Proof. Let us prove that (i) implies (ii). It suffices to show that $J^\perp \cap \ell ^\infty (S,\mathbb {R})^n=\{0\}$ . Let $\mathbf {f}\in \ell ^\infty (S,\mathbb {R})^n$ be such that $\langle \mathbf {f}, \mathbf {a}\rangle =0$ for all $\mathbf {a}\in J$ . Then, $\langle \unicode{x3bb} \mathbf {f}, \mathbf {a}\rangle =0$ for every $\unicode{x3bb} \in \mathbb {R}$ and $\mathbf {a}\in J$ , so Proposition 3.3 allows us to conclude that $\pi (\unicode{x3bb} \mathbf {f})\in X_{J}$ for every $\unicode{x3bb} \in \mathbb {R}$ .

By expansivity, there exists an open neighborhood U of $\mathbf {0}\in X_J$ such that $\bigcap _{t\in S}t^{-1}U=\{0\}$ (take, for instance, $U=B(0,\epsilon )$ with $\epsilon>0$ the expansivity constant). Since the shift action $S\curvearrowright \ell ^\infty (S,\mathbb {R})^n$ is by isometries, and the map $\pi \colon \ell ^\infty (S,\mathbb {R}^n)\to (\mathbb {T}^n)^S$ is continuous and S-equivariant, there exists $\unicode{x3bb} _0\in \mathbb {R}$ such that $S\pi (\unicode{x3bb} \mathbf {f})\subseteq U$ for all $0<\unicode{x3bb} <\unicode{x3bb} _0$ . Hence,

$$ \begin{align*}\pi(\unicode{x3bb} \mathbf{f})\in\bigcap_{t\in S}t^{-1}(S\pi(\unicode{x3bb} \mathbf{f}))\subseteq \bigcap_{t\in S}t^{-1}U=\{0\},\end{align*} $$

which means $\pi (\unicode{x3bb} \mathbf {f})=0$ for all $0<\unicode{x3bb} < \unicode{x3bb} _0$ . In other words, $\unicode{x3bb} \mathbf {f}\in \ell ^{\infty }(S,\mathbb {Z})^n$ for all $0<\unicode{x3bb} < \unicode{x3bb} _0$ , so we must have $\mathbf {f}=0$ . Thus, we have property (ii) as desired.

Now, we prove that (ii) implies (iii). Let $K\subseteq S$ be a finite subset satisfying $S=KS$ and consider $\Omega =\{\omega \leq J:\omega \text { is a finitely generated right }\mathbb {Z}[S]\text {-submodule}\}$ ordered by inclusion. Assume that $\omega ^\perp \neq \{0\}$ for all $\omega \in \Omega $ . Then, we can find for each $\omega \in \Omega $ a non-zero element $\mathbf {f}^\omega =(f^\omega _1,\ldots ,f^\omega _n)\in \omega ^\perp $ and, by renormalizing, we may assume that $\|\mathbf {f}^\omega \|_\infty =1$ . Moreover, since $S=KS$ and $s\cdot \mathbf {f}^\omega \in \omega ^{\perp }$ for all $s\in S$ , by replacing $\mathbf {f}^\omega $ with $s\cdot \mathbf {f}^\omega $ for a suitable $s\in S$ , we may assume that $\|\mathbf {f}^\omega \rvert _K\|_\infty \geq 1/2$ . By compactness of $([-1,1]^n)^S$ , we may take a limit point $\mathbf {f}=(f_1,\ldots ,f_n)\in \ell ^\infty (S)^n$ of the net $(\mathbf {f}^\omega )_{\omega \in \Omega }$ , where $\Omega $ is ordered by inclusion. Note that it must be the case that $\|\mathbf {f}\rvert _K\|_\infty \geq 1/2$ , so in particular, $\mathbf {f}\neq 0$ . However, since $\langle \mathbf {f}^\omega , \mathbf {a}\rangle =0$ for all $\mathbf {a}\in \omega $ and $\omega \in \Omega $ , if $\mathbf {a}\in J$ and $\epsilon>0$ , then taking $\omega \in \Omega $ large enough so that $\mathbf {a}\in \omega $ and $\sum _{i=1}^{n}\|(f_i-f^{\omega }_i)\mathbf {1}_{\mathrm {supp}(a_i)}\|_\infty <\epsilon $ , we have

$$ \begin{align*} |\langle\mathbf{f}, \mathbf{a}\rangle|=|\langle\mathbf{f}-\mathbf{f}^\omega, \mathbf{a}\rangle|\leq \sum_{i=1}^n\sum_{t\in \mathrm{supp}(a_i)}|(f_i-f_i^\omega)(t)a_i(t)|\leq \epsilon\|\mathbf{a}\|_1. \end{align*} $$

Thus, $\langle \mathbf {f}, \mathbf {a}\rangle =0$ for all $\mathbf {a}\in J$ , which implies $\mathbf {f}\in J^\perp =\{0\}$ , and this is a contradiction.

Let us see that (iii) implies (i). Since the metric $\rho $ is invariant under translations by elements of $\mathbb {R}/\mathbb {Z}$ , we have that $d(\mathbf {x},\mathbf {y})=d(\mathbf {x}-\mathbf {y},0)$ for all $\mathbf {x},\mathbf {y}\in X_\omega $ , and so it suffices to show that there exists $\epsilon>0$ such that, for an arbitrary non-zero element $\mathbf {x}=(x_1,\ldots ,x_n)$ of $X_\omega $ , $\sup _{s\in S}d(s\cdot \mathbf {x},0)\geq \epsilon $ . Let $\mathbf {x}$ be such an element. There exists a non-zero $\mathbf {f}=(f_1,\ldots ,f_n)\in ([-1/2,1/2]^S)^n$ with $\pi (\mathbf {f})=\mathbf {x}$ and such that

$$ \begin{align*}\|\mathbf{f}\|_\infty=\sup_{s\in S}\max_{1\leq j\leq n}\rho(x_j(s),0_{\mathbb{T}}).\end{align*} $$

Now, since $\omega ^\perp =\{0\}$ and $\mathbf {f}$ is non-zero, we must have that $\langle \mathbf {f}, \mathbf {a}\rangle \neq 0$ for some $\mathbf {a}\in \omega $ . Writing $\omega =\mathbf {A}\mathbb {Z}[S]^k$ for some $k\in \mathbb {N}$ and $\mathbf {A}=(a_{ij})_{i,j}\in \mathrm {M}_{n\times k}(\mathbb {Z}[S])$ , we have $\mathbf {a}=\mathbf {A}\ast \mathbf {b}$ for some $\mathbf {b}=(b_1,\ldots ,b_k)\in \mathbb {Z}[S]^k$ . Therefore,

$$ \begin{align*}0\neq \langle\mathbf{f}, \mathbf{a}\rangle=\langle\mathbf{f}, \mathbf{A}\ast \mathbf{b}\rangle=\bigg{\langle}\mathbf{f},\mathbf{A}\ast \sum_{j=1}^{k}\sum_{t\in S}b_j(t)\delta_t\mathbf{e}_j\bigg{\rangle}=\sum_{j=1}^{k}\sum_{t\in S}b_j(t)\langle\mathbf{f},\mathbf{A}\ast \delta_t\mathbf{e}_j\rangle,\end{align*} $$

so there must exist $t\in S$ and $1\leq j\leq k$ such that $\langle \mathbf {f}, \mathbf {A}\ast \delta _{t}\mathbf {e}_j\rangle \neq 0$ . Since $\mathbf {x}\in X_\omega $ , we have that $\pi (\langle \mathbf {f}, \mathbf {a}\rangle )=\langle \mathbf {x}, \mathbf {a}\rangle =0$ for all $\mathbf {a}\in \omega $ , by Proposition 3.3. Hence, $\langle \mathbf {f},\mathbf {a}\rangle \in \mathbb {Z}$ for every $\mathbf {a}\in \omega $ , and in particular, $\langle \mathbf {f},\mathbf {A}\ast \delta _{t}\mathbf {e}_j\rangle \in \mathbb {Z}$ , yielding $|\langle \mathbf {f},\mathbf {A}\ast \delta _{t}\mathbf {e}_j\rangle |\geq 1$ . Therefore, using the fact that $\|a_{ij}\ast \delta _t\|_1\leq \|a_{ij}\|_1\|\delta _t\|_1=\|a_{ij}\|_1$ , we get

$$ \begin{align*} 1&\leq |\langle\mathbf{f},\mathbf{A}\ast \delta_{t}\mathbf{e}_j\rangle|= \bigg|\sum_{i=1}^{n}\langle f_i,a_{ij}\ast \delta_{t}\rangle\bigg|\leq \sum_{i=1}^{n}\|f_i\|_\infty\|a_{ij}\ast \delta_t\|_1\leq \|\mathbf{f}\|_\infty\|\mathbf{A}\|_1. \end{align*} $$

With this, since $KS=S$ , we have

$$ \begin{align*} \sup_{s\in S}\sup_{t\in K}\max_{1\leq j\leq n}\rho((s\cdot x_j) (t),0)&=\sup_{s\in S}\sup_{t\in K}\max_{1\leq j\leq n}\rho(x_j (ts),0)\\ &=\sup_{s\in S}\max_{1\leq j\leq n}\rho(x_j(s),0)=\|\mathbf{f}\|_\infty\geq \frac{1}{\|\mathbf{A}\|_1}. \end{align*} $$

Taking an enumeration $\{s_1,s_2,\ldots \}$ of S, there is some $r\geq 1$ such that $K\subseteq \{s_1,\ldots ,s_r\}$ . Since $\rho $ is bounded by $1$ , the metric $\rho /(1+\rho )$ is bounded below by $\rho /2$ , and we have

$$ \begin{align*}d(\mathbf{y},\mathbf{y}')\geq \max_{1\leq j\leq n}\frac{\rho(y_j(t),y_j'(t))}{2^{r}(1+\rho(y_j(t),y_j'(t)))} \geq\frac{1}{2^{r+1}}\max_{1\leq j\leq n}\rho(y_j(t),y_j'(t))\end{align*} $$

for all $\mathbf {y}=(y_1,\ldots ,y_n),\mathbf {y}'=(y_1',\ldots ,y_n')\in (\mathbb {T}^S)^n$ and $t\in K$ , which implies

$$ \begin{align*}\sup_{s\in S}d(s\cdot \mathbf{x},0)\geq\frac{1}{2^{r+1}}\sup_{s\in S}\sup_{t\in K}\max_{1\leq j\leq n}\rho((s\cdot x_j) (t),0)\geq\frac{1}{2^{r+1}\|\mathbf{A}\|_1}.\end{align*} $$

Therefore, $\epsilon := (2^{r+1}\|\mathbf {A}\|_1)^{-1}$ is an expansivity constant for $S\curvearrowright X_\omega $ . Since $\omega \leq J$ , by duality, we have $X_J\subseteq X_\omega $ , yielding the expansivity of $S\curvearrowright X_J$ .

Proof. It follows from Proposition 4.1 that (i) implies (ii), so let us prove that (ii) implies (i). Assume statement (ii) and suppose that $KS\subsetneq S$ for every finite $K\subseteq S$ . The algebraic semigroup action of S upon the closed subgroup $X\leq \mathbb {T}^S$ given by

$$ \begin{align*}X=\{x\in \mathbb{T}^S:x(s)\in\{0,1/2\}\text{ for all }s\in S\}\end{align*} $$

cannot be expansive by [Reference Ceccherini-Silberstein, Coornaert and PhungCSCP26, Proposition 4.2], but we include a brief proof for completeness. For any given $\epsilon>0$ , there exists a finite $K\subseteq S$ such that $d(x,y)<\epsilon $ whenever x and y coincide on K. Choose an element $s_K\in S-KS$ , and let $x,y\in X$ be equal over $S-\{s_K\}$ and such that $x(s_K)\neq y(s_K)$ . Then, $ts\neq s_K$ for every $s\in S$ and $t\in K$ , so

$$ \begin{align*}(s\cdot x)(t)=x(ts)=y(ts)=(s\cdot y)(t).\end{align*} $$

Hence, $s\cdot x$ and $s\cdot y$ coincide on K for all $s\in S$ , implying $d(s\cdot x,s\cdot y)<\epsilon $ for all $s\in S$ . As this has been done for arbitrary $\epsilon>0$ , we conclude that $\sup _{s\in S} d(s\cdot x,s\cdot y)=0$ but $x\neq y$ , so that $S\curvearrowright X$ is not expansive.

Observe that $x(s)\in \{0,1/2\}$ if and only if $\langle x, 2\delta _s\rangle =2x(s)=0$ . Hence, we have that $X={[\mathbb {Z}[S]/2\mathbb {Z}[S]]}^{\wedge }$ . Thus, X is the algebraic action associated with the right $\mathbb {Z}[S]$ -module $J=2\mathbb {Z}[S]$ . Now, if $f\in J^\perp $ , then for every $t\in S$ , we have

$$ \begin{align*}0=\langle f, 2\delta_t\rangle=2\sum_{r\in S}f(r)\delta_t(r)=2f(t),\end{align*} $$

so $f=0$ . Hence, $J^\perp =\{0\}$ , but $S\curvearrowright X_J$ is non-expansive.

Proof. Let $e\in \ell ^1(S)$ be a left identity. Then, $\mathrm {Re}\{e\}\in \ell ^1(S)$ is a left identity too. Let $K\subseteq S$ be a finite subset such that $\|\mathrm {Re}\{e\}\mathbf {1}_{K^{\text {c}}}\|_1<1$ . Note that, for every $s\in S$ , we have

$$ \begin{align*}1=\delta_s(s)=(\delta_s\ast \mathrm{Re}\{e\})(s)=\sum_{t: st=s}\mathrm{Re}\{e\}(t),\end{align*} $$

and hence, $\{t\in S:st=s\}\cap K\neq \varnothing $ . Indeed, if it were not the case, we would have

$$ \begin{align*}1=\sum_{t\in S:st=s}\mathrm{Re}\{e\}(t)\leq \sum_{t\in K^{\text{c}}}|\mathrm{Re}\{e\}(t)|=\|\mathrm{Re}\{e\}\mathbf{1}_{K^{\text{c}}}\|_1<1,\end{align*} $$

which is absurd. Therefore, for every $s\in S$ , there exists $t\in K$ such that $s=ts\in KS$ and we conclude that $S=KS$ .

Proof. For all $\mathbf {f}=(f_1,\ldots ,f_n)\in \ell ^\infty (S)^n$ , $\mathbf {A}=(a_{ij})_{i,j}\in \mathrm {M}_{n\times k}(\ell ^1(S))$ and $\mathbf {b}=(b_1,\ldots ,b_k)\in \ell ^1(S)^k$ , it holds that

$$ \begin{align*} \langle\mathbf{f},\mathbf{A}\ast \mathbf{b}\rangle &= \sum_{j=1}^{k}\sum_{i=1}^{n}\langle f_i,a_{ij}\ast b_j\rangle = \sum_{j=1}^{k}\sum_{i=1}^{n} \sum_{t\in S}f_i(t)\sum_{rs=t}a_{ij}(r)b_j(s)\\ &= \sum_{j=1}^{k}\sum_{i=1}^{n} \sum_{r,s\in S}f(rs)a_{ij}(r)b(s)=\sum_{j=1}^{k}\sum_{i=1}^{n} \sum_{s\in S}(f\star a_{ij})(s)b_j(s)\\ &= \sum_{j=1}^{k} \bigg{\langle}\sum_{i=1}^{n}f_i\star a_{ij}, b_j\bigg{\rangle}=\langle \mathbf{f}\star \mathbf{A},\mathbf{b}\rangle. \end{align*} $$

Hence, the first claim is proved. The left $\mathrm {M}_n(\ell ^1(S))$ -module structure $\ast \colon \mathrm {M}_n(\ell ^1(S)) \times \ell ^1(S)^n\to \ell ^1(S)^n$ is determined by the maps $\psi _{\mathbf {A}}\colon \ell ^1(S)^n\to \ell ^1(S)^n$ , $\mathbf {b}\mapsto \mathbf {A}\ast \mathbf {b}$ . What we have just proven implies that for all $\mathbf {A}\in \mathrm {M}_n(\ell ^1(S))$ , the map $\varphi _{\mathbf {A}}\colon \ell ^\infty (S)^n\to \ell ^\infty (S)$ , $\mathbf {f}\mapsto \mathbf {f}\star \mathbf {A}$ is exactly the dual map of $\psi _{\mathbf {A}}$ . Hence, the maps $\varphi _{\mathbf {A}}$ determine a right $\mathrm {M}_n(\ell ^1(S))$ -module structure, dual to $\ast \colon \mathrm {M}_n(\ell ^1(S))\times \ell ^1(S)^n\to \ell ^1(S)^n$ . Finally, the fact that

$$ \begin{align*}\|\mathbf{f}\star \mathbf{A}\|_\infty & =\sup_{\|\mathbf{b}\|_1\leq 1}|\langle\mathbf{f}\star \mathbf{A},\mathbf{b}\rangle| \\ &=\sup_{\|\mathbf{b}\|_1\leq 1}|\langle\mathbf{f},\mathbf{A}\ast\mathbf{b}\rangle|\leq \sup_{\|\mathbf{b}\|_1\leq 1}\|\mathbf{f}\|_\infty\|\mathbf{A}\|_1\|\mathbf{b}\|_1\leq \|\mathbf{f}\|_\infty\|\mathbf{A}\|_1\end{align*} $$

implies that said right $\mathrm {M}_n(\ell ^1(S))$ -module structure is Banach.

Proof. Note that, for every $\mathbf {A}\in \mathrm {M}_{n\times k}(\mathbb {Z}[S])$ , we have $(\mathbf {A}\mathbb {Z}[S]^k)^\perp =\mathrm {ker}(\varphi _{\mathbf {A}})$ . Indeed, we know from Proposition 4.5 that $\langle \mathbf {f},\mathbf {A}\ast \mathbf {b}\rangle =\langle \mathbf {f}\star \mathbf {A},\mathbf {b}\rangle $ for all $\mathbf {f}\in \ell ^\infty (S)^n$ and $\mathbf {b}\in \ell ^1(S)^m$ . Hence, if $\mathbf {f}\star \mathbf {A}=0$ , then $\langle \mathbf {f},\mathbf {A}\ast \mathbf {b}\rangle =0$ for all $\mathbf {b}\in \mathbb {Z}[S]^k$ , so $\mathbf {f}\in (\mathbf {A}\mathbb {Z}[S]^k)^\perp $ . However, if $\mathbf {f}\in (\mathbf {A}\mathbb {Z}[S]^k)^\perp $ , we have that $\langle \mathbf {f}\star \mathbf {A},\mathbf {b}\rangle =0$ for all $\mathbf {b}\in \mathbb {Z}[S]^k$ , which immediately implies $\mathbf {f}\star \mathbf {A}=0$ . The equivalence between statements (i) and (ii) is now a direct consequence of Theorem A.

Proof. We first claim that $\mathbf {f}\star \mathbf {I}= \mathbf {f}$ for every $\mathbf {f}\in \ell ^\infty (S)^n$ . To show this, we first prove that for a left identity $e\in \ell ^1(S)$ , we have $f\star e= f$ for all $f\in \ell ^\infty (S)$ . Indeed, for all $r\in S$ , it holds that $e\ast \delta _r= \delta _r$ , so $(f\star e)\star \delta _r=f\star (e\ast \delta _r)=f\star \delta _r$ . By Lemma 4.3, S is expansive, thus, for every $s\in S$ , we can write $s=rr'$ for some $r,r'\in S$ . We then have

$$ \begin{align*} (f\star e-f)(s)&=((f\star e-f)\star \delta_r)(r')=(f\star\delta_r-f\star \delta_r)(r')=0, \end{align*} $$

and we conclude that $f\star e=f$ , as desired. Next, note that if $a\in \ell ^1(S)$ is such that $a\ast b= 0$ for all $b\in \ell ^1(S)$ , then it holds that $f\star a= 0$ for all $f\in \ell ^\infty (S)$ . Indeed, for all $r\in S$ , we have $a\ast \delta _r= 0$ , implying

$$ \begin{align*}(f\star a)(s)=((f\star a)\star\delta_r)(r')=(f\star (a\ast \delta_r))(r')=0.\end{align*} $$

Finally, write $\mathbf {I}=(e_{ij})_{i,j}$ . By considering the products $\mathbf {I}\ast \mathbf {A}_{ij}$ , where $\mathbf {A}_{ij}$ is the matrix with all entries equal to zero except for the entry $(i,j)$ , which is an arbitrary $a\in \ell ^1(S)$ , it is readily checked that $e_{jj}$ is a left identity in $\ell ^1(S)$ for all $1\leq j\leq n$ , and that $e_{ij}$ satisfies $e_{ij}\ast b= 0$ for every $b\in \ell ^1(S)$ and $i\neq j$ . Hence, for every $\mathbf {f}=(f_1,\ldots ,f_n)\in \ell ^\infty (S)^n$ , we get

$$ \begin{align*}\mathbf{f}\star\mathbf{I}=\sum_{i=1}^n\sum_{j=1}^n(f_j\star e_{ij})\mathbf{e}_i=\sum_{i=1}^nf_i\mathbf{e}_i=\mathbf{f},\end{align*} $$

and the claim is proven.

The result now follows directly from the claim. Let $\mathbf {f}\in (\mathbf {A}\mathbb {Z}[S]^n)_\infty ^\perp $ and $\mathbf {B}\in \mathrm {M}_n(\ell ^1(S))$ such that $\mathbf {A}\ast \mathbf {B}=\mathbf {I}$ . Then, as $\mathbf {f}\star \mathbf {I}= \mathbf {f}$ , we have $\mathbf {f}=\mathbf {f}-(\mathbf {f}\star \mathbf {A})\star \mathbf {B}=\mathbf {f}-\mathbf {f}\star (\mathbf {A}\ast \mathbf {B}) =0$ . Therefore, $(\mathbf {A}\mathbb {Z}[S]^n)_\infty ^\perp $ is trivial and so is $J^\perp \subseteq (\mathbf {A}\mathbb {Z}[S]^n)^\perp $ . Since S is expansive, we conclude by Theorem A that $S\curvearrowright X_J$ is an expansive action.

Proof. Note that e is clearly right e-invertible, so that the set of right e-invertible elements is non-empty. Let $a\in \mathfrak {A}$ be right e-invertible and let $a^{-1}\in \mathfrak {A}$ be such that $aa^{-1}=e$ . Given $x\in \mathfrak {A}$ with $\|x\|<\|a^{-1}\|^{-1}$ , define $S_n := a^{-1}+\sum _{k=1}^{n}a^{-1}(xa^{-1})^{k}$ and note that $\|xa^{-1}\|< 1$ , so the series $\sum _{k=1}^{\infty }\|a^{-1}(xa^{-1})^{k}\|$ converges. Since $\mathfrak {A}$ is a complete normed space, every absolutely convergent series converges (see [Reference BollobásBol99, Theorem 2.8]), so we can define $S_\infty :=a^{-1}+\sum _{k=1}^{\infty }a^{-1}(xa^{-1})^{k}$ . Thus,

$$ \begin{align*}(a-x)S_\infty=\lim_{n\to\infty}\bigg(e+\sum_{k=1}^{n}(xa^{-1})^k-\sum_{k=1}^{n+1}(xa^{-1})^k\bigg)=\lim_{n\to\infty}(e-(xa^{-1})^{n+1})=e,\end{align*} $$

showing that $a-x$ is right e-invertible. Hence, every element at a distance less than $\|a^{-1}\|^{-1}$ from a is right e-invertible and we get that the set of right e-invertible elements is open.

Proof. The fact that (ii) implies (i) directly follows from Proposition 4.8.

We prove now that (i) implies (ii). By Corollary 4.6, the map $\varphi _{\mathbf {A}}\colon \ell ^\infty (S)^n\to \ell ^\infty (S)^k$ given by $\mathbf {f}\mapsto \mathbf {f}\star \mathbf {A}$ is injective. Consider now the $\mathbb {C}$ -linear map $\psi _{\mathbf {A}}\colon \ell ^1(S^1)^k\to \ell ^1(S)^n$ given by $\mathbf {a}\mapsto \mathbf {A}\ast \mathbf {a}$ . By Proposition 4.5, we know that $\varphi _{\mathbf {A}}$ is the dual map of $\psi _{\mathbf {A}}$ . Now, $\psi _{\mathbf {A}}$ must have dense image in $\ell ^1(S)^n$ , because otherwise, there exists an element $\mathbf {a}\in \ell ^1(S)^n-\overline {\mathrm {im}(\psi )}$ and by the Hahn–Banach theorem, we may extend the zero functional on $\overline {\mathrm {im}(\psi )}$ to obtain an element $\mathbf {f}\in \overline {\mathrm {im}(\psi _{\mathbf {A}})}^\perp \subseteq \ell ^\infty (S)^n$ such that $\langle \mathbf {f},\mathbf {a}\rangle \neq 0$ . Thus, $\mathbf {f}$ is non-zero but $\langle \mathbf {f},\psi _{\mathbf {A}}(\mathbf {b})\rangle =0$ for all $\mathbf {b}\in \ell ^1(S)^k$ , so that $\varphi _{\mathbf {A}}(\mathbf {f})=\psi _{\mathbf {A}}^*(\mathbf {f})=0$ , which contradicts the injectivity of $\varphi _{\mathbf {A}}$ .

Let $\epsilon>0$ and let $\mathbf {e}_1,\ldots ,\mathbf {e}_n$ be the rows of $\mathbf {I}$ . Since $\mathrm {im}(\psi _{\mathbf {A}})$ is dense in $\ell ^1(S)^n$ , we may choose elements $\mathbf {a}_1,\ldots ,\mathbf {a}_n \in \ell ^1(S^1)^k$ such that for all $1\leq i\leq n$ , we have $\|\psi _{\mathbf {A}}(\mathbf {a}_{i})-\mathbf {e}_i\|_1<\epsilon /n$ . Observe that, for all $1\leq i\leq n$ ,

$$ \begin{align*} \|\psi_{\mathbf{A}}(\mathbf{a}_{i})-\mathbf{e}_i\|_1&= \|\psi_{\mathbf{A}}(\mathrm{Re}\{\mathbf{a}_{i}\})-\mathrm{Re}\{\mathbf{e}_i\}+\mathrm{i}\,(\psi_{\mathbf{A}}(\mathrm{Im}\{\mathbf{a}_{i}\})-\mathrm{Im}\{\mathbf{e}_i\})\|_1\\ &\geq \|\psi_{\mathbf{A}}(\mathrm{Re}\{\mathbf{a}_{i}\})-\mathrm{Re}\{\mathbf{e}_i\}\|_1. \end{align*} $$

Hence, $\|\mathbf {A}\ast \mathrm {Re}\{\mathbf {a}_i\}-\mathrm {Re}\{\mathbf {e}_i\}\|_1<\epsilon /n$ . Since $\mathbb {Q}[S]^k$ is dense in $\ell ^1(S^1,\mathbb {R})^k$ and $\psi _{\mathbf {A}}$ is continuous, we can choose elements $\mathbf {b}_{i}\in \mathbb {Q}[S]^k$ such that $\|\mathbf {A}\ast \mathbf {b}_i-\mathrm {Re}\{\mathbf {e}_i\}\|_1<\epsilon /n$ for all $1\leq i\leq n$ . Hence, defining $\mathbf {B}_\epsilon \in M_{k\times n}(\mathbb {Q}[S])$ to be the matrix whose ith column is $\mathbf {b}_i$ , we get

$$ \begin{align*}\|\mathbf{A}\ast\mathbf{B}_\epsilon-\mathrm{Re}\{\mathbf{I}\}\|_1=\sum_{i=1}^{n}\|\mathbf{A}\ast\mathbf{b}_i-\mathrm{Re}\{\mathbf{e}_i\}\|_1<\epsilon.\end{align*} $$

By Lemma 4.10, the set of right $\mathrm {Re}\{\mathbf {I}\}$ -invertible elements is an open subset of $\mathrm {M}_n(\ell ^1(S))$ . Hence, taking $\epsilon $ sufficiently small, we can ensure that $\mathbf {A}\ast \mathbf {B}_\epsilon $ is right $\mathrm {Re}\{\mathbf {I}\}$ -invertible. Furthermore, since the rows of $\mathbf {B}_\epsilon $ are elements from $\mathbb {Q}[S]^k$ , we may take $m\in \mathbb {N}$ such that $m\mathbf {B}_\epsilon \in \mathrm {M}_{k\times n}(\mathbb {Z}[S])$ . Therefore, the matrix $\mathbf {B}:=\mathbf {A}\ast m\mathbf {B}_\epsilon $ belongs to $\mathrm {M}_n(\mathbb {Z}[S])$ , is right $\mathrm {Re}\{\mathbf {I}\}$ -invertible, and satisfies $\mathbf {B}\mathbb {Z}[S]^n\subseteq \mathbf {A}\mathbb {Z}[S]^k$ .

Finally, assume that $\ell ^1(S)$ has an identity e. It is already clear that (iii) implies (ii). To see that (i) implies (iii), first note that for all $s\in S$ , we have

$$ \begin{align*}\delta_s\ast e=\delta_s\ast \mathrm{Re}\{e\}+\mathrm{i}(\delta_s\ast \mathrm{Im}\{e\})=\delta_s\in \ell^1(S,\mathbb{R}).\end{align*} $$

Thus, $\delta _s\ast \mathrm {Im}\{e\}=0$ for all $s\in S$ , which implies $\mathrm {Im}\{e\}=e\ast \mathrm {Im}\{e\}=0$ . Hence, $e=\mathrm {Re}\{e\}$ and we conclude that $\mathbf {I}=\mathrm {Re}\{\mathbf {I}\}$ as well. The same argument of the proof of the fact that (i) implies (ii) allows us to construct a matrix $\mathbf {B}\in \mathrm {M}_n(\mathbb {Z}[S])$ arbitrarily close to $\mathbf {I}$ such that $\mathbf {B}\mathbb {Z}[S]^n\subseteq \mathbf {A}\mathbb {Z}[S]^k$ . Since the set of invertible elements of $\mathrm {M}_n(\ell ^1(S))$ is open, we can assume that such $\mathbf {B}$ is invertible.

Proof. By the equivalence between statements (i) and (iii) from Theorem B, we just need to show that $\mathbf {A}$ is invertible in $\mathrm {M}_n(\ell ^1(S))$ if and only if there exists $\mathbf {B}\in \mathrm {M}_n(\mathbb {Z}[S])$ that is invertible in $\mathrm {M}_n(\ell ^1(S))$ satisfying $\mathbf {B}\mathbb {Z}[S]^n\subseteq \mathbf {A}\mathbb {Z}[S]^n$ . Clearly, the invertibility of $\mathbf {A}$ implies this last condition. For the other direction, assume such $\mathbf {B}$ exists. It then follows that $\mathbf {B}\mathrm {M}_n(\ell ^1(S))\subseteq \mathbf {A}\mathrm {M}_n(\ell ^1(S))$ for every $n\geq 1$ . Since the identity element $\mathbf {I}\in \mathrm {M}_n(\ell ^1(S))$ belongs to $\mathbf {B}\mathrm {M}_n(\ell ^1(S))$ , there must exist $\mathbf {C}\in \mathrm {M}_n(\ell ^1(S))$ such that $\mathbf {A}\ast \mathbf {C}=\mathbf {I}$ , so $\mathbf {A}$ is right invertible in $\mathrm {M}_n(\ell ^1(S))$ . By the stable finiteness of $\ell ^1(S)$ , we conclude that $\mathbf {A}$ is invertible in $\mathrm {M}_n(\ell ^1(S))$ .

Proof. It is shown in [Reference Dales, Lau and StraussDLS10, Proposition 4.3] that if $\ell ^1(S)$ has a left approximate identity and $S=FS$ for some finite $F\subseteq E_S$ , then $\ell ^1(S)$ has a left identity. We will show that if S is expansive and $\ell ^1(S)$ admits a left approximate identity, then such F exists. Let $(e_\alpha )_{\alpha }$ be a left approximate identity in $\ell ^1(S)$ . Note that for every $s\in S$ , we have $e_\alpha \ast \delta _s\to \delta _s$ , so that

$$ \begin{align*}\sum_{r\in S: rs=s}e_\alpha(r)=(e_\alpha\ast \delta_s)(s)\to \delta_s(s)=1.\end{align*} $$

In particular, for every $s\in S$ , there exists $r\in S$ with $rs=s$ . Let $K\subseteq S$ be a finite subset with $S=KS$ . Choose, for every $t\in K$ , an element $r_t\in S$ with $r_tt=t$ , and let $K'=\{r_t:t\in K\}$ . Then, for every $s\in S$ , we can write $s=ts'$ with $t\in K$ and $s'\in S$ , and it follows that $r_ts=r_tts'=ts'=s$ . Hence, $K'$ is finite and for every $s\in S$ , there is $t\in K'$ with $ts=s$ (in particular, $K'S=S$ ).

Let $n:= |K'|$ and set $K^{\prime }_n:= K'$ . If every element $t\in K'$ satisfies $t^2=t$ , we are done by taking $F=K^{\prime }_n$ . Otherwise, if $t\in K'$ is not an idempotent, there is $t'\in K'$ with $t't=t$ and $t'\neq t$ . It follows that every element $s\in S$ satisfying $ts=s$ also satisfies $t's=t'ts=ts=s$ , so we can define $K_{n-1}':= K^{\prime }_n-\{t\}$ and still for every $s\in S$ , there is $t\in K^{\prime }_{n-1}$ satisfying $ts=s$ . By iterating this process, we see that for some $1\leq k\leq n$ , $F:= K^{\prime }_k$ is a subset of $E_S$ and satisfies $S=FS$ .

Proof. The equivalence between assertions (ii) and (iv) is proven in [Reference GrønbækGrø88, Corollary 1.3 and Remark 1.4], and since (iv) implies (iii) and (iii) implies (ii), the equivalences between assertions (ii), (iii), and (iv) follow. That (iii) implies (i) was proven in Lemma 4.3, so it just remains to prove that (i) implies (iii).

Let $K\subseteq S$ be finite such that $S=KS$ , and note that if $t=t's$ for some $t,t'\in K$ with $t\neq t'$ and $s\in S$ , then every $r\in S$ satisfying $r\in tS$ also satisfies $r\in tS=t'sS\subseteq t'S$ . Hence, by removing such t from K and iterating the process, we may assume that every $t\in K$ satisfies $t=ts_t$ for some $s_t\in S$ . With this, we get by left cancellativity that

$$ \begin{align*}t=ts_t=t^2s_t\implies s_t=ts_t=t,\end{align*} $$

so that $t^2=t$ for every $t\in K$ . Now, assume $|K|\geq 2$ and let $t_1,t_2\in K$ be different elements. Then, $t_1t_2=t_1^2t_2=t_1t_2^2$ , so left and right cancellativity yield that $t_1=t_1t_2=t_2$ . Hence, $|K|=1$ and we write $K=\{t\}$ . Note that for every element $s\in S$ , it holds that $s=ts'$ for some $s'\in S$ , so $ts=tts'=ts'=s$ and we conclude that t is a left identity element. Finally, by right cancellativity and the fact that $st=st^2$ , we get $s=st$ for every $s\in S$ and we obtain that S is a monoid, as desired.

Proof. If $\mathfrak {A}$ admits a faithful trace $\tau '$ , then the trace $\tau \colon \mathrm {M}_n(\mathfrak {A})\to \mathbb {C}$ defined for every $\mathbf {A}=(a_{ij})_{i,j}\in \mathrm {M}_n(\mathfrak {A})$ by $\tau (\mathbf {A})=\sum _{j=1}^n\tau '(a_{jj})$ is faithful, since

$$ \begin{align*}\tau(\mathbf{A}^*\mathbf{A})=\sum_{j=1}^n\tau'\bigg(\sum_{i=1}^{n}a_{ij}^*a_{ij}\bigg)=\sum_{1\leq i,j\leq n}\tau'(a_{ij}^* a_{ij})>0\end{align*} $$

whenever $\mathbf {A}\neq \mathbf {0}$ .

The assertion that every right invertible element of $\mathrm {M}_n(\mathfrak {A})$ is also left invertible is equivalent to the following statement: for all projections $\mathbf {0}\leq \mathbf {Q}\leq \mathbf {P}$ admitting an element $\mathbf {V}\in \mathrm {M}_n(\mathfrak {A})$ with $\mathbf {P}=\mathbf {V}^*\mathbf {V}$ and $\mathbf {Q}=\mathbf {V}\mathbf {V}^*$ , we have $\mathbf {P}=\mathbf {Q}$ [Reference Rørdam, Larsen and LaustsenRLL00, Lemma 5.1.2]. If $\mathbf {P},\mathbf {Q},\mathbf {V}\in \mathrm {M}_n(\mathfrak {A})$ are as above, we get

$$ \begin{align*}\tau(\mathbf{P}-\mathbf{Q})=\tau(\mathbf{V}^*\mathbf{V})-\tau(\mathbf{V}\mathbf{V}^*)=0,\end{align*} $$

and since $\mathbf {P}-\mathbf {Q}$ is positive, we get that $\mathbf {P}=\mathbf {Q}$ . Hence, every right invertible element from $\mathrm {M}_n(\mathfrak {A})$ is invertible and we conclude that $\mathfrak {A}$ is stably finite.

Proof. Let $\tau \colon C^*_r(S)\to \mathbb {C}$ be given by

$$ \begin{align*}\tau(a)=\sum_{p\in E_S}\langle a(\delta_p),\delta_p\rangle_{\ell^2(S)}\quad \text{for all }a\in C^*_r(S).\end{align*} $$

To prove that $\tau $ satisfies $\tau (ab)=\tau (ba)$ for all $a,b\in C^*_r(S)$ , it suffices to show $\tau (\unicode{x3bb} (s)\unicode{x3bb} (t))=\tau (\unicode{x3bb} (t)\unicode{x3bb} (s))$ for all $s,t\in S$ . Note that, by the definitions of $\unicode{x3bb} (s)$ and $\unicode{x3bb} (t)$ , we have

$$ \begin{align*}\tau(\unicode{x3bb}(s)\unicode{x3bb}(t))=\sum_{p\in E_S}\langle (\unicode{x3bb}(s)\unicode{x3bb}(t))(\delta_p),\delta_p\rangle_{\ell^2(S)}=\sum_{p:p\leq t^*t,tpt^*\leq s^*s}\delta_{stp}(p)=|\Omega_{s,t}|,\end{align*} $$

where $\Omega _{s,t}:=\{p\in E_s:p\leq t^*t,tpt^*\leq s^*s\text { and }stp=p\}$ . It is thus enough to show that $|\Omega _{s,t}|=|\Omega _{t,s}|$ . Consider the map $\theta \colon \Omega _{s,t}\to \Omega _{t,s}$ given by $p\mapsto tpt^*$ for $p\in \Omega _{s,t}$ . We first note that it is well defined. Indeed, if $p\in \Omega _{s,t}$ , we want to show that $\theta (p)\in E_S$ , and satisfies $\theta (p)\leq s^*s$ , $s\theta (p)s^*\leq t^*t$ and $ts\theta (p)=\theta (p)$ . Note that, for such p, we have that:

• • $(tpt^*)^2=tp(t^*t)pt^*=tt^*tp^2t^*=tpt^*$ , so $\theta (p)\in E_S$ ;

• • the condition $tpt^*\leq s^*s$ by definition yields $\theta (p)\leq s^*s$ ;

• • the fact that $stp=p$ implies $pt^*s^*=p$ because $p\in E_S$ , and since $pt^*t=p$ , we get

  $$ \begin{align*} (s\theta(p)s^*)(t^*t)&=(stp)t^*s^*t^*t=(pt^*s^*)t^*t=pt^*t\\ &=p=stp=st(pt^*s^*)=s\theta(p)s^*, \end{align*} $$
  and so $s\theta (p)s^*\leq t^*t$ ; and

• • $ts\theta (p)=t(stp)t^*=tpt^*=\theta (p)$ .

Hence, $\theta $ is well defined. Now, if $tpt^*=tqt^*$ for $p,q\in \Omega _{s,t}$ , then

$$ \begin{align*}p=t^*(tpt^*)t=t^*(tqt^*)t=q,\end{align*} $$

so $\theta \colon \Omega _{s,t}\to \Omega _{t,s}$ is an injection. By symmetry, the map $\eta \colon \Omega _{t,s}\to \Omega _{s,t}$ given by $e\mapsto ses^*$ is an injection as well, so we have proven that $|\Omega _{s,t}|=|\Omega _{t,s}|$ .

Let us see that $\tau $ is faithful. For every $a\in C^*_r(S)$ , the element $a^*a$ is self-adjoint with non-negative spectrum and so the continuous functional calculus (see, e.g., [Reference Reed and SimonRS72, Theorem VII.1]) gives a self-adjoint element $|a|\in C^*_r(S)$ such that $|a|^2=a^*a$ . Also, if $s\in S$ , then

$$ \begin{align*} a^*a-|a|\unicode{x3bb}(s)^*\unicode{x3bb}(s)|a| & =|a|(e-\unicode{x3bb}(s)^*\unicode{x3bb}(s))|a| \\ & =((e-\unicode{x3bb} bda(s)^*\unicode{x3bb}(s))|a|)^*(e-\unicode{x3bb}(s)^*\unicode{x3bb}(s))|a|, \end{align*} $$

so the element $a^*a-|a|\unicode{x3bb} (s)^*\unicode{x3bb} (s)|a|$ is positive. Thus, $\tau (|a|\unicode{x3bb} (s)^*\unicode{x3bb} (s)|a|)\leq \tau (a^*a)$ . With this, if $\tau (a^*a)=0$ , then for every $s\in S$ , we have

$$ \begin{align*}\sum_{p\in E_S}\|(a\unicode{x3bb}(s))(\delta_p)\|^2_{\ell^2(S)}=\tau(\unicode{x3bb}(s)^*a^*a\unicode{x3bb}(s))=\tau(|a|\unicode{x3bb}(s)^*\unicode{x3bb}(s)|a|)\leq\tau(a^*a)=0,\end{align*} $$

which implies $(a\unicode{x3bb} (s))(\delta _p)=0$ for all $p\in E_S$ . Taking $p=s^*s$ , we have $s^*s\leq s^*s$ so that $\unicode{x3bb} (s)(\delta _{s^*s})=\delta _{ss^*s}=\delta _s$ , yielding $a(\delta _s)=0$ for all $s\in S$ . Hence, $\tau $ is faithful.

Question 5.10. In [Reference Ceccherini-Silberstein, Coornaert and PhungCSCP25], the authors prove that for every surjunctive monoid S and field K, the ring $K[S]$ is stably finite. Is it true that $\ell ^1(S)$ is stably finite for this class of monoids?

Proof. Write $\mathbf {P}=(p_{\unicode{x3bb} i})_{i,\unicode{x3bb} }$ . Let us prove statement (i). Since S contains a non-zero element, it is clear that if $S=KS$ , then $\mathbf {P}$ is non-zero. Conversely, let $i_0\in I$ and $\unicode{x3bb} _0\in \Lambda $ be such that $p_{\unicode{x3bb} _0 i_0}\in G$ , and note that $K:=\{(i,1_G,\unicode{x3bb} _0):i\in I\}$ is finite and satisfies

$$ \begin{align*}(i,g,\unicode{x3bb})=(i,1_G,\unicode{x3bb}_0)(i_0,p_{\unicode{x3bb}_0 i_0}^{-1}g,\unicode{x3bb})\end{align*} $$

for all $(i,g,\unicode{x3bb} )\in S$ . Hence, $S=KS$ .

To see statement (ii), note that for $g\in G$ , we have that $(i,g,\unicode{x3bb} )$ is an idempotent if and only if $p_{\unicode{x3bb} i}\in G$ and $g=p_{\unicode{x3bb} i}^{-1}$ . Thus, if $S=KS$ for some $K\subseteq E_S$ , then for all $i\in I$ , there must be some $\unicode{x3bb} \in \Lambda $ with $p_{\unicode{x3bb} i}\in G$ . Conversely, assume for each $i\in I$ , there is $\unicode{x3bb} (i)\in \Lambda $ such that $p_{\unicode{x3bb} (i) i}\in G$ . Then, $K:=\{(i,p_{\unicode{x3bb} (i)i}^{-1},\unicode{x3bb} (i)):i\in I\}\subseteq E_S$ is finite and

$$ \begin{align*}(i,g,\unicode{x3bb})=(i,p_{\unicode{x3bb}(i)i}^{-1},\unicode{x3bb}(i))(i,g,\unicode{x3bb})\quad\text{for all }(i,g,\unicode{x3bb})\in S,\end{align*} $$

yielding $S=KS$ .

Proof. To see statement (i), note that the ‘only if’ part follows by taking $K_i:=K\cap \Sigma _i$ for each $1\leq i\leq n$ , where $K\subseteq S$ is finite such that $S=KS$ , and the ‘if’ part by taking $K:=K_1\sqcup \ldots \sqcup K_n$ , where each $K_i\subseteq \Sigma _i$ is finite such that $\Sigma _i=K_i\Sigma _i$ .

Now we prove statement (ii). Consider, for $1\leq i\leq n$ , the character $\varphi _{\Sigma _i}\colon \ell ^1(\Sigma _i)\to \mathbb {C}$ given by

$$ \begin{align*}\varphi_{\Sigma_i}(a):= \sum_{t\in S}a(t)\quad\text{for all }a\in \ell^1(\Sigma_i).\end{align*} $$

It can be verified that $\ell ^1(S)\simeq \mathbb {C}\oplus \bigoplus _{i=1}^{n}\ell ^1(\Sigma _i)$ , where the product of $(\alpha ,a_1,\ldots ,a_n)$ and $(\beta ,b_1,\ldots ,b_n)$ is given in the first coordinate by

(1) $$ \begin{align} \sum_{i=1}^{n}\bigg[\varphi_{\Sigma_i}(b_i)\bigg(\alpha+\sum_{j\neq i}\varphi_{\Sigma_j}(a_j)\bigg)\bigg]+\beta\bigg(\alpha+\sum_{i=1}^{n}\varphi_{\Sigma_i}(a_i)\bigg), \end{align} $$

and defined coordinate-wise in the last n coordinates. It is then clear that if $\ell ^1(S)$ has a left identity $(\varepsilon ,e_1,\ldots ,e_n)$ , then each $e_i$ is a left identity in $\ell ^1(\Sigma _i)$ .

Conversely, if for each $1\leq i\leq n$ there is a left identity $e_i\in \ell ^1(\Sigma _i)$ , then the element $e:= (1-n,e_1,\ldots ,e_n)$ is a left identity in $\ell ^1(S)$ . Indeed, for all $(\beta ,b_1,\ldots ,b_n)\in \ell ^1(S)$ , we have, by (1), that the first coordinate of the product $(1-n,e_1,\ldots ,e_n)\cdot (\beta ,b_1,\ldots ,b_n)$ is

$$ \begin{align*} &\sum_{i=1}^{n}\bigg[\varphi_{\Sigma_i}(b_i)\bigg(1-n+\sum_{j\neq i}\varphi_{\Sigma_j}(e_j)\bigg)\bigg]+\beta\bigg(1-n+\sum_{i=1}^{n}\varphi_{\Sigma_i}(e_i)\bigg)\\ &\quad=\sum_{i=1}^{n}\varphi_{\Sigma_i}(b_i)(1-n+n-1)+\beta(1-n+n)=\beta, \end{align*} $$

where we use the fact that $\varphi _{\Sigma _i}$ is a character, so $\varphi _{\Sigma _i}(e_i)=1$ for all $1\leq i\leq n$ . This yields $(1-n,e_1,\ldots ,e_n)\cdot (\beta ,b_1,\ldots ,b_n)=(\beta ,b_1,\ldots ,b_n)$ ; hence, $\ell ^1(S)$ has a left identity.