Proof. It is enough to prove that if $(\alpha _i)$ is a net in $\operatorname{Aut}(M)$ converging to the identity, then so does $(\hat {\alpha }_i)$ . We have, for $\chi \in \hat {M}$ and $x\in M$ ,

\begin{equation*}\hat {\alpha }_i(\chi )(x)=\chi (\alpha _i^{-1}(x))\end{equation*}

and thus

\begin{equation*}\hat {\alpha }_i(\chi )(x)-\chi (x)=\chi (\alpha _i^{-1}(x)-x).\end{equation*}

Then $\alpha _i^{-1}(x)-x$ converges to $0$ , uniformly for $x$ in compact subsets. Hence $\chi (\alpha _i^{-1}(x)-x)$ converges to $0$ uniformly for $(x,\chi )$ in compact subsets. This means that $\hat {\alpha }_i(\chi )$ converges to $\chi$ uniformly for $\chi$ in compact subsets. In turn, this means that $\hat {\alpha }_i$ converges to the identity uniformly on compact subsets. The same argument with $\alpha _i^{-1}$ yields that ${\hat {\alpha }_i}^{-1}$ also converges to the identity uniformly on compact subsets. So $\hat {\alpha }_i$ tends to the identity in the topology of $\operatorname{Aut}(\hat {M})$ .

Proof. Consider all pairs $(N,f)$ consisting of an $A$ -module $N$ and a homomorphism $f:M\to N$ such that for every compact $A$ -module $K$ and homomorphism $g:K\to M$ , the composition $f\circ g$ is zero. It is understood that homomorphisms in this categories are homomorphisms between the given modules so that the obvious triangle with $M$ commutes. Then the initial object in this category is the quotient map $p:M\to M/\overline {{W}(M)}$ , and its kernel is the inclusion $i:\overline {{W}(M)}\to M$ .

Write $L=M/\overline {{W}(M)}$ . Let $\hat {p}:\hat {L}\to \hat {M}$ be the Pontryagin dual map of $p$ . Consider the category of pairs $(N,f)$ consisting of an $A$ -module $N$ and a homomorphism $N\to \hat {M}$ such that for every discrete $A$ -module $D$ and homomorphism $g:\hat {M}\to D$ , we have $g\circ f=0$ . Then, by Pontryagin duality and the property of the previous paragraph, in this category, $\hat {p}$ is a terminal object, and its cokernel is $\hat {i}:\hat {M}\to \widehat {\overline {{W}(M)}}$ .

But it follows from the definition of $\Omega (\cdot )$ that a terminal object in the latter category is also given by the inclusion $\Omega (\hat {M})\to \hat {M}$ , whose cokernel is the projection $\hat {M}\to \hat {M}/\Omega (\hat {M})$ . By uniqueness of the terminal object and of taking the cokernel, we deduce that the Pontryagin dual of the inclusion $i$ is naturally the projection $\hat {M}\to \hat {M}/\Omega (\hat {M})$ .

Recall that the orthogonal of a submodule $N$ of $M$ is $N^\bot =\{\chi \in \hat {M}:\chi |_N=0\}$ . There is a canonical isomorphism $\hat {N}\simeq \hat {M}/N^\bot$ . The second fact therefore follows from the first. Alternatively, a submodule is compact if and only if its orthogonal is open. Orthogonality is an order-reversing isomorphism between the lattices of submodules of $M$ and $\hat {M}$ . Then ${W}(M)$ , as supremum (in this lattice) of the set of all compact submodules of $M$ , is mapped to the infimum of the set of all open submodules of $\hat {M}$ , and this is $\Omega (\hat {M})$ .

Proof. This is [Reference CornulierCor15, Theorem 2.8]. Let us point out that the proof in [Reference CornulierCor15], published in 2015, has been slightly corrected in the 2017 arXiv version at https://arxiv.org/abs/1306.4194. Let us also mention that the main case, that of $G$ totally disconnected, was essentially due to Trofimov. More precisely, he obtained the case of cocompact subgroups of automorphisms of locally finite connected graphs. The totally disconnected case follows by the ‘Cayley–Abels graph’ construction [Reference Cornulier and de la HarpeCH16, Proposition 2.E.9].

Proof. Let $N$ be the $A$ -submodule generated by some compact subset of $M$ with nonempty interior. Thus $N$ is an open compactly generated submodule of $M$ . So $N\rtimes A$ is a compactly generated LC group, and hence ${W}_{\mathrm{grp}}(N\rtimes A)$ is closed by Theorem2.4. We see directly from the definition that ${W}_{\mathrm{grp}}(N\rtimes A)\cap N$ equals ${W}(N)$ . Thus ${W}(N)$ is closed in $N$ . It also follows from the definition that ${W}(M)\cap N={W}(N)$ . Thus the subgroup ${W}(M)$ is locally closed, and hence it is closed.

For the additional statement, let $K$ be a compact subset of ${W}(M)$ . Above, we can choose $N$ to contain $K$ . Then the above reference also ensures that $K$ is contained in a compact normal subgroup $L$ of $N\rtimes A$ . Then $L\cap N$ is a compact submodule of $M$ containing $K$ .

Proof. We actually check the stronger facts that ${W}^\flat (M)\subseteq \Omega (M)$ and ${W}(M)\subseteq \Omega ^\sharp (M)$ . If $N_1$ is a compact parafinite submodule and $N_2$ is an open submodule, the projection of $N_1$ on $M/N_2$ is compact discrete, hence finite; since $N_1$ is parafinite, it is trivial. This shows that $N_1\subseteq N_2$ . Since this holds for all $N_1,N_2$ , we deduce the inclusion ${W}^\flat (M)\subseteq \Omega ^\sharp (M)$ . (The other inclusion is similar.)

Now assume that $A$ is compactly generated, and let us show that the reduced polycompact radical is compact for every LC $A$ -module. By Pontryagin duality, openness of the augmented codiscrete radical follows for every LC $A$ -module.

Define $N$ as the closure of ${W}^\flat (M)$ . Let $K$ be the closure of the submodule generated by some compact subset of $N$ with nonempty interior. By Proposition2.5, $K$ is compact; by construction, $K$ is open in $N$ . Since $N$ is generated by its compact parafinite submodules, this is also true for the discrete module $N/K$ , which is thus generated by its finite parafinite submodules. Since a finite parafinite module is trivial, this means that $N=K$ . Thus $N$ is compact, and being topologically generated by parafinite submodules, it is parafinite. Thus $N={W}^\flat (M)$ . Hence ${W}^\flat (M)$ is compact; it is clearly parafinite. By Pontryagin duality (applied to $\hat {M}$ ), we deduce that $\Omega ^\sharp (M)$ is open, with purely discrete quotient.

Given that ${W}^\flat (M)$ is compact parafinite and the extension of two compact parafinite modules is compact parafinite, we see that $M/{W}^\flat (M)$ is lower-sheer, and hence so is $\Omega ^\sharp (M)/{W}^\flat (M)$ . By Pontryagin duality, the latter is therefore also upper-sheer.

Proof. Upper-sheer means that every homomorphism to a purely discrete module is zero, and this clearly passes to extensions. The lower-sheer case follows by Pontryagin duality. The sheer case follows from the other two cases.

Proof. Let $N$ be an upper-sheer closed submodule of $M$ . Then the image of $N$ into every purely discrete quotient of $M$ is zero, so $N\subseteq \Omega ^\sharp (M)$ . Then $M/\Omega ^\sharp (\Omega ^\sharp (M))$ is an extension of $\Omega ^\sharp (M)/\Omega ^\sharp (\Omega ^\sharp (M))$ by $M/\Omega ^\sharp (M)$ , both of which are purely discrete (by Theorem2.6), and hence is purely discrete. So $\Omega ^\sharp (\Omega ^\sharp (M)\subseteq \Omega ^\sharp (M)$ , whence the two are equal, which means that $\Omega ^\sharp (M)$ itself is upper-sheer.

The second statement follows by Pontryagin duality.

Proof. By Proposition 2.8, $\Omega ^\sharp (N)$ is upper-sheer. Hence the image of $\Omega ^\sharp (N)$ in the purely discrete module $M/\Omega ^\sharp (M)$ is zero, which means that $\Omega ^\sharp (N)\subseteq \Omega ^\sharp (M)$ .

Proof. To obtain a contradiction, let $X\neq \{0\}$ be an orbit accumulating at zero. Let $W$ be the span of the given orbit, and $W'$ an irreducible quotient of $W$ . Then the image $X'$ of $X$ in $W'$ is also an orbit accumulating at zero, and spans $W'\neq \{0\}$ . Since $G$ has a compact closure in ${GL}(W')$ by Proposition 2.10, a nonzero orbit cannot accumulate at zero. This is a contradiction.

Proof. It is no restriction to assume that $C$ is contractible.

$\operatorname{Hom}(V,C)=0$ . Let $f:V\to C$ be a homomorphism, and $H$ the closure of the image; it is also contractible. Also, $H$ is connected. So, by the general structure of LCA groups, it has a maximal compact subgroup $W$ and the quotient $H/W$ is isomorphic to $\mathbf{R}^k$ for some $k$ . Then $H/W$ is a quotient module of $V$ and hence is distal as well; since it is contractible we deduce $H/W=\{0\}$ , so $H=W$ is compact. Also, $W$ is compact and contractible, hence trivial. So $H=\{0\}$ .

$\operatorname{Hom}(V,E)=0$ . This is clear since $V$ is connected and amorphic, which implies it is totally disconnected.

$\operatorname{Hom}(C,V)=0$ , $\operatorname{Hom}(C,E)=0$ . This is clear since no $A$ -orbit in $V$ or $E$ accumulates at zero (for $V$ , this is Corollary 2.11).

$\operatorname{Hom}(E,V)=0$ , $\operatorname{Hom}(E,C)=0$ . This is clear since $V$ and $C$ have no nonzero compact submodule, and $E$ is the union of its compact submodules.

Proof. Since being amorphic is invariant under Pontryagin duality, it is enough to consider submodules. Namely, let $M$ be amorphic, $\Omega$ a compact open submodule, and $N$ a closed submodule. Since $\Omega$ is profinite, so is $N\cap \Omega$ , and the discrete quotient $N/(N\cap \Omega )$ being a submodule of the locally finite discrete module $M/\Omega$ , it is locally finite as well. Hence $N$ is amorphic.

Proof. Let us first check that it has an open image, that is, that $M^\circ +{W}(M)$ is open. This property is trivial when $M$ is connected. It is inherited from open subgroups, and from quotients by compact subgroups. Since $M$ is compact-by-Lie, it is therefore inherited by $M$ .

For properness, first assume that $M$ is $\sigma$ -compact. The kernel is compact (namely, the diagonal of ${W}(M^\circ )$ ) and its image is open by the previous paragraph, hence is closed. So properness follows. In general, properness follows from being proper on every $\sigma$ -compact open subgroup.

The second assertion (on the kernel) is straightforward.

Let us now prove the third assertion; we can suppose that ${W}(M^\circ )=\{0\}$ . First assume that $M/M^\circ$ is elliptic, in which case we have to show $M=M^\circ +{W}(M)$ . Modding out by a compact subgroup, we can suppose that $M$ is Lie. Then the divisible group $M^\circ$ has a complement in $M$ as abstract group. This complement is a discrete subgroup, hence is a complement, and is actually locally finite, thus equals ${W}(M)$ .

Now, let us prove the second assertion in general. If by contradiction $M/M^\circ {W}(M)$ is not torsion-free, it has a nontrivial finite subgroup. Let $N$ be its inverse in $M$ . Then $N/N^\circ {W}(N)$ is finite and nontrivial. So $N$ is connected-by-elliptic, implying, by the previous paragraph, $N=N^\circ {W}(N)$ , contradiction.

Finally, clearly every homomorphism to a torsion-free discrete group is trivial on both $M^\circ$ and ${W}(M)$ , whence the ‘largest’ assertion.

Proof. This is a particular case of Lemma 2.14, in this case the map $M^\circ \times {W}_{\{0\}}(M)\to M$ of Lemma 2.14 is an isomorphism: injectivity holding because the kernel ${W}(M^\circ )$ is trivial, and surjectivity holding because $M$ is upper-sheer. If $M=C\times T$ is another decomposition ( $C$ connected and $T$ totally disconnected), clearly $C=M^\circ$ . Also $T$ being totally disconnected and upper-sheer, we have $T={W}_{\{0\}}(T)$ , so $T\subseteq {W}_{\{0\}}(M)$ and then $M=C+T$ ensures $T={W}_{\{0\}}(M)$ .

Proof.

1. (1) If $G$ is connected and Lie, then it has a f.g. dense subgroup.

2. (2) If $G$ is totally disconnected, let $X$ be Cayley–Abels for $G$ , that is, a nonempty connected graph of finite valency with a vertex-transitive action of $G$ with compact vertex stabilizers. It exists by [Reference Cornulier and de la HarpeCH16, Proposition 2.E.9]. Fix a vertex $x_0$ of $X$ and let $S$ be a finite subset of $G$ such that the neighbors of $x_0$ are the $sx_0$ for $s\in S$ . Let $\Gamma$ be the subgroup generated by $S$ . Then $\Gamma$ acts transitively on $X$ . Indeed, if $x\in \Gamma x_0$ , writing $x=\gamma x_0$ , each neighbor of $x$ has the form $\gamma s x_0$ for some $s\in S$ , and hence also belongs to $\Gamma x_0$ . If $K$ is the stabilizer of $x_0$ , we then have $G=\Gamma K$ , so $\overline {\Gamma }$ is cocompact.

3. (3) If ${W}(G^\circ )$ is trivial, then $G^\circ$ is Lie and hence has a dense f.g. subgroup, and since $G/G^\circ$ has a f.g. subgroup with cocompact closure, so does $G$ .

4. (4) In general, the previous case applies to $G/{W}(G^\circ )$ . Lifting a f.g. subgroup of the latter, we obtain a f.g. subgroup $\Gamma$ such that $\overline {\Gamma }{W}(G^\circ )$ is cocompact. Thus $\overline {\Gamma }$ is cocompact.

Proof. We can assume that ${C}_\alpha (M)$ is dense and we have to show that $M$ is $\alpha$ -contractive.

Let $V$ be a compact open subgroup, write $V^-=\bigcap _{n\ge 0}\alpha ^{-n}(V)$ (so $\alpha (V^-)\subseteq V^-$ ) and $V^{-}=\bigcup _{n\ge 0}\alpha ^{-n}(V^-)$ . For each $x\in {C}_\alpha (M)$ , the element $\alpha ^n(x)$ belongs to $V$ for $n$ large enough, say $n\ge n_0$ . Hence $\alpha ^{n_0}(x)\in V^-$ , and therefore $x\in V^{-}$ . This proves that ${C}_\alpha (M)\subseteq V^{-}$ .

Willis’s theorem (Theorem3.1) ensures that there exists $V$ such that $V^{-}$ is closed; we assume $V$ has this property. Then the density of ${C}_\alpha (M)$ ensures that $V^{-}=M$ . Note that by Baire’s theorem, this implies that $V^-$ has nonempty interior, hence is open.

Define $F=\bigcap _{n\ge 0}\alpha ^n(V^-)$ . Then $F$ is a compact submodule and $\alpha$ induces a contracting automorphism of $M/F$ . To conclude, it remains to show that $F=\{0\}$ . Since ${W}^\flat (M)=0$ , to show that $F=\{0\}$ it is enough to check that $F$ has no proper open submodule. Otherwise, modding out, we can suppose that $F$ is finite, and let us deduce that $F=\{0\}$ .

Let $\Omega$ be a compact open subgroup of $M$ with $\Omega \cap F=\{0\}$ . Since $\alpha$ is contracting modulo $F$ there exists a compact open subgroup $K$ of $\Omega$ with $\alpha (K)\subseteq K+F$ , and $K$ can be chosen arbitrary small. In particular, $K$ can be chosen to be contained in $\alpha ^{-1}(\Omega )$ . So $\alpha (K)\subseteq (K+F)\cap \Omega =K$ . Write $F'=F\smallsetminus \{0\}$ . Since $\alpha (F')=F'$ , we deduce $\alpha (K+F')\subseteq K+F'$ . It follows that $K+F'\cap {C}_\alpha (M)=\emptyset$ . By density of ${C}_\alpha (M)$ , we deduce that the open subset $K+F'$ is empty, and hence $F=\{0\}$ .

Proof. If $N$ is a discrete quotient of $M$ , then $N$ is also $\alpha$ -compactive, and clearly this implies that $N$ is a locally finite module. Thus $M$ is upper-sheer, and hence sheer. In particular, it is sheer as LCA group (as mentioned at the beginning of § 2.9).

By Theorem2.5, ${W}(M)$ is closed, and has a compact open submodule $\Omega$ . Then ${W}(M)/\Omega$ is $\alpha$ -compactive and discrete, hence finite. Hence ${W}(M)$ is compact.

Now let us prove the decomposition statement. By Corollary 2.15, we can suppose that $M$ is either Euclidean or totally disconnected. The Euclidean case is clear: in this case, since $M$ has no nonzero compact submodule, it is $\alpha$ -contractive.

Assume $M$ is totally disconnected. By Proposition 3.2, ${C}_\alpha (M)$ is closed. Consider the natural homomorphism $f:{C}_\alpha (M)\times {W}(M)\to M$ . It is proper, by compactness of ${W}(M)$ . Its kernel is therefore compact, and hence its projection to ${C}_\alpha (M)$ is a compact $\alpha$ -contractive module, hence trivial. Thus the kernel is contained in ${W}(M)$ on which the homomorphism is clearly injective. Finally, surjectivity of $f$ is given by [Reference Baumgartner and WillisBW04, Corollary 3.17].

Proof. It is enough to assume that $M$ is either $\alpha$ -contractive or $\alpha ^{-1}$ -contractive. If the $\alpha$ -contractive case works, the other case follows since $\alpha ^{-1}-1=-\alpha ^{-1}(\alpha -1)$ . So assume that $M$ is $\alpha$ -contractive.

By Siebert’s theorem [Reference SiebertSie86, Proposition 4.2] (or Corollary 2.15), we can suppose that $M$ is either Euclidean or totally disconnected. The Euclidean case being straightforward from linear algebra, assume that $M$ is totally disconnected.

Let $\Omega$ be a compact open $\mathbf{Z}[\alpha ]$ -submodule. Observe that for every $v\in M$ , we have

(3.1) \begin{equation} v\in \Omega \quad \Leftrightarrow \quad \alpha v-v\in \Omega . \end{equation}

The implication $\Rightarrow$ is clear. For the sake of a contradiction, let $v$ be a counterexample to $\Leftarrow$ , namely $\alpha v-v\in \Omega$ and $v\notin \Omega$ . Considering $n$ maximal such that $\alpha ^nv\notin \Omega$ (which exists by contractivity), and define $w=\alpha ^nv$ . Then $w\notin \Omega$ and $\alpha w\in \Omega$ . Since $\alpha v-v\in \Omega$ and $\alpha \Omega \subseteq \Omega$ , we have $\alpha ^n(\alpha v-v)=\alpha w-w\in \Omega$ . Hence $w\in \Omega$ , a contradiction.

Consider the ultrametric on $M$ defined by $d(x,y)=\exp (\inf \{n:\alpha ^n(x-y)\in \Omega \})$ . It is complete, addition-invariant and defines the topology of $M$ . Moreover, by (3.1), multiplication by $\alpha -1$ is an isometric embedding of $(M,d)$ into itself; in particular its image is closed, by completeness. It remains to check that it is surjective: indeed, if we mod out by the image, we obtain a module on which $\alpha$ is contracting but acts as the identity, so we obtain the trivial module.

Proof. For a direct product of an Euclidean and a totally disconnected module, all given subspaces split accordingly. Hence Corollary 2.15 reduces to the cases when $M$ is either Euclidean or totally disconnected (to apply the corollary, note that $M$ being a sheer $A$ -module, it is a sheer LCA group).

The case when $M$ is Euclidean is just plain linear algebra: the subspaces in the decomposition are then as follows. The sum of characteristic subspaces of $\alpha$ with respect to complex eigenvalues of modulus greater than $1$ (respectively, less than $1$ , equal to $1$ ) is ${C}_\alpha (M)$ (respectively, ${C}_{\alpha ^{-1}}(M)$ , ${\mathcal{E}}(M)$ ), and ${W}(M)=\{0\}$ .

Now assume that $M$ is totally disconnected. We then have to prove that $M={C}_\alpha (M)\oplus {C}_{\alpha ^{-1}}(M)\oplus {W}(M)$ . By Willis’s theorem (Theorem3.1), there are closed submodules $N_+$ , $N_-$ such that $N_+$ is $\alpha$ -compactive, $N_-$ is $\alpha ^{-1}$ -compactive, and $N_++N_-$ is open. Therefore, by Lemma 3.4, there are closed submodules $C_+$ , $C_-$ , and a compact submodule $\Omega$ such that $C_+$ is $\alpha$ -contractive, $C_-$ is $\alpha ^{-1}$ -contractive, $\Omega$ is compact, and $C_++C_-+\Omega$ is open. Note that $C_\pm ={C}_\pm (M)$ , and $\Omega \subseteq {W}(M)$ .

Consider the sum homomorphism $f:{C}_\alpha (M)\oplus {C}_{\alpha ^{-1}}(M)\oplus {W}(M)\to M$ . Let $K$ be its kernel. Let $s=(c_+,c_-,z)$ be an element of $K$ in this decomposition. Then $c_++c_-+z=0$ in $M$ . Thus, for every $n$ , $0=\alpha ^n(c_+)+\alpha ^n(c_-)+\alpha ^n(z)$ . Then, for $n\to \infty$ , $\alpha ^n(c_++z)$ is bounded, while $\alpha ^n(c_-)$ tends to infinity unless $c_-=0$ . This forces $c_-=0$ . Similarly, letting $n$ tend to $-\infty$ ensures $c_+=0$ . Hence $c_+=c_-=0$ , so $z=-c_+-c_-=0$ showing that $K=\{0\}$ , that is, $f$ is injective. All this argument showing that $f$ is injective with open image is true for some $\sigma$ -compact submodule of $M$ . Hence $f$ is proper.

It remains to see that $f$ is surjective. Write $T={C}_\alpha (M)+{C}_{\alpha ^{-1}}(M)+{W}(M)$ , which is the image of $f$ and which we already proved to be open; let us show that $M=T$ . Since $M$ is sheer, the discrete quotient $M/T$ is locally finite as an $A$ -module.

Take $x\in M$ . Since the discrete module $M/T$ is a locally finite module (because $M$ is sheer), for some $k\ge 1$ we have $\alpha ^kx=x$ in $M/T$ , that is, $(\alpha ^k-1)x\in T$ . Write $P={C}_\alpha (M)\oplus {C}_{\alpha ^{-1}}(M)$ , so that $T=P\oplus {W}(M)$ . We can decompose $(\alpha ^k-1)x\in T$ according to this decomposition. By Lemma 3.5, we can write $(\alpha ^k-1)x=(\alpha ^k-1)u+v$ with $u\in P$ and $v\in {W}(T)$ . Set $y=x-u$ , so that $(\alpha ^k-1)y=v$ . Modulo the closed submodule $N$ generated by $v$ (which is compact), we have $\alpha ^ky=y$ , and therefore, modulo $N$ , the submodule generated by $y$ is actually generated, as a group, by the finite set $\{y,\ldots ,\alpha ^{k-1}y\}$ . Since $M$ is elliptic as an LCA group, we deduce that the closed submodule of $M$ generated by $y$ is compact, and therefore $y\in {W}(M)$ . So $x=u+y\in P+{W}(M)\subseteq T$ .

Proof. Using Corollary 2.15, we can assume that $M$ is either totally disconnected or Euclidean (since the decomposition is preserved at each step).

We start with the easier Euclidean case: we have to prove that $B_n$ is distal. Complexify and triangulate the $A$ -action on the real vector space $B_n$ . Consider a diagonal coefficient, given by a continuous homomorphism $\omega :A\to {\mathbf{C}}^*$ . Then by construction, for each $\alpha \in S$ we have $|\omega (\alpha )|\le 1$ . Since $S=S^{-1}$ , we deduce that $S\subseteq \operatorname{Ker}(|\omega |)$ . Hence the homomorphism $\log |\omega |:A\to \mathbf{R}$ factors through the compact group $A/ \overline {\langle S\rangle }$ , and hence it is zero. Thus $|\omega |=1$ , that is, $B_n$ is distal.

Now suppose that $M$ is totally disconnected. For every $\alpha \in S$ , the module $B_n$ is amorphic over $\mathbf{Z}[\alpha ^{\pm 1}]$ . For $T\subseteq S$ symmetric, we prove by induction on $|T|$ that there is a compact open $\langle T\rangle$ -submodule; the case $T$ empty is clear. More precisely, suppose that $T$ has maximal cardinal such that this holds, with a compact open $\langle T\rangle$ -submodule $\Omega$ , and by contradiction choose $\alpha \in S\smallsetminus T$ . Since $B_n$ is amorphic over $\mathbf{Z}[\alpha ^{\pm 1}]$ , the union $\bigcup _{n\in \mathbf{Z}}\alpha ^nT$ has a compact closure. Hence the closure of the additive subgroup by this union is compact and invariant under $T\cup \{s,-s\}$ , contradiction.

Thus $M$ is an amorphic $\langle S\rangle$ -module. Let $\Omega$ be a compact open $\langle S\rangle$ -submodule. Its stabilizer in $A$ is open and contains the subgroup $\langle S\rangle$ with cocompact closure, and hence is a finite-index subgroup $B$ of $A$ . Therefore the sum $\sum _{a\in A}a\Omega$ is a finite sum, and thus is a compact open $A$ -submodule.

The existence of $n$ and $(\alpha _i)$ is ensured by Proposition 2.17: find a finite family $(\alpha _1,\ldots ,\alpha _m)$ generating a dense subgroup of a cocompact subgroup. Then set $n=2m$ , $\alpha _{m+i}=\alpha _i^{-1}$ for $1\le i\le m$ , and $\alpha _i=1$ for $i\gt 2m$ .

Proof of Theorem 4.1. Lemma 2.12 ensures uniqueness: if $P_1\oplus E_1\oplus V_1$ is another decomposition, considering projections $P_1\to E$ , $P_1\to V$ , the lemma then implies these are zero, so $P_1\subseteq P$ and similarly $E_1\subseteq E$ , $V_1\subseteq V$ , whence equality holds.

For the existence, apply Lemma 4.2 (for a suitable sequence $(\alpha _i)$ , which exists since $A$ is compactly generated). Then, in the notation preceding the lemma, $M=C_1\oplus \cdots \oplus C_n\oplus B_n$ with each $C_i$ contractible, and the lemma says that $B_n$ is the direct sum of an amorphic and a distal Euclidean module.

Proof. Modding out by the compact submodule ${W}^\flat (M)$ , we can suppose that $M$ is lower-sheer. By Theorem4.1, write $N=\Omega ^\sharp (M)=P\oplus E\oplus V$ with $E={W}(N)$ , $P$ polycontractible and $V$ distal Euclidean. Let $E'$ be a compact open submodule of $E$ . Then $Q=M/(P\oplus E'\oplus V)$ is a discrete $A$ -module, with the locally finite submodule $E/E'$ .

Let $B$ be a cocompact lattice in $A$ . Then since $Q\rtimes A$ is a compactly generated LC group, so is its cocompact lattice $Q\rtimes B$ (see Proposition 2.16). Hence $Q$ is a finitely generated $B$ -module. By noetherianity, we deduce that $E/E'$ is also a finitely generated $B$ -module, and since it is also locally finite as a $B$ -module, it is finite. Thus $E$ is compact.

Since the projection of ${W}(M)$ in the purely discrete group $M/N$ is trivial, we deduce that ${W}(M)={W}(N)=E$ is compact.

Proof. Let $N=P_2\oplus E_2\oplus V_2$ be the decomposition provided by Theorem4.1. By Lemma 2.12, $P_2\subseteq P$ , and similarly for $E_2$ and $V_2$ . So $P_2\subset P_1$ , etc. Thus $N=P_2\oplus E_2\oplus V_2\subseteq P_1\oplus E_1\oplus V_1 N$ . Thus all inclusions are equalities and the corollary is proved.

Proof. Because of Pontryagin duality, it is enough to prove one direction. Suppose that $N$ is sheer.

We apply Lemma 4.2 simultaneously to $M$ and $N$ . Namely, $M=C_1\oplus \cdots \oplus C_n\oplus B_n$ , $B_n={W}(B_n)\oplus B_n^\circ$ with ${W}(B_n)={W}(M)$ amorphic and $B_n^\circ$ distal Euclidean, and $C_i={C}_{\alpha _i}(C_i\oplus \cdots \oplus C_n\oplus B_n)$ . Consider the same decomposition for $N$ , namely $N=C'_1\oplus \cdots \oplus C'_n\oplus B'_n$ . Then from the definition it follows that $C'_i=C_i\cap N$ . Also ${W}(B'_n)\subseteq {W}(M)=B_n$ and $(B'_n)^\circ \subseteq B_n^\circ$ . So $M/N=C''_1\oplus \cdots \oplus C''_n\oplus B''_n$ with $C''_i=C_i/C'_i$ contracted by $\alpha _i$ and $B''_n=B_n/B'_n$ . Clearly $B_n^\circ /(B'_n)^\circ$ is distal Euclidean, and ${W}(B_n)/{W}(B'_n)$ , being a quotient of the amorphic module ${W}(B_n)$ , has to be amorphic as well (Lemma 2.13). Hence $M/N$ is sheer.

Proof of Theorem 4.6. It is enough to prove all assertions about the sheer core, since those about the sheer envelope follow by Pontryagin duality (using Theorem4.5 in some cases).

The assertion $N^\boxminus \subseteq N$ is trivial. $N^\boxminus$ is sheer: it is lower-sheer as submodule of the sheer module $M$ , and is upper-sheer by Proposition 2.8. That $N/N^\boxminus$ is purely discrete is part of Theorem2.6. If $N_1\subseteq N_2$ , then by Proposition 2.8 applied successively in $N_1$ and $N_2$ , the submodule $N_1^\boxminus$ is upper-sheer, and hence contained in $N_2^\boxminus$ .

Proof. First case: $G$ is totally disconnected. This case is actually covered by [Reference Glockner and WillisGW10, Theorem 3.3]; let us provide a much shorter argument. Let $\alpha ^{-1}$ multiply the Haar measure of $G$ by $n_G$ . Then $n_G$ is a positive integer, namely, it equals the index $[\Omega :\alpha (\Omega )]$ for some/every compact open subgroup $\Omega$ of $G$ such that $\alpha \Omega \subseteq \Omega$ . Now let us check that if $H$ is an $\langle \alpha \rangle$ -invariant closed subgroup then $n_H\le n_G$ , and if $n_H=n_G$ then $H=G$ . Then $n_G=[\Omega :\alpha (\Omega )]$ and $n_H=[\Omega \cap H:\alpha (\Omega \cap H)]$ . The inequality immediately follows; if it is an equality we deduce that $\Omega =\alpha (\Omega )(\Omega \cap H)$ . It follows that the image of $\Omega$ in $G/H$ is a compact $\langle \alpha \rangle$ -invariant subset. Since $\alpha$ acts as a contraction (to the base-point) on the coset space $G/H$ , we deduce that this is a point, that is, that $\Omega \subseteq H$ . Thus $H$ is open, and since the discrete coset space $G/H$ admits a contraction, we deduce that it is reduced to a point and thus $H=G$ . The bound immediately follows (e.g., $\log _2(n_G)$ is such a bound).

Second case: $G$ is connected, namely a $d$ -dimensional simply connected nilpotent Lie group [Reference SiebertSie86, § 2]. Then since every contractible subspace is a real subspace, we have the bound $d$ .

The general case then follows from the Siebert’s decomposition [Reference SiebertSie86, § 4].

Proof. Given Lemma 4.2, this follows from the contractible case, Lemma 4.8.

Proof. By assumption, $M\neq \{0\}$ . Clearly, $M$ is either connected or totally disconnected.

If $M$ is connected, let us conclude (we will not need the assumption on $A$ ). Then ${W}(M)$ is a compact characteristic subgroup, hence is either $M$ or trivial. In the first case, $M$ is compact (hence case (2)), and otherwise, $M$ is Euclidean. If $M$ is not distal (case (5)), let us check that $M$ is contractible (case (4)). Indeed, this means that there is some $\alpha$ acting with an eigenvalue $z$ of modulus not equal to $1$ . If $z$ is real (respectively, nonreal), then $\operatorname{Ker}(\alpha -z)$ (respectively, $\operatorname{Ker}((\alpha -z)(\alpha -\bar {z}))$ is a nontrivial subspace, invariant under $\operatorname{Aut}(M)$ , hence equal to $M$ which is then contractible.

Now suppose that $M$ is totally disconnected. Then $\Omega ^\sharp (M)$ is either zero or $M$ , and in the first case, Theorem2.6 (which uses that $A$ is compactly generated) ensures that $M$ is purely discrete (case (3)). Similarly ${W}^\flat (M)$ is either zero or $M$ , and in the second case, $M$ is compact paracompact (case (2)). Otherwise, $M$ is sheer. Using that $A$ is compactly generated (and that $M$ is totally disconnected), Theorem4.1 then ensures that $M$ is either contractible (case (6)) or amorphic.

If $M$ is amorphic, then it is compact (case (2)), discrete (case (3)), or neither (case (7)). In the latter case, it has a proper open submodule that is infinite, so is not topologically simple.

Proof. Take the second assertion for granted: write $N=W+M_n$ for $n\ge n_0$ . Then, for $n\ge n_0$ , $M_n$ is cocompact in $M_n+W=N$ , and hence is also cocompact in the smaller submodule $\overline {\bigcup _i M_i}$ . The first assertion is therefore proved.

Let us now prove the second assertion. By Corollary 4.3, ${W}(M)$ is compact. Hence, there is no restriction in assuming that ${W}(M)$ is trivial.

The module $\Omega ^\sharp (M)$ is sheer. Let us use the notation $\boxplus$ (see Theorem4.6) relative to the ambient module $\Omega ^\sharp (M)$ . Since ${W}(M)=\{0\}$ , we can write $\Omega ^\sharp (M)=P\oplus V$ with $P$ polycontractible and $V$ distal Euclidean. Hence, by Corollary 4.9, the sequence $(M_n\cap \Omega ^\sharp (M))^\boxplus$ is stationary, say, equal to a certain submodule $L$ . There is no restriction in assuming that $(M_n\cap \Omega ^\sharp (M))^\boxplus =L$ for all $n$ . By Theorem4.6, $(M_n\cap \Omega ^\sharp (M))^\boxplus$ contains $M_n\cap \Omega ^\sharp (M)$ as a closed cocompact subgroup. Hence $L\cap M_n$ is closed cocompact in $L$ . It follows that $M'_n=L+M_n$ is closed and contains $M_n$ as a closed cocompact subgroup.

We claim that $(M'_n)$ is stationary. We first observe that $M'_n\cap \Omega ^\sharp (M)=L$ . The inclusion $\subseteq$ is trivial. If $x\in M'_n\cap \Omega ^\sharp (M)$ , write it as $y+z$ , $y\in L$ , $z\in M_n$ . Since $x\in \Omega ^\sharp (M)$ , we have $z\in \Omega ^\sharp (M)$ , so $z\in M_n\cap \Omega ^\sharp (M)\subseteq L$ and hence $x\in L$ . The observation ensures that $M'_n\cap \Omega ^\sharp (M)$ is stationary. Then the claim follows because the projection of $M'_n$ in the discrete quotient is ascending, hence stationary by noetherianity.

Proof. Let $M$ be a compactly generated LC module. Let $N$ be a closed submodule. Let $N_0$ be an open, compactly generated closed submodule of $N$ . Since $M$ is $\sigma$ -compact, so is $N$ , and hence the discrete quotient $N/N_0$ is countable. Hence there is an ascending sequence $(N_n)$ of compactly generated submodules of $N$ with $N_0$ as previously defined, and $\bigcup N_i=N$ . By Theorem4.13, for some $n$ , $N_n$ is cocompact in $N$ . Since $N_n$ is compactly generated and cocompact in $N$ , it follows that $N$ is compactly generated (Proposition 2.16).