noncommutative in this case. We prove it by using representations of the ${\varepsilon}$-sphere which factor through the group $C^*$-algebras of certain Coxeter groups (see also [@BanCox; @BanSurv]).

\[DefCox\] Let $\mathbf F_n$ be the free group with $n$ generators $a_1,\ldots,a_n$ and denote by $\mathbf F_n^{\varepsilon}$ the quotient of $\mathbf F_n$ by the relations $a_ia_j=a_ja_i$ if ${\varepsilon}_{ij}=1$. Denote by $\mathbb Z_2^{\varepsilon}$ the quotient of $\mathbf F_n^{\varepsilon}$ by the relations $a_i^2=e$.

By $\mathbb Z_2=\mathbb Z/2\mathbb Z$ we denote the cyclic group of order two. We may view $\mathbb Z_2^{\varepsilon}$ as the quotient of the $n$-fold free product $\mathbb Z_2^{*n}$ by the relations $a_ia_j=a_ja_i$ if ${\varepsilon}_{ij}=1$. It is a Coxeter group with the presentation: $$\mathbb Z_2^{\varepsilon}=\langle a_1,\ldots,a_n\;|\; (a_ia_j)^{m_{ij}}=e\rangle,\qquad 
m_{ij}=\begin{cases} 1 &\textnormal{ if } i=j\\2 &\textnormal{ if } {\varepsilon}_{ij}=1\\\infty &\textnormal{ if }{\varepsilon}_{ij}=0\end{cases}$$

We have:

-   ${ {\mathbb Z} }_2^{\varepsilon}={ {\mathbb Z} }_2\times\ldots\times{ {\mathbb Z} }_2$ for ${\varepsilon}={\varepsilon}_{\operatorname{comm}}$ as in Example \[ExEps\](a)

-   ${ {\mathbb Z} }_2^{\varepsilon}={ {\mathbb Z} }_2*\ldots*{ {\mathbb Z} }_2$ for ${\varepsilon}={\varepsilon}_{\operatorname{free}}$ as in Example \[ExEps\](b)

-   ${ {\mathbb Z} }_2^{\varepsilon}={ {\mathbb Z} }_2^{\times n}*{ {\mathbb Z} }_2^{*m}$ for ${\varepsilon}$ as in Example \[ExEps\](c)

-   ${ {\mathbb Z} }_2^{\varepsilon}=({ {\mathbb Z} }_2\times{ {\mathbb Z} }_2)*({ {\mathbb Z} }_2\times{ {\mathbb Z} }_2)$ for ${\varepsilon}$ 