{2,2}=\left\{\begin{pmatrix}a&b&0&0\\ c&d&0&0\\ 0&0&e&f\\ 0&0&g&h\end{pmatrix}\right\},$	$\displaystyle D_{2,3}=\left\{\begin{pmatrix}a&b&c&d\\ e&f&g&h\\ -h&g&f&-e\\ d&-c&-b&a\end{pmatrix}\right\},$
	$\displaystyle D_{2,4}=\left\{\begin{pmatrix}a&b&c&d\\ e&f&g&h\\ c&d&a&b\\ g&h&e&f\end{pmatrix}\right\},$	$\displaystyle D_{3,2}=\left\{\begin{pmatrix}a&b&c&d\\ e&f&g&h\\ h&g&f&e\\ d&c&b&a\end{pmatrix}\right\},$
	$\displaystyle D_{3,3}=\left\{\begin{pmatrix}a&0&b&0\\ 0&c&0&d\\ e&0&f&0\\ 0&g&0&h\end{pmatrix}\right\},$	$\displaystyle D_{3,4}=\left\{\begin{pmatrix}a&b&c&d\\ b&a&-d&-c\\ e&f&g&h\\ -f&-e&h&g\end{pmatrix}\right\},$
	$\displaystyle D_{4,2}=\left\{\begin{pmatrix}a&b&c&d\\ e&f&g&h\\ c&-d&a&-b\\ -g&h&-e&f\end{pmatrix}\right\},$	$\displaystyle D_{4,3}=\left\{\begin{pmatrix}a&b&c&d\\ b&a&d&c\\ e&f&g&h\\ f&e&h&g\end{pmatrix}\right\},$
	$\displaystyle D_{4,4}=\left\{\begin{pmatrix}a&0&0&b\\ 0&c&d&0\\ 0&e&f&0\\ g&0&0&h\end{pmatrix}\right\},$

where $a,b,c,d,e,f,g,h$ run through $\mathbb{C}$ .

Definition 11.3.

For each $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ , define $D_{(i_{2}i_{3}i_{4})}\subset\mathbb{R}P^{3}$ by

D_{(i_{2}i_{3}i_{4})}\coloneqq D_{i_{2},2}\cap D_{i_{3},3}\cap D_{i_{4},4}.

Proposition 11.4.

For each $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ , we have

D_{(i_{2}i_{3}i_{4})}=D_{i_{2},2}\cap D_{i_{3},3}=D_{i_{2},2}\cap D_{i_{4},4}=D_{i_{3},3}\cap D_{i_{4},4},

and $D_{(i_{2}i_{3}i_{4})}$ is isomorphic to $\mathbb{C}^{4}$ . More precisely, we have

	$\displaystyle D_{(234)}=\left\{\begin{pmatrix}a&0&0&0\\ 0&b&0&0\\ 0&0&c&0\\ 0&0&0&d\end{pmatrix}\right\}$	$\displaystyle D_{(423)}=\left\{\begin{pmatrix}a&b&c&d\\ b&a&-d&-c\\ c&-d&a&-b\\ d&-c&-b&a\end{pmatrix}\right\}$
	$\displaystyle D_{(342)}=\left\{\begin{pmatrix}a&b&c&d\\ b&a&d&c\\ c&d&a&b\\ d&c&b&a\end{pmatrix}\right\}$	$\displaystyle D_{(243)}=\left\{\begin{pmatrix}a&b&0&0\\ b&a&0&0\\ 0&0&c&d\\ 0&0&d&c\end{pmatrix}\right\}$
	$\displaystyle D_{(432)}=\left\{\begin{pmatrix}a&0&b&0\\ 0&c&0&d\\ b&0&a&0\\ 0&d&0&c\end{pmatrix}\right\}$	$\displaystyle D_{(324)}=\left\{\begin{pmatrix}a&0&0&d\\ 0&b&c&0\\ 0&c&b&0\\ d&0&0&a\end{pmatrix}\right\}$

where $a,b,c,d$ run through $\mathbb{C}$ .

Definition 11.5.

We set $B^{\circ}\coloneqq M_{4}(C_{0}(\widetilde{F}^{\circ}))^{\beta}$ and $B^{\text{\textbullet}}\coloneqq M_{4}(C(\widetilde{F}^{\text{\textbullet}}))^{\beta}$ . We also set $B^{\circ}_{i,j}\coloneqq M_{4}(C_{0}(\widetilde{F}^{\circ}_{i,j}))^{\beta}$ for $i,j=2,3,4$ and $B_{(i_{2}i_{3}i_{4})}\coloneqq M_{4}(C_{0}(\widetilde{F}_{(i_{2}i_{3}i_{4})}))^{\beta}$ for $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ .

From the discussion up to here, we have the following proposition.

Proposition 11.6.

We have

\displaystyle B^{\circ}

\displaystyle\cong\bigoplus_{i,j=2}^{4}B^{\circ}_{i,j},

\displaystyle B^{\text{\textbullet}}

\displaystyle\cong\bigoplus_{\{i_{2},i_{3},i_{4}\}=\{2,3,4\}}B_{(i_{2}i_{3}i_{4})}.

We also have

\displaystyle B^{\circ}_{i,j}

\displaystyle\cong C_{0}(F^{\circ}_{i,j},D_{i,j})\cong C_{0}\big{(}(0,1),M_{2}(\mathbb{C})\oplus M_{2}(\mathbb{C})\big{)},

for $i,j=2,3,4$ and

\displaystyle B_{(i_{2}i_{3}i_{4})}

\displaystyle\cong C(F_{(i_{2}i_{3}i_{4})},D_{(i_{2}i_{3}i_{4})})\cong\mathbb{C}^{4}

for $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ .

From this proposition, we get

\displaystyle B^{\circ}

\displaystyle\cong C_{0}\big{(}(0,1),M_{2}(\mathbb{C})\oplus M_{2}(\mathbb{C})\big{)}^{9}\cong C_{0}\big{(}(0,1),M_{2}(\mathbb{C}))^{18},

\displaystyle B^{\text{\textbullet}}

\displaystyle\cong(\mathbb{C}^{4})^{6}\cong\mathbb{C}^{24}.

12. K-groups of the quotient $B$

From the short exact sequence

0\longrightarrow B^{\circ}\longrightarrow B\longrightarrow B^{\text{\textbullet}}\longrightarrow 0,

we get a six-term exact sequence

Next we compute $\delta\colon K_{0}(B^{\text{\textbullet}})\to K_{1}(B^{\circ})$ .

Proposition 12.1.

Under the isomorphism $\varPhi\colon A(4)\to A$ , the $C^{*}$ -algebra $A^{\text{ab}}(4)$ is canonically isomorphic to $B^{\text{\textbullet}}$ .

Proof.

Since $B^{\text{\textbullet}}\cong\mathbb{C}^{24}$ is commutative, the surjection $A(4)\cong A\twoheadrightarrow B\twoheadrightarrow B^{\text{\textbullet}}$ factors through the surjection $A(4)\twoheadrightarrow A^{\text{ab}}(4)$ . The induced surjection $A^{\text{ab}}(4)\twoheadrightarrow B^{\text{\textbullet}}$ is an isomorphism because $A^{\text{ab}}(4)\cong\mathbb{C}^{24}$ . ∎

For $i,j=1,2,3,4$ , we denote the image of $P_{i,j}\in A$ under a surjection by the same symbol $P_{i,j}$ . By Proposition 1.7 and Proposition 12.1, the 24 minimal projections of $B^{\text{\textbullet}}$ are

P_{(i_{1}i_{2}i_{3}i_{4})}\coloneqq P_{i_{1},1}P_{i_{2},2}P_{i_{3},3}P_{i_{4},4}\in B^{\text{\textbullet}}

for $(i_{1}i_{2}i_{3}i_{4})\in\mathfrak{S}_{4}$ .

Definition 12.2.

For $\sigma\in\mathfrak{S}_{4}$ , we define $q_{\sigma}\coloneqq[P_{\sigma}]_{0}\in K_{0}(B^{\text{\textbullet}})$ .

Note that $\{q_{\sigma}\}_{\sigma\in\mathfrak{S}_{4}}$ is a basis of $K_{0}(B^{\text{\textbullet}})\cong\mathbb{Z}^{24}$ .

Proposition 12.3.

For each $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ , the $4$ minimal projections of $\mathbb{C}^{4}\cong B_{(i_{2}i_{3}i_{4})}\subset B^{\text{\textbullet}}$ are $P_{\sigma t_{k}}$ for $k=1,2,3,4$ where $\sigma\coloneqq(1i_{2}i_{3}i_{4})$ .

Proof.

Take $i_{2},i_{3},i_{4}$ with $\{i_{2},i_{3},i_{4}\}=\{2,3,4\}$ . Since the 4 points in $\widetilde{F}_{(i_{2}i_{3}i_{4})}$ are fixed by $\sigma_{i_{2},2}$ , $\sigma_{i_{3},3}$ and $\sigma_{i_{4},4}$ , we have $P_{k,l}=P_{t_{i_{j}}(k),t_{j}(l)}$ in $B_{(i_{2}i_{3}i_{4})}$ for $k,l=1,2,3,4$ and $j=2,3,4$ by Lemma 9.3. More concretely we have

	$\displaystyle P_{1,1}=P_{i_{2},2}=P_{i_{3},3}=P_{i_{4},4},$
	$\displaystyle P_{i_{2},1}=P_{1,2}=P_{i_{4},3}=P_{i_{3},4},$
	$\displaystyle P_{i_{3},1}=P_{i_{4},2}=P_{1,3}=P_{i_{2},4},$
	$\displaystyle P_{i_{4},1}=P_{i_{3},2}=P_{i_{2},3}=P_{1,4}$

in $B_{(i_{2}i_{3}i_{4})}$ . These four projections are mutually orthogonal, and their sum equals to $1$ . Thus the 4 minimal projections of $B_{(i_{2}i_{3}i_{4})}$ are $P_{(1i_{2}i_{3}i_{4})}$ , $P_{(i_{2}1i_{4}i_{3})}$ , $P_{(i_{3}i_{4}1i_{2})}$ and $P_{(i_{4}i_{3}i_{2}1)}$ . ∎

Take $i,j=2,3,4$ , and fix them for a while. Let $(1m_{2}m_{3}m_{4})\in\mathfrak{S}_{4}$ be the unique even permutation with $m_{j}=i$ , and $(1n_{2}n_{3}n_{4})\in\mathfrak{S}_{4}$ be the unique odd permutation with $n_{j}=i$ . We set $\sigma=(1m_{2}m_{3}m_{4})$ and $\tau=(1n_{2}n_{3}n_{4})$ . Then we have the following commutative diagram with exact rows;

By Lemma 9.3, we have $P_{k,l}=P_{t_{i}(k),t_{j}(l)}$ in $B_{i,j}$ for $k,l=1,2,3,4$ . Let $\omega=(1342)\in\mathfrak{S}_{4}$ . Note that we have $t_{i}(\omega(i))=\omega^{2}(i)$ and $t_{i}(\omega^{2}(i))=\omega(i)$ . One can see that $B_{i,j}$ is a direct sum of two $C^{*}$ -subalgebras $B_{i,j}^{\cap}$ and $B_{i,j}^{\cup}$ where $B_{i,j}^{\cap}$ is generated by

\displaystyle P_{1,1}

\displaystyle=P_{i,j},

\displaystyle P_{1,j}

\displaystyle=P_{i,1},

\displaystyle P_{\omega(i),\omega(j)}

\displaystyle=P_{\omega^{2}(i),\omega^{2}(j)},

\displaystyle P_{\omega(i),\omega^{2}(j)}

\displaystyle=P_{\omega^{2}(i),\omega(j)}

and $B_{i,j}^{\cup}$ is generated by

\displaystyle P_{1,\omega(j)}

\displaystyle=P_{i,\omega^{2}(j)},

\displaystyle P_{1,\omega^{2}(j)}

\displaystyle=P_{i,\omega(j)},

\displaystyle P_{\omega(i),1}

\displaystyle=P_{\omega^{2}(i),j},

\displaystyle P_{\omega(i),j}

\displaystyle=P_{\omega^{2}(i),1}.

Note that $P_{1,1}+P_{1,j}=P_{\omega(i),\omega(j)}+P_{\omega(i),\omega^{2}(j)}$ is the unit of $B_{i,j}^{\cap}$ , and $P_{1,\omega(j)}+P_{1,\omega^{2}(j)}=P_{\omega(i),1}+P_{\omega(i),j}$ is the unit of $B_{i,j}^{\cup}$ . It turns out that both $B_{i,j}^{\cap}$ and $B_{i,j}^{\cup}$ are isomorphic to the universal unital $C^{*}$ -algebra generated by two projections, which is isomorphic to

\Big{\{}f\in C([0,1],M_{2}(\mathbb{C}))\ \Big{|}\ f(0)=\begin{pmatrix}*&0\\ 0&*\end{pmatrix},f(1)=\begin{pmatrix}*&0\\ 0&*\end{pmatrix}\Big{\}}.

This fact can be proved directly, but we do not prove it here because we do not need it. The image of $B_{i,j}^{\cap}$ under the surjection $B_{i,j}\to B_{(m_{2}m_{3}m_{4})}\oplus B_{(n_{2}n_{3}n_{4})}$ is $(\mathbb{C}p_{\sigma}+\mathbb{C}p_{\sigma t_{j}})\oplus(\mathbb{C}p_{\tau}+\mathbb{C}p_{\tau t_{j}})$ . Therefore, the image of $B_{i,j}^{\cup}$ under the surjection $B_{i,j}\to B_{(m_{2}m_{3}m_{4})}\oplus B_{(n_{2}n_{3}n_{4})}$ is $(\mathbb{C}p_{\sigma t_{\omega(j)}}+\mathbb{C}p_{\sigma t_{\omega^{2}(j)}})\oplus(\mathbb{C}p_{\tau t_{\omega(j)}}+\mathbb{C}p_{\tau t_{\omega^{2}(j)}})$ . We set $v_{i,j}^{\cap},v_{i,j}^{\cup}\in K_{1}(B^{\circ}_{i,j})$ by $v_{i,j}^{\cap}\coloneqq\delta^{\prime}(q_{\sigma})$ and $v_{i,j}^{\cup}\coloneqq\delta^{\prime}(q_{\sigma t_{\omega(j)}})$ where

\delta^{\prime}\colon K_{0}(B_{(m_{2}m_{3}m_{4})}\oplus B_{(n_{2}n_{3}n_{4})})\to K_{1}(B^{\circ}_{i,j})

is the exponential map. Then we have the following.

Lemma 12.4.

The set $\{v_{i,j}^{\cap},v_{i,j}^{\cup}\}$ is a generator of $K_{1}(B^{\circ}_{i,j})\cong\mathbb{Z}^{2}$ , and we have

	$\displaystyle\delta^{\prime}(q_{\sigma})$	$\displaystyle=\delta^{\prime}(q_{\sigma t_{j}})=v_{i,j}^{\cap},$	$\displaystyle\delta^{\prime}(q_{\sigma t_{\omega(j)}})$	$\displaystyle=\delta^{\prime}(q_{\sigma t_{\omega^{2}(j)}})=v_{i,j}^{\cup},$
	$\displaystyle\delta^{\prime}(q_{\tau})$	$\displaystyle=\delta^{\prime}(q_{\tau t_{j}})=-v_{i,j}^{\cap},$	$\displaystyle\delta^{\prime}(q_{\tau t_{\omega(j)}})$	$\displaystyle=\delta^{\prime}(q_{\tau t_{\omega^{2}(j)}})=-v_{i,j}^{\cup}.$

Proof.

Choose a closed interval $Z\subset\mathbb{R}P^{3}$ such that $\pi\colon Z\to F_{i,j}$ is a homeomorphism (see Figure 13.2 and the remark around it for an example of such a space). Let $z_{0},z_{1}\in Z$ be the point such that $\pi(z_{0})=v_{(m_{2}m_{3}m_{4})}$ $\pi(z_{1})=v_{(n_{2}n_{3}n_{4})}$ . Then we have $B^{\circ}_{i,j}\cong C_{0}(Z\setminus\{z_{0},z_{1}\},D_{i,j})$ . Let $B_{i,j}^{\prime}$ be the inverse image of $B_{(m_{2}m_{3}m_{4})}$ under the surjection $B_{i,j}\to B_{(m_{2}m_{3}m_{4})}\oplus B_{(n_{2}n_{3}n_{4})}$ . Then we have the following commutative diagram with exact rows;

Let us denote by $\varphi$ the homomorphism from $K_{0}(B_{(m_{2}m_{3}m_{4})})$ to $K_{0}(D_{i,j})$ induced by the vertical map from $B_{(m_{2}m_{3}m_{4})}\cong D_{(m_{2}m_{3}m_{4})}$ to $D_{i,j}$ . Then $K_{0}(D_{i,j})\cong\mathbb{Z}^{2}$ is spanned by $\varphi(q_{\sigma})=\varphi(q_{\sigma t_{j}})$ and $\varphi(q_{\sigma t_{\omega(j)}})=\varphi(q_{\sigma t_{\omega^{2}(j)}})$ . Since $K_{l}(C_{0}(Z\setminus\{z_{0}\},D_{i,j}))=0$ for $l=0,1$ , $K_{0}(D_{i,j})\to K_{1}(B^{\circ}_{i,j})$ is an isomorphism. This shows that $\{v_{i,j}^{\cap},v_{i,j}^{\cup}\}$ is a generator of $K_{1}(B^{\circ}_{i,j})\cong\mathbb{Z}^{2}$ . We also have $\delta^{\prime}(q_{\sigma})=\delta^{\prime}(q_{\sigma t_{j}})$ and $\delta^{\prime}(q_{\sigma t_{\omega(j)}})=\delta^{\prime}(q_{\sigma t_{\omega^{2}(j)}})$ . Similarly, we have $\delta^{\prime}(q_{\tau})=\delta^{\prime}(q_{\tau t_{j}})$ and $\delta^{\prime}(q_{\tau t_{\omega(j)}})=\delta^{\prime}(q_{\tau t_{\omega^{2}(j)}})$ .

Since the image of the projection $P_{1,1}\in B_{i,j}$ under the surjection $B_{i,j}\to B_{(m_{2}m_{3}m_{4})}\oplus B_{(n_{2}n_{3}n_{4})}$ is $P_{\sigma}+P_{\tau}$ , we have $\delta^{\prime}(q_{\sigma}+q_{\tau})=0$ . Hence $\delta^{\prime}(q_{\tau})=-v_{i,j}^{\cap}$ . Similarly we have $\delta^{\prime}(q_{\sigma t_{\omega(j)}}+q_{\tau t_{\omega(j)}})=0$ because the image of $P_{1,\omega(j)}\in B_{i,j}$ under the surjection $B_{i,j}\to B_{(m_{2}m_{3}m_{4})}\oplus B_{(n_{2}n_{3}n_{4})}$ is $P_{\sigma t_{\omega(j)}}+P_{\tau t_{\omega(j)}}$ . We are done. ∎

From these computation, we get the following proposition.

Proposition 12.5.

The exponential map $\delta\colon K_{0}(B^{\text{\textbullet}})\to K_{1}(B^{\circ})$ is as Table 12.1.

Table 12.1. Computation of the exponential map

\delta

	2,2		3,3		4,4		4,3		2,4		3,2		3,4		4,2		2,3
$q\diagdown v$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$
(1234)	1	0	1	0	1	0	0	0	0	0	0	0	0	0	0	0	0	0
(2143)	1	0	0	1	0	1	0	0	0	0	0	0	0	0	0	0	0	0
(3412)	0	1	1	0	0	1	0	0	0	0	0	0	0	0	0	0	0	0
(4321)	0	1	0	1	1	0	0	0	0	0	0	0	0	0	0	0	0	0
(1342)	0	0	0	0	0	0	1	0	1	0	1	0	0	0	0	0	0	0
(2431)	0	0	0	0	0	0	0	1	1	0	0	1	0	0	0	0	0	0
(3124)	0	0	0	0	0	0	0	1	0	1	1	0	0	0	0	0	0	0
(4213)	0	0	0	0	0	0	1	0	0	1	0	1	0	0	0	0	0	0
(1423)	0	0	0	0	0	0	0	0	0	0	0	0	1	0	1	0	1	0
(2314)	0	0	0	0	0	0	0	0	0	0	0	0	0	1	0	1	1	0
(3241)	0	0	0	0	0	0	0	0	0	0	0	0	1	0	0	1	0	1
(4132)	0	0	0	0	0	0	0	0	0	0	0	0	0	1	1	0	0	1
(1243)	$-1$	0	0	0	0	0	$-1$	0	0	0	0	0	$-1$	0	0	0	0	0
(2134)	$-1$	0	0	0	0	0	0	$-1$	0	0	0	0	0	$-1$	0	0	0	0
(3421)	0	$-1$	0	0	0	0	0	$-1$	0	0	0	0	$-1$	0	0	0	0	0
(4312)	0	$-1$	0	0	0	0	$-1$	0	0	0	0	0	0	$-1$	0	0	0	0
(1432)	0	0	$-1$	0	0	0	0	0	$-1$	0	0	0	0	0	$-1$	0	0	0
(2341)	0	0	0	$-1$	0	0	0	0	$-1$	0	0	0	0	0	0	$-1$	0	0
(3214)	0	0	$-1$	0	0	0	0	0	0	$-1$	0	0	0	0	0	$-1$	0	0
(4123)	0	0	0	$-1$	0	0	0	0	0	$-1$	0	0	0	0	$-1$	0	0	0
(1324)	0	0	0	0	$-1$	0	0	0	0	0	$-1$	0	0	0	0	0	$-1$	0
(2413)	0	0	0	0	0	$-1$	0	0	0	0	0	$-1$	0	0	0	0	$-1$	0
(3142)	0	0	0	0	0	$-1$	0	0	0	0	$-1$	0	0	0	0	0	0	$-1$
(4231)	0	0	0	0	$-1$	0	0	0	0	0	0	$-1$	0	0	0	0	0	$-1$

We will see that $K_{1}(B)\cong\operatorname{coker}\delta$ is isomorphic to $\mathbb{Z}^{4}\oplus\mathbb{Z}/2\mathbb{Z}$ in Proposition 15.5. This implies $K_{0}(B)\cong\ker\delta$ is isomorphic to $\mathbb{Z}^{10}$ because $\ker\delta$ is a free abelian group with dimension $24-18+4=10$ . Below, we examine the generator of $K_{0}(B)\cong\ker\delta$ .

For $i,j=1,2,3,4$ , we have

\displaystyle P_{i,j}=P_{i,j}\sum_{k\neq i}\sum_{l=1}^{n}P_{k,l}=\sum_{i=\sigma(j)}P_{\sigma}

in $B$ . Hence $[P_{i,j}]_{0}=\sum_{i=\sigma(j)}q_{\sigma}$ in $K_{0}(B)$ .

Proposition 12.6.

The group $\ker\delta$ is generated by $\{[P_{i,j}]_{0}\mid i,j=1,2,3,4\}$ .

Proof.

It is straightforward to check that $[P_{i,j}]_{0}$ is in $\ker\delta$ for $i,j=1,2,3,4$ .

Take $x\in\ker\delta$ , and we will show that $x$ is in the subgroup generated by $\{[P_{i,j}]_{0}\mid i,j=1,2,3,4\}$ . Write $x=\sum_{\sigma\in\mathfrak{S}_{4}}n_{\sigma}q_{\sigma}$ with $n_{\sigma}\in\mathbb{Z}$ . Subtracting $n_{(4213)}[P_{2,2}]_{0}+n_{(4132)}[P_{1,2}]_{0}$ from $x$ , we may assume $n_{(4213)}=n_{(4132)}=0$ without loss of generality. Subtracting $n_{(4312)}[P_{3,2}]_{0}+n_{(4123)}[P_{2,3}]_{0}+n_{(4231)}[P_{1,4}]_{0}$ from $x$ , we may further assume $n_{(4312)}=n_{(4123)}=n_{(4231)}=0$ without loss of generality. Subtracting $n_{(2341)}[P_{2,1}]_{0}+n_{(3142)}[P_{3,1}]_{0}$ from $x$ , we may further assume $n_{(2341)}=n_{(3142)}=0$ without loss of generality. Subtracting $n_{(2413)}[P_{4,2}]_{0}+n_{(3214)}[P_{4,4}]_{0}+n_{(1324)}[P_{1,1}]_{0}$ from $x$ , we may further assume $n_{(2413)}=n_{(3214)}=n_{(1324)}=0$ without loss of generality. Now we will show $x=0$ using $x\in\ker\delta$ .

Since $n_{(3241)}+n_{(4132)}=n_{(3142)}+n_{(4231)}$ , we have $n_{(3241)}=0$ .

Since $n_{(2314)}+n_{(3241)}=n_{(2341)}+n_{(3214)}$ , we have $n_{(2314)}=0$ .

Since $n_{(1423)}+n_{(2314)}=n_{(1324)}+n_{(2413)}$ , we have $n_{(1423)}=0$ .

Since $n_{(1423)}+n_{(4132)}=n_{(1432)}+n_{(4123)}$ , we have $n_{(1432)}=0$ .

Since $n_{(3124)}+n_{(4213)}=n_{(3214)}+n_{(4123)}$ , we have $n_{(3124)}=0$ .

Since $n_{(2431)}+n_{(4213)}=n_{(2413)}+n_{(4231)}$ , we have $n_{(2431)}=0$ .

Since $n_{(1342)}+n_{(2431)}=n_{(1432)}+n_{(2341)}$ , we have $n_{(1342)}=0$ .

Since $n_{(2314)}+n_{(4132)}=n_{(2134)}+n_{(4312)}$ , we have $n_{(2134)}=0$ .

Since $n_{(2431)}+n_{(3124)}=n_{(2134)}+n_{(3421)}$ , we have $n_{(3421)}=0$ .

Since $n_{(1423)}+n_{(3241)}=n_{(1243)}+n_{(3421)}$ , we have $n_{(1243)}=0$ .

Since $n_{(1234)}+n_{(2143)}=n_{(1243)}+n_{(2134)}=0$ , $n_{(1234)}+n_{(3412)}=n_{(1432)}+n_{(3214)}=0$ and $n_{(2143)}+n_{(3412)}=n_{(2413)}+n_{(3142)}=0$ , we have $2n_{(1234)}=0$ . Hence $n_{(1234)}=0$ . This implies $n_{(2143)}=n_{(3412)}=0$ . Finally, since $n_{(1234)}+n_{(4321)}=n_{(1324)}+n_{(4231)}$ , we have $n_{(4321)}=0$ . We have shown that $x=0$ . This completes the proof. ∎

From Proposition 12.6 (or its proof), we see that $K_{0}(B)\cong\ker\delta$ is isomorphic to $\mathbb{Z}^{n}$ with $n\leq 10$ . Note that the group generated by $\{[P_{i,j}]_{0}\mid i,j=1,2,3,4\}$ is in fact generated by $10$ elements

[P_{1,1}]_{0},[P_{1,2}]_{0},[P_{1,3}]_{0},[P_{1,4}]_{0},[P_{2,1}]_{0},[P_{2,2}]_{0},[P_{2,3}]_{0},[P_{3,1}]_{0},[P_{3,2}]_{0},[P_{3,3}]_{0}.

We will show that $K_{0}(B)\cong\ker\delta$ is isomorphic to $\mathbb{Z}^{10}$ in Proposition 15.5.

Table 12.2. Computation of

[P_{i,j}]_{0}

$i$	1				2				3				4
$q\diagdown j$	1	2	3	4	1	2	3	4	1	2	3	4	1	2	3	4
(1234)	1	0	0	0	0	1	0	0	0	0	1	0	0	0	0	1
(2143)	0	1	0	0	1	0	0	0	0	0	0	1	0	0	1	0
(3412)	0	0	1	0	0	0	0	1	1	0	0	0	0	1	0	0
(4321)	0	0	0	1	0	0	1	0	0	1	0	0	1	0	0	0
(1342)	1	0	0	0	0	0	0	1	0	1	0	0	0	0	1	0
(2431)	0	0	0	1	1	0	0	0	0	0	1	0	0	1	0	0
(3124)	0	1	0	0	0	0	1	0	1	0	0	0	0	0	0	1
(4213)	0	0	1	0	0	1	0	0	0	0	0	1	1	0	0	0
(1423)	1	0	0	0	0	0	1	0	0	0	0	1	0	1	0	0
(2314)	0	0	1	0	1	0	0	0	0	1	0	0	0	0	0	1
(3241)	0	0	0	1	0	1	0	0	1	0	0	0	0	0	1	0
(4132)	0	1	0	0	0	0	0	1	0	0	1	0	1	0	0	0
(1243)	1	0	0	0	0	1	0	0	0	0	0	1	0	0	1	0
(2134)	0	1	0	0	1	0	0	0	0	0	1	0	0	0	0	1
(3421)	0	0	0	1	0	0	1	0	1	0	0	0	0	1	0	0
(4312)	0	0	1	0	0	0	0	1	0	1	0	0	1	0	0	0
(1432)	1	0	0	0	0	0	0	1	0	0	1	0	0	1	0	0
(2341)	0	0	0	1	1	0	0	0	0	1	0	0	0	0	1	0
(3214)	0	0	1	0	0	1	0	0	1	0	0	0	0	0	0	1
(4123)	0	1	0	0	0	0	1	0	0	0	0	1	1	0	0	0
(1324)	1	0	0	0	0	0	1	0	0	1	0	0	0	0	0	1
(2413)	0	0	1	0	1	0	0	0	0	0	0	1	0	1	0	0
(3142)	0	1	0	0	0	0	0	1	1	0	0	0	0	0	1	0
(4231)	0	0	0	1	0	1	0	0	0	0	1	0	1	0	0	0

The positive cone $K_{0}(B^{\text{\textbullet}})_{+}$ of $K_{0}(B^{\text{\textbullet}})$ is the set of sums of $q_{\sigma}$ ’s. In other words, we have

K_{0}(B^{\text{\textbullet}})_{+}=\Big{\{}\sum_{\sigma\in\mathfrak{S}_{4}}n_{\sigma}q_{\sigma}\ \Big{|}\ n_{\sigma}=0,1,2,\ldots\Big{\}}

Proposition 12.7.

The intersection $K_{0}(B^{\text{\textbullet}})_{+}\cap\ker\delta$ is the set of sums of $[P_{i,j}]_{0}$ ’s.

Proof.

It is clear that $[P_{i,j}]_{0}$ is in $K_{0}(B^{\text{\textbullet}})_{+}\cap\ker\delta$ for $i,j=1,2,3,4$ . Thus the set of sums of $[P_{i,j}]_{0}$ ’s is contained in $K_{0}(B^{\text{\textbullet}})_{+}\cap\ker\delta$ .

Take $x\in K_{0}(B^{\text{\textbullet}})_{+}\cap\ker\delta$ . By Proposition 12.6, we have $n_{i,j}\in\mathbb{Z}$ for $i,j=1,2,3,4$ such that $x=\sum_{i,j=1}^{4}n_{i,j}[P_{i,j}]_{0}$ . We set $n\coloneqq\sum_{n_{i,j}<0}(-n_{i,j})$ . If $n=0$ , then $x$ is in the set of sums of $[P_{i,j}]_{0}$ ’s. If $n>0$ , then we will show that there exists $n^{\prime}_{i,j}\in\mathbb{Z}$ for $i,j=1,2,3,4$ such that $x=\sum_{i,j=1}^{4}n^{\prime}_{i,j}[P_{i,j}]_{0}$ and that $n^{\prime}\coloneqq\sum_{n^{\prime}_{i,j}<0}(-n^{\prime}_{i,j})$ satisfies $0\leq n^{\prime}<n$ . Repeating this argument at most $n$ times, we will find $n^{\prime\prime}_{i,j}\in\mathbb{Z}$ for $i,j=1,2,3,4$ such that $x=\sum_{i,j=1}^{4}n^{\prime\prime}_{i,j}[P_{i,j}]_{0}$ and that $n^{\prime\prime}\coloneqq\sum_{n^{\prime\prime}_{i,j}<0}(-n^{\prime\prime}_{i,j})$ satisfies $n^{\prime\prime}=0$ . This shows that $x$ is in the set of sums of $[P_{i,j}]_{0}$ ’s.

Since $n>0$ we have $i_{0},j_{0}\in\{1,2,3,4\}$ such that $n_{i_{0},j_{0}}<0$ . To simplify the notation, we assume $i_{0}=3$ and $j_{0}=1$ . The other $15$ cases can be shown similarly. Since $x\in K_{0}(B^{\text{\textbullet}})_{+}$ , the coefficient of $v_{\sigma}$ is non-negative for all $\sigma\in\mathfrak{S}_{4}$ . In particular, so is for $\sigma\in\mathfrak{S}_{4}$ with $i_{0}=\sigma(j_{0})$ . Since the coefficient of $v_{(3,1,2,4)}$ is non-negative we have $n_{3,1}+n_{1,2}+n_{2,3}+n_{4,4}\geq 0$ . Since $n_{3,1}<0$ , we have $n_{1,2}+n_{2,3}+n_{4,4}>0$ . Hence either $n_{1,2}$ , $n_{2,3}$ or $n_{4,4}$ is positive. Similarly, since the coefficients of

v_{(3,1,4,2)},v_{(3,2,1,4)},v_{(3,2,4,1)},v_{(3,4,1,2)},v_{(3,4,2,1)}

are non-negative, we obtain that either $n_{1,2}$ , $n_{4,3}$ or $n_{2,4}$ is positive etc. Then by Lemma 12.8 below we have either

(i)

$n_{i_{1},2}$ $n_{i_{1},3}$ and $n_{i_{1},4}$ are positive for some $i_{1}\in\{1,2,4\}$ ,
(ii)

$n_{1,j_{1}}$ $n_{2,j_{1}}$ and $n_{4,j_{1}}$ are positive for some $j_{1}\in\{2,3,4\}$ , or
(iii)

$n_{i_{1},j_{1}}$ , $n_{i_{1},j_{2}}$ , $n_{i_{2},j_{1}}$ and $n_{i_{2},j_{2}}$ are positive for some distinct $i_{1},i_{2}\in\{1,2,4\}$ and distinct $j_{1},j_{2}\in\{2,3,4\}$ .

In the case (i), we set $n^{\prime}_{i,j}$ by

n^{\prime}_{i,j}=\begin{cases}n_{i,j}+1&\text{for $i\in\{1,2,3,4\}\setminus\{i_{1}\}$ and $j=1$,}\\ n_{i,j}-1&\text{for $i=i_{1}$ and $j=2,3,4$}\\ n_{i,j}&\text{otherwise.}\end{cases}

Then since $n^{\prime}_{3,1}=n_{3,1}+1$ , $n^{\prime}\coloneqq\sum_{n^{\prime}_{i,j}<0}(-n^{\prime}_{i,j})$ satisfies $0\leq n^{\prime}<n$ . We also have $x=\sum_{i,j=1}^{4}n^{\prime}_{i,j}[P_{i,j}]_{0}$ because $\sum_{i=1}^{4}[P_{i,1}]_{0}=\sum_{j=1}^{4}[P_{i_{1},j}]_{0}$ . In the case (ii), we get the same conclusion for $n^{\prime}_{i,j}$ defined by

n^{\prime}_{i,j}=\begin{cases}n_{i,j}+1&\text{for $i=3$ and $j\in\{1,2,3,4\}\setminus\{j_{1}\}$,}\\ n_{i,j}-1&\text{for $i=1,2,4$ and $j=j_{1}$}\\ n_{i,j}&\text{otherwise.}\end{cases}

In the case (iii), we define $n^{\prime}_{i,j}$ by

n^{\prime}_{i,j}=\begin{cases}n_{i,j}+1&\text{for $i\in\{1,2,3,4\}\setminus\{i_{1},i_{2}\}$ and $j\in\{1,2,3,4\}\setminus\{j_{1},j_{2}\}$,}\\ n_{i,j}-1&\text{for $i=i_{1},i_{2}$ and $j=j_{1},j_{2}$}\\ n_{i,j}&\text{otherwise.}\end{cases}

Since $n^{\prime}_{3,1}=n_{3,1}+1$ , $n^{\prime}\coloneqq\sum_{n^{\prime}_{i,j}<0}(-n^{\prime}_{i,j})$ satisfies $0\leq n^{\prime}<n$ . We also have $x=\sum_{i,j=1}^{4}n^{\prime}_{i,j}[P_{i,j}]_{0}$ because

\sum_{i=1}^{4}[P_{i,j_{1}}]_{0}+\sum_{i=1}^{4}[P_{i,j_{2}}]_{0}=\sum_{j=1}^{4}[P_{i_{3},j}]_{0}+\sum_{j=1}^{4}[P_{i_{4},j}]_{0}.

where $\{i_{3},i_{4}\}=\{1,2,3,4\}\setminus\{i_{1},i_{2}\}$ . This completes the proof. ∎

Lemma 12.8.

Let $a,b,c$ and $d,e,f$ are distinct three numbers, respectively. Suppose $n_{i,j}\in\mathbb{Z}$ for $i=a,b,c$ and $j=d,e,f$ satisfy that either $n_{\omega(d),d}$ , $n_{\omega(e),e}$ or $n_{\omega(f),f}$ is positive for all bijection $\omega\colon\{d,e,f\}\to\{a,b,c\}$ . Then we have either

(i)

$n_{i_{1},d}$ $n_{i_{1},e}$ and $n_{i_{1},f}$ are positive for some $i_{1}\in\{a,b,c\}$ ,
(ii)

$n_{a,j_{1}}$ $n_{b,j_{1}}$ and $n_{c,j_{1}}$ are positive for some $j_{1}\in\{d,e,f\}$ , or
(iii)

$n_{i_{1},j_{1}}$ , $n_{i_{1},j_{2}}$ , $n_{i_{2},j_{1}}$ and $n_{i_{2},j_{2}}$ are positive for some distinct $i_{1},i_{2}\in\{a,b,c\}$ and distinct $j_{1},j_{2}\in\{d,e,f\}$ .

Proof.

To the contrary, assume that the conclusion does not hold. Then for $j=d,e,f$ , either $n_{a,j}$ , $n_{b,j}$ or $n_{c,j}$ is non-positive. Thus we obtain a map $\omega\colon\{d,e,f\}\to\{a,b,c\}$ such that $n_{\omega(j),j}$ is non-positive for $j=d,e,f$ . If the cardinality of the image of $\omega$ is three, then $\omega$ is a bijection and it contradicts the assumption. If the cardinality of the image of $\omega$ is two, let $i_{1}$ be the element in $\{a,b,c\}$ which is not in the image of $\omega$ . Then we have either $n_{i_{1},d}$ $n_{i_{1},e}$ or $n_{i_{1},f}$ is non-positive. Let $j_{1}\in\{d,e,f\}$ be an element such that $n_{i_{1},j_{1}}$ is non-positive. If the cardinality of $\omega^{-1}(\omega(j_{1}))$ is two, we get a bijection $\omega^{\prime}\colon\{d,e,f\}\to\{a,b,c\}$ such that $n_{\omega(d),d}$ , $n_{\omega(e),e}$ and $n_{\omega(f),f}$ are non-positive. This is a contradiction. If the cardinality of $\omega^{-1}(\omega(j_{1}))$ is one, we have either $n_{i_{1},j_{2}}$ , $n_{i_{1},j_{3}}$ , $n_{i_{2},j_{2}}$ or $n_{i_{2},j_{3}}$ is non-positive where $i_{2}=\omega(j_{1})$ and $\{j_{2},j_{3}\}=\{d,e,f\}\setminus\{j_{1}\}$ . In this case, we can find a bijection $\omega^{\prime}\colon\{d,e,f\}\to\{a,b,c\}$ such that $n_{\omega(d),d}$ , $n_{\omega(e),e}$ and $n_{\omega(f),f}$ are non-positive. This is a contradiction. Finally, if the cardinality of the image of $\omega$ is one, let $i_{1}$ be the unique element of the image of $\omega$ , and $i_{2}$ and $i_{3}$ be the other two elements in $\{a,b,c\}$ . We have $j_{2},j_{3}\in\{d,e,f\}$ such that $n_{i_{2},j_{2}}$ and $n_{i_{3},j_{3}}$ are non-positive. If $j_{2}\neq j_{3}$ , then we can find a bijection $\omega^{\prime}\colon\{d,e,f\}\to\{a,b,c\}$ such that $n_{\omega(d),d}$ , $n_{\omega(e),e}$ and $n_{\omega(f),f}$ are non-positive. This is a contradiction. If $j_{2}=j_{3}$ , then we have either $n_{i_{2},j_{1}}$ , $n_{i_{2},j^{\prime}_{1}}$ , $n_{i_{3},j^{\prime}_{1}}$ or $n_{i_{3},j_{1}}$ is non-positive where $\{j_{1},j^{\prime}_{1}\}=\{d,e,f\}\setminus\{j_{2}\}$ . In this case, we can find a bijection $\omega^{\prime}\colon\{d,e,f\}\to\{a,b,c\}$ such that $n_{\omega(d),d}$ , $n_{\omega(e),e}$ and $n_{\omega(f),f}$ are non-positive. This is a contradiction. We are done. ∎

13. The Structure of the Ideal $I$

Definition 13.1.

Define a subspace $V$ of $\mathbb{R}P^{3}$ by

V\coloneqq\{[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}\mid a_{1},a_{2},a_{3}>|a_{4}|\}.

The next proposition gives us a motivation to compute the subspace $V$ and its closure $\overline{V}$ in $\mathbb{R}P^{3}$ .

Proposition 13.2.

We have the following facts.

(i)

For each $i,j=1,2,3,4$ with $(i,j)\neq(1,1)$ , we have $\sigma_{i,j}(V)\cap V=\emptyset$
(ii)

The restriction of $\pi$ to $V$ is a homeomorphism onto $\pi(V)\subset X$ .
(iii)

$\overline{V}=\{[a_{1},a_{2},a_{3},a_{4}]\in\mathbb{R}P^{3}\mid a_{1},a_{2},a_{3}\geq|a_{4}|\}$ and $\pi(\overline{V})=X$ .

Proof.

(i) and (iii) can be checked directly, and (ii) follows from (i). ∎

In the next proposition, when we write $[a_{1},a_{2},a_{3},a_{4}]\in\overline{V}$ , we mean $(a_{1},a_{2},a_{3},a_{4})$ satisfies $a_{1},a_{2},a_{3}\geq|a_{4}|$ .

Proposition 13.3.

The map

h\colon\overline{V}\ni[a_{1},a_{2},a_{3},a_{4}]\mapsto(3a_{1}^{2}+a_{4}^{2}+4a_{4}|a_{4}|,3a_{2}^{2}+a_{4}^{2}+4a_{4}|a_{4}|,3a_{3}^{2}+a_{4}^{2}+4a_{4}|a_{4}|)\in\mathbb{R}^{3}

is a homeomorphism onto the hexahedron whose 6 faces are isosceles right triangles and whose vertices are $(0,0,0)$ , $(3,0,0)$ , $(0,3,0)$ , $(0,0,3)$ and $(2,2,2)$ . This map sends $V$ onto the interior of the hexahedron.

Proof.

First note that we have $|a_{4}|\leq 1/2$ for $[a_{1},a_{2},a_{3},a_{4}]\in\overline{V}$ . When $|a_{4}|=1/2$ , we have $a_{1}=a_{2}=a_{3}=1/2$ . We have $h([1/2,1/2,1/2,1/2])=(2,2,2)$ and $h([1/2,1/2,1/2,-1/2])=(0,0,0)$ . When $|a_{4}|=0$ , we have $a_{1},a_{2},a_{3}\geq 0$ and $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=1$ . Thus

\{h([a_{1},a_{2},a_{3},0])\mid[a_{1},a_{2},a_{3},0]\in\overline{V}\}=\{(x,y,z)\in\mathbb{R}^{3}\mid x,y,z\geq 0,\ x+y+z=3\}

which is the equilateral triangle whose vertices are $(3,0,0)$ , $(0,3,0)$ and $(0,0,3)$ . For each $t$ with $-1/2<t<0$ , we have

	$\displaystyle\{h([a_{1},a_{2},a_{3},t])$	$\displaystyle\mid[a_{1},a_{2},a_{3},t]\in\overline{V}\}$
		$\displaystyle=\{(x,y,z)\in\mathbb{R}^{3}\mid x,y,z\geq 0,\ x+y+z=3(1-4t^{2})\}$

which is the equilateral triangle whose vertices are $(3(1-4t^{2}),0,0)$ , $(0,3(1-4t^{2}),0)$ and $(0,0,3(1-4t^{2}))$ . Thus

\{h([a_{1},a_{2},a_{3},a_{4}])\mid[a_{1},a_{2},a_{3},a_{4}]\in\overline{V},a_{4}\leq 0\}

is the tetrahedron whose vertices are $(0,0,0)$ , $(3,0,0)$ , $(0,3,0)$ and $(0,0,3)$ . Note that for each $[a_{1},a_{2},a_{3},a_{4}]\in\overline{V}$ with $a_{4}\geq 0$ , the point $h([a_{1},a_{2},a_{3},a_{4}])$ is the reflection point of $h([a_{1},a_{2},a_{3},-a_{4}])$ with respect to the plane $x+y+z=3$ because the vector $(8a_{4}^{2},8a_{4}^{2},8a_{4}^{2})$ is orthogonal to the plane $x+y+z=3$ and the point $(3a_{1}^{2}+a_{4}^{2},3a_{2}^{2}+a_{4}^{2},3a_{3}^{2}+a_{4}^{2})$ is on the plane $x+y+z=3$ . Thus

\{h([a_{1},a_{2},a_{3},a_{4}])\mid[a_{1},a_{2},a_{3},a_{4}]\in\overline{V},a_{4}\geq 0\}

is the reflection of the tetrahedron above with respect to the plane $x+y+z=3$ , which in turn is the tetrahedron whose vertices are $(3,0,0)$ , $(0,3,0)$ , $(0,0,3)$ and $(2,2,2)$ . From the discussion above, we see that $h$ is injective. Therefore we see that $h$ is a homeomorphism from $\overline{V}$ onto the hexahedron whose vertices are $(0,0,0)$ , $(3,0,0)$ , $(0,3,0)$ , $(0,0,3)$ and $(2,2,2)$ . We can also see that the map $h$ sends $V$ onto the interior of the hexahedron. ∎

Figure 13.1.

\overline{V}

Definition 13.4.

Define $O_{0}\coloneqq\pi(V)\subset O$ .

By Proposition 13.3, $O_{0}\cong V$ is homeomorphic to $\mathbb{R}^{3}$ .

Definition 13.5.

We set $E\coloneqq\widetilde{F}\cap\overline{V}$ and $E_{i,j}\coloneqq\widetilde{F}_{i,j}\cap\overline{V}$ for $i,j=2,3,4$ .

We have $E=\bigcup_{i,j=2}^{4}E_{i,j}$ . For $i,j=2,3,4$ with $i\neq j$ , the map $\pi\colon E_{i,j}\to F_{i,j}$ is a homeomorphism. For $i=2,3,4$ the map $\pi\colon E_{i,i}\to F_{i,i}$ is a $2$ -to- $1$ map except the middle point.

Figure 13.2.

\pi\colon E\to F

(

t=1/\sqrt{2}

)

We have

	$\displaystyle E_{2,2}=\{[a,b,0,0]\in\overline{V}\mid a,b\geq 0,\ a^{2}+b^{2}=1\},$
	$\displaystyle E_{2,3}=\{[a,b,b,-a]\in\overline{V}\mid 0\leq a\leq b,\ 2(a^{2}+b^{2})=1\},$
	$\displaystyle E_{2,4}=\{[a,b,a,b]\in\overline{V}\mid 0\leq b\leq a,\ 2(a^{2}+b^{2})=1\},$
	$\displaystyle E_{3,2}=\{[a,b,b,a]\in\overline{V}\mid 0\leq a\leq b,\ 2(a^{2}+b^{2})=1\},$
	$\displaystyle E_{3,3}=\{[a,0,b,0]\in\overline{V}\mid a,b\geq 0,\ a^{2}+b^{2}=1\},$
	$\displaystyle E_{3,4}=\{[a,a,b,-b]\in\overline{V}\mid 0\leq b\leq a,\ 2(a^{2}+b^{2})=1\},$
	$\displaystyle E_{4,2}=\{[a,b,a,-b]\in\overline{V}\mid 0\leq b\leq a,\ 2(a^{2}+b^{2})=1\},$
	$\displaystyle E_{4,3}=\{[a,a,b,b]\in\overline{V}\mid 0\leq b\leq a,\ 2(a^{2}+b^{2})=1\},$
	$\displaystyle E_{4,4}=\{[0,a,b,0]\in\overline{V}\mid a,b\geq 0,\ a^{2}+b^{2}=1\}.$

Definition 13.6.

We set $R^{+}_{x},R^{+}_{y},R^{+}_{z},R^{-}_{x},R^{-}_{y},R^{-}_{z}\subset\overline{V}$ by

	$\displaystyle R^{\pm}_{x}$	$\displaystyle\coloneqq\{[\sqrt{1-3t^{2}},t,t,\pm t]\in\overline{V}\mid 0<t<1/2\}$
	$\displaystyle R^{\pm}_{y}$	$\displaystyle\coloneqq\{[t,\sqrt{1-3t^{2}},t,\pm t]\in\overline{V}\mid 0<t<1/2\}$
	$\displaystyle R^{\pm}_{z}$	$\displaystyle\coloneqq\{[t,t,\sqrt{1-3t^{2}},\pm t]\in\overline{V}\mid 0<t<1/2\}$

We see that $R^{+}_{x}\cup R^{+}_{y}\cup R^{+}_{z}\cup R^{-}_{x}\cup R^{-}_{y}\cup R^{-}_{z}$ is the space obtained by subtracting $E$ from the “edges” of $\overline{V}$ .

Definition 13.7.

We set $R^{+},R^{-}\subset O$ by

\displaystyle R^{\pm}

\displaystyle\coloneqq\pi(R^{\pm}_{x})=\pi(R^{\pm}_{y})=\pi(R^{\pm}_{z})

Note that $\pi$ induces a homeomorphism from $R^{\pm}_{x}$ (or $R^{\pm}_{y}$ , $R^{\pm}_{z}$ ) to $R^{\pm}$ . Hence both $R^{+}$ and $R^{-}$ are homeomorphic to $\mathbb{R}$ .

Definition 13.8.

We set

	$\displaystyle\widehat{T}_{2,3}$	$\displaystyle\coloneqq\{[t,a,b,-t]\in\overline{V}\mid 0<t<1/2,\ a,b>t,\ a^{2}+b^{2}=1-2t^{2}\},$
	$\displaystyle\widehat{T}_{3,4}$	$\displaystyle\coloneqq\{[a,b,t,-t]\in\overline{V}\mid 0<t<1/2,\ a,b>t,\ a^{2}+b^{2}=1-2t^{2}\},$
	$\displaystyle\widehat{T}_{4,2}$	$\displaystyle\coloneqq\{[b,t,a,-t]\in\overline{V}\mid 0<t<1/2,\ a,b>t,\ a^{2}+b^{2}=1-2t^{2}\},$
	$\displaystyle\widehat{T}_{3,2}$	$\displaystyle\coloneqq\{[t,a,b,t]\in\overline{V}\mid 0<t<1/2,\ a,b>t,\ a^{2}+b^{2}=1-2t^{2}\},$
	$\displaystyle\widehat{T}_{4,3}$	$\displaystyle\coloneqq\{[a,b,t,t]\in\overline{V}\mid 0<t<1/2,\ a,b>t,\ a^{2}+b^{2}=1-2t^{2}\},$
	$\displaystyle\widehat{T}_{2,4}$	$\displaystyle\coloneqq\{[b,t,a,t]\in\overline{V}\mid 0<t<1/2,\ a,b>t,\ a^{2}+b^{2}=1-2t^{2}\}.$

These 6 spaces are the interiors of the 6 “faces” of $\overline{V}$ .

Definition 13.9.

We set

$\displaystyle\widehat{T}_{2,3}^{r}$	$\displaystyle\coloneqq\{[t,a,b,-t]\in\widehat{T}_{2,3}\mid a>b\},$	$\displaystyle\widehat{T}_{2,3}^{l}$	$\displaystyle\coloneqq\{[t,a,b,-t]\in\widehat{T}_{2,3}\mid a<b\}$
$\displaystyle\widehat{T}_{3,4}^{r}$	$\displaystyle\coloneqq\{[a,b,t,-t]\in\widehat{T}_{3,4}\mid a>b\},$	$\displaystyle\widehat{T}_{3,4}^{l}$	$\displaystyle\coloneqq\{[a,b,t,-t]\in\widehat{T}_{3,4}\mid a<b\}$
$\displaystyle\widehat{T}_{4,2}^{r}$	$\displaystyle\coloneqq\{[b,t,a,-t]\in\widehat{T}_{4,2}\mid a>b\},$	$\displaystyle\widehat{T}_{4,2}^{l}$	$\displaystyle\coloneqq\{[b,t,a,-t]\in\widehat{T}_{4,2}\mid a<b\}$
$\displaystyle\widehat{T}_{3,2}^{r}$	$\displaystyle\coloneqq\{[t,a,b,t]\in\widehat{T}_{3,2}\mid a>b\},$	$\displaystyle\widehat{T}_{3,2}^{l}$	$\displaystyle\coloneqq\{[t,a,b,t]\in\widehat{T}_{3,2}\mid a<b\}$
$\displaystyle\widehat{T}_{4,3}^{r}$	$\displaystyle\coloneqq\{[a,b,t,t]\in\widehat{T}_{4,3}\mid a>b\},$	$\displaystyle\widehat{T}_{4,3}^{l}$	$\displaystyle\coloneqq\{[a,b,t,t]\in\widehat{T}_{4,3}\mid a<b\}$
$\displaystyle\widehat{T}_{2,4}^{r}$	$\displaystyle\coloneqq\{[b,t,a,t]\in\widehat{T}_{2,4}\mid a>b\},$	$\displaystyle\widehat{T}_{2,4}^{l}$	$\displaystyle\coloneqq\{[b,t,a,t]\in\widehat{T}_{2,4}\mid a<b\}.$

For $i,j=2,3,4$ with $i\neq j$ , the set $\widehat{T}_{i,j}\setminus(\widehat{T}_{i,j}^{r}\cup\widehat{T}_{i,j}^{l})$ is the interior of $E_{i,j}$ .

Definition 13.10.

For $i,j=2,3,4$ with $i\neq j$ , we set

\displaystyle T_{i,j}\coloneqq\pi(\widehat{T}_{i,j}^{r})=\pi(\widehat{T}_{i,j}^{l}).

Note that $\pi$ induces a homeomorphism from $\widehat{T}_{i,j}^{r}$ (or $\widehat{T}_{i,j}^{l}$ ) to $T_{i,j}$ . Hence $T_{i,j}$ is homeomorphic to $\mathbb{R}^{2}$ .

The space $O$ is a disjoint union (as a set) of

O_{0},\ T_{2,3},\ T_{3,4},\ T_{4,2},\ R^{-},\ T_{3,2},\ T_{4,3},\ T_{2,4},\ R^{+}.

We use these spaces to compute the K-groups of $I=M_{4}(C_{0}(\widetilde{O}))^{\beta}$ .

14. K-groups of the ideal $I$

Definition 14.1.

We set $I_{0}\coloneqq M_{4}\big{(}C_{0}(\pi^{-1}(O_{0}))\big{)}^{\beta}$ and $I^{\star}\coloneqq M_{4}\big{(}C_{0}(\pi^{-1}(O\setminus O_{0}))\big{)}^{\beta}$ .

We have a short exact sequence

0\longrightarrow I_{0}\longrightarrow I\longrightarrow I^{\star}\longrightarrow 0.

We have $I_{0}\cong M_{4}(C_{0}(V))\cong M_{4}(C_{0}(O_{0}))\cong M_{4}(C_{0}(\mathbb{R}^{3}))$ .

Definition 14.2.

We set $T\coloneqq T_{2,3}\cup T_{3,4}\cup T_{4,2}\cup T_{3,2}\cup T_{4,3}\cup T_{2,4}$ and $R\coloneqq R^{-}\cup R^{+}$ . We set $I^{\circ}\coloneqq M_{4}\big{(}C_{0}(\pi^{-1}(T))\big{)}^{\beta}$ and $I^{\text{\textbullet}}\coloneqq M_{4}\big{(}C_{0}(\pi^{-1}(R))\big{)}^{\beta}$ .

We have $I^{\circ}\cong M_{4}(C_{0}(T))\cong\bigoplus_{i,j}M_{4}(C_{0}(T_{i,j}))$ and

I^{\text{\textbullet}}\cong M_{4}(C_{0}(R))\cong M_{4}(C_{0}(R^{-}))\oplus M_{4}(C_{0}(R^{+})).

We have a short exact sequence

0\longrightarrow I^{\circ}\longrightarrow I^{\star}\longrightarrow I^{\text{\textbullet}}\longrightarrow 0.

This induces a six-term exact sequence

We set $r^{-}\in K_{1}\big{(}M_{4}(C_{0}(R^{-}))\big{)}$ and $r^{+}\in K_{1}\big{(}M_{4}(C_{0}(R^{+}))\big{)}$ to be the images of $v_{(1234)}\in K_{0}(B_{(234)})\subset K_{0}(B^{\text{\textbullet}})$ under the exponential maps coming from the exact sequences

0\longrightarrow M_{4}(C_{0}(R^{\pm}))\longrightarrow M_{4}\big{(}C_{0}(\pi^{-1}(R^{\pm}\cup\{x_{(234)}\}))\big{)}^{\beta}\longrightarrow B_{(234)}\longrightarrow 0.

Then similarly as the proof of Lemma 12.4, we see that $r^{-}$ and $r^{+}$ are the generators of $K_{1}\big{(}M_{4}(C_{0}(R^{-}))\big{)}\cong\mathbb{Z}$ and $K_{1}\big{(}M_{4}(C_{0}(R^{+}))\big{)}\cong\mathbb{Z}$ , respectively.

Let $\omega=(1342)\in\mathfrak{S}_{4}$ . For $i=2,3,4$ , we set $w_{i,\omega(i)}\in K_{0}\big{(}M_{4}(C_{0}(T_{i,\omega(i)}))\big{)}$ to be the image of the generator $r^{-}$ of $K_{1}\big{(}M_{4}(C_{0}(R^{-}))\big{)}$ under the index map coming from the exact sequences

0\longrightarrow M_{4}(C_{0}(T_{i,\omega(i)}))\longrightarrow M_{4}\big{(}C_{0}(\pi^{-1}(T_{i,\omega(i)}\cup R^{-}))\big{)}^{\beta}\longrightarrow M_{4}(C_{0}(R^{-}))\longrightarrow 0.

Since

M_{4}\big{(}C_{0}(\pi^{-1}(T_{2,3}\cup R^{-}))\big{)}^{\beta}\cong M_{4}\big{(}C_{0}(\widehat{T}_{2,3}^{r}\cup R_{y}^{-})\big{)}\cong M_{4}\big{(}C_{0}((0,1)\times(0,1])\big{)}

whose K-groups are $0$ , $w_{2,3}$ is a generator of $K_{0}\big{(}M_{4}(C_{0}(T_{2,3}))\big{)}\cong\mathbb{Z}$ . Similarly, $w_{3,4}$ and $w_{4,2}$ are generators of $K_{0}\big{(}M_{4}(C_{0}(T_{3,4}))\big{)}\cong\mathbb{Z}$ and $K_{0}\big{(}M_{4}(C_{0}(T_{4,2}))\big{)}\cong\mathbb{Z}$ , respectively.

Similarly for $i=2,3,4$ , we set the generator $w_{\omega(i),i}$ of $K_{0}\big{(}M_{4}(C_{0}(T_{\omega(i),i}))\big{)}\cong\mathbb{Z}$ to be the image of the generator $r^{+}$ of $K_{1}\big{(}M_{4}(C_{0}(R^{+}))\big{)}$ under the index map coming from the exact sequences

0\longrightarrow M_{4}(C_{0}(T_{\omega(i)},i))\longrightarrow M_{4}\big{(}C_{0}(\pi^{-1}(T_{\omega(i),i}\cup R^{+}))\big{)}^{\beta}\longrightarrow M_{4}(C_{0}(R^{+}))\longrightarrow 0.

Then the index map from

K_{1}(I^{\text{\textbullet}})\cong K_{1}\big{(}M_{4}(C_{0}(R^{-}))\big{)}\oplus K_{1}\big{(}M_{4}(C_{0}(R^{+}))\big{)}\cong\mathbb{Z}^{2}

to

	$\displaystyle K_{0}(I^{\circ})\cong K_{0}\big{(}$	$\displaystyle M_{4}(C_{0}(T_{2,3}))\big{)}\oplus K_{0}\big{(}M_{4}(C_{0}(T_{3,4}))\big{)}\oplus K_{0}\big{(}M_{4}(C_{0}(T_{4,2}))\big{)}$
		$\displaystyle\oplus K_{0}\big{(}M_{4}(C_{0}(T_{3,2}))\big{)}\oplus K_{0}\big{(}M_{4}(C_{0}(T_{4,3}))\big{)}\oplus K_{0}\big{(}M_{4}(C_{0}(T_{2,4}))\big{)}\cong\mathbb{Z}^{6}$

becomes $\mathbb{Z}^{2}\ni(a,b)\mapsto(a,a,a,b,b,b)\in\mathbb{Z}^{6}$ . Thus we have the following.

Proposition 14.3.

We have $K_{0}(I^{\star})\cong\mathbb{Z}^{4}$ and $K_{1}(I^{\star})=0$ .

We denote by $s_{1},s_{2},s_{3},s_{4}\in K_{0}(I^{\star})$ the images of $w_{2,3},w_{3,4},w_{3,2},w_{4,3}\in K_{0}(I^{\circ})$ . Then $\{s_{1},s_{2},s_{3},s_{4}\}$ becomes a basis of $K_{0}(I^{\star})\cong\mathbb{Z}^{4}$ . Note that the images of $w_{4,2},w_{2,4}\in K_{0}(I^{\circ})$ are $-s_{1}-s_{2}\in K_{0}(I^{\star})$ and $-s_{3}-s_{4}\in K_{0}(I^{\star})$ , respectively.

We have a six-term exact sequence

(14.1)

To compute the index map $K_{0}(I^{\star})\to K_{1}(I_{0})$ , we need the following lemma.

Lemma 14.4.

The index map from $K_{0}(I^{\circ})\cong\mathbb{Z}^{6}$ to $K_{1}(I_{0})\cong\mathbb{Z}$ coming from the short exact sequence

0\longrightarrow I_{0}\longrightarrow M_{4}\big{(}C_{0}(\pi^{-1}(O_{0}\cup T))\big{)}^{\beta}\longrightarrow I^{\circ}\longrightarrow 0.

is $0$ .

Proof.

We set $\widehat{T}\coloneqq\bigcup_{i,j}(\widehat{T}_{i,j}^{r}\cup\widehat{T}_{i,j}^{l})$ where $i,j$ run $2,3,4$ with $i\neq j$ . We have the following commutative diagram with exact rows;

Note that $V\cup\widehat{T}=\pi^{-1}(O_{0}\cup T)\cap\overline{V}$ . From this diagram, we see that the index map $K_{0}(I^{\circ})\to K_{1}(I_{0})$ factors through $K_{0}(M_{4}(C_{0}(\widehat{T})))$ .

Take $i,j=2,3,4$ with $i\neq j$ . Let $a_{i,j}^{r}\in K_{0}\big{(}M_{4}(C_{0}(\widehat{T}_{i,j}^{r}))\big{)}$ and $a_{i,j}^{l}\in K_{0}\big{(}M_{4}(C_{0}(\widehat{T}_{i,j}^{l}))\big{)}$ be the images of the generator $w_{i,j}$ of $K_{0}\big{(}M_{4}(C_{0}(T_{i,j}))\big{)}$ under the homomorphism induced by $\pi$ . Under the map $K_{0}(I^{\circ})\to K_{0}(M_{4}(C_{0}(\widehat{T})))$ , the generator $w_{i,j}$ of $K_{0}\big{(}M_{4}(C_{0}(T_{i,j}))\big{)}$ goes to $a_{i,j}^{r}+a_{i,j}^{l}$ . Under the index map $K_{0}(M_{4}(C_{0}(\widehat{T})))\to K_{1}\big{(}M_{4}(C_{0}(V))\big{)}$ the element $a_{i,j}^{r}+a_{i,j}^{l}$ goes to $0$ because the side to $V$ from $\widehat{T}_{i,j}^{r}$ and the one from $\widehat{T}_{i,j}^{l}$ differ if $\widehat{T}_{i,j}^{r}$ and $\widehat{T}_{i,j}^{l}$ are identified through the map $\pi$ to $T_{i,j}$ . Thus we see that the map $K_{0}(I^{\circ})\to K_{1}\big{(}M_{4}(C_{0}(V))\big{)}\cong K_{1}(I_{0})$ is $0$ . ∎

By this lemma, the composition of the map $K_{0}(I^{\circ})\to K_{0}(I^{\star})$ and the index map $K_{0}(I^{\star})\to K_{1}(I_{0})$ is $0$ . Since the map $\mathbb{Z}^{6}\cong K_{0}(I^{\circ})\to K_{0}(I^{\star})\cong\mathbb{Z}^{4}$ is a surjection, we see that the index map $K_{0}(I^{\star})\to K_{1}(I_{0})$ is $0$ . Thus we have the following.

Proposition 14.5.

We have $K_{0}(I)\cong K_{0}(I^{\star})\cong\mathbb{Z}^{4}$ and $K_{1}(I)\cong K_{1}(I_{0})\cong\mathbb{Z}$ .

15. K-groups of $A$

Recall the six-term exact sequence

In this section, we calculate the exponential map $\delta_{0}\colon K_{0}(B)\to K_{1}(I)$ and the index map $\delta_{1}\colon K_{1}(B)\to K_{0}(I)$ .

Proposition 15.1.

The exponential map $\delta_{0}\colon K_{0}(B)\to K_{1}(I)$ is $0$ .

Proof.

Since $K_{0}(B)$ is generated by 16 elements $\{[P_{i,j}]_{0}\}_{i,j=1}^{4}$ , the map $K_{0}(A)\to K_{0}(B)$ is surjective. Hence the exponential map $\delta_{0}\colon K_{0}(B)\to K_{1}(I)$ is $0$ . ∎

By the definitions of the generators of $K$ -groups we did so far, we have the following. (see Figure 13.2 for the relation between $T$ and $F$ .)

Proposition 15.2.

The index map $\delta^{\prime\prime}\colon K_{1}(B^{\circ})\cong\mathbb{Z}^{18}\to K_{0}(I^{\circ})\cong\mathbb{Z}^{6}$ coming from the short exact sequence

0\longrightarrow I^{\circ}\longrightarrow M_{4}\big{(}C_{0}(\pi^{-1}(T\cup F^{\circ}))\big{)}^{\beta}\longrightarrow B^{\circ}\longrightarrow 0.

is as Table 15.1.

Table 15.1. Computation of the index map

\delta^{\prime\prime}

	2,2		3,3		4,4		2,3		3,4		4,2		3,2		4,3		2,4
$w\diagdown v$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$
2,3	0	0	0	0	$-1$	$-1$	1	1	0	0	0	0	0	0	0	0	0	0
3,4	$-1$	$-1$	0	0	0	0	0	0	1	1	0	0	0	0	0	0	0	0
4,2	0	0	$-1$	$-1$	0	0	0	0	0	0	1	1	0	0	0	0	0	0
3,2	0	0	0	0	$-1$	$-1$	0	0	0	0	0	0	1	1	0	0	0	0
4,3	$-1$	$-1$	0	0	0	0	0	0	0	0	0	0	0	0	1	1	0	0
2,4	0	0	$-1$	$-1$	0	0	0	0	0	0	0	0	0	0	0	0	1	1

Definition 15.3.

The composition of the index map $\delta^{\prime\prime}\colon K_{1}(B^{\circ})\to K_{0}(I^{\circ})$ and the map $K_{0}(I^{\circ})\to K_{0}(I^{\star})$ is denoted by $\eta\colon K_{1}(B^{\circ})\to K_{0}(I^{\star})$

We set $\widetilde{\eta}\colon K_{1}(B^{\circ})\to K_{0}(I^{\star})\oplus\mathbb{Z}/2\mathbb{Z}$ by $\widetilde{\eta}(w_{i,j}^{\cap})=(\eta(w_{i,j}^{\cap}),0)$ and $\widetilde{\eta}(w_{i,j}^{\cup})=(\eta(w_{i,j}^{\cup}),1)$ for $i,j=2,3,4$ .

We denote the generator of $\mathbb{Z}/2\mathbb{Z}$ in $K_{0}(I^{\star})\oplus\mathbb{Z}/2\mathbb{Z}$ by $s_{5}$ .

Table 15.2. Computation of

\widetilde{\eta}

	2,2		3,3		4,4		2,3		3,4		4,2		3,2		4,3		2,4
$s\diagdown v$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$	$\cap$	$\cup$
1	0	0	1	1	$-1$	$-1$	1	1	0	0	$-1$	$-1$	0	0	0	0	0	0
2	$-1$	$-1$	1	1	0	0	0	0	1	1	$-1$	$-1$	0	0	0	0	0	0
3	0	0	1	1	$-1$	$-1$	0	0	0	0	0	0	1	1	0	0	$-1$	$-1$
4	$-1$	$-1$	1	1	0	0	0	0	0	0	0	0	0	0	1	1	$-1$	$-1$
5	0	1	0	1	0	1	0	1	0	1	0	1	0	1	0	1	0	1

Proposition 15.4.

The map $\widetilde{\eta}\colon K_{1}(B^{\circ})\to K_{0}(I^{\star})\oplus\mathbb{Z}/2\mathbb{Z}$ is surjective, and its kernel coincides with the image of $\delta\colon K_{0}(B^{\text{\textbullet}})\to K_{1}(B^{\circ})$ .

Proof.

Since

\displaystyle\widetilde{\eta}(w_{2,3}^{\cap})=s_{1},\quad\widetilde{\eta}(w_{3,4}^{\cap})=s_{2},\quad\widetilde{\eta}(w_{3,2}^{\cap})=s_{3},\quad\widetilde{\eta}(w_{4,3}^{\cap})=s_{4},

$s_{1},s_{2},s_{3},s_{4}$ are in the image of $\widetilde{\eta}$ . Since $\widetilde{\eta}(w_{2,2}^{\cup}+w_{3,3}^{\cup}+w_{4,4}^{\cup})=s_{5}$ , $s_{5}$ is also in the image of $\widetilde{\eta}$ . Thus $\widetilde{\eta}$ is surjective.

It is straightforward to check $\widetilde{\eta}\circ\delta=0$ Hence the image of $\delta$ is contained in the kernel of $\widetilde{\eta}$ . Suppose

x=\sum_{i,j=2}^{4}n_{i,j}^{\cap}w_{i,j}^{\cap}+\sum_{i,j=2}^{4}n_{i,j}^{\cup}w_{i,j}^{\cup}

is in the kernel of $\widetilde{\eta}$ where $n_{i,j}^{\cap},n_{i,j}^{\cup}\in\mathbb{Z}$ for $i,j=2,3,4$ . We will show that $x$ is in the image of $\delta$ . By adding

n_{2,3}^{\cup}\delta(q_{(3142)})+n_{3,4}^{\cup}\delta(q_{(4312)})+n_{4,2}^{\cup}\delta(q_{(2341)})+n_{3,2}^{\cup}\delta(q_{(2413)})+n_{4,3}^{\cup}\delta(q_{(3421)})+n_{2,4}^{\cup}\delta(q_{(4123)})

we may assume

n_{2,3}^{\cup}=n_{3,4}^{\cup}=n_{4,2}^{\cup}=n_{3,2}^{\cup}=n_{4,3}^{\cup}=n_{2,4}^{\cup}=0

without loss of generality. By subtracting $n_{3,3}^{\cup}\delta(q_{(4321)})+n_{4,4}^{\cup}\delta(q_{(3412)})$ , we may further assume $n_{3,3}^{\cup}=n_{4,4}^{\cup}=0$ without loss of generality. Then $n_{2,2}^{\cup}$ is even since the coefficient of $c_{5}$ for $\widetilde{\eta}(x)$ is $0$ . Hence by adding

\frac{n_{2,2}^{\cup}}{2}\big{(}\delta(q_{(2143)})-\delta(q_{(3412)})-\delta(q_{(4321)})\big{)}

we may further assume $n_{2,2}^{\cup}=0$ without loss of generality. Thus we may assume $x=\sum_{i,j=2}^{4}n_{i,j}^{\cap}w_{i,j}^{\cap}$ . By adding $n_{2,2}^{\cap}\delta(q_{(1243)})+n_{3,3}^{\cap}\delta(q_{(1432)})+n_{4,4}^{\cap}\delta(q_{(1324)})$ , we may further assume $n_{2,2}^{\cap}=n_{3,3}^{\cap}=n_{4,4}^{\cap}=0$ without loss of generality. By subtracting $n_{4,2}^{\cap}\delta(q_{(1423)})+n_{2,4}^{\cap}\delta(q_{(1342)})$ , we may further assume $n_{4,2}^{\cap}=n_{2,4}^{\cap}=0$ without loss of generality. Thus we may assume

x=n_{2,3}^{\cap}w_{2,3}^{\cap}+n_{3,4}^{\cap}w_{3,4}^{\cap}+n_{3,2}^{\cap}w_{3,2}^{\cap}+n_{4,3}^{\cap}w_{4,3}^{\cap}.

Then we have $n_{2,3}^{\cap}=n_{3,4}^{\cap}=n_{3,2}^{\cap}=n_{4,3}^{\cap}=0$ because

\widetilde{\eta}(x)=n_{2,3}^{\cap}s_{1}+n_{3,4}^{\cap}s_{2}+n_{3,2}^{\cap}s_{3}+n_{4,3}^{\cap}s_{4}.

Thus $x=0$ . We have shown that $x$ is in the image of $\delta$ . Hence the image of $\delta$ coincides with the kernel of $\widetilde{\eta}$ . ∎

As a corollary of this proposition, we have the following as predicted.

Proposition 15.5.

We have $K_{0}(B)\cong\mathbb{Z}^{10}$ and $K_{1}(B)\cong\mathbb{Z}^{4}\oplus\mathbb{Z}/2\mathbb{Z}$ .

Proof.

By Proposition 15.4, we see that $K_{1}(B)\cong\operatorname{coker}\delta$ is isomorphic to $\mathbb{Z}^{4}\oplus\mathbb{Z}/2\mathbb{Z}$ . This implies $K_{0}(B)\cong\ker\delta$ is isomorphic to $\mathbb{Z}^{10}$ because $\ker\delta$ is a free abelian group with dimension $24-18+4=10$ . ∎

We also have the following.

Proposition 15.6.

The index map $\delta_{1}\colon K_{1}(B)\to K_{0}(I)$ is as $K_{1}(B)\cong\mathbb{Z}^{4}\oplus\mathbb{Z}/2\mathbb{Z}\ni(n,m)\mapsto n\in\mathbb{Z}^{4}\cong K_{0}(I)$ .

Proof.

From the commutative diagram with exact rows

the index map $\delta_{1}\colon K_{1}(B)\to K_{0}(I)$ coincides with the map $K_{1}(B)\to K_{0}(I^{\star})$ if we identify $K_{0}(I)\cong K_{0}(I^{\star})$ as we did in Proposition 14.5.

From the commutative diagram with exact rows

we have the commutative diagram

From this diagram, we see that the map $K_{1}(B)\to K_{0}(I^{\star})$ is as $K_{1}(B)\cong\mathbb{Z}^{4}\oplus\mathbb{Z}/2\mathbb{Z}\ni(n,m)\mapsto n\in\mathbb{Z}^{4}\cong K_{0}(I^{\star})$ . This completes the proof. ∎

Definition 15.7.

Define a unitary $w\in C(S^{3},M_{2}(\mathbb{C}))$ by

	$\displaystyle w(a_{1},a_{2},a_{3},a_{4})$	$\displaystyle=a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}+a_{4}c_{4}$
		$\displaystyle=\begin{pmatrix}a_{1}+a_{2}\sqrt{-1}&a_{3}+a_{4}\sqrt{-1}\\ -a_{3}+a_{4}\sqrt{-1}&a_{1}-a_{2}\sqrt{-1}\end{pmatrix}$

for $(a_{1},a_{2},a_{3},a_{4})\in S^{3}$ .

Then $[w]_{1}$ is the generator of $K_{1}\big{(}C(S^{3},M_{2}(\mathbb{C}))\big{)}\cong K_{1}\big{(}M_{4}(C(S^{3}))\big{)}\cong\mathbb{Z}$ .

Let $\varphi\colon A\to M_{4}(C(S^{3}))$ be the composition of the embedding $A\to M_{4}(C(\mathbb{R}P^{3}))$ and the map $M_{4}(C(\mathbb{R}P^{3}))\to M_{4}(C(S^{3}))$ induced by $[\cdot]\colon S^{3}\to\mathbb{R}P^{3}$ . Let $\widetilde{\pi}\colon S^{3}\to X$ be the composition of $[\cdot]\colon S^{3}\to\mathbb{R}P^{3}$ and $\pi\colon\mathbb{R}P^{3}\to X$ . We set $V^{\prime}$ of $S^{3}$ by

V^{\prime}\coloneqq\{(a_{1},a_{2},a_{3},a_{4})\in S^{3}\mid a_{1},a_{2},a_{3}>|a_{4}|\}.

Then $V^{\prime}$ is homeomorphic to $V$ via $[\cdot]$ , and hence to $O_{0}$ via $\widetilde{\pi}$ . Note that the map $M_{4}(C_{0}(V^{\prime}))\hookrightarrow M_{4}(C(S^{3}))$ induces the isomorphism

K_{1}\big{(}M_{4}(C_{0}(V^{\prime}))\big{)}\to K_{1}\big{(}M_{4}(C(S^{3}))\big{)}.

Since $I_{0}\cong M_{4}(C_{0}(O_{0}))\cong M_{4}(C_{0}(V^{\prime}))$ canonically, we set a generator $y$ of $K_{1}(I_{0})$ which corresponds to the generator $[w]_{1}$ of $K_{1}\big{(}M_{4}(C(S^{3}))\big{)}$ via the isomorphism $K_{1}\big{(}M_{4}(C_{0}(V^{\prime}))\big{)}\to K_{1}\big{(}M_{4}(C(S^{3}))\big{)}$ . We denote by the same symbol $y$ the generator of $K_{1}(I)\cong K_{1}(I_{0})$ corresponding $y\in K_{1}(I_{0})$ .

Proposition 15.8.

The image of $y\in K_{1}(I)$ under the map $K_{1}(I)\to K_{1}(A)\to K_{1}\big{(}M_{4}(C(S^{3}))\big{)}$ is $32[w]_{1}$ .

Proof.

The map $I_{0}\to I\to A\to M_{4}(C(S^{3}))$ is induced by $\widetilde{\pi}\colon\widetilde{\pi}^{-1}(O_{0})\to O_{0}$ when we identify $I_{0}$ with $M_{4}(C_{0}(O_{0}))$ . We have

\widetilde{\pi}^{-1}(O_{0})=\coprod_{i,j=1}^{4}\sigma^{\prime}_{i,j,+}(V^{\prime})\amalg\coprod_{i,j=1}^{4}\sigma^{\prime}_{i,j,-}(V^{\prime})

where $\sigma^{\prime}_{i,j,\pm}\colon S^{3}\to S^{3}$ is induced by the unitary $\pm U_{i,j}$ similarly as $\sigma_{i,j}\colon\mathbb{R}P^{3}\to\mathbb{R}P^{3}$ for $i,j=1,2,3,4$ . These $32$ homeomorphisms preserve the orientation of $S^{3}$ . Therefore, the image of $y\in K_{1}(I_{0})$ , and hence the one of $y\in K_{1}(I)$ , in $K_{1}\big{(}M_{4}(C(S^{3}))\big{)}$ is $32[w]_{1}$ . ∎

Definition 15.9.

Define the linear map $\xi:M_{2}(\mathbb{C})\to\mathbb{C}^{4}$ by

\displaystyle\xi\left(\begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}\right)=\frac{1}{\sqrt{2}}(a_{11},a_{12},a_{21},a_{22}).

Definition 15.10.

Define unital $*$ -homomorphisms $\iota,\iota^{\prime}\colon M_{2}(\mathbb{C})\to M_{4}(\mathbb{C})$ by

	$\displaystyle\iota\left(\begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}\right)=\begin{pmatrix}a_{11}&a_{12}&0&0\\ a_{21}&a_{22}&0&0\\ 0&0&a_{11}&a_{21}\\ 0&0&a_{21}&a_{22}\end{pmatrix},$
	$\displaystyle\iota^{\prime}\left(\begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}\right)=\begin{pmatrix}a_{11}&0&a_{12}&0\\ 0&a_{11}&0&a_{12}\\ a_{21}&0&a_{22}&0\\ 0&a_{21}&0&a_{22}\end{pmatrix}.$

Lemma 15.11.

For each $M,N\in M_{2}(\mathbb{C})$ , we have

\displaystyle\xi(M)\iota(N)

\displaystyle=\xi(MN),

\displaystyle\iota^{\prime}(M)\xi(N)^{\mathrm{T}}

\displaystyle=\xi(MN)^{\mathrm{T}}.

Proof.

It follows from a direct computation. ∎

Definition 15.12.

Define $U\in M_{4}(A)$ by

U=\begin{pmatrix}P_{11}&P_{12}&P_{13}&P_{14}\\ P_{21}&P_{22}&P_{23}&P_{24}\\ P_{31}&P_{32}&P_{33}&P_{34}\\ P_{41}&P_{42}&P_{43}&P_{44}\end{pmatrix}.

It can be easily checked that $U$ is a unitary.

Proposition 15.13.

The image of $[U]_{1}\in K_{1}(A)$ under the map $K_{1}(A)\to K_{1}\big{(}M_{4}(C(S^{3}))\big{)}$ is $16[w]_{1}$ .

Proof.

Let $\varphi_{4}\colon M_{4}(A)\to M_{4}\big{(}M_{4}(C(S^{3}))\big{)}$ be the $*$ -homomorphism induced by $\varphi$ . Set $\mathbb{U}\coloneqq\varphi_{4}(U)$ . For $i,j=1,2,3,4$ , the $(i,j)$ -entry $\mathbb{U}_{i,j}\in C(S^{3},M_{4}(\mathbb{C}))$ of $\mathbb{U}$ is given by

\mathbb{U}_{i,j}(a_{1},a_{2},a_{3},a_{4})=U_{i,j}(a_{1},a_{2},a_{3},a_{4})^{\mathrm{T}}(a_{1},a_{2},a_{3},a_{4})U_{i,j}^{*}

for each $(a_{1},a_{2},a_{3},a_{4})\in S^{3}$ .

Let $W\in M_{4}(\mathbb{C})$ be

W=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-\sqrt{-1}&0&0\\ 0&0&1&-\sqrt{-1}\\ 0&0&-1&-\sqrt{-1}\\ 1&\sqrt{-1}&0&0\end{pmatrix}.

Then $W$ is a unitary.

Take $(a_{1},a_{2},a_{3},a_{4})\in S^{3}$ and $i,j=1,2,3,4$ . We set $(b_{1},b_{2},b_{3},b_{4})=(a_{1},a_{2},a_{3},a_{4})U_{i,j}^{*}$ . By Proposition 5.2, we have $\sum_{k=1}^{4}b_{k}c_{k}=c_{i}\Big{(}\sum_{k=1}^{4}a_{k}c_{k}\Big{)}c_{j}^{*}$ . We also have

	$\displaystyle\xi\Big{(}\sum_{k=1}^{4}b_{k}c_{k}\Big{)}W$	$\displaystyle=\frac{1}{\sqrt{2}}(b_{1}+b_{2}\sqrt{-1},b_{3}+b_{4}\sqrt{-1},-b_{3}+b_{4}\sqrt{-1},b_{1}-b_{2}\sqrt{-1})W$
		$\displaystyle=(b_{1},b_{2},b_{3},b_{4})$

Hence we get

	$\displaystyle(a_{1},a_{2},a_{3},a_{4})U_{i,j}^{*}$	$\displaystyle=\xi\left(c_{i}\Big{(}\sum_{k=1}^{4}a_{k}c_{k}\Big{)}c_{j}^{*}\right)W$
		$\displaystyle=\xi(c_{i})\iota\left(\Big{(}\sum_{k=1}^{4}a_{k}c_{k}\Big{)}c_{j}^{*}\right)W$
		$\displaystyle=\xi(c_{i})\iota(w(a_{1},a_{2},a_{3},a_{4}))\iota(c_{j}^{*})W$

by Lemma 15.11. Similarly, we get

	$\displaystyle U_{i,j}(a_{1},a_{2},a_{3},a_{4})^{\mathrm{T}}$	$\displaystyle=W^{\mathrm{T}}\xi\left(c_{i}\Big{(}\sum_{k=1}^{4}a_{k}c_{k}\Big{)}c_{j}^{*}\right)^{\mathrm{T}}$
		$\displaystyle=W^{\mathrm{T}}\iota^{\prime}\left(c_{i}\Big{(}\sum_{k=1}^{4}a_{k}c_{k}\Big{)}\right)\xi(c_{j}^{*})^{\mathrm{T}}$
		$\displaystyle=W^{\mathrm{T}}\iota^{\prime}(c_{i})\iota^{\prime}(w(a_{1},a_{2},a_{3},a_{4}))\xi(c_{j}^{*})^{\mathrm{T}}$

by Lemma 15.11. Define $\mathbb{V},\mathbb{W},\mathbb{W}^{\prime}\in M_{4}(M_{4}(\mathbb{C}))$ by

	$\displaystyle\mathbb{V}=(\xi(c_{j}^{*})^{\mathrm{T}}\xi(c_{i}))_{i,j=1}^{4},$
	$\displaystyle\mathbb{W}=\begin{pmatrix}\iota(c_{1}^{})W&0&0&0\\ 0&\iota(c_{2}^{})W&0&0\\ 0&0&\iota(c_{3}^{})W&0\\ 0&0&0&\iota(c_{4}^{})W\end{pmatrix},$
	$\displaystyle\mathbb{W}^{\prime}=\begin{pmatrix}W^{\mathrm{T}}\iota^{\prime}(c_{1})&0&0&0\\ 0&W^{\mathrm{T}}\iota^{\prime}(c_{2})&0&0\\ 0&0&W^{\mathrm{T}}\iota^{\prime}(c_{3})&0\\ 0&0&0&W^{\mathrm{T}}\iota^{\prime}(c_{4})\end{pmatrix}.$

One can check that these are unitaries. If we consider these as constant functions in $M_{4}\big{(}C(S^{3},M_{4}(\mathbb{C}))\big{)}$ , we have

\mathbb{U}=\mathbb{W}^{\prime}\iota^{\prime}_{4}(w)\mathbb{V}\iota_{4}(w)\mathbb{W},

where $\iota_{4}(w),\iota^{\prime}_{4}(w)\in M_{4}(C(S^{3},M_{4}(\mathbb{C}))$ are defined as

	$\displaystyle\iota_{4}(w)=\begin{pmatrix}\iota(w(\cdot))&0&0&0\\ 0&\iota(w(\cdot))&0&0\\ 0&0&\iota(w(\cdot))&0\\ 0&0&0&\iota(w(\cdot))\end{pmatrix},$
	$\displaystyle\iota^{\prime}_{4}(w)=\begin{pmatrix}\iota^{\prime}(w(\cdot))&0&0&0\\ 0&\iota^{\prime}(w(\cdot))&0&0\\ 0&0&\iota^{\prime}(w(\cdot))&0\\ 0&0&0&\iota^{\prime}(w(\cdot))\end{pmatrix}.$

Since $[\iota_{4}(w)]_{1}=[\iota^{\prime}_{4}(w)]_{1}=8[w]_{1}$ , we obtain $[\mathbb{U}]_{1}=16[w]_{1}$ . ∎

Proposition 15.14.

We have $K_{0}(A)\cong\mathbb{Z}^{10}$ and $K_{1}(A)\cong\mathbb{Z}$ . More specifically, $K_{0}(A)$ is generated by $\{[P_{i,j}]_{0}\}_{i,j=1}^{4}$ , and $K_{1}(A)$ is generated by $[U]_{1}$ . Moreover, the positive cone $K_{0}(A)_{+}$ of $K_{0}(A)$ is generated by $\{[P_{i,j}]_{0}\}_{i,j=1}^{4}$ as a monoid.

Proof.

We have already seen that $K_{0}(A)\to K_{0}(B)$ is isomorphic, and we have a short exact sequence

0\longrightarrow K_{1}(I)\longrightarrow K_{1}(A)\longrightarrow\mathbb{Z}/2\mathbb{Z}\longrightarrow 0.

From this, we see that $K_{1}(A)$ is isomorphic to either $\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}$ . If $K_{1}(A)$ is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ , one can choose an isomorphism so that $y\in K_{1}(I)$ goes to $(1,0)\in\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ . Then the image of the map $K_{1}(A)\to K_{1}\big{(}M_{4}(C(S^{3}))\big{)}\cong\mathbb{Z}$ is $32\mathbb{Z}$ by Proposition 15.8. This is a contradiction because the image of $[U]_{1}\in K_{1}(A)$ is $16$ by Proposition 15.13. Hence $K_{1}(A)$ is isomorphic to $\mathbb{Z}$ so that $y\in K_{1}(I)$ goes to $2$ . By Proposition 15.8 and Proposition 15.13, $[U]_{1}\in K_{1}(A)$ corresponds to $1\in\mathbb{Z}$ . Thus $[U]_{1}$ is a generator of $K_{1}(A)\cong\mathbb{Z}$ .

It is clear that the monoid generated by $\{[P_{i,j}]_{0}\}_{i,j=1}^{4}$ is contained in the positive cone $K_{0}(A)_{+}$ . The positive cone $K_{0}(A)_{+}$ maps into the positive cone $K_{0}(B^{\text{\textbullet}})_{+}$ under the surjection $A\to B^{\text{\textbullet}}$ . Hence by Proposition 12.7, $K_{0}(A)_{+}$ is contained in the monoid generated by $\{[P_{i,j}]_{0}\}_{i,j=1}^{4}$ . Thus $K_{0}(A)_{+}$ is the monoid generated by $\{[P_{i,j}]_{0}\}_{i,j=1}^{4}$ . ∎

Definition 15.15.

Define $u\in M_{4}(A(4))$ by

u=\begin{pmatrix}p_{11}&p_{12}&p_{13}&p_{14}\\ p_{21}&p_{22}&p_{23}&p_{24}\\ p_{31}&p_{32}&p_{33}&p_{34}\\ p_{41}&p_{42}&p_{43}&p_{44}\end{pmatrix}.

It can be easily checked that $u$ is a unitary. This unitary $u$ is called the defining unitary of the magic square C*-algebra $A(4)$ .

By Proposition 15.14, we get the third main theorem.

Theorem 15.16.

We have $K_{0}(A(4))\cong\mathbb{Z}^{10}$ and $K_{1}(A(4))\cong\mathbb{Z}$ . More specifically, $K_{0}(A(4))$ is generated by $\{[p_{i,j}]_{0}\}_{i,j=1}^{4}$ , and $K_{1}(A(4))$ is generated by $[u]_{1}$ .

The positive cone $K_{0}(A(4))_{+}$ of $K_{0}(A(4))$ is generated by $\{[p_{i,j}]_{0}\}_{i,j=1}^{4}$ as a monoid.

As mentioned in the introduction, the computation $K_{0}(A(4))\cong\mathbb{Z}^{10}$ and $K_{1}(A(4))\cong\mathbb{Z}$ and that $K_{0}(A(4))$ is generated by $\{[p_{i,j}]_{0}\}_{i,j=1}^{4}$ were already obtained by Voigt in [V]. We give totally different proofs of these facts. That $K_{1}(A(4))$ is generated by $[u]_{1}$ and the computation of the positive cone $K_{0}(A(4))_{+}$ of $K_{0}(A(4))$ are new.

On the magic square C*-algebra of size 4

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

0. Introduction

1. Definitions of and basic facts on magic square C*-algebras

Definition 1.1.

Remark 1.2.

Definition 1.3.

Definition 1.4.

Lemma 1.5.

Proposition 1.6.

Proof.

Proposition 1.7.

Proof.

Corollary 1.8.

2. General results on magic square C*-algebras

Proposition 2.1.

Proof.

Proposition 2.2.

Proof.

Corollary 2.3.

Proof.

Corollary 2.4.

Proof.

Proposition 2.5.

Proof.

3. Twisted crossed product

Definition 3.1.

Definition 3.2.

Lemma 3.3.

Lemma 3.4.

Proof.

Definition 3.5.

Theorem 3.6.

4. Real projective space ℝ​P3\mathbb{R}P^{3}

Definition 4.1.

Definition 4.2.

Lemma 4.3.

Proof.

Definition 4.4.

Proposition 4.5.

Proof.

5. Unitaries

Definition 5.1.

Proposition 5.2.

Proof.

Proposition 5.3.

Proof.

6. Projections

Definition 6.1.

Proposition 6.2.

Proof.

Proposition 6.3.

Proof.

Proposition 6.4.

Proof.

7. The inverse map

Definition 7.1.

Definition 7.2.

Lemma 7.3.

Proof.

Proposition 7.4.

Proof.

Proposition 7.5.

Proof.

Proposition 7.6.

Proof.

Proposition 7.7.

Proof.

Proposition 7.8.

Proof.

Corollary 7.9 (cf. [BC, Theorem 4.1]).

Proof.

8. Action

Proposition 8.1.

Proof.

Theorem 8.2.

Proof.

9. Quotient Space ℝ​P3/(K×K)\mathbb{R}P^{3}/(K\times K)

4. Real projective space $\mathbb{R}P^{3}$

9. Quotient Space $\mathbb{R}P^{3}/(K\times K)$

11. The Structure of the Quotient $B$

12. K-groups of the quotient $B$

13. The Structure of the Ideal $I$

14. K-groups of the ideal $I$

15. K-groups of $A$