3.1 General theory

We define the following model-theoretic connected components of ${\bar {R}}$ in a way analogous to model-theoretic connected components of groups.

The existence of all these components (over the fixed set A) is clear and in fact the index of the smallest one (i.e., ${\bar {R}}^{000}_{A,ring}$ ) in $\bar R$ is bounded by $2^{|\mathcal {L}| +|A|}$ , as the relation of lying in the same coset of ${\bar {R}}^{000}_{A,ring}$ is coarser than the finest bounded, A-invariant equivalence relation (i.e., the relation of having the same Lascar strong type over A) which is well-known to have at most $2^{|\mathcal {L}| +|A|}$ classes.

The following inclusions are obvious:

In fact, we prove in Proposition 3.6(i)–(iii) that the components of the top row of the diagram coincide with the respective components of the bottom row. That is, there is no need to distinguish between the ring components and ideal components, which justifies item (v) of Definition 3.1. We moreover prove in Proposition 3.6(iv) that the components ${\bar {R}}^{0}_{A,ring}$ and ${\bar {R}}^{00}_{A,ring}$ also coincide in any (unital) ring. This means that among the defined components there are only at most two distinct ones, and we leave as a question whether they coincide (see Question 3.8). We will keep distinguishing the components from the diagram until after Proposition 3.6 is proven.

The following example shows that (similarly to the group 00-component) the component ${\bar {R}}^{00}_{R,ideal}$ can be thought of as a generalization of the kernel of the standard part map in the sense that it coincides with this kernel in a certain class of compact rings.

Consider the action of the monoid $({\bar {R}}, \cdot )$ on ${\bar {R}}$ by left multiplication. For any $r \in {\bar {R}}$ , the map $f_r \colon {\bar {R}} \to {\bar {R}}$ given by $f_r(x) := r \cdot x$ is an endomorphism of the additive group $({\bar {R}},+)$ . For $X \subseteq {\bar {R}}$ , define its setwise stabilizer as

$$ \begin{align*}{\mathrm {Stab}}_{\bar{R}}(X): = \left\{r \in {\bar{R}}: r \cdot X \subseteq X\right\}.\end{align*} $$

For any X, ${\mathrm {Stab}}_{\bar {R}}(X)$ is a submonoid of $({\bar {R}}, \cdot )$ . Moreover, if $X=G \leq ({\bar {R}},+)$ is a subgroup, then ${\mathrm {Stab}}_{\bar {R}}(G)$ is a subring of ${\bar {R}}$ . Further, if $X=S \subseteq {\bar {R}}$ is a subring, then ${\mathrm {Stab}}_{\bar {R}}(S)$ is the largest subring of ${\bar {R}}$ in which S is a left ideal; it is known as the left idealizer of S in ${\bar {R}}$ [Reference Goodearl7, p. 121].

We similarly consider the action of $({\bar {R}}, \cdot )$ on ${\bar {R}}$ by right multiplication and denote the setwise stabilizer of X under this action by ${\mathrm {Stab}}^{\prime }_{\bar {R}}(X):=\left \{r \in {\bar {R}}: X \cdot r \subseteq X\right \}$ .

A key point in what follows is the trivial observation below that the assumption that the index of the stabilizer is bounded is always satisfied when G is a bounded index subring of ${\bar {R}}$ .

A standard observation about the connected components of groups is that each component has only boundedly many conjugates, so it must contain their intersection. In Lemma 3.3, we instead used the assumption on the index of the stabilizer. Interestingly, the assumption that the index of the left [or right] stabilizer is bounded is sufficient to find a two-sided ideal instead of just one-sided one, as proved in the proposition below.

We are now able to prove that some of the connected components introduced in Definition 3.1 are actually equal.

Proof. Items (i) and (ii) follow from Remark 3.4 and Proposition 3.5 applied to $G=S:={\bar {R}}^{000}_{A,ring}$ and $G=S:={\bar {R}}^{00}_{A,ring}$ , respectively.

We prove (iii) and (iv). Since the quotient ring ${\bar {R}}/{\bar {R}}^{00}_{A,ideal}$ is compact, it is profinite by Fact 2.9, so there is a basis of neighborhoods of $0$ that consists of clopen two-sided ideals. Let $\pi \colon {\bar {R}} \to {\bar {R}}/{\bar {R}}^{00}_{A,ideal}$ be the quotient map. We have

$$ \begin{align*}{\bar{R}}^{00}_{A,ideal} = \bigcap\left\{\pi^{-1}[\mathcal{I}]: \mathcal{I} \text{ is a clopen two-sided ideal of } {\bar{R}}/{\bar{R}}^{00}_{A,ideal}\right\}.\end{align*} $$

Consider a clopen two-sided ideal $\mathcal {I}$ of ${\bar {R}}/{\bar {R}}^{00}_{A,ideal}$ . Both $\mathcal {J} :=\pi ^{-1}[\mathcal {I}]$ and its complement are type-definable, hence definable. Also, $\mathcal {J}$ has finite index. Since ${\bar {R}}^{00}_{A,ideal} \leq \mathcal {J}$ , the orbit of $\mathcal {J}$ under ${\mathrm {Aut}}({\bar {R}}/A)$ is bounded and so finite by definability of $\mathcal {J}$ . Thus, $\bigcap _{f \in {\mathrm {Aut}}({\bar {R}}/A)} f[\mathcal {J}]$ is A-definable with finite index. This shows that ${\bar {R}}^{00}_{A,ideal}$ is an intersection of A-definable two-sided ideals with finite index, and therefore ${\bar {R}}^{00}_{A,ideal} = {\bar {R}}^{0}_{A,ideal}$ . In particular, ${\bar {R}}^{00}_{A,ring} \subseteq {\bar {R}}^{0}_{A,ring} \subseteq {\bar {R}}^{0}_{A,ideal} = {\bar {R}}^{00}_{A,ideal} = {\bar {R}}^{00}_{A,ring}$ (where the last equality holds by (ii)), and so we get (iii) and (iv).⊣

We now adopt the notation from item (v) of Definition 3.1; that is, we write ${\bar {R}}^{000}_A$ for ${\bar {R}}^{000}_{A,ideal} (= {\bar {R}}^{000}_{A,ring})$ , ${\bar {R}}^{00}_A$ for ${\bar {R}}^{00}_{A,ideal} (= {\bar {R}}^{00}_{A,ring})$ , and ${\bar {R}}^{0}_A$ for ${\bar {R}}^{0}_{A,ideal} (= {\bar {R}}^{0}_{A,ring})$ .

Proposition 3.6 establishes that ${\bar {R}}^{00}_A = {\bar {R}}^{0}_A$ regardless of the first-order structure of R. This is in stark contrast to the case of groups. The key difference is that due to Pontryagin duality, every (unital) compact topological ring is necessarily profinite, hence totally disconnected (which forces ${\bar {R}}/{\bar {R}}^{00}_A$ and ${\bar {R}}/{\bar {R}}^{0}_A$ to be the same object). The analogous statement is not true for groups. In particular, by Corollary 2.7, given an abelian group G with infinite exponent considered with the full structure, the (compact) quotient $\bar G / \bar {G}^{00}_G$ is not profinite; it follows that $\bar {G}^{00}_G \neq \bar {G}^{0}_G$ . A concrete instance of this case is $({\mathbb {Z}}, +)$ , discussed in more detail in Example 3.18. Another counterexample is the circle group $S := S^1(\mathbb {R})$ defined in an o-minimal expansion of $\mathbb {R}$ . We have $\bar {S}^{0}_\emptyset = \bar {S}$ , but $\bar {S}^{00}_\emptyset \neq \bar {S}^{0}_\emptyset $ as it consists of the infinitesimal elements of $\bar {S}$ .

Regarding the components ${\bar {R}}^{00}_{A}$ and ${\bar {R}}^{000}_{A}$ , let us write explicitly what we have observed in the first paragraph of the proof of Proposition 3.5.

This question is strongly related to some problems concerning our computation of the type-definable connected component of unitriangular groups, which will be discussed in Section 4.2 after Question 4.6. In particular, see Lemma 4.8 for equivalent statements.

We conclude this subsection with a discussion on what happens if we drop the assumption that R is unital. First, observe that this assumption is not needed in Lemma 3.3, Remark 3.4 and Proposition 3.5 (working with $J(G) \cap G$ in place of $J(G)$ in the proof). However, the assumption that R is unital was used in the proofs of Proposition 3.6 (iii) and (iv). Nevertheless, it turns out that (iii) holds also for non-unital rings, which is explained below, whereas (iv) fails in general: to see it, start from any abelian group $(R,+,\dots )$ for which $(\bar R,+)^{00}_A \ne (\bar R,+)^0_A$ and turn it into a (non-unital) ring with the trivial multiplication. Then the above additive group components coincide with the respective ring components, so ${\bar {R}}^{00}_{A} = (\bar R,+)^{00}_A \ne (\bar R,+)^0_A= {\bar {R}}^{0}_{A}$ .

The proofs of Lemma 3.3, Remark 3.4 and Proposition 3.5 can be easily adapted to yield the following lemma.

3.2 Characterization of the ring components

We now give a characterization of the ring components in terms of subgroups of the additive group. For convenience, the following result is stated in two parts, even though the components ${\bar {R}}^{00}_A$ and ${\bar {R}}^{0}_A$ are equal.

Proof. If G is a subgroup of $({\bar {R}},+)$ with bounded index such that ${\mathrm {Stab}}_{{\bar {R}}}(G)$ has bounded index, then, by Proposition 3.5(i), ${\bar {R}}^{00}_A \subseteq G$ , and so:

$$ \begin{align*} {\bar{R}}^{00}_A & \subseteq \bigcap \left\{G \leq ({\bar{R}},+) : G \text{ is } A\text{-type-definable}, [{\bar{R}} : G] < \kappa, [{\bar{R}} : {\mathrm {Stab}}_{{\bar{R}}}(G)] < \kappa\right\} \\ & \subseteq \bigcap \left\{G \leq ({\bar{R}},+) : G \text{ is } A\text{-definable}, [{\bar{R}} : G] < \omega, [{\bar{R}} : {\mathrm {Stab}}_{{\bar{R}}}(G)] < \omega\right\} \\ & \subseteq \bigcap \left\{I \leq ({\bar{R}},+) : I \text{ is an } A\text{-definable two-sided ideal with finite index}\right\} \\ & = {\bar{R}}^{0}_A = {\bar{R}}^{00}_A. \\[-38pt] \end{align*} $$

⊣

Conversely, we have the following lemma.

Likewise the lemma below for A-definable groups.

3.3 Ring components vs. additive group components

Our goal is to compare the connected components of ${\bar {R}}$ to the connected components of the additive group $({\bar {R}},+)$ . We start with an immediate observation.

Therefore, if ${\bar {R}}^{00}_A = ({\bar {R}},+)^{00}_A$ , then $({\bar {R}},+)^{0}_A = ({\bar {R}},+)^{00}_A$ . It is natural to ask under which conditions ${\bar {R}}^{00}_A$ is equal to one of the group components. Namely,

Our objective is now to find a characterization of when $({\bar {R}},+)^{0}_A = ({\bar {R}},+)^{00}_A$ . This equality means exactly that the group quotient ${\bar {R}}/({\bar {R}},+)^{00}_A$ is profinite (this equivalence is well-known and can be justified by an argument as in the proof of Proposition 3.6). Below is an immediate consequence of Corollary 2.7 for additive groups of rings.

A fundamental example of a ring whose additive group has infinite exponent is the ring of integers. Regardless of the structure on ${\mathbb {Z}}$ , every subgroup of $({\bar {{\mathbb {Z}}}},+)$ with finite index is of the form $n{\bar {{\mathbb {Z}}}}$ for some $n \neq 0$ , and so it is $0$ -definable. Hence, for any structure on ${\mathbb {Z}}$ , $({\bar {{\mathbb {Z}}}},+)^{0} = \bigcap _{n \neq 0}n{\bar {{\mathbb {Z}}}}$ exists and is an ideal, so it coincides with $\bar {\mathbb {Z}}^0$ (which therefore exists).

Using more explicit arguments, in [Reference Bowler, Chen and Gismatullin2, Example 4.5] the same conclusion was obtained working with the pure ring structure $({\mathbb {Z}},+,\cdot )$ .

The core argument behind Corollary 2.7 relies on harmonic analysis and the description of the Bohr compactification which it provides. On the other hand, both this corollary as well as the corollaries which we derive from it are stated in algebraic and model-theoretic terms. This leads to a question whether they can be proved by means of model-theory, e.g.:

We have already seen that ${\bar {R}}^{00}_A = {\bar {R}}^{0}_A$ may be strictly bigger than $({\bar {R}},+)^{00}_A$ . Now, we give examples where ${\bar {R}}^{00}_A$ is strictly bigger than $({\bar {R}},+)^{0}_A$ .

Lemmas 3.12 and 3.13 give us the following straightforward criteria for when the type-definable connected component of ${\bar {R}}$ differs from the connected components of $({\bar {R}},+)$ .

Observe that if $A \subseteq R$ , then on the right-hand side of the second criterion the ring ${\bar {R}}$ can be replaced by R.

Now, we give an application of the second criterion. The example ${\mathbb {Z}}_2[X]$ was suggested to us by Światosław Gal.

Example 3.23. (1) Let $R:= {\mathbb {Z}}_2[X]$ be equipped with the full structure. We will show that it satisfies the right hand side of Corollary 3.22(ii) for any $A\subseteq R$ , so ${\bar {R}}^{0}_A \neq ({\bar {R}},+)^{0}_A$ . Let $h\colon R \to {\mathbb {Z}}_2$ be given by

$$ \begin{align*}h\left(\sum a_i X^i\right) := \sum_{ k \in \omega} a_{2^k}.\end{align*} $$

Then h is an epimorphism of groups and $G := \ker h$ is a subgroup of R of index 2. We will check that $f \in {\mathrm {Stab}}_{R}(G)$ iff f is constant, which directly implies that $[({\bar {R}},+) : {\mathrm {Stab}}_{{\bar {R}}}(\bar G)]$ is infinite.

Clearly $0, 1 \in {\mathrm {Stab}}_{R}(G)$ . Now, take $f \in {\mathbb {Z}}_2[X]$ with $\deg (f) = k> 0$ . Fix some natural $n> 1$ such that $2^n - k> 2^{n-1}$ . Let $g := X^{2^n - k}$ . Then $g \in G$ , but $h(f \cdot g) = a_{2^n} = 1$ , so $f\cdot g \notin G$ .

(2) The above example generalizes to any $R:=K[\bar X]$ equipped with the full structure, where K is a field of characteristic $p>0$ and $\bar X =(X_i)_{i<\lambda }$ is a (possibly infinite) tuple of variables. Namely, let $h\colon R \to {\mathbb {Z}}_p$ be given by

$$ \begin{align*}h\left(\sum a_{\bar i} X^{\bar i}\right) := \sum_{k \in \omega} \pi(a_{2^k}),\end{align*} $$

where $\pi \colon (K,+) \to ({\mathbb {Z}}_p,+)$ is any group homomorphism which is the identity on ${\mathbb {Z}}_p$ , and $a_{2^k}$ is $a_{\bar i}$ for the tuple $\bar i$ with $2^k$ on the first position and $0$ elsewhere. As in (1), $G:=\ker (h)$ has finite index in $(R,+)$ , whereas ${\mathrm {Stab}}_R(G)$ has infinite index, because each polynomial in $K[X_0] \setminus {\mathbb {Z}}_p$ is not in ${\mathrm {Stab}}_R(G)$ . So ${\bar {R}}^{0}_A \neq ({\bar {R}},+)^{0}_A$ for any $A\subseteq R$ .

Example 3.23(1) implies that for $R :={\mathbb {Z}}[X]$ we also have ${\bar {R}}^{0}_A \neq ({\bar {R}},+)^{0}_A$ , but in order to see this, we need to make a few general remarks which may be useful in other situations, too.

Notice that whenever R is a ring definable in a structure M, then each of the components $\bar R^*_A$ and $(\bar R,+)^*_A$ (where $*\in \{0,00,000\}$ and $A \subseteq R$ ) computed with respect to the language $\mathcal {L}_{set,R}$ coincides with the one with respect to the language $\mathcal {L}_{set,M}$ .

In Example 3.18, the left hand side of the criterion in 3.22(i) holds, so the right hand side holds as well. But can one see directly that the RHS of (i) holds in this example? Also, the left hand of the criterion in 3.22(ii) fails in this example, so the right hand side fails as well, but this is trivially seen directly, as each subgroup of finite index of $({\mathbb {Z}},+)$ is an ideal.

Below we show a positive result for the case of a group component which does not depend on the parameters (which is for example always the case under NIP).

3.4 Definable compactifications of rings

We now turn our attention to the notion of definable compactifications of rings. Let us recall the notion of definable compactification.

As in the context of groups, if a ring R is considered in the full set-theoretic language $\mathcal {L}_{set,{R}}$ , then a definable [Bohr] compactification is the same thing as a classical [resp. Bohr] compactification of R considered with the discrete topology.

An essential result of [Reference Gismatullin, Penazzi and Pillay6] shows the existence and uniqueness of the definable Bohr compactification of a definable group by means of its connected components. We state an analogous result for rings.

Proposition 3.28. The definable Bohr compactification of R is ${\bar {R}}/{\bar {R}}^{00}_R$ with the natural map

$$ \begin{align*}R \ni r \mapsto r/{\bar{R}}^{00}_R \in {\bar{R}}/{\bar{R}}^{00}_R.\end{align*} $$

The above definition and proposition are valid also for non-unital rings. Let us note that in contrast with groups, by Fact 2.9, the definable Bohr compactification of a unital ring R coincides with the universal definable, profinite compactification of R.

3.5 Model-theoretic connected components of topological rings and the classical Bohr compactification

In [Reference Gismatullin, Penazzi and Pillay6] and in Section 2 of [Reference Krupiński and Pillay14], the classical Bohr compactification of a topological group G was described as $\bar G/{\bar G}^{00}_{\mathrm {top}}$ , where G is equipped with a structure in which all open sets are 0-definable (e.g., with the full structure), where ${\bar G}^{00}_{\mathrm {top}}$ can be described as the smallest 0-type-definable, bounded index subgroup of $\bar G$ containing the infinitesimals. In fact, several equivalent definitions of ${\bar G}^{00}_{\mathrm {top}}$ are given in Section 2 of [Reference Krupiński and Pillay14].

Here, we want to present an analog for topological rings, describing their Bohr compactifications (which coincide with the universal profinite compactifications for unital rings) in terms of a suitable component, where the Bohr compactification of a topological ring R is, of course, defined as the unique universal (ring) compactification of R.

Let R be a ring 0-definable in a structure M so that all open subsets of R are 0-definable (e.g., $M=R$ is equipped with the full structure). Let $\mu $ be the ring of infinitesimal elements in $\bar R$ . Then $R\mu \subseteq \mu $ , but $\mu $ is not necessarily a left ideal of $\bar R$ .

Since ${\bar {R}}^{00}_M$ is the smallest M-type-definable two-sided ideal [ring] of bounded index, we get the following corollary.

4.1 Some linear algebra over rings

First, we analyze the structure of the group ${\mathrm {UT}}_n(R)$ for a unital ring R. The following belongs to standard linear algebra. A matrix $B \in {\mathrm {UT}}_{n+1}(R)$ can be written as $ \begin {pmatrix} \begin {array}{c|c} {A} & {v} \\ \hline {0} & {1} \end {array}\end {pmatrix}$ for some $A \in {\mathrm {UT}}_{n}(R)$ and $v \in R^n$ . The map $\psi \colon {\mathrm {UT}}_{n+1}(R) \to {\mathrm {UT}}_{n}(R)$ given by sending B to its upper-left $n \times n$ submatrix A is a group epimorphism. Its kernel consists of all matrices of the form $ \begin {pmatrix} \begin {array}{c|c} {I} & {v} \\ \hline {0} & {1} \end {array}\end {pmatrix}$ , $v \in R^n$ , and is naturally isomorphic to $(R,+)^n$ . The short exact sequence

$$ \begin{align*}1 \to (R,+)^n \to {\mathrm{UT}}_{n+1}(R) \xrightarrow{\psi} {\mathrm{UT}}_{n}(R) \to 1\end{align*} $$

splits via the map $s \colon {\mathrm {UT}}_{n}(R) \to {\mathrm {UT}}_{n+1}(R)$ which sends A to $ \begin {pmatrix} \begin {array}{c|c} {A} & {0} \\ \hline {0} & {1} \end {array}\end {pmatrix}$ . Hence, ${\mathrm {UT}}_{n+1}(R)$ becomes a semidirect product ${\mathrm {UT}}_{n}(R) \ltimes _\phi (R,+)^n$ . With a direct calculation, we verify that the action $\phi \colon {\mathrm {UT}}_{n}(R) \to \mathrm {Aut}((R,+)^n)$ is just the standard action of ${\mathrm {UT}}_{n}(R)$ on the R-module $R^n$ :

$$ \begin{align*} \begin{pmatrix} \begin{array}{c|c} {A} & {0} \\ \hline {0} & {1} \end{array}\end{pmatrix} \begin{pmatrix} \begin{array}{c|c} {I} & {v} \\ \hline {0} & {1} \end{array}\end{pmatrix} \begin{pmatrix} \begin{array}{c|c} {A} & {0} \\ \hline {0} & {1} \end{array}\end{pmatrix}^{-1} = \begin{pmatrix} \begin{array}{c|c} {A} & {Av} \\ \hline {0} & {1} \end{array}\end{pmatrix} \begin{pmatrix} \begin{array}{c|c} {A^{-1}} & {0} \\ \hline {0} & {1} \end{array}\end{pmatrix} = \begin{pmatrix} \begin{array}{c|c} {I} & {Av} \\ \hline {0} & {1} \end{array}\end{pmatrix}. \end{align*} $$

Thus, the group operation in ${\mathrm {UT}}_{n}(R) \ltimes _\phi (R,+)^n$ is just $(A,v)\cdot (A',v') = (AA', v + Av')$ .

We now perform a similar analysis for ${\mathrm {T}}_n(R)$ . First, consider the following variant of the semidirect product of groups. Suppose that $K, H$ and N are groups, and that there are: a left action $\phi _1 \colon K \to {\mathrm {Aut}}(N)$ and a right action $\phi _2 \colon H \to {\mathrm {Aut}}(N)$ . For $k \in K, h \in H, n \in N$ , write $kn$ and $nh$ in place of $\phi _1(k)(n)$ and $\phi _2(h)(n)$ , respectively. The set $K \times H \times N$ can be equipped with the following operation:

$$ \begin{align*}(k, h, n) \cdot (k', h', n') = \left(k k', h h', ( k n')(n h')\right).\end{align*} $$

It is easy to see that this is a group operation if and only if both actions commute, that is if $k(nh) = (kn)h$ for all $k \in K, h \in H, n \in N$ . In that case, we will denote such a group as $(K, H) \ltimes _{\phi _1}^{\phi _2} N$ . The groups $K \times H$ and N are naturally embedded in $(K, H) \ltimes _{\phi _1}^{\phi _2} N$ as $K \times H \times \{1\}$ , and $\{1\} \times \{1\} \times N$ respectively. The subgroup $N \leq (K, H) \ltimes _{\phi _1}^{\phi _2} N$ is normal. The action of $K \times H$ on N by conjugation is as follows:

$$ \begin{align*}(k, h, 1) \cdot (1, 1, n) \cdot (k, h, 1)^{-1} = (k, h, 1) \cdot (1, 1, n) \cdot (k^{-1}, h^{-1}, 1) = (1,1,knh^{-1}).\end{align*} $$

Note that if either of the actions $\phi _1, \phi _2$ is trivial, then $(K, H) \ltimes _{\phi _1}^{\phi _2} N$ is just a semidirect product of $K \times H$ and N.

Now, consider a matrix $B \in {\mathrm {T}}_{n+1}(R)$ . It can be written as $ \begin {pmatrix} \begin {array}{c|c} {A} & {v} \\ \hline {0} & {r} \end {array}\end {pmatrix}$ for some $A \in {\mathrm {T}}_{n}(R), v \in R^n$ , and $r \in R^*$ . We consider a product of two matrices represented this way:

$$ \begin{align*} \begin{pmatrix} \begin{array}{c|c} {A} & {v} \\ \hline {0} & {r} \end{array}\end{pmatrix} \begin{pmatrix} \begin{array}{c|c} {A'} & {v'} \\ \hline {0} & {r'} \end{array}\end{pmatrix} = \begin{pmatrix} \begin{array}{c|c} {AA'} & {Av' + vr'} \\ \hline {0} & {rr'} \end{array}\end{pmatrix}. \end{align*} $$

From the calculation above, it follows that ${\mathrm {T}}_{n+1}(R)$ is isomorphic to the group $\left ({\mathrm {T}}_n(R), R^*\right ) \ltimes _{\phi _1}^{\phi _2} (R^n, +)$ with $A \in {\mathrm {T}}_n(R)$ acting on $R^n$ by $v \mapsto Av$ , and $R^*$ acting on $R^n$ by $v \mapsto vr$ . Hence, the conjugate of $v \in R^n$ by $(A, r) \in {\mathrm {T}}_n(R) \times R^*$ is $(Av)r^{-1} = A(vr^{-1})$ .

4.2 Discrete triangular groups

Recall that R is a unital ring, and $A\subset {\bar {R}}$ is a small set of parameters.

Our first goal is to describe ${\mathrm {UT}}({\bar {R}})^{00}_A$ , the A-type-definable connected component of ${\mathrm {UT}}_n({\bar {R}})$ , along with the quotient ${\mathrm {UT}}({\bar {R}})/{\mathrm {UT}}({\bar {R}})^{00}_A$ . In particular, for $A :=R$ , we get a description of the definable Bohr compactification of ${\mathrm {UT}}_n(R)$ ; working in $\mathcal {L}_{set,{R}}$ , this compactification coincides with the classical Bohr compactification of the discrete group ${\mathrm {UT}}_n(R)$ .

A natural candidate for the component is ${\mathrm {UT}}_n({\bar {R}}^{00}_A)$ . However, we will see that in general it may happen that ${\mathrm {UT}}({\bar {R}})^{00}_A \lneq {\mathrm {UT}}_n({\bar {R}}^{00}_A)$ .

Define a sequence $I_{i,A}({\bar {R}})$ , $i\in {\mathbb {N}}_{> 0}$ , of A-type-definable subgroups of $({\bar {R}},+)$ as follows:

• • $I_{1,A}({\bar {R}}) = ({\bar {R}},+)^{00}_A$ and

• • for $i> 0$ let $I_{i+1,A}({\bar {R}})$ be the smallest A-type-definable subgroup of $({\bar {R}},+)$ containing the set ${\bar {R}} \cdot I_{i,A}({\bar {R}})$ .

We clearly have

$$ \begin{align*}({\bar{R}},+)^{00}_A = I_{1,A}({\bar{R}}) \leq I_{2,A}({\bar{R}}) \leq \cdots \leq I_{i,A}({\bar{R}}) \leq \cdots \leq {\bar{R}}^{00}_A.\end{align*} $$

Moreover, if for some $i \in {\mathbb {N}}_{>0}$ the group $I_{i,A}({\bar {R}})$ is a two-sided ideal (or just left ideal) of ${\bar {R}}$ , then $I_{j,A}({\bar {R}}) = {\bar {R}}^{00}_A$ for all $j \geq i$ . Conversely, if $I_{j,A}({\bar {R}})$ is constant for $j \geq i$ , then $I_{i,A}({\bar {R}}) = {\bar {R}}^{00}_A$ is an ideal. Indeed, since $I_{i,A}({\bar {R}}) = I_{i+1,A}({\bar {R}})$ , it is a bounded index, A-type-definable left ideal contained in ${\bar {R}}^{00}_A$ , and so it coincides with ${\bar {R}}^{00}_A$ by Corollary 3.7.

When ${\bar {R}}$ and A are fixed, we will omit the parameters and write $I_i$ to denote $I_{i,A}({\bar {R}})$ .

Proposition 4.1.

$$ \begin{align*}{\mathrm{UT}}_n({\bar{R}})^{00}_A = \begin{pmatrix} 1 & I_1 & I_2 & \ldots & I_{n-2} & I_{n-1} \\ 0 & 1 & I_1 & \ldots & I_{n-3} & I_{n-2} \\ 0 & 0 & 1 & \ldots & I_{n-4} & I_{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 1 & I_1 \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ \end{pmatrix}.\end{align*} $$

While the groups $I_i$ need not be (two-sided) ideals, if $i, j < k$ , then for any coset $a + I_i \in ({\bar {R}},+)/I_i$ and $b + I_j \in ({\bar {R}},+)/I_j$ we have $(a + I_i)(b + I_j) \subseteq ab + I_k$ ; that is, the coset $ab + I_k$ is well-defined. Consequently, if $S = \sum _s v_s w_s$ where each $w_s$ and each $v_s$ is a coset of $I_{i_s}$ and $I_{j_s}$ , respectively, then S can be unambiguously considered as an element of $({\bar {R}},+)/I_{k}$ for any k such that $i_s, j_s < k$ for all s. In the result below, the group operation on the set of matrices is defined using this identification.

Proposition 4.2. The definable Bohr compactification of the $($ discrete $)$ group ${\mathrm {UT}}_n(R)$ is

$$ \begin{align*}{\mathrm{UT}}_n({\bar{R}})/{\mathrm{UT}}_n({\bar{R}})^{00}_R \cong \begin{pmatrix} 1 & B/I_1 & B/I_2 & \ldots & B/I_{n-2} & B/I_{n-1} \\ 0 & 1 & B/I_1 & \ldots & B/I_{n-3} & B/I_{n-2} \\ 0 & 0 & 1 & \ldots & B/I_{n-4} & B/I_{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 1 & B/I_1 \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ \end{pmatrix},\end{align*} $$

where $B := ({\bar {R}},+)$ and $\cong $ is a topological group isomorphism, with the right hand side equipped with the product topology induced from the logic topologies on the quotients $B/I_i$ . The quotient $B/I_1$ is exactly the definable Bohr compactification of $(R,+)$ .

More precisely, the definable Bohr compactification of ${\mathrm {UT}}_n(R)$ is the homomomorphism from ${\mathrm {UT}}_n(R)$ to the above group of matrices given coordinatewise as the quotients by the appropriate $I_i$ ’s.

To state the analogous results for the group ${\mathrm {T}}_n({\bar {R}})$ , we need to define another non-decreasing sequence $I_{i,A}'({\bar {R}})$ , $i\in {\mathbb {N}}_{> 0}$ , of A-type-definable subgroups of $({\bar {R}},+)$ as follows:

• • $I_{1,A}'({\bar {R}})$ is the smallest A-type-definable subgroup of $({\bar {R}},+)$ which contains $({\bar {R}},+)^{00}_A$ and which is closed under multiplication by ${\bar {R}}^*$ from both left and right.

• • for $i> 0$ let $I_{i+1,A}'({\bar {R}})$ be the smallest A-type-definable subgroup of $({\bar {R}},+)$ that contains the set ${\bar {R}} \cdot I_{i,A}'({\bar {R}}) \cdot {\bar {R}}^*$ and that is closed under multiplication by ${\bar {R}}^*$ from both left and right.

By definition and induction, we have $ I_{i,A}({\bar {R}}) \subseteq I_{i,A}'({\bar {R}}) \subseteq {\bar {R}}^{00}_A$ for all ${\bar {R}}, A, i$ . Hence, if $I_{j,A}({\bar {R}})$ is constant for $j \geq i$ , then $I_{i,A}'({\bar {R}}) = {\bar {R}}^{00}_A$ . Also, as before, if $I_{j,A}'({\bar {R}})$ is constant for $j \geq i$ , then $I_{i,A}'({\bar {R}}) = {\bar {R}}^{00}_A$ .

Again, when ${\bar {R}}$ and A are fixed, we write $I_i'$ to denote $I_{i,A}'({\bar {R}})$ .

Proposition 4.3.

$$ \begin{align*}{\mathrm{T}}_n({\bar{R}})^{00}_A = \begin{pmatrix} ({\bar{R}}^*,\cdot)_A^{00} & I_1' & I_2' & \ldots & I_{n-2}' & I_{n-1}' \\ 0 & ({\bar{R}}^*,\cdot)_A^{00} & I_1' & \ldots & I_{n-3}' & I_{n-2}' \\ 0 & 0 & ({\bar{R}}^*,\cdot)_A^{00} & \ldots & I_{n-4}' & I_{n-3}' \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & ({\bar{R}}^*,\cdot)_A^{00} & I_1' \\ 0 & 0 & 0 & \ldots & 0 & ({\bar{R}}^*,\cdot)_A^{00} \\ \end{pmatrix}.\end{align*} $$

The group operation in the result below uses the identifications analogous to those discussed before Proposition 4.2:

Proposition 4.4. The definable Bohr compactification of the $($ discrete $)$ group ${\mathrm {T}}_n(R)$ is

$$ \begin{align*}{\mathrm{T}}_n({\bar{R}})/{\mathrm{T}}_n({\bar{R}})^{00}_R \cong \begin{pmatrix} P & B/I_1' & B/I_2' & \ldots & B/I_{n-2}' & B/I_{n-1}' \\ 0 & P & B/I_1' & \ldots & B/I_{n-3}' & B/I_{n-2}' \\ 0 & 0 & P & \ldots & B/I_{n-4}' & B/I_{n-3}' \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & P & B/I_1' \\ 0 & 0 & 0 & \ldots & 0 & P \\ \end{pmatrix},\end{align*} $$

where $P := ({\bar {R}}^*,\cdot )/({\bar {R}}^*,\cdot )_R^{00}$ is the definable Bohr compactification of $(R^*,\cdot )$ , $B := ({\bar {R}},+)$ , and $\cong $ is a topological group isomorphism, with the right hand side equipped with the product topology induced from the logic topologies on the quotients $B/I_i'$ .

More precisely, the definable Bohr compactification of ${\mathrm {T}}_n(R)$ is the homomomorphism from ${\mathrm {T}}_n(R)$ to the above group of matrices given coordinatewise as the quotients by $({\bar {R}}^*,\cdot )_R^{00}$ or by the appropriate $I_i'$ ’s.

We will prove Propositions 4.1–4.4 later in this subsection.

From now on, when we compute Bohr compactications, we will be describing them only as compact groups, skipping the information about the actual homomorphisms from the groups in question to these compact groups, since these homomorphisms are always as in the last parts of Propositions 4.2 and 4.4.

The descriptions of the definable Bohr compactifications of ${\mathrm {UT}}_n(R)$ and ${\mathrm {T}}_n(R)$ given by Propositions 4.2 and 4.4 can be significantly improved under the following condition on the ring R:

(‡) $$ \begin{align}I_{i,R}({\bar{R}}) = {\bar{R}}^{00}_R \text{ for all } i \geq 2.\end{align} $$

The condition asserts exactly that the sequence $I_{i,R}({\bar {R}})$ stabilizes after (at most) two steps. Assuming $(\dagger )$ , each quotient $({\bar {R}},+)/I_{i,R}({\bar {R}})$ and $({\bar {R}},+)/I_{i,R}'({\bar {R}})$ for $i \geq 2$ is the ring definable Bohr compactification ${R}^{\mathrm {dBohr}}$ of R. Hence, by Propositions 4.2 and 4.4, we get:

Corollary 4.5. Assume R satisfies $(\dagger )$ . Then the definable Bohr compactification of the group ${\mathrm {UT}}_n(R)$ is

$$ \begin{align*}\begin{pmatrix} 1 & {(R, +)}^{\mathrm {dBohr}} & {R}^{\mathrm {dBohr}} & \ldots & {R}^{\mathrm {dBohr}} & {R}^{\mathrm {dBohr}} \\ 0 & 1 & {(R, +)}^{\mathrm {dBohr}} & \ldots & {R}^{\mathrm {dBohr}} & {R}^{\mathrm {dBohr}} \\ 0 & 0 & 1 & \ldots & {R}^{\mathrm {dBohr}} & {R}^{\mathrm {dBohr}} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 1 & {(R, +)}^{\mathrm {dBohr}} \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ \end{pmatrix}\end{align*} $$

considered with the product topology. Also, the definable Bohr compactification of the group ${\mathrm {T}}_n(R)$ is

$$ \begin{align*}\begin{pmatrix} {(R^*,\cdot)}^{\mathrm {dBohr}} & ({\bar{R}},+)/I_1' & {R}^{\mathrm {dBohr}} & \ldots & {R}^{\mathrm {dBohr}} & {R}^{\mathrm {dBohr}} \\ 0 & {(R^*,\cdot)}^{\mathrm {dBohr}} & ({\bar{R}},+)/I_1' & \ldots & {R}^{\mathrm {dBohr}} & {R}^{\mathrm {dBohr}} \\ 0 & 0 & {(R^*,\cdot)}^{\mathrm {dBohr}} & \ldots & {R}^{\mathrm {dBohr}} & {R}^{\mathrm {dBohr}} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & {(R^*,\cdot)}^{\mathrm {dBohr}} & ({\bar{R}},+)/I_1' \\ 0 & 0 & 0 & \ldots & 0 & {(R^*,\cdot)}^{\mathrm {dBohr}} \\ \end{pmatrix}\end{align*} $$

considered with the product topology.

In the next subsection, we will consider several classes of rings, each time showing that they satisfy $(\dagger )$ . This motivates the following:

Condition $(\dagger )$ is strongly related to Question 3.8, which is explained in the next two lemmas.

A positive answer to Question 3.8 is the assertion that condition (iii) of Lemma 4.8 holds, yielding $(\dagger )$ and the descriptions of the definable Bohr compactifications of ${\mathrm {UT}}_n(R)$ and ${\mathrm {T}}_n(R)$ given by Corollary 4.5. So Question 3.8 can be restated in the following enriched form.

We expect a positive answer to this question (so also to Question 4.6). This will be dealt with in a forthcoming paper of the third author and Tomasz Rzepecki. In the next subsection, we will give a positive answer in several concrete examples.

Let us only argue here that in order to answer Questions 3.8 and 4.6 for commutative, unital rings R in the full language $\mathcal {L}_{set,{R}}$ , we can restrict the context to polynomial rings over ${\mathbb {Z}}$ in possibly infinitely many variables. This essentially follows from the fact that for each commutative, unital ring there is a ring of polynomials ${\mathbb {Z}}[\bar X]$ (where $\bar X$ is a tuple of possibly infinitely many variables) and an epimorphism $f \colon {\mathbb {Z}}[\bar X] \to R$ . Indeed, let us work in the two-sorted structure M with sorts R and ${\mathbb {Z}}[\bar X]$ in the language $\mathcal {L}_{set,{M}}$ . Then all the relevant “components” associated with R computed in $\mathcal {L}_{set,{R}}$ coincide with the ones computed in $\mathcal {L}_{set,{M}}$ , and similarly for the ring ${\mathbb {Z}}[\bar X]$ ; hence, we can work in $\mathcal {L}_{set,{M}}$ . Put $P:={\mathbb {Z}}[\bar X]$ , and let $\bar P$ be the interpretation of P in the monster model $\bar M$ . Finally, since f is a 0-definable ring epimorphism, by Remark 3.24, we have $f[\bar P^{000}_\emptyset ] = {\bar {R}}^{000}_\emptyset $ , $f[\bar P^{00}_\emptyset ] = {\bar {R}}^{00}_\emptyset $ , and we easily check that $f[I_{i,\emptyset }(\bar P)]= I_{i,\emptyset }(\bar R)$ for all i.

The same holds for non-commutative rings, using free rings in non-commuting variables in place of polynomial rings.

We now show a number of lemmas needed in the proofs of Propositions 4.1–4.4. We will be using notations and observations from Section 4.1.

Proof. For $k \leq n$ , let $S_k$ be the image of the embedding of $({\bar {R}},+)^k$ into $({\bar {R}},+)^n$ by the map $(v_k,\ldots ,v_1) \mapsto (0,\ldots ,0,v_k,\ldots ,v_1)$ . We freely identify $S_k$ with $({\bar {R}},+)^k$ .

First we show (i). We prove the following by induction on k:

$$ \begin{align*}N' \cap S_k \supseteq I_k \times \cdots \times I_1.\end{align*} $$

For $k=1$ , simply observe that $N' \cap S_1$ is an A-type-definable subgroup of $({\bar {R}},+)$ of bounded index so it must contain $I_1$ . Now, suppose the statement holds for some $k> 0$ . Let

$$ \begin{align*}C = \{c \in {\bar{R}}: (c,0,0,\ldots,0) \in N' \cap S_{k+1}\}.\end{align*} $$

We have $N' \cap S_{k+1} \supseteq C \times I_k \times \cdots \times I_1$ . We conclude the induction by showing that $C \supseteq I_{k+1}$ . Let $x \in I_k$ be arbitrary and take $v := (0,x,0,\ldots ,0) \in N' \cap S_{k+1}$ . For any $r \in {\bar {R}}$ there is an $A \in {\mathrm {UT}}_n({\bar {R}})$ such that $\phi (A)(v) = (rx,x,0,\ldots ,0) \in S_{k+1}$ . As $N'$ is invariant under ${\mathrm {UT}}_n({\bar {R}})$ , we have $\phi (A)(v) \in N'$ and also $\phi (A)(v) - v = (rx,0,0,\ldots ,0) \in N' \cap S_{k+1}$ . This shows that C contains the set ${\bar {R}} \cdot I_k$ . Therefore, since C is an A-type-definable subgroup of $({\bar {R}},+)$ of bounded index, it contains $I_{k+1}$ .

We have that $N'= N' \cap S_n \supseteq I_n \times \cdots \times I_1$ . As $I_n \times \cdots \times I_1$ is A-type-definable with bounded index, it remains to show that it is invariant under the action of ${\mathrm {UT}}_n({\bar {R}})$ . Take $v = (v_n, v_{n-1}, \ldots , v_1) \in {\bar {R}}^n$ . For a unitriangular matrix A, $Av$ is of the form $(v_n + v_n', v_{n-1} + v_{n-1}', \ldots , v_2 + v_2', v_1)$ , where each $v_i'$ is an ${\bar {R}}$ -linear combination of $\{v_j : j < i\}$ . So $v \in I_n \times \cdots \times I_1$ implies $Av \in I_n \times \cdots \times I_1$ , since $I_i \supseteq I_i + {\bar {R}} I_{i-1} + \cdots + {\bar {R}} I_1$ .

We now prove (ii). First, let $r, r' \in {\bar {R}}^*$ and let I denote the $n \times n$ identity matrix. Since $N'$ is closed under the actions $\phi _1, \phi _2$ , we have $rN'r'=(rI)N'r' \subseteq N'$ , so $N'$ is closed under multiplication by ${\bar {R}}^*$ from both left and right.

Now, similarly to (i), we prove by induction on k that

$$ \begin{align*}N' \cap S_k \supseteq I_k' \times \cdots \times I_1'.\end{align*} $$

For $k=1$ , we again observe that $N' \cap S_1$ is an A-type-definable subgroup of $({\bar {R}},+)$ of bounded index. It is closed under multiplication by ${\bar {R}}^*$ from both sides, so it must contain $I_1'$ . Now, suppose the statement holds for some $k> 0$ . Define C as in the proof of item (i). Then $N' \cap S_{k+1} \supseteq C \times I_k' \times \cdots \times I_1'$ and we need to show $C \supseteq I_{k+1}'$ .

Let $x \in I_k'$ be arbitrary and take $v := (0,x,0,\ldots ,0) \in N' \cap S_{k+1}$ . For any $r \in {\bar {R}}$ and $r' \in {\bar {R}}^*$ there is an $A \in {\mathrm {T}}_n({\bar {R}})$ such that $Avr' = (rxr',xr',0,\ldots ,0) \in S_{k+1}$ . We have $Avr' \in N'$ and also $Avr' - vr' = (rxr',0,0,\ldots ,0) \in N' \cap S_{k+1}$ . This shows that C contains the set ${\bar {R}} \cdot I_k' \cdot {\bar {R}}^*$ . Since C is an A-type-definable subgroup of $({\bar {R}},+)$ of bounded index, closed under multiplication by ${\bar {R}}^*$ from both left and right, it contains $I_{k+1}'$ .

We now have that $N'= N' \cap S_n \supseteq I_n' \times \cdots \times I_1'$ . As $I_n' \times \cdots \times I_1'$ is A-type-definable with bounded index, and clearly invariant under multiplication by ${\bar {R}}^*$ from the right, it remains to show that it is invariant under the action of ${\mathrm {T}}_n({\bar {R}})$ . Take $v = (v_n, v_{n-1}, \ldots , v_1) \in {\bar {R}}^n$ . For a triangular matrix A, $Av$ is of the form $(r_n v_n + v_n', r_{n-1}v_{n-1} + v_{n-1}', \ldots , r_2 v_2 + v_2', r_1 v_1)$ , where for each i, $v_i'$ is an ${\bar {R}}$ -linear combination of $\{v_j : j < i\}$ and $r_i \in {\bar {R}}^*$ . So $v \in I_n' \times \cdots \times I_1'$ implies $Av \in I_n' \times \cdots \times I_1'$ , since $I_i' \supseteq {\bar {R}}^* I_i' + {\bar {R}} I_{i-1}' + \cdots + {\bar {R}} I_1'$ .⊣

We are now ready to prove the previously stated results.

Proof of Proposition 4.1. By Corollary 4.11 and Lemma 4.12(i), we have

$$ \begin{align*} {\mathrm{UT}}_{n+1}({\bar{R}})^{00}_A & \cong {\mathrm{UT}}_{n}({\bar{R}})^{00}_A \ltimes (I_n \times \cdots \times I_1)\\ & \cong \begin{pmatrix} \begin{array}{c|c} {{\mathrm{UT}}_{n}({\bar{R}})^{00}_A} & {\begin{array}{c}I_n \\ \vdots \\ I_1 \end{array}} \\ {0} & {1} \end{array}\end{pmatrix}, \end{align*} $$

where the isomorphisms are the obvious ones so that the first and the last group are in fact equal. Hence, the result follows by induction on n.⊣

Proof of Proposition 4.2. Write H for the group of matrices on the right hand side of the formula in Proposition 4.2. Let $F \colon {\mathrm {UT}}_n({\bar {R}}) \to H$ be the map sending a matrix $[a_{ij}] \in {\mathrm {UT}}_n({\bar {R}})$ to the matrix $[b_{ij}] \in H$ defined by

$$ \begin{align*} b_{ij} := \begin{cases} a_{ij} + I_{j-i}, & \text{if } i < j, \\ 1, & \text{if } i = j, \\ 0, & \text{otherwise.} \end{cases} \end{align*} $$

As the group operation in H is the ordinary matrix multiplication after the identification of the product of cosets $a + I_i$ , $b + I_j$ with $ab + I_k$ for all $a,b \in {\bar {R}}$ and $i,j < k$ , the map F is a group homomorphism. It is clearly onto, and Proposition 4.1 implies that $\ker (F) = {\mathrm {UT}}_n({\bar {R}})^{00}_A$ . Hence, ${\mathrm {UT}}_n({\bar {R}})/{\mathrm {UT}}_n({\bar {R}})^{00}_A \cong H$ as an abstract group. By compactness of the logic topologies, in order to see that this isomorphism is a homeomorphism, it is enough to check that it is continuous. But this is clear, as the preimage by F of a subbasic closed set S in the product topology on H (i.e., S consists of all matrices in H whose fixed $(i,j)$ -th entry belongs to a fixed closed subset of $B/I_{j-i}$ ) is type-definable.⊣

The proofs of Propositions 4.3 and 4.4 are similar to the two previous ones.

4.3 Connected components of abelian groups via characters

In this subsection, we give a description in terms of characters of the type-definable connected component of any abelian group. It will be needed to get descriptions of some Bohr compactifications in the next subsection, and may prove to be useful in future studies.

Let G be an abelian group definable in a structure M. Recall that ${\mathrm {Hom}}(G, S^1)$ is the group of all homomorphisms from G to the compact group $S^1={\mathbb {R}}/{\mathbb {Z}}=[-\frac {1}{2},\frac {1}{2})$ . By ${{\mathrm {Hom}}}_{{\mathrm {{def}}}}(G,S^1)$ we denote the subgroup consisting of all definable homomorphisms in the sense explained before Fact 2.1. Note that if all subsets of G are definable, then ${{\mathrm {Hom}}}_{{\mathrm {{def}}}}(G,S^1) = {\mathrm {Hom}}(G,S^1)$ .

The next fact follows from the proofs of Lemma 3.2 and Proposition 3.4 of [Reference Gismatullin, Penazzi and Pillay6].

Lemma 3.2 of [Reference Gismatullin, Penazzi and Pillay6] provides the following construction of $\bar {\chi }$ from the fact above. Let $g \in \bar {G}$ and let $p(x) := {\mathrm {{tp}}}(g/M)$ . For a formula $\phi (x) \in p$ , let ${\mathrm {{cl}}}\left (\chi [\phi (G)]\right )$ denote the closure of $\chi [\phi (G)]$ in $S^1$ . The set $\bigcap _{\phi \in p} {\mathrm {{cl}}}\left (\chi [\phi (G)]\right )$ is shown to be a singleton in $S^1$ , and $\bar {\chi }(g)$ is defined to be the unique element of this singleton.

Proposition 4.14. Suppose that G is an abelian group $0$ -definable in M. Then

$$ \begin{align*}(\bar{G},+)^{00}_M &= \bigcap_{\chi\in{{\mathrm{Hom}}}_{{\mathrm{{def}}}}(G,S^1)} \bigcap_{m\in{\mathbb{N}}_{>0}} \bar{\chi}^{-1}\left[\left(-\frac{1}{m},\frac{1}{m}\right)\right] \\&= \bigcap_{\chi\in{{\mathrm{Hom}}}_{{\mathrm{{def}}}}(G,S^1)} \bigcap_{m\in{\mathbb{N}}_{>0}} \bar{\chi}^{-1}\left[\left[-\frac{1}{m},\frac{1}{m}\right]\right].\end{align*} $$

Proof. The second equality is obvious. So we focus on the first equality.

( $\subseteq $ ) Observe that for every $\chi \in {{\mathrm {Hom}}}_{{\mathrm {{def}}}}(G,S^1)$ ,

$$ \begin{align*}\ker (\bar{\chi}) = \bigcap_{m\in{\mathbb{N}}_{>0}} \bar{\chi}^{-1}\left[\left(-\frac{1}{m},\frac{1}{m}\right)\right]=\bigcap_{m\in{\mathbb{N}}_{>0}} \bar{\chi}^{-1}\left[\left[-\frac{1}{m},\frac{1}{m}\right]\right]\end{align*} $$

is an M-type-definable subgroup of $\bar {G}$ of bounded index. Hence, $\ker (\bar {\chi })$ contains $(\bar {G},+)^{00}_M$ .

( $\supseteq $ ) Take $a\in \bar {G}\setminus (\bar {G},+)^{00}_M$ . Let

$$ \begin{align*}i\colon \bar{G} \to \bar{G}/ (\bar{G},+)^{00}_M \end{align*} $$

be the (M-definable) quotient map. Since $i(a)$ is not the neutral element and $\bar {G}/ (\bar {G},+)^{00}_M$ is a compact abelian group, the second part of Fact 2.2 yields $\varphi \in {\mathrm {Hom}}_c(\bar {G}/ (\bar {G},+)^{00}_M,S^1)$ with $\varphi (i(a)) \ne 0$ . Then $\chi ':= \varphi \circ i \colon \bar G \to S^1$ is a character which is definable over M. Hence, by Fact 4.13, $ \chi : = \chi ' |_{G} \colon G \to S^1$ is a definable character with $\bar \chi =\chi '$ . We get that $\bar {\chi }(a)\neq 0$ , so $a\not \in \bar {\chi }^{-1}\left [\left (-\frac {1}{m},\frac {1}{m}\right )\right ]$ for any $m\in {\mathbb {N}}$ such that $\frac {1}{m}<|\bar {\chi }(a)|$ .⊣

By Proposition 4.14 and Remark 4.15, we get

Corollary 4.16. Let G be an abelian group equipped with the full structure. Then

$$ \begin{align*}(\bar{G},+)^{00}_\emptyset = \bigcap_{\chi\in{\mathrm{Hom}}(G,S^1)} \bigcap_{m\in{\mathbb{N}}_{>0}} \overline{{\chi}^{-1}\left[\left(-\frac{1}{m},\frac{1}{m}\right)\right]}.\end{align*} $$

4.4 Triangular groups over some classical rings

We apply Propositions 4.2 and 4.4 (more precisely, Corollary 4.5) to compute definable (so also classical by equipping the ring of coefficients with the full structure) Bohr compactifications of ${\mathrm {UT}}_n(R)$ and ${\mathrm {T}}_n(R)$ for the following classical rings R: fields, ${\mathbb {Z}}$ , $K[\bar X]$ or even $K[G]$ (where K is a field and G is a group or semigroup).

For each of the above classes of rings, we first consider the group ${\mathrm {UT}}_n(R)$ . We show that the set $\bar {R} \cdot (\bar {R}, +)^{00}_R$ generates a group in finitely many steps, whence condition (ii) of Lemma 4.8 is satisfied. This shows that $(\dagger )$ holds for each of the considered rings, so we can apply Corollary 4.5 to compute the definable Bohr compactification of ${\mathrm {UT}}_n(R)$ . In fact, in these examples, the set $\bar {R} \cdot (\bar {R}, +)^{00}_R$ generates ${\bar {R}}^{00}_R$ in one step, i.e., $\bar {R} \cdot (\bar {R}, +)^{00}_R={\bar {R}}^{00}_R$ . (On the other hand, one can show that the case of ${\mathbb {Z}}[X]$ equipped with the full structure requires exactly two steps, which will be shown in the aforementioned forthcoming paper of the third author with Tomasz Rzepecki). For each R, after dealing with the compactification of ${\mathrm {UT}}_n(R)$ , we follow with the computation of the compactification of ${\mathrm {T}}_n(R)$ .

We begin with the case of an infinite field $R = K$ . For any A, $\bar {K} \cdot (\bar {K}, +)^{00}_A = \bar {K}$ and so for all $i \geq 2$ we have $I_{i,A}(\bar {K}) = \bar {K}$ , the only non-trivial ideal of $\bar {K}$ . Corollary 4.5 gives us that the definable Bohr compactification ${{\mathrm {UT}}_n(K)}^{\mathrm {dBohr}}$ of ${\mathrm {UT}}_n(K)$ is

$$ \begin{align*} \begin{pmatrix} 1 & {(K, +)}^{\mathrm {dBohr}} & 0 & \ldots & 0 & 0 \\[-1pt] 0 & 1 & {(K, +)}^{\mathrm {dBohr}} & \ldots & 0 & 0 \\[-1pt] 0 & 0 & 1 & \ldots & 0 & 0 \\[-1pt] \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\[-1pt] 0 & 0 & 0 & \ldots & 1 & {(K, +)}^{\mathrm {dBohr}} \\[-1pt] 0 & 0 & 0 & \ldots & 0 & 1 \\[-1pt] \end{pmatrix} \cong \left({(K, +)}^{\mathrm {dBohr}}\right)^{n-1}.\end{align*} $$

We similarly note that for any A we have $I_{i,A}'(\bar {K}) = \bar {K}$ for all $i \geq 1$ , so, by Corollary 4.5, the definable Bohr compactification ${{\mathrm {T}}_n(K)}^{\mathrm {dBohr}}$ of ${\mathrm {T}}_n(K)$ is

$$ \begin{align*} \begin{pmatrix} {(K^*, \cdot)}^{\mathrm {dBohr}} & 0 & \ldots & 0 \\[-1pt] 0 & {(K^*, \cdot)}^{\mathrm {dBohr}} & \ldots & 0 \\[-1pt] \vdots & \vdots & \ddots & \vdots \\[-1pt] 0 & 0 & \ldots & {(K^*, \cdot)}^{\mathrm {dBohr}} \\ \end{pmatrix} \cong \left({(K^*, \cdot)}^{\mathrm {dBohr}}\right)^{n}.\end{align*} $$

We now work with $R := {\mathbb {Z}}$ .

Proof. Since $({\bar {{\mathbb {Z}}}},+)^0 ={\mathbb {Z}}^0$ is an ideal, we clearly have $ {\bar {{\mathbb {Z}}}} \cdot ({\bar {{\mathbb {Z}}}},+)^{00}_{\mathbb {Z}} \subseteq ({\bar {{\mathbb {Z}}}},+)^{0}$ , so it is remains to prove $({\bar {{\mathbb {Z}}}},+)^{0} \subseteq {\bar {{\mathbb {Z}}}} \cdot ({\bar {{\mathbb {Z}}}},+)^{00}_{\mathbb {Z}}$ . The group $({\bar {{\mathbb {Z}}}},+)^{00}_{\mathbb {Z}}$ is the intersection of a downward directed by inclusion family $\{P_i({\bar {{\mathbb {Z}}}})\}_{i\in I}$ of 0-definable sets. For every $i \in I$ we can find $n_i \in P_i({\mathbb {Z}}) \setminus \{0\}$ . Then $n_i \cdot {\bar {{\mathbb {Z}}}} \subseteq P_i({\bar {{\mathbb {Z}}}}) \cdot {\bar {{\mathbb {Z}}}} = {\bar {{\mathbb {Z}}}} \cdot P_i({\bar {{\mathbb {Z}}}})$ . Thus, $({\bar {{\mathbb {Z}}}},+)^0 = \bigcap _{n\in {\mathbb {N}}_{>0}}n\cdot {\bar {{\mathbb {Z}}}} \subseteq {\bar {{\mathbb {Z}}}}\cdot P_i({\bar {{\mathbb {Z}}}})$ . By compactness, we conclude that

$$ \begin{align*}({\bar{{\mathbb{Z}}}},+)^0\subseteq {\bar{{\mathbb{Z}}}} \cdot \bigcap_{i \in I} P_i({\bar{{\mathbb{Z}}}}) = {\bar{{\mathbb{Z}}}} \cdot ({\bar{{\mathbb{Z}}}},+)^{00}_{\mathbb{Z}}. \\[-42pt] \end{align*} $$

⊣

This lemma implies that $(\dagger )$ holds for $R={\mathbb {Z}}$ . The quotient $({\bar {{\mathbb {Z}}}},+)/I_{1,{\mathbb {Z}}}({\bar {{\mathbb {Z}}}})$ is the definable Bohr compactification ${({\mathbb {Z}},+)}^{\mathrm {dBohr}}$ of $({\mathbb {Z}},+)$ , whereas $({\bar {{\mathbb {Z}}}},+)/I_{2,{\mathbb {Z}}}({\bar {{\mathbb {Z}}}}) = ({\bar {{\mathbb {Z}}}},+)/({\bar {{\mathbb {Z}}}},+)^{0}$ is $\hat {{\mathbb {Z}}}$ , i.e., the profinite completion of ${\mathbb {Z}}$ . So, by Corollary 4.5, we get that the definable Bohr compactification ${{\mathrm {UT}}_n({\mathbb {Z}})}^{\mathrm {dBohr}}$ of ${\mathrm {UT}}_n({\mathbb {Z}})$ is

$$ \begin{align*}\begin{pmatrix} 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} & \hat{{\mathbb{Z}}} & \ldots & \hat{{\mathbb{Z}}} & \hat{{\mathbb{Z}}} \\ 0 & 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} & \ldots & \hat{{\mathbb{Z}}} & \hat{{\mathbb{Z}}} \\ 0 & 0 & 1 & \ldots & \hat{{\mathbb{Z}}} & \hat{{\mathbb{Z}}} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ \end{pmatrix}.\end{align*} $$

One easily gets a description of the (topological) connected component of ${{\mathrm {UT}}_n({\mathbb {Z}})}^{\mathrm {dBohr}}$ . For a topological group G, we will denote its topological connected component as $G^{t}$ in order to avoid confusions with its model-theoretic components.

Proof. As the ring $\hat {{\mathbb {Z}}}$ is totally disconnected, we have

$$ \begin{align*} \left({{\mathrm{UT}}_n({\mathbb{Z}})}^{\mathrm {dBohr}}\right)^t & \cong \begin{pmatrix} 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} & 0 & \ldots & 0 & 0 \\ 0 & 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} & \ldots & 0 & 0 \\ 0 & 0 & 1 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ \end{pmatrix}^t \\ & \cong \left(\left({({\mathbb{Z}},+)}^{\mathrm {dBohr}}\right)^{n-1}\right)^t = \left(\left({({\mathbb{Z}},+)}^{\mathrm {dBohr}}\right)^t\right)^{n-1}.\\[-38pt] \end{align*} $$

⊣

Moving to ${\mathrm {T}}_n({\mathbb {Z}})$ , observe that ${\bar {{\mathbb {Z}}}}^* = {\mathbb {Z}}^* = \{1, -1\}$ , and hence $I_i' = I_i$ for all i. Then, by Proposition 4.4 or Corollary 4.5, the definable Bohr compactification ${{\mathrm {T}}_n({\mathbb {Z}})}^{\mathrm {dBohr}}$ of ${\mathrm {T}}_n({\mathbb {Z}})$ is

$$ \begin{align*}\begin{pmatrix} \pm 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} & \hat{{\mathbb{Z}}} & \ldots & \hat{{\mathbb{Z}}} & \hat{{\mathbb{Z}}} \\ 0 & \pm 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} & \ldots & \hat{{\mathbb{Z}}} & \hat{{\mathbb{Z}}} \\ 0 & 0 & \pm 1 & \ldots & \hat{{\mathbb{Z}}} & \hat{{\mathbb{Z}}} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & \pm 1 & {({\mathbb{Z}},+)}^{\mathrm {dBohr}} \\ 0 & 0 & 0 & \ldots & 0 & \pm 1 \\ \end{pmatrix}.\end{align*} $$

We are now interested in the rings of polynomials $R:=K[\bar X]$ , where K is an arbitrary infinite field and $\bar X$ is a (possibly infinite) tuple of variables. We show that ${\bar {R}} \cdot ({\bar {R}},+)^{00}_R =\bar R ={\bar {R}}^{0}_R$ (i.e., again ${\bar {R}} \cdot ({\bar {R}},+)^{00}_R$ generates ${\bar {R}}^{0}_R$ in a single step). In fact, we will work more generally with any ring R containing an infinite subfield K, covering also rings of the form $K[G]$ , where G is a group or semigroup.

Recall the notion of a thick set from [Reference Gismatullin5, Definition 3.1].

By compactness, it is clear that for any A-definable group G, each definable superset $\bar D$ of $\bar G^{00}_A$ contains a definable superset of $\bar G^{00}_A$ that is thick in $\bar G$ , namely $\bar D \cap \bar D^{-1}$ . Hence $\bar G^{00}_A$ is the intersection of some directed family of A-definable thick subsets of $\bar G$ .

Note that for an arbitrary (unital) ring R, ${\bar {R}} \cdot ({\bar {R}},+)^{00}_A \subseteq {\bar {R}}^{00}_A = {\bar {R}}^{0}_A$ . Hence, by compactness, we get

As a corollary, we extend the description of the definable Bohr compactification of ${\mathrm {UT}}_n(K)$ from the beginning of the subsection to ${\mathrm {UT}}_n(R)$ for any R containing an infinite field K. By the last proposition, $I_{i,R}({\bar {R}}) = {\bar {R}}$ for all $i> 1$ , so Corollary 4.5 yields the following description of ${{\mathrm {UT}}_n(R)}^{\mathrm {dBohr}}$ :

$$ \begin{align*} \begin{pmatrix} 1 & {(R,+)}^{\mathrm {dBohr}} & 0 & \ldots & 0 & 0 \\ 0 & 1 & {(R,+)}^{\mathrm {dBohr}} & \ldots & 0 & 0 \\ 0 & 0 & 1 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 1 & {(R,+)}^{\mathrm {dBohr}} \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ \end{pmatrix} \cong \left({(R,+)}^{\mathrm {dBohr}}\right)^{n-1}.\end{align*} $$

We now turn to the group ${\mathrm {T}}_n(R)$ for R containing an infinite field K. First, we need to show the following strengthening of Proposition 4.21 whose proof is less elementary, as it uses Proposition 4.14.

Proof. Without loss of generality, we can assume that $A =R$ (enlarging R and A if necessary).

We need to check that the right hand side of item (iii) from Lemma 4.20 is satisfied. For this, take any symmetric R-definable set $\bar D$ containing $({\bar {R}},+)^{00}_A$ . Note that $\bar D$ is thick in ${\bar {R}}$ and that its realization D in R is also thick in R. We need to show that $R =R^*D$ .

Choose a basis $\{b_i\}_{i \in I}$ for R treated as a linear space over K.

Claim. For every finite $J \subseteq I$ there is a thick subset $D_J$ of K such that $\sum _{j \in J} D_Jb_{j} \subseteq D$ .

Proof of Claim. By Proposition 4.14 and compactness, there are finitely many R-definable characters $\chi _0, \dots ,\chi _{k-1} \colon (R,+) \to S^1$ and $m \in \mathbb {N}_{>0}$ such that $\chi _0^{-1}[[-\frac {1}{m}, \frac {1}{m}]] \cap \dots \cap \chi _{k-1}^{-1}[[-\frac {1}{m}, \frac {1}{m}]]\subseteq D$ . Let $\chi _{ij} \colon (K,+) \to S^1$ be the character defined as the composition $\chi _i \circ e_j$ , where $e_j \colon (K,+) \to R$ is given by $e_j(a) :=ab_j$ . Put $n:=|J|$ . Then, for $D_{ij}:=\chi _{ij}^{-1}[[-\frac {1}{mn}, \frac {1}{mn}]]$ (where $i=0,\dots ,k-1$ and $j \in J$ ) we have

$$ \begin{align*}\chi_i\left[\sum_{j \in J} D_{ij}b_j\right] = \sum_{j \in J}\chi_{ij}\left[D_{ij}\right] \subseteq \left[-\frac{1}{m}, \frac{1}{m}\right].\end{align*} $$

Hence,

$$ \begin{align*}\sum_{j \in J} D_{ij}b_j \subseteq \chi_i^{-1}\left[\left[ -\frac{1}{m}, \frac{1}{m}\right]\right].\end{align*} $$

Since each $D_{ij}$ is thick (as the preimage of a thick set by a homomorphism), the set $D_J$ defined as $\bigcap _{i<k,j \in J} D_{ij}$ is also thick (see [Reference Gismatullin4, Lemma 1.2]). By the last displayed formula, we have

$$ \begin{align*}\sum_{j \in J} D_Jb_j \subseteq \bigcap_{i<k} \chi_i^{-1}\left[\left[ -\frac{1}{m}, \frac{1}{m}\right]\right] \subseteq D.\\[-38pt] \end{align*} $$

⊣

Claim. For every $n \in \omega $ and thick subset $D_n$ of K we have $K\cdot D_n^{\times n} = K^{\times n}$ , where $X^{\times n}$ denotes the n-fold Cartesian power, and $\cdot $ coordinatewise multiplication.

Proof of Claim. We need to show that for every $a_0,\dots , a_{n-1} \in K$ there exists $a \in K$ such that $(a_0,\dots ,a_{n-1}) \in a \cdot D_n^{\times n}$ . Since $0 \in D_n$ , we can assume that all the $a_i$ ’s are non-zero. Then the last statement is equivalent to the condition $a_0^{-1}D_n \cap \dots \cap a_{n-1}^{-1}D_n \ne \{ 0\}$ . Now, since $D_n$ is thick and each $a_i^{-1} \cdot $ is an automorphism of $(K,+)$ , each $a_i^{-1}D_n$ is thick, so the intersection of all of them is also thick by [Reference Gismatullin4, Lemma 1.2], so contains a non-zero element, because K is infinite.⊣

Now take any $x \in R$ . Then there are a finite $J \subseteq I$ and $k_j \in K$ for $j \in J$ such that $x = \sum _{j \in J}k_j b_j$ . By the first claim, there is a thick subset $D_J$ of K such that $\sum _{j \in J}D_J b_j \subseteq D$ . Hence, we have $x \in \sum _{j \in J}K b_j = K\sum _{j \in J}D_J b_j \subseteq KD$ , where the equality $\sum _{j \in J}K b_j = K\sum _{j \in J}D_J b_j$ is provided by the second claim. Thus, we have shown that $R \subseteq KD \subseteq R^*D$ , so $R=R^*D$ .⊣

From Proposition 4.22, we obtain that $I_i' = {\bar {R}}$ for all i. So, by Proposition 4.4 or Corollary 4.5, ${{\mathrm {T}}_n(R)}^{\mathrm {dBohr}}$ is

$$ \begin{align*} &\begin{pmatrix} {(R^*,\cdot)}^{\mathrm {dBohr}} & 0 & 0 & \ldots & 0 & 0 \\[1pt] 0 & {(R^*,\cdot)}^{\mathrm {dBohr}} & 0 & \ldots & 0 & 0 \\[1pt] 0 & 0 & {(R^*,\cdot)}^{\mathrm {dBohr}} & \ldots & 0 & 0 \\[1pt] \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\[1pt] 0 & 0 & 0 & \ldots & {(R^*,\cdot)}^{\mathrm {dBohr}} & 0 \\[1pt] 0 & 0 & 0 & \ldots & 0 & {(R^*,\cdot)}^{\mathrm {dBohr}} \\[1pt] \end{pmatrix}\\[1pt] &\quad \cong \left({(R^*,\cdot)}^{\mathrm {dBohr}}\right)^{n}.\end{align*} $$

4.5 Topological triangular groups

We describe how our approach can be adapted to compute the classical Bohr compactification of ${\mathrm {UT}}_n(R)$ and ${\mathrm {T}}_n(R)$ treated as topological groups with the product topology induced from the topology on R, where we assume that R is a (unital) topological ring.

In order to do that, we need first to recall how to present model-theoretically the Bohr compactification of a topological group. So let G be a topological group 0-definable in a first order structure M in such a way that all open subsets of G are 0-definable (e.g., we can work in $\mathcal {L}_{set,{M}}$ ). Following [Reference Krupiński and Pillay14, Definition 2.3], we define $\bar G^{00}_{\mathrm {top}}$ to be the smallest bounded index subgroup of $\bar G$ which is an intersection of some sets of the form $\bar U$ for U open in G. Let $\mu $ denote the intersection of the $\bar U$ ’s for U ranging over all open neighborhoods of the neutral element of G; $\mu $ is the group of infinitesimal elements of $\bar G$ . Proposition 2.1 of [Reference Gismatullin, Penazzi and Pillay6] or Fact 2.4 of [Reference Krupiński and Pillay14] says that $\bar G^{00}_{\mathrm {top}}$ is a normal subgroup of $\bar G$ , and the quotient mapping $\pi \colon G \to \bar G/\bar G^{00}_{\mathrm {top}}$ is the Bohr compactification of G (treated as a topological group). Proposition 2.5 of [Reference Krupiński and Pillay14] describes $\bar G^{00}_{\mathrm {top}}$ as the smallest M-type-definable [or 0-type-definable], bounded index subgroup of $\bar G$ which contains $\mu $ . We will be using this description rather than the original definition.

We need the following variant of Lemma 4.10.