Proof Since $\tau $ is a linear NUCA with finite memory, there exists a finite subset $M \subset G$ and $s \in S^G$ , where $S= \mathcal {L}(V^M, V)$ such that $\tau = \sigma _s$ . By hypothesis, s is asymptotic to a constant configuration $c \in S^G$ . Up to enlarging M, we can also suppose that $s\vert _{G \setminus M}= c\vert _{G \setminus M}$ and that $1_G \in M$ . Since the group G is finitely generated, thus countable, it admits an increasing sequence of finite subsets $M=E_0 \subset \dots \subset E_n \dots $ such that $G=\cup _{n\in \mathbb {N}} E_n$ .

Suppose on the contrary that there does not exist a finite subset $N\subset G$ which satisfies condition $\mathrm {(C)}$ . Then, by linearity, there exist, for each $n \in \mathbb {N}$ , a configuration $d_n \in \Gamma $ and an element $g_n \in G$ such that for $c_n=\tau ^{-1}(d_n)$ (which is well-defined since $\tau $ is injective), we have

$$ \begin{align*} d_n\vert_{g_nE_n}=0^{g_nE_n} \quad \text{and} \quad c_n(g_n)\neq 0. \end{align*} $$

Consequently, by letting $x_n=g_n^{-1}c_n$ and $y_n= g_n^{-1}d_n$ , we infer from [Reference Phung33, Lemma 5.1] that $\sigma _{g_n^{-1}s}(x_n)=y_n$ and

$$ \begin{align*} y_n\vert_{E_n}=0^{E_n} \quad \text{and} \quad x_n(1_G)\neq 0. \end{align*} $$

Since s is asymptotic to a constant configuration $c \in S^G$ by hypothesis, the set ${T=\{s(g) \mathop{\colon} g \in G\}}$ is actually a finite subset of $S= \mathcal {L}(V^M, V)$ . It follows that $\Sigma (s) \subset T^G$ is a compact subspace. Therefore, up to restricting to a subsequence, we can suppose without loss of generality that the sequence $(g_n^{-1}s)_{n \in \mathbb {N}}$ converges to a configuration $t \in T^G \subset S^G$ with respect to the prodiscrete topology.

By [Reference Phung33, Lemma 8.1], we know that $\Sigma (s) = \{gs \mathop{\colon} g \in G\} \cup \{c\}$ . Note that if s is constant then the lemma results from [Reference Ceccherini-Silberstein and Coornaert5]. Hence, we can suppose in the sequel that s is not a constant configuration. In particular, T and $\Sigma (s)$ are not singletons. We distinguish two cases according to whether $t = c$ or not.

Case 1: $t=gs$ for some $g \in G$ . Then, since G is infinite and s is asymptotic but not equal to c, we can, up to restricting to a subsequence again, assume without loss of generality that $g_n^{-1}=g$ for all $n \in \mathbb {N}$ . Up to replacing s by $gs$ , we can also suppose that $g=1_G$ so that $\sigma _s(x_n)=y_n$ for all $n \in \mathbb {N}$ . For each $n\in \mathbb {N}$ , consider the following linear subspace of $V^{E_nM}$ :

Observe that $x_n\vert _{E_nM} \in I_n \setminus \{0^{E_nM} \}$ . Note also that, for all $n \leq m \leq k$ , the projection $p_{nm} \mathop{\colon} A^{E_m} \to A^{E_n}$ induces a linear map $\pi _{nm} \mathop{\colon} I_m \to I_n$ and $\pi _{nk}(I_{k}) \subset \pi _{nm}(I_m)$ . Hence, for each $n\in \mathbb {N}$ , we obtain a decreasing sequence of linear subspaces $(\pi _{nm}(I_m))_{m \geq n}$ of $I_n$ . Hence, $(\pi _{nm}(I_m))_{m \geq n}$ is stationary and there exists a linear subspace $J_n \subset I_n\subset A^{E_nM}$ such that $\pi _{nm}(I_m)=J_n$ for all m large enough.

Observe that $\pi _{nm}(J_m) \subset J_n$ for all $m \geq n$ . We claim that the restriction linear map $q_{nm} \mathop{\colon} J_m \to J_n$ is surjective for all $m\geq n$ . Indeed, let $y\in J_n$ and let $k \geq m$ be sufficiently large such that $q_{n k}(I_k)=J_n$ and $q_{m k}(I_k)=J_m$ . Thus, $q_{nk}(x)=y$ for some $x\in I_k$ . As $q_{nk}= q_{nm} \circ q_{mk}$ , we have $q_{nm}(y')=y$ , where $y'=q_{mk}(x) \in J_m$ . The claim is proved.

We choose $k\in \mathbb {N}$ large enough such that $\pi _{0k}(I_k)=J_0$ . Let $z_0=\pi _{0k}(x_k) \in J_0$ then $z_0(1_G) \neq 0$ . We define by induction a sequence $(z_n)_{n \in \mathbb {N}}$ , where $z_n \in J_n$ for all $n\in \mathbb {N}$ as follows. Given $z_n \in J_n$ for some $n\in \mathbb {N}$ , there exists by the surjectivity of the map $q_{n,n+1}$ an element

$$\begin{align*}z_{n+1}\in q^{-1}_{n,n+1}(z_n) \subset J_{n+1}\subset A^{E_{n+1}M}. \end{align*}$$

We thus obtain a configuration $c \in V^G$ defined by $z\vert _{E_nM}=z_n$ for all $n \in \mathbb {N}$ . Since $G=\cup _{n \in \mathbb {N}} E_n M$ , the configuration z is well-defined.

By construction, we have for all $n \in \mathbb {N}$ that

$$\begin{align*}\tau(z)\vert_{E_n}=f^+{E_n, s\vert_{E_n}}(z\vert_{E_nM})=f^+{E_n, s\vert_{E_n}}(z_n)=0^{E_n}. \end{align*}$$

Therefore, $\tau (z)= 0^{G}$ but $z(1_G)\neq 0$ which then contradicts the injectivity of the linear NUCA $\tau $ .

Case 2: $t = c$ . Then, since $\lim _{n \to \infty } g_n^{-1}s=t$ and $s \neq c$ , we deduce immediately that $g_n \to \infty $ when $n \to \infty $ , i.e., for every finite subset $E \subset G$ , there exists $N \in \mathbb {N}$ such that $g_n \notin E$ for all $n \geq N$ . Consequently, by restricting to a suitable subsequence, we can suppose without loss of generality that $g_nE_nM \cap M = \varnothing $ for all $n \in \mathbb {N}$ . As $s\vert _{G \setminus M}= c\vert _{G \setminus M}$ , it follows that $(g_n^{-1}s)\vert _{E_nM} = c\vert _{E_nM}$ for all $ n \in \mathbb {N}$ . Since c is constant, we deduce that $\sigma _c(x_n)\vert _{E_n}=0^{E_n}$ and $x_n(1_G)\neq 0$ . We infer from the stable injectivity of $\sigma _s$ that $\sigma _c$ is injective. Therefore, a similar argument as in Case 1 applied for $\sigma _c$ and the sequence $(x_n)_{n \in \mathbb {N}}$ leads to a contradiction.

Consequently, there must exist a finite subset $N \subset G$ which satisfies condition (C) and the proof is thus complete.

Proof As the linear NUCA $\tau $ has finite memory, we can find a finite subset $M \subset G$ and $s \in S^G$ where $S= \mathcal {L}(V^M, V)$ such that $\tau = \sigma _s$ . By hypothesis, the configuration s is asymptotic to a constant configuration $c \in S^G$ . Hence, we can, up to enlarging M, suppose that $s\vert _{G \setminus M}= c\vert _{G \setminus M}$ and that $1_G \in M$ .

Assume first that G is a finitely generated infinite group. Then we infer from Lemma 4.1 that there exists a finite subset $N \subset G$ such that for any $d\in \tau (V^G)$ and $g \in G$ , the element $\tau ^{-1}(d)(g)\in V$ depends only on the restriction $d \vert _{gN}$ . Up to enlarging M and N, we can clearly suppose that $M=N$ . Consequently, for each $g \in G$ , we have a well-defined map $\varphi _g\mathop{\colon} \tau (V^G)_{gM} \to V$ given by $d\vert _{gM} \mapsto \tau ^{-1}(d)(g)$ for every $d \in V^G$ .

Since $\tau $ is linear and $\tau (V^G)_{gM}$ is a linear subspace of $V^{gM}$ , it follows that $\varphi _g$ is also a linear map and we can extend $\varphi _g$ to a linear map $\tilde {\varphi }_g \mathop{\colon} V^{gM} \to V$ which coincides with $\varphi _g$ on $\tau (V^G)_{gM}$ . Let $\phi _g \mathop{\colon} V^M \to V^{gM}$ be the canonical automorphism induced by the bijection $M \simeq gM$ , $h \mapsto gh$ . Let us define an configuration $t \in S^G$ where $S= \mathcal {L}(V^M, V)$ by setting $t(g)= \tilde {\varphi }_g \circ \phi _g \mathop{\colon} V^M \to M$ for every $g \in G$ . It is immediate from the construction that for every $c \in V^G$ , $g \in G$ , and $d = \tau (c) \in \tau (V^G)$ , we have

$$ \begin{align*} \sigma_t(\sigma_s(c))(g)= \sigma_t(d)(g)=t(g)((g^{-1}d)\vert_{M})= \tau^{-1}(d)(g)=c(g). \end{align*} $$

Therefore, $\sigma _t \circ \sigma _s = \operatorname {\mathrm {Id}}$ and we conclude that $\tau =\sigma _s$ is left-invertible. In fact, since $s\vert _{G \setminus M}= c\vert _{G \setminus M}$ , the linear spaces $W=\phi _g^{-1}(\tau (V^G)_{gM}) =\phi ^{-1}( f^+{gM,s\vert _{gM}}(V^{gM^2}))$ coincide as linear subspaces of $V^M$ for all $g \in G \setminus M M^{-1}$ . Let us fix a direct sum decomposition $V^M = W \oplus U$ of $V^M$ . Thus, if we define $\tilde {\varphi }_g$ by setting $\tilde {\varphi }_g(v)=0$ for all $v \in \phi _g(U)$ and $\tilde {\varphi }_g(v)=\varphi _g(v)$ if $v \in \phi _g(W)$ and extend by linearity on the whole space $V^{gM}$ , then it is clear that t is also asymptotically constant, which completes the proof of the theorem in the case when G is a finitely generated infinite group.

The case when G is a finite group is trivial since every injective endomorphism of a finite-dimensional vector space is an automorphism. Let us consider the general case where G is an infinite group. Let H be the subgroup of G generated by M. Let $G/H=\{gH \mathop{\colon} g \in G\}$ be the set of all right cosets of H in G. By identifying $x \in A^G$ with $(x\vert _{u})_{u \in G/H}$ , we obtain a factorization $A^G = \prod _{u \in G/H} A^u$ . Moreover, $\sigma _s= \prod _{u \in G/H} \sigma _s^u$ , where $\sigma _s^u\mathop{\colon} A^u \to A^u$ is given by $\sigma _{s}^u (y) = \sigma _s(x)\vert _u$ for all $y \in A^u$ and any $x \in A^G$ extending y. Similarly, we have $\sigma _c= \prod _{u \in G/H} \sigma _c^u$ .

For every coset $u \in G/H$ , let us choose $g_u\in G$ such that $g_H=1_G$ . Then, if $u \neq H$ , we have $s\vert _u =c\vert _u$ and $\sigma _s^u=\sigma _c^u$ is conjugate to the restriction CA $ \sigma _{c\vert _H} = \sigma _{c}^H \mathop{\colon} A^H \to A^H$ by the uniform homeomorphism $\phi _u \mathop{\colon} A^u \to A^H$ given by $\phi _u(y)(h)=y(g_uh)$ for all $y \in A^u$ and $h \in H$ (cf. the discussion following [Reference Ceccherini-Silberstein, Coornaert and Phung11, Lemma 2.8]). Hence, $\sigma _s$ and $\sigma _c$ are left-invertible (resp. injective) if and only if so are $\sigma _{s\vert _H}$ and $\sigma _{c \vert _H}$ (see also [Reference Ceccherini-Silberstein and Coornaert7, Theorem 1.2]). Consequently, the general case follows from the case when G is finite or when G is a finitely generated infinite group. The proof is thus complete.

Proof As in the proof of Theorem 4.2, we can suppose without loss of generality that G is a finitely generated infinite group. Since $\tau $ is a linear NUCA with finite memory and left-invertible, we can find a finite subset $M\subset G$ and $s,t \in S^G$ , where $S= \mathcal {L}(V^M, V)$ such that $\tau =\sigma _s$ and $\sigma _t \circ \sigma _s= \operatorname {\mathrm {Id}}$ . In particular, we deduce immediately that $\sigma _s$ is injective.

As s is asymptotically constant, we infer from [Reference Phung33, Lemma 8.1] that $\Sigma (s)=\{g s \mathop{\colon} g \in G\}\cup \{c\}$ for some constant configuration $c \in S^G$ . Note that by [Reference Phung33, Lemma 5.1], the injectivity of $\sigma _{gs}$ for all $g \in G$ follows from the injectivity of $\sigma _s$ . We must show that $\sigma _c$ is injective. For this, we can suppose, up to enlarging M, that $s\vert _{G \setminus M}= c\vert _{G \setminus M}$ . Since G is infinite, there exists $g \in G$ such that $g M \cap M= \varnothing $ . It follows that $s\vert _{gM}=c\vert _{gM}$ . On the other hand, we infer from the identity $\sigma _t \circ \sigma _s= \operatorname {\mathrm {Id}}$ that

$$\begin{align*}t(g) \circ f^+_{gM, s\vert_{gM}}= \pi_{gM^2, g}, \end{align*}$$

where $\pi _{F,E} \mathop{\colon} V^F \to V^E$ denotes the canonical projection induced by any inclusion of sets $E \subset F$ . Consequently, $t(g) \circ f^+_{gM, c\vert _{gM}}= \pi _{gM^2, g}$ . Since c is constant, we deduce that $\sigma _d \circ \sigma _c= \operatorname {\mathrm {Id}}$ , where $d \in S^G$ is the constant configuration defined by $d(h)=t(g)$ for all $h \in G$ . In particular, $\sigma _c$ is injective and we conclude that $\sigma _s$ is stably injective. The proof is thus complete.

Proof Since the addition is component-wise and $k[G]$ and $(k[G])[G]$ are abelian groups, $D^1(k[G])$ is also an abelian group. It is clear that $(\alpha , \beta )*(1_G, 0)=(1_G, 0)*(\alpha , \beta )=(\alpha , \beta )$ for all $(\alpha , \beta ) \in D^1(k[G])$ . Moreover, the associativity of the multiplication is satisfied since for all $(\alpha _i, \beta _i) \in D^1(k[G])$ ( $i=1,2,3$ ), we find that

$$ \begin{align*} & \left( (\alpha_1, \beta_1) * (\alpha_2, \beta_2)\right) * (\alpha_3, \beta_3) = (\alpha_1 \alpha_2, \alpha_1 \beta_2 + \beta_1 \alpha_2 + \beta_1 \beta_2) * (\alpha_3, \beta_3) \\ & = (\alpha_1 \alpha_2 \alpha_3, \alpha_1 \alpha_2 \beta_3+ \alpha_1 \beta_2 \alpha_3 + \beta_1 \alpha_2 \alpha_3+ \beta_1 \beta_2 \alpha_3+ \alpha_1 \beta_2 \beta_3+ \beta_1 \alpha_2\beta_3 + \beta_1 \beta_2\beta_3) \\ &= (\alpha_1 \alpha_2 \alpha_3, \alpha_1 \alpha_2 \beta_3+ \alpha_1 \beta_2 \alpha_3 + \alpha_1 \beta_2 \beta_3+ \beta_1 \alpha_2 \alpha_3+ \beta_1 \alpha_2\beta_3 + \beta_1 \beta_2 \alpha_3+ \beta_1 \beta_2\beta_3)\\ & = (\alpha_1, \beta_1) * (\alpha_2 \alpha_3, \alpha_2 \beta_3 + \beta_2 \alpha_3 + \beta_2 \beta_3)\\ & = (\alpha_1, \beta_1) * \left( (\alpha_2, \beta_2) * (\alpha_3, \beta_3) \right). \end{align*} $$

Finally, we see without difficulty that the distributivity of $D^1(k[G])$ follows from the distributivity of $k[G]$ and $(k[G])[G]$ . Hence, we conclude that $D^1(k[G])$ is a ring with unit $(1_G, 0)$ and neutral element $(0,0)$ .

Proof The map $\varphi $ is trivially injective. Moreover, it is a direct consequence of the definition of the addition of multiplication operations of $D^1(k[G])$ that $\varphi (\alpha _1) + \varphi (\alpha _2) = \varphi (\alpha _1+\alpha _2)$ and $\varphi (\alpha _1 \alpha _2) =\varphi (\alpha _1\alpha _2)$ for all $\alpha _1, \alpha _2 \in k[G]$ .

Proof Since $\alpha (g) , \beta (g)(h) \in M_n(k)$ for all $g,h \in G$ , it follows from (6.1) that $\tau ^\omega $ is a linear map. Let $M = \operatorname {\mathrm {supp}}(\omega ) \subset G$ (see (6.2)).

We define a configuration of local defining maps $s \in S^G$ , where $S= \mathcal {L}(V^G, V)$ as follows. For every $g \in G$ , let $s(g) \in S$ be the linear map determined for all $w \in V^M$ by

$$\begin{align*}s(g)(w) = \sum_{h \in M} \alpha(h)w + \sum_{h \in M} \beta(g)(h)w. \end{align*}$$

Let $x \in V^G$ and $g \in G$ . Then we infer from the definition (6.1) and the choice of M that

$$ \begin{align*} \sigma_s(x)(g) &= s(g)((g^{-1}x)\vert_M) \\ & = \sum_{h \in M} \alpha(h)x(gh) + \sum_{h \in M} \beta(g)(h)x(gh)\\ & = \sum_{h \in G} \alpha(h)x(gh) + \sum_{h \in G} \beta(g)(h)x(gh)\\ & = \tau^\omega(x)(g). \end{align*} $$

We deduce that $\tau ^\omega = \sigma _s$ is indeed a linear NUCA with finite memory. On the other hand, if we denote $E= \operatorname {\mathrm {supp}} \beta $ then E is a finite subset of G and we have $s(g)= \alpha (g)$ for all $g \in G \setminus E$ by construction. Consequently, s is asymptotic to the constant configuration $\alpha ^G$ . Thus, $\tau ^\omega \in \mathrm {LNUCA}_c(G, k^n)$ and the proof is complete.

Proof Let $V= k^n$ . We claim that $\Psi $ is injective. Indeed, let $\omega =(\alpha , \beta ) \in D^1(M_n(k)[G])$ be an element such that $\tau ^\omega = 0$ as a map from $V^G$ to itself. Let $M = \operatorname {\mathrm {supp}}(\omega )$ (see (6.2)) then M is a finite subset of G. Since G is infinite, we can choose some $g_0 \in G \setminus M$ . In particular, $\beta (g_0)=0$ by the choice of M. Then for every $x \in V^G$ , we find that $\tau ^\omega (x)(g_0)=0$ and it follows from (6.1) that

$$\begin{align*}\sum_{h \in M} \alpha(h)x(g_0h) = \sum_{h \in G} \alpha(h)x(g_0h)=\tau^\omega(x)(g_0)=0. \end{align*}$$

Since x is arbitrary, we deduce that $\alpha (h)=0$ for all $h \in M$ and thus $\alpha =0$ since $\operatorname {\mathrm {supp}} (\alpha ) \subset \operatorname {\mathrm {supp}}(\omega )=M$ . Consequently, we infer again from (6.1) that for all $g \in G$ :

$$\begin{align*}\sum_{h \in M} \beta(g)(h)x(gh)=\sum_{h \in G} \beta(g)(h)x(gh)=0. \end{align*}$$

Thus, $\beta (g)(h)=0$ for all $g,h \in G$ . In other words, $\beta =0$ and we conclude that $\omega =0$ . Hence, $\Psi $ is indeed injective as claimed.

To check that $\Psi $ is surjective, let $\sigma _s \in \mathrm {LNUCA}_c(G, V)$ where $s \in S^G$ for some $S=\mathcal {L}(V^M, V)$ , where $M \subset G$ is a finite subset, such that s is asymptotic to a constant configuration $c \in S^G$ . Up to enlarging M, we can also suppose that $s\vert _{G \setminus M} = c\vert _{G \setminus M}$ .

Since $c(1_G) \in \mathcal {L}(V^M, V)$ , there exist $\gamma _h \in \mathcal {L}(V,V)= M_n(k)$ for every $h \in M$ such that for all $w \in V^M$ , we have $c(1_G)(w) = \sum _{h \in M} \gamma _h w$ . Let us denote $\alpha = \sum _{h \in M}\gamma _h h \in M_n(k)[G]$ .

For each $g \in M$ , we define $ \delta _g=s(g) - c(g) \in \mathcal {L}(V^M,V)$ . By linearity, there exists uniquely $\delta _g(h) \in M_n(k)= \mathcal {L}(V,V)$ for $h \in M$ such that for all $w \in V^M$ , we have

$$\begin{align*}\delta_g(w) = \sum_{h \in M} \delta_g(h)w(h). \end{align*}$$

Hence, we obtain an element $\mu _g = \sum _{h \in M} \delta _g(h) \in M_n(k)[G]$ for every $g \in M$ . Let us denote $\beta = \sum _{g \in M} \mu _g g \in (M_n(k)[G])[G]$ and $\omega =(\alpha , \beta ) \in D^1(M_n(k)[G])$ . We claim that $\tau ^\omega = \sigma _s$ . Indeed, for every $x \in V^G$ and $g \in G$ , we find that

$$ \begin{align*} \tau^\omega(x)(g) & = \sum_{h \in G} \alpha(h) x(gh) + \sum_{h \in G} \beta(g)(h)x(gh) \\ & = \sum_{h \in M} \gamma_h x(gh) + \sum_{h \in M} \mu_g(h)x(gh)\\ &= c(1_G)((g^{-1}x)\vert_M) + \sum_{h \in M} \delta_g(h)x(gh)\\ & = c(g)((g^{-1}x)\vert_M) + \delta_g((g^{-1}x)\vert_M)\\ & = s(g)((g^{-1}x)\vert_M)\\ &=\sigma_s(x)(g). \end{align*} $$

We conclude that $\sigma _s = \tau ^\omega =\Psi (\omega )$ from which the surjectivity of the map $\Psi $ follows. It is immediate from (6.1) that $\Psi $ is a k-linear homomorphism of groups and $\Psi $ sends the unit of $D^1(M_n(k)[G])$ to the unit of $\mathrm {LNUCA}_c(G, V)$ .

To finish the proof, we have to check that for all elements $\omega =(\alpha , \beta )$ and $\omega '=(\alpha ', \beta ')$ of $D^1(M_n(k)[G])$ , we have $\Psi (\omega * \omega ') = \Psi (\omega ) \circ \Psi (\omega ')$ . Indeed, using the formula $\omega * \omega ' = (\alpha \alpha ', \alpha \beta ' + \beta \alpha '+ \beta \beta ')$ and (5.2), we can compute for all $x \in V^G$ , $g \in G$ , and $y = \tau ^{\omega '}(x)$ that

$$ \begin{align*} & \Psi(\omega * \omega') (x)(g) = \sum_{h \in G} \left( \alpha\alpha' + (\alpha \beta' + \beta \alpha'+ \beta \beta')(g)\right)(h) x(gh) \\ &\ \ = \sum_{h, t \in G} \left( \alpha(t) ( \alpha'(t^{-1}h)\kern1.2pt{+}\kern1.2pt \beta'(gt)(t^{-1}h)) \kern1.2pt{+}\kern1.2pt \beta(g)(t) (\alpha'(t^{-1}h) \kern1.2pt{+} \beta'(gt)(t^{-1}h)\right)x(gh) \\ &\ \ = \sum_{t\in G} ( \alpha(t) + \beta(g)(t))\sum_{ r \in G}\left( \alpha'(r) + \beta'(gt)(r) \right) x(gtr) \qquad \qquad (r=t^{-1}h) \\ &\ \ = \sum_{t \in G} \left( \alpha(t) + \beta(g)(t)\right) y(gt) \\ &\ \ = \tau^\omega(y)(g) \\ &\ \ = (\tau^\omega \circ \tau^{\omega'})(x)(g) \\ &\ \ = \left( \Psi(\omega)\circ \Psi(\omega')\right) (x)(g). \end{align*} $$

It follows that $\Psi (\omega * \omega ')=\Psi (\omega )\circ \Psi (\omega ')$ . The proof is thus complete.

Proof By [Reference Phung28, Lemma 9.4], there exists a canonical isomorphism of rings $ M_n(k)[G] \simeq M_n(k[G])$ given by $\sum _{g\in G}A(g) g \mapsto (\sum _{g \in G} A(g)_{ij}g)_{1 \leq i,j \leq n}$ . Consider the map $F\mathop{\colon} D^1(M_n(k)[G]) \to M_n(D^1(k[G]))$ defined as follows. For $ x= (\alpha , \beta ) \in D^1(M_n(k)[G])$ , we can write $\beta = \sum _{g \in G} \beta (g)g \in (M_n(k)[G])[G]$ where $\beta (g) = (\beta (g)_{ij})_{1 \leq i,j \leq n} \in M_n(k)[G]$ for all $g \in G$ . Then we define $F(x) \in M_n(D^1(k[G]))$ by setting for all $1 \leq i,j \leq n$ :

$$\begin{align*}F(x)_{ij} = (\alpha_{ij}, \sum_{g \in G} \beta(g)_{ij} g) \in D^1(k[G]). \end{align*}$$

It is clear that F is a bijective homomorphism of groups and that $F((I_n, 0))= J_n$ , where $I_n \in M_n(k)[G]$ and $J_n \in M_n(D^1(k[G]))$ are identity matrices of $M_n(k)[G]$ and $M_n(D^1(k[G]))$ , respectively.

Now, let $x_i=(\alpha _i, \beta _i)\in D^1(M_n(k)[G])$ for $i=1,2$ . Then $x_1*x_2=(\alpha _1 \alpha _2, \alpha _1 \beta _2+ \beta _1 \alpha _2+\beta _1 \beta _2)$ and thus

(7.2) $$ \begin{align} F(x_1*x_2)_{ij} = ((\alpha_1\alpha_2)_{ij}, \sum_{g \in G} ((\alpha_1 \beta_2+ \beta_1 \alpha_2+\beta_1 \beta_2)(g))_{ij} g ). \end{align} $$

On the other hand, we find that

$$ \begin{align*} (F(x_1)F(x_2))_{ij} = \sum_{r=1}^n F(x_1)_{ir}*F(x_2)_{rj} = \sum_{r=1}^n ((\alpha_1)_{ir}, (\beta_1)_{ir})*((\alpha_2)_{rj}, (\beta_2)_{rj}). \end{align*} $$

Therefore, if we denote $(F(x_1)F(x_2))_{ij}= (u,v)$ then

(7.3) $$ \begin{align} u= \sum_{r=1}^n (\alpha_1)_{ir} (\alpha_2)_{rj} = (\alpha_1\alpha_2)_{ij} \end{align} $$

by the definition of matrix multiplication. Moreover, we deduce from the definition of the operation $*$ that $v= \sum _{g \in G} v(g) g$ , where

$$ \begin{align*} v(g) & = \sum_{r=1}^n ((\alpha_1)_{ir}(\beta_2)_{rj})(g) + ((\beta_1)_{ir}(\alpha_2)_{rj})(g)+ ((\beta_1)_{ir}(\beta_2)_{rj})(g). \end{align*} $$

We infer from (5.2) that

$$ \begin{align*} \sum_{r=1}^n ((\alpha_1)_{ir}(\beta_2)_{rj})(g) & = \sum_{r=1}^n \sum_{h \in G}((\alpha_1)_{ir}(\beta_2)_{rj})(g)(h) h \\ & = \sum_{r=1}^n \sum_{h, t \in G} \alpha_1(t)_{ir} \beta_2(gt)(t^{-1}h)_{rj} h\\ & = \sum_{h\in G} \sum_{t\in G}\sum_{r=1}^n \alpha_1(t)_{ir} \beta_2(gt)(t^{-1}h)_{rj} h\\ & = \sum_{h \in G} (\alpha_1 \beta_2)(g)(h)_{ij} h \\ & = (\alpha_1 \beta_2)(g)_{ij}. \end{align*} $$

Similarly, we have the equalities $ \sum _{r=1}^n ((\beta _1)_{ir}(\alpha _2)_{rj})(g) = (\beta _1 \alpha _2)(g)_{ij}$ and also $ \sum _{r=1}^n ((\beta _1)_{ir}(\beta _2)_{rj})(g) = (\beta _1 \beta _2)(g)_{ij}$ . Comparing to (7.2), it follows that v is equal to the singular part of $F(x_1*x_2)_{ij}$ . Consequently, we deduce from (7.3) that for all $1 \leq i,j \leq n$ , we have

$$\begin{align*}(F(x_1)F(x_2))_{ij} = F(x_1*x_2)_{ij}. \end{align*}$$

Hence, $F(x_1)F(x_2)= F(x_1*x_2)$ and we can finally conclude that F is a ring isomorphism. The proof is thus complete.

Proof The equivalence between (ii) and (iii) results directly from Proposition 7.1 and Theorem 6.2 which imply that $\mathrm {LNUCA}_c(G,k^n) \simeq M_n(D^1(k[G]))$ .

Suppose that (i) holds and let $\tau , \sigma \in \mathrm {LNUCA}_c(G,k^n)$ be two linear NUCA such that $\tau \circ \sigma = \operatorname {\mathrm {Id}}$ . Then Theorem 4.3 implies that $\sigma $ is stably injective. Consequently, we infer from (i) that $\sigma $ is surjective. In particular, $\sigma $ is bijective and thus so is $\tau $ . It follows from $\tau \circ \sigma = \operatorname {\mathrm {Id}}$ that $\sigma \circ \tau = \operatorname {\mathrm {Id}}$ as well. This shows that the ring $\mathrm {LNUCA}_c(G,k^n)$ is directly finite. Therefore, we have shown that (i) implies (ii).

Suppose now that (ii) holds and let $\tau \in \mathrm {LNUCA}_c(G,k^n)$ be a stably injective linear NUCA. Then we deduce from Theorem 4.2 that $\tau $ is left-invertible, i.e., there exists $\sigma \in \mathrm {LNUCA}_c(G,k^n)$ such that $\sigma \circ \tau = \operatorname {\mathrm {Id}}$ . Hence, (ii) implies that $\tau \circ \sigma = \operatorname {\mathrm {Id}}$ and it follows at once that $\tau $ is surjective. Therefore, we also have that (ii) implies (i). The proof is thus complete.

Proof It is a direct consequence of Theorem 7.2.

Proof Let V be a finite-dimensional vector space, and let $\tau \in \mathrm {LNUCA}_c(G,V)$ . Then we obtain a dual linear NUCA $\tau ^* \in \mathrm {LNUCA}_c(G,V)$ whose dual is exactly $\tau $ , that is, $(\tau ^*)^* = \tau $ (see [Reference Phung37]). We infer from [Reference Phung37, Theorem A] that $\tau $ is pre-injective if and only if $\tau ^*$ is surjective and that $\tau ^*$ is stably injective if and only if $\tau $ is stably post-surjective. Hence, we deduce immediately the equivalences (i) $\iff $ (v) and (ii) $\iff $ (vi).

The equivalence (i) $\iff $ (ii) is the content of Corollary 7.3. Similarly, the exact same proof of Theorem 7.2 shows that (ii) $\iff $ (iv). Finally, the equivalence (i) $\iff $ (ii) results from Theorem 8.2. The proof is thus complete.

Proof Since the case when G is finite is clear and (ii) $\implies $ (i) trivially, we suppose in the rest of the proof that G is an infinite group which satisfies (i). Let V be a finite-dimensional vector space over a field k (not necessarily finite), and let $\tau \in \mathrm {LNUCA}_c(G, V)$ . Suppose that $\tau $ is stably injective. Then, by definition, we can choose a finite subset $M \subset G$ with $1_G \in M=M^{-1}$ and a configuration $s \in \mathcal {L}(V^M,V)^G$ which is asymptotic to a constant configuration $c \in \mathcal {L}(V^M,V)^G$ such that $\tau =\sigma _s$ and $s_{G \setminus M}=c\vert _{G \setminus M}$ . We infer from Theorem 4.2 that $\tau $ is left-invertible. Hence, we can find $\sigma \in \mathrm {LNUCA}_c(G,V)$ such that $\sigma \circ \tau =\operatorname {\mathrm {Id}}$ . Moreover, up to enlarging the finite set M, we can find $t \in \mathcal {L}(V^M, V)^G$ asymptotic to a constant configuration $d\in \mathcal {L}(V^M, V)^G$ such that $\sigma = \sigma _t$ and $t\vert _{G \setminus M}=d\vert _{G \setminus M}$ .

Let us denote $\Gamma = \tau (V^G)$ . As $\sigma _t \circ \sigma _s= \operatorname {\mathrm {Id}}$ , we deduce for all $g \in G$ that

(8.1) $$ \begin{align} f^+_{\{g\}, t(g)} \circ f^+_{gM, s \vert_{gM}} = \pi_{gM^2, \{g\}}, \end{align} $$

where $\pi _{F, E} \mathop{\colon} V^F\to V^E$ denotes the canonical projection for all sets $E \subset F$ . Consider the similar condition where we switch the role of s and t:

(8.2) $$ \begin{align} f^+_{\{g\}, s(g)} \circ f^+_{gM, t \vert_{gM}} = \pi_{gM^2, \{g\}}. \end{align} $$

Since G is infinite, we can choose a finite subset $M^*\subset G$ such that $M^2 \subsetneq M^*$ . Then observe that $\sigma _t \circ \sigma _s= \operatorname {\mathrm {Id}}$ , resp. $\sigma _s \circ \sigma _t= \operatorname {\mathrm {Id}}$ , if and only if (8.1), resp. (8.2), holds for all $g \in M^*$ (see [Reference Phung34, Lemma 2.2] for the case of CA). Hence, up to making the base change to $k'$ (replacing V, $s(g)$ , $t(g)$ resp. by $V \otimes _k k'$ , $s(g) \otimes _k k'$ , $t(g) \otimes _k k'$ etc.), where $k'$ is an algebraically closed field which contains k, we can suppose without loss of generality that k is algebraically closed.

We obtain from [Reference Phung36, Lemma 2.1] a finitely generated $\mathbb {Z}$ -algebra $R \subset k$ and R-modules of finite type $V_R$ and

$$\begin{align*}s_R, t_R \in \mathrm{Hom}_{R-mod}((V_R)^M, V_R)^G \end{align*}$$

such that for some fixed $g_0 \in M^*\setminus M^2$ , $s_R(g) = s_R(g_0)$ , $t_R(g)=t_R(g_0)$ for all $g \in G \setminus M^*$ and the following hold:

1. I. $V= V_R \otimes _R k$ ,

2. II. $s(g)= s_R(g) \otimes _R k$ and $t(g)= t_R(g) \otimes _R k$ for all $g \in M^*$ ,

where $\pi _R \mathop{\colon} (V_R)^{E_n M} \to (V_R)^{\{1_G\}}$ is the canonical projection. Essentially, we can take $R=\mathbb {Z}[\Omega ]$ , where $\Omega \subset k$ is a finite subset consisting of the entries of the matrices which represent the linear maps $s(g)$ , $t(g)$ for all $g \in M^*$ .

Let us denote $S=\operatorname {\mathrm {Spec}} R$ which is a $\mathbb {Z}$ -scheme of finite type. Then we infer from [Reference Phung36, Lemma 2.2] that the set of closed points of $V_R^{E_n}$ is given by $\Delta = \cup _{p \in \mathcal {P}, a \in S_p, d\in \mathbb {N}} H_{p,a,d}^{E_n}$ . Here, $\mathcal {P}$ denotes the set of prime numbers. By $a \in S_p= S \otimes _{\mathbb {Z}} \mathbb {F}_p$ , we mean a is a closed point of $S_p$ . In particular, $\kappa (a)$ is a finite field. The set $H_{p,a, d}$ is defined by

(8.3) $$ \begin{align} H_{p,a, d}= \{x \in V_a \mathop{\colon} \vert \kappa(x) \vert=p^r, 1 \leq r \leq d\}, \end{align} $$

which is a finite linear subspace of the finite-dimensional $\kappa (a)$ -vector space ${V_a=V_R \otimes _R \kappa (a)}$ .

Let us fix $p \in \mathcal {P}$ , $a \in S_p$ , $d \in \mathbb {N}$ and consider the configurations of local defining maps $s_a, t_a \in \mathcal {L}(V_a^M, V_a)^G$ where for all element $g \in G$ , we define $s_a(g) = s_R(g)\otimes _R \kappa (a)$ and $t_a(g) = t_R(g)\otimes _R \kappa (a)$ . Observe that $s_{a}(g)(H_{p,a,d}^M)$ and $t_{a}(g)(H_{p,a,d}^M)$ are subsets of $H_{p,a,d}$ for all $g \in G$ (cf. e.g., [Reference Phung35, Lemma 3.1]). Consequently, we can define $s_{p,a,d}, t_{p,a,d} \in \mathcal {L}(H_{p,a,d}^M, H_{p,a,d})$ by setting $s_{p,a,d} = s_a\vert _{H_{p,a,d}}$ and $t_{p,a,d} = t_a\vert _{H_{p,a,d}}$ for all $g \in G$ . Thus, we obtain well-defined linear NUCA $\sigma _{s_{p,a,d}}, \sigma _{t_{p,a,d}} \mathop{\colon} H_{p,a,d}^G \to H_{p,a,d}^G$ .

From (8.1), it is clear from our construction that for all $g \in M^*$ , we have

(8.4) $$ \begin{align} f^+_{\{g\}, t_{p,a,d}(g)} \circ f^+_{gM, s_{p,a,d} \vert_{gM}} = \pi^{p,a,d}_{gM^2, \{g\}}, \end{align} $$

where $\pi ^{p,a,d}_{F, E} \mathop{\colon} H_{p,a,d}^F\to H_{p,a,d}^E$ denotes the canonical projection for all sets $E \subset F$ . It follows that $\sigma _{t_{p,a,d}}\circ \sigma _{s_{p,a,d}}=\operatorname {\mathrm {Id}}$ . In particular, $\sigma _{s_{p,a,d}}$ is left-invertible and we deduce from Theorem 4.3 that $\sigma _{s_{p,a,d}}$ is stably injective. Since (i) holds by hypothesis and $H_{p,a,d}$ is a finite $\kappa (a)$ -vector space, $\sigma _{s_{p,a,d}}$ is surjective. It follows at once that ${\sigma _{s_{p,a,d}}\circ \sigma _{t_{p,a,d}}=\operatorname {\mathrm {Id}}}$ .

Therefore, we deduce that for every $g \in M^*$ , the equality

(8.5) $$ \begin{align} f^+_{\{g\}, s_R(g)} \circ f^+_{gM, t_R \vert_{gM}} = \pi^R_{gM^2, \{g\}}, \end{align} $$

where $\pi ^R_{gM^2, \{g\}} \mathop{\colon} V_R^{gM^2} \to V_R^{\{g\}}$ , holds over the set $\Delta = \cup _{p \in \mathcal {P}, s \in S_p, d\in \mathbb {N}} H_{p,s,d}^{gM^2}$ of all closed points of $(V_R)^{gM^2}$ . Since $V_R$ is a Jacobson scheme (cf., e.g., [Reference Phung35, Section 3]), an argument using the equalizer as in [Reference Ceccherini-Silberstein, Coornaert and Phung8, Lemma 7.2] shows that $ f^+_{\{g\}, s_R(g)} \circ f^+_{gM, t_R \vert _{gM}} = \pi ^R_{gM^2, \{g\}} $ as R-morphisms $V_R^{gM^2} \to V_R^{\{g\}}$ . Consequently, we obtain the relation (8.2) for all $g \in M^*$ by making the base change (8.5) $\otimes _R k$ . It follows that $\sigma _s \circ \sigma _t=\operatorname {\mathrm {Id}}$ and we can finally conclude that $\tau =\sigma _s$ is surjective. Therefore, we also have (i) $\implies $ (ii) and the proof is complete.

Proof Let G be an initially subamenable group, and let V be a finite vector space. Suppose that $\tau \in \mathrm {LNUCA}_c(G,V)$ is stably injective. Then we can infer without difficulty from [Reference Phung33, Theorem A] or [Reference Phung37, Theorem B] that there exist a large enough finite subset $M \subset G$ and two configurations $s,t \in \mathcal {L}(V^M, V)^G$ and another configuration ${c \in \mathcal {L}(V^M, V)^G}$ such that $\tau = \sigma _s$ , $s\vert _{G \setminus M} = t\vert _{G \setminus M} = c\vert _{G \setminus M}$ , and $\sigma _t \circ \sigma _s= \operatorname {\mathrm {Id}}$ . Up to enlarging M, we can suppose without loss of generality that $1_G \in M$ and $M=M^{-1}$ .

If $G=M^4$ then G is finite and the theorem is trivial since every injective endomorphism of $V^G$ is surjective. Consider the case $M^4 \subsetneq G$ . Let $E \subset G$ be any finite subset which contains strictly $M^4$ , that is, $M^4 \subsetneq E$ . Since the group G is initially subamenable, we can find an amenable group H and an injective map $\varphi \mathop{\colon} E \to H$ such that $\varphi (gh)=\varphi (g) \varphi (h)$ for all $g,h \in E$ such that $gh \in E$ . In particular, $\varphi (gh)=\varphi (g) \varphi (h)$ for all $g,h \in M$ . Since $M=M^{-1}$ and $1_G \in M$ , we deduce that $1_H \in \varphi (M)$ and $\varphi (M)=\varphi (M)^{-1}$ .

Up to replacing H by the subgroup generated by $\varphi (E)$ , we can suppose that H is generated by $\varphi (E)$ . As $\sigma _t \circ \sigma _s= \operatorname {\mathrm {Id}}$ , we deduce for all $g \in E$ that

(9.1) $$ \begin{align} f^+_{\{g\}, t(g)} \circ f^+_{gM, s \vert_{gM}} = \pi_{gM^2, \{g\}}, \end{align} $$

where we denote by $\pi _{F, Q} \mathop{\colon} V^F\to V^Q$ the canonical linear projection for all sets ${Q \subset F}$ . The bijection $\varphi \vert _E \mathop{\colon} E \to \varphi (E)$ induces, in particular, an isomorphism $\phi \mathop{\colon} V^{\varphi (M)} \to V^M$ . Let us fix $g_0 \in E \setminus M$ . The patterns $s\vert _{E}$ , $t\vert _E$ in turn induce the configurations $\tilde {s}, \tilde {t}\in \mathcal {L}(V^{\varphi (M)}, V)^H$ defined by $\tilde {s}(h) = s(h) \circ \phi $ , $\tilde {t}(h) = t(h) \circ \phi $ for all $h \in \varphi (E)$ and $\tilde {s}(h) = s(g_0) \circ \phi $ , $\tilde {t}(h) = t(g_0) \circ \phi $ for all $h \in H \setminus \varphi (E)$ .

Since $\varphi $ is injective, it follows from (9.1) that for all $h \in \varphi (E)$ , we have

(9.2) $$ \begin{align} f^+_{\{h\}, \tilde{t}(h)} \circ f^+_{h\varphi(M), \tilde{s} \vert_{h\varphi(M)}} = \pi_{h\varphi(M^2), \{h\}}. \end{align} $$

Consequently, we deduce that $\sigma _{\tilde {t}}\circ \sigma _{\tilde {s}}=\operatorname {\mathrm {Id}}$ . In particular, $\sigma _{\tilde {s}}$ is injective. Since H is amenable and $\tilde {s}$ is asymptotically constant by construction, we infer from [Reference Phung33, Theorem B(i)] that $\tilde {s}$ is surjective. Hence, it follows from $\sigma _{\tilde {t}}\circ \sigma _{\tilde {s}}=\operatorname {\mathrm {Id}}$ that $\sigma _{\tilde {s}}\circ \sigma _{\tilde {t}}=\operatorname {\mathrm {Id}}$ . We deduce that for every $h \in \varphi (E)$ , we have

(9.3) $$ \begin{align} f^+_{\{h\}, \tilde{s}(h)} \circ f^+_{h\varphi(M), \tilde{t} \vert_{h\varphi(M)}} = \pi_{h\varphi(M^2), \{h\}}. \end{align} $$

Therefore, via the injection $\varphi $ , we obtain for all $g \in E$ that

(9.4) $$ \begin{align} f^+_{\{g\}, s(g)} \circ f^+_{gM, t \vert_{gM}} = \pi_{gM^2, \{g\}}. \end{align} $$

By the choice of E and $\varphi $ , we can thus conclude that $\sigma _s \circ \sigma _t= \operatorname {\mathrm {Id}}$ which implies in particular that $\sigma _s$ is surjective. The proof is thus complete.

Proof of Theorem C

Thanks to Theorem B, we infer, respectively, from Theorem 9.1 and [Reference Phung33, Theorem B(ii)] that all initially amenable groups and all residually finite groups are $L^1$ -surjunctive. We can thus conclude the proof of the theorem since dual $L^1$ -surjunctivity is equivalent to $L^1$ -surjunctivity also by Theorem B.