# Omega-categorical groups and rings of finite dimension

Moreno Invitti

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Moreno Invitti. Omega-categorical groups and rings of finite dimension. 2025. hal-05355829

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# On $\omega$-categorical groups of finite dimension

Moreno Invitti

## Abstract

We prove that a finite-dimensional $\omega$-categorical group is finite-by-abelian-by-finite and that a finite-dimensional $\omega$-categorical ring is virtually finite-by-null.

## Introduction

This article analyzes the structure of $\omega$-categorical groups and rings of finite dimension.

**Definition.** A structure $\mathfrak{M}$ is said to be $\omega$-categorical if its theory $T = T(\mathfrak{M})$ has a unique countable model up to isomorphism.

Some examples of $\omega$-categorical structures are: dense linear ordered set without endpoints (this is the theory of $(\mathbb{Q}, <)$ that is $\omega$-categorical by Cantor's Theorem), random graphs, infinite vector spaces over finite fields, and boolean atomless algebras. The fundamental result that holds for these structures, proved independently by Ryll-Nardzewski [8], Svenonius [12], and Engeler [13], states that an $\omega$-categorical structure has only finitely many parameter-free definable sets, up to equivalence.

?(RyllNard)? **Theorem 0.1.** Let $\mathfrak{M}$ be a first-order structure. Then, TFAE:

1. $\mathfrak{M}$ is $\omega$-categorical;

2. In any arity, there exist finitely many parameter-free definable sets;

3. For any $n < \omega$, there exists only finitely many orbits in $M^n$ for the action of $\text{Aut}_\emptyset(\mathfrak{M})$.

From Theorem 0.1(3), we can derive the following easy corollary.

**Corollary 0.2.** Let $X$ be a definable subset of an $\omega$-categorical structure, $\{c_i\}_{i<n_C}$ a finite set of elements in $X$, $\{R_i\}_{i\in I}$ a family of relations definable over a finite set $R$ and $\{f_i\}_{j\in J}$ a family of functions definable over a finite set $F$. Then, the structure $(X, \{c_i\}_{i\le n_F}, \{R_i\}_{i\in I}, \{f_j\}_{j\in J})$ is $\omega$-categorical.

*Proof.* Let $D$ the set of definition of $X$ and $n_D = |D|$. Let $n_R = |R|$ and $n_F = |F|$ and $N = n_D + n_C + n_F + n_R$.

It is sufficient to observe that, for any $n < \omega$, the action of $\text{Aut}(\mathfrak{M})$ on the set $M^{n+N}$ has finitely many orbits. In particular, there are finitely many orbits for the subset $(d_1, \dots, d_{n_D}, c_1, \dots, c_{n_C}, f_1, \dots, f_{n_F}, r_1, \dots, r_{n_R}) \times X^n$. Since all the automorphisms of $\mathfrak{M}$ fixing $(d_1, \dots, d_{n_D}, c_1, \dots, c_{n_C}, f_1, \dots, f_{n_F}, r_1, \dots, r_{n_R})$ are automorphisms of $X$, the proof is completed. □
The study of $\omega$-categorical groups has a long history. The following results are a straightforward application of Theorem 0.1.

**Lemma 0.3.** An $\omega$-categorical group is locally finite and of finite exponent.

?(ChaSimSer)?**Lemma 0.4.** Let $G$ be an $\omega$-categorical group. Then, there exists a chain of subgroups $G = G_0 \ge G_1 \ge \dots \ge G_n = \{0\}$ such that

• $G_i$ is characteristic and definable in $G$ for any $i \le n$;

• $G_i/G_{i+1}$ is characteristically simple for any $i \le n-1$.

The main theorem about $\omega$-categorical groups is the classification of $\omega$-categorical characteristically simple groups obtained by Wilson in [3].

?(CharSimGro)?**Fact 0.5.** For each infinite, countable, $\omega$-categorical, characteristically simple group $H$, one of the following holds:

• For some prime number $p$, $H$ is an elementary abelian $p$-group;

• $H \simeq B(F)$ or $H \simeq B^{-}(F)$ for some non-abelian finite simple group $F$. $B(F)$ denotes the group of all the continuous functions from the Cantor space $C$ to $F$ (with the discrete topology) while $B^{-}(F)$ is the subgroup consisting of the functions $f \in B(F)$ such that $f(x_0) = e$ for a fixed element $x_0 \in C$.

• $H$ is a perfect $p$-group for some prime $p$.

The important question is whether, under an assumption of "tameness", we can classify $\omega$-categorical groups and $\omega$-categorical rings. In particular, the two big meta-conjectures are the following.

**Theorem 0.6.** 1. A tame $\omega$-categorical group is virtually nilpotent. A tame $\omega$-categorical ring is virtually nilpotent.

2. A supertame $\omega$-categorical group is finite-by-abelian-by-finite. A supertame $\omega$-categorical ring is finite-by-null-by-finite.

These have been proven under several conditions of tameness. If tame is substituted by stable, (1) has been proved by Felgner [10] for groups and by Baldwin and Rose [14] for rings, while (2) by Bauer, Cherlin, and Macintyre in [11]. In the simple case, (2) has been verified by Evans and Wagner in [7]). (1) holds also for NSOP and NIP theories as proved by Macpherson in [9] and Krupinski in [4] (under the assumption of f.s.g. for groups) respectively. Finally, (2) has been proved under the hypothesis of finite burden by Wagner and Dobrowolski in [1].

This article proves (2) under the assumption of finite-dimensionality, an important model-theoretic notion introduced by Wagner in [5]. In the first section, we introduce finite-dimensional theories and describe the properties of finite-dimensional groups. Moreover, we introduce the notion of dimensional-generic of a group. In the second section, we define bilinear quasi-forms. These, as in [1], are fundamental in the analysis of $\omega$-categorical groups. In the third section, we define principal indiscernible dimensional-generic sequences. These are used in section 4 to prove the finite-dimensional version of [1, Theorem 7.5].
?⟨VirtTriv⟩? **Theorem 0.7.** A bilinear quasi-form definable in a finite-dimensional ω-categorical structure is virtually almost trivial.

In the fifth section, we apply Theorem 0.7 to prove the following result.

**Theorem 0.8.** An ω-categorical group of finite dimension is finite-by-abelian-by-finite, and an ω-categorical ring of finite dimension is virtually null-by-finite.

# 1 Finite-dimensional groups

We define finite-dimensional theories.

**Definition.** A theory $T$ is finite-dimensional if there exists a function $\dim$ from the class of all the interpretable subsets in any model $\mathcal{M}$ of $T$ into $\omega \cup \{-\infty\}$ such that for any $\phi(x, y)$ formula, $X, Y$ interpretable sets in $T$ and $f$ interpretable function from $X$ to $Y$, then

* If $a, a'$ have the same type over $\emptyset$, $\dim(\phi(x, a)) = \dim(\phi(x, a'))$;

* $\dim(\emptyset) = -\infty$ and $\dim(X) = 0$ if and only if $X$ is finite;

* $\dim(X \cup Y) = \max\{\dim(X), \dim(Y)\}$;

* If $\dim(f^{-1}(y)) \ge k$ for any $y \in Y$, then $\dim(X) \ge \dim(Y) + k$;

* If $\dim(f^{-1}(y)) \le k$ for any $y \in Y$, then $\dim(X) \le \dim(Y) + k$.

Examples of finite-dimensional theories are superstable theories of finite Lascar rank, supersimple theories of finite Lascar rank, and $o$-minimal theories.

We now focus our attention on finite-dimensional groups. An important property of finite-dimensional groups is the $\omega$-DCC.

**Definition.** Let $G$ be a definable group. Then, $G$ has the $\omega$-DCC property if there exists no infinite strictly descending chain of definable subgroups $\{G_i\}_{i<\omega}$ such that $|G_i : G_{i+1}| > \omega$.

Groups definable in a finite-dimensional theory respect the $\omega$-DCC (Corollary 2.4 of [5]). Moreover, finite-dimensional groups have the ucc, as a consequence of the following fundamental Lemma [15, Lemma 1.1].

?⟨boundedind⟩? **Lemma 1.1.** Let $G$ be a definable group of finite dimension and $\{H_i\}_{i \in I}$ a family of uniformly definable subgroups. Then, there exist $n, d < \omega$ such that there is no $J = \{j_1, \dots, j_n\} \subseteq I$ of cardinality $n$ with $|\bigcap_{i=1}^k H_{j_i} : \bigcap_{i=1}^{k+1} H_{j_i}| \ge d$ for any $k \le n-1$.

If we apply these results to the centralisers of elements in $G$, we obtain that a finite-dimensional definable group has the hereditarily $\tilde{\mathfrak{M}}_c$-property.
**Definition.** A group $G$ is hereditarily-$\tilde{\mathfrak{M}}_c$ if, for any definable subgroups $H, N$ such that $N$ is normalised by $H$, there exist natural numbers $n_{HN}$ and $d_{HN}$ such that any sequence of centralisers

$$C_H(a_0/N) \geq C_H(a_0, a_1/N) \geq \dots \geq C_H(a_0, a_1, \dots, a_n/N) \geq \dots$$

with $|C_H(a_0, \dots, a_n/N) : C_H(a_0, \dots, a_{n+1}/N)| \geq d_{HN}$ has length at most $n_{HN}$.

The notions of almost containment and commensurability are fundamental in the study of groups definable in finite-dimensional theories.

**Definition.** • Let $G$ be a group and $A, B \leq G$ subgroups. A almost contains $B$, denoted $A \geq B$, if $|B : B \cap A|$ is finite.

• If either $A \leq B$ or $B \leq A$, we say that $A$ and $B$ are comparable;

• If $A \leq B$ and $B \leq A$, $A$ and $B$ are commensurable, denoted $A \sim B$. $\sim$ is clearly an equivalent relation on the class of subgroups of $G$.

• Given a family of subgroups $\{G_i\}_{i \in I}$ in $G$, the family $\{G_i\}_{i \in I}$ is uniformly commensurable if there exists $n < \omega$ such that $|G_i/G_i \cap G_j| \leq n$ for every $i, j \in I$.

We also define the notion of almost direct sum.

**Definition.** Let $G$ be a group and $H, K$ subgroups of $G$ that normalize each other. The sum $H + K$ is almost direct, denoted by $H \oplus K$, if $H \cap K$ is finite.

A consequence of the hereditarily $\tilde{\mathfrak{M}}_c$-condition is that the almost centralisers are definable.

**Definition.** Let $G$ be a group, $K, H \leq G$ and $N$ a subgroup of $G$ normalized by both $H$ and $K$. The almost centraliser in $K$ of $H/N$, denoted $\tilde{C}_K(H/N)$ is the subgroup

$$\{k \in K : C_H(k/N) \geq H\}.$$

If we assume $N = 0$ and $K = H = G$, we have the notion of almost center.

**Definition.** Let $G$ be a group. We define iteratively the $n$-almost center of $G$, denoted by $\tilde{Z}^n(G)$, as follows:

• $\tilde{Z}^1(G) = \tilde{C}_G(G)$;

• $\tilde{Z}^{n+1}(G)$ is such that $\tilde{Z}(G/\tilde{Z}^n(G)) = \tilde{Z}^{n+1}(G)/\tilde{Z}^n(G)$.

The following Lemma is proven in [2, Proposition 3.3].

**Lemma 1.2.** Let $G$ be a hereditarily $\tilde{\mathfrak{M}}_c$-group and $K, H, N$ as before. If $K, H, N$ are definable, $\tilde{C}_K(H/N)$ is definable.

An easy result for the almost center is the following.

?(Zfin)? Lemma 1.3. Let $G$ be a group such that $\tilde{Z}(G)$ is finite. Then, $\tilde{Z}(G/\tilde{Z}(G)) = \tilde{Z}(G)$.
Proof. Let $g \in \tilde{Z}(G/\tilde{Z}(G))$. Then, $[g, G/\tilde{Z}(G)] + \tilde{Z}(G)$ is finite. Since $\tilde{Z}(G)$ is finite, $[g, G]$ is finite i.e. $g \in \tilde{Z}(G)$. $\square$

Another important property of almost centralisers for definable groups is symmetry [2, Theorem 2.10].

?(sym)?**Lemma 1.4.** Let $G$ be a definable group and $H, K, N$ definable subgroups such that both $H$ and $K$ normalize $N$. Then, $H \leq \tilde{C}_G(K/N)$ iff $K \leq \tilde{C}_G(H/N)$.

In the last section of this article, we will use a fundamental theorem of linearization for the action of an almost abelian group definable in a finite-dimensional theory.

**Definition.** Let $G$ be a group. $G$ is almost abelian if $\tilde{Z}(G) = G$.

We introduce the notion of absolutely minimal $G$-group.

**Definition.** Let $G$ be a definable group acting on the definable group $A$. $A$ is absolutely $G$-minimal if, for any definable subgroup $H$ of finite index in $G$, $A$ is $H$-minimal.

We introduce the notion of almost centralisers of an action.

**Definition.** Let $G$ be a group acting on a group $A$. The almost centraliser of the action in $G$, denoted by $\tilde{C}_G(A)$, is the subgroup $\{g \in G : C_A(g) \sim A\}$. The almost centraliser of the action in $A$, denoted by $\tilde{C}_A(G)$, is the subgroup $\{a \in A : C_G(a) \sim G\}$.

It follows from Lemma 1.1 that the two almost centralisers are definable. We introduce almost trivial actions.

**Definition.** Let $G$ be a group acting on a group $A$. The action of $G$ on $A$ is almost trivial if either $\tilde{C}_G(A)$ is of finite index in $G$ or $\tilde{C}_A(G)$ is of finite index in $A$.

In the finite-dimensional case, these two conditions are equivalent by [15, Lemma 12.4]. The following result is fundamental in the analysis of almost trivial actions.

?(lemma 4)?**Lemma 1.5.** Let $G$ be a definable almost abelian group acting on the definable almost abelian group $A$. Then,

$$[\tilde{C}_G(A), \tilde{C}_A(G)] = \langle [g, a] = ga - a : g \in \tilde{C}_G(A), a \in \tilde{C}_A(G)\rangle$$

is finite.

Proof. By Lemma 1.1, the almost centralisers $H = \tilde{C}_G(A)$, $B = \tilde{C}_A(G)$ are definable subgroups such that $B$ is $G$-invariant and $H$ is $G$-normal. Since $A$ is almost abelian, $B'$ is finite and characteristic, and $H$ acts on the abelian group $B/B'$. Since $B$ is definable and every element has finite $H$-orbit, there exists $b \in B$ such that $Hb$ is of maximal cardinality. Let $h_1, \dots, h_n$ be a left transversal of $C_H(b)$ in $H$. Since every element $h \in H$ has centraliser in $A$ of finite index, $C = C_B(h_1, \dots, h_n)$ is of finite index in $B$. Let $b_1, \dots, b_m$ be a transversal of $C$ in $B$. Denote by $F$ the $H$-invariant finitely generated subgroup $\langle Hb, Hb_1, \dots, Hb_m \rangle$. Clearly, $F$ has countable cardinality. Moreover, by definition, $B = F + C$. On the other hand, let $D = \{[h, b] : h \in H, b \in B\}$. $D$ is clearly a definable subset of $B$ and therefore either finite or unbounded. We verify that
$D \leq F$. Given $b \in B$, there exist $c \in C$ and $f \in F$ such that $b = c + f$. Therefore, $[h, b] = h(c + f) - c - f = h(c) - c + h(f) - f$. Since $h(f) \in F$, $[h, b]$ is contained in $F$ iff $h(c) - c \in F$. Define $w = c + a$, then $h_i(w) = h_i(c) + h_i(b)$ but $c \in C$ and so this is equal to $c + h_i(b)$. Since $h_i(b) \neq h_j(b)$ when $i \neq j$ and by maximality of the cardinality of the orbit of $b$, the orbit of $w$ is $c + g_i(b)$ for $i = 1, ..., n$. Therefore, for any $h \in H$, $h(w) = c + h_i(b)$ for a certain $i$. Consequently,

$$h(c) - c = h(w) - h(b) - c = c + h_i(b) - h_j(b) - c = h_i(b) - h_j(b) \leq F.$$

Being $D$ definable and not unbounded, it is finite. Therefore, the subgroup $[H, B]$ is finitely generated and, being $B$ abelian, it is finite iff the order of any generator is finite. But $n(g - 1)(b) = g - 1(b) \leq D$ that is finite so also $o(g - 1(b))$ is finite for every $g \in H$ and $b \in B$. Since $H'$ is finite, this proves the theorem. $\square$

Finally, if $G$ and $A$ are two almost abelian groups such that $G$ acts on $A$ not almost trivially and $A$ is absolutely $G$-minimal, then the action can be linearized [15, Theorem 12.6].

#### Theorem 1.6. Let $G$ be a definable almost abelian group acting on a definable infinite almost abelian group $A$. Assume that:

* $A$ is absolutely $G$-minimal;

* the action is not almost trivial.

Then, $G/\tilde{C}_G(A)$ is abelian and there exists a definable field $K$ such that $G/\tilde{C}_G(A)$ definably embeds in $K^\times$ and $A/\tilde{C}_A(G)$ is isomorphic to $K^+$.

## 1.1 Dimensional-generics

We introduce the notion of dimensional-generics of a definable group of finite dimension. We also recall the notion of subgroup-generic used in [1].

**Definition.** Given $G$ a group definable in a theory $T$, a definable subset $X$ of $G$ is:

* subgroup-generic if it is not contained in the union of finitely many translates of interpretable subgroups of infinite index;

* dimensional-generic if $\dim(G) = \dim(X)$ with $T$ a finite-dimensional theory.

$A$ (possibly partial) type $p$ is *-generic if every $\emptyset \neq p$ is *-generic (with * equal to either subgroup or dimensional).

We prove the existence of dimensional-generics.

#### Lemma 1.7. Let $G$ be a definable finite-dimensional group and $A$ a set of parameters. Then, there always exists a dimensional-generic type $p \in S(A)$. Moreover, we can construct infinite indiscernible sequences $\{a_i \in G\}_{i \in I}$ over $A$ such that $tp(a_i/\{a_j\}_{j<i} \cup A)$ is dimensional-generic.
*Proof.* Assume that $G$ is $\mathfrak{0}$-definable.

For the first part, let $q$ be the partial type over $A$ given by $\{X \subseteq G : \dim(G - X) < \dim(G)\}$. We verify that it is finitely consistent. Assume not, then there exists a finite subfamily $\{X_i\}_{i \le n}$ of $q$ such that $\bigcap_{i \le n} X_i$ is empty. This implies that $G - \bigcap_{i \le n} X_i = \bigcup_{i \le n} G - X_i = G$. Therefore, $\dim(G) = \dim(\bigcup_{i \le n} G - X_i) = \max_{i \le n} \dim(G - X_i)$ by union. By hypothesis, $\dim(G) = \max_{i \le n} \dim(G - X_i) < \dim(G)$, a contradiction. Therefore, let $p$ be a completion of $q$ in $S_A(G)$. Then, this type is dimensional-generic. Assume not, then there exists an $A$-definable set $X$ in $p$ such that $\dim(X) < \dim(G)$. By definition of $q$, $G - X \in q \subseteq p$ and so $X, G - X \in p$, contradicting the finite satisfiability of $p$.

For the second part, apply iteratively the first part to obtain an infinite sequence $\{a_i\}_{i \in I}$ of elements in $G$ such that $tp(a_i/A \cup \{a_j\}_{j<i})$ is dimensional-generic. This sequence is also $A$-dimensional independent since $\dim.tp(a_i/A \cup \{a_j\}_{j<i}) = \dim(G) \ge \dim.tp(a_i/A))$ since $a_i \in G$. The conclusion follows by Ramsey's Theorem. $\square$

We verify that dimensional-generics are subgroup-generics.

?⟨dimgen⟩? Lemma 1.8. A dimensional-generic is subgroup-generic.

*Proof.* Let $p$ be a dimensional-generic for a definable group $G$ of finite dimension. Assume, for a contradiction, that $p$ is not subgroup-generic. Then, by definition, there exists $\phi(x) \in p$ such that $\phi(G)$ is contained in a translate of an interpretable subgroup $H$ of infinite index in $G$. By fibration, $\dim(H) < \dim(G)$. Therefore, by invariance, $\dim(\phi(G)) \le \dim(H) < \dim(G)$, contradicting the dimensional-genericity. $\square$

## 2 Bilinear quasi-forms

We introduce bilinear quasi-forms, which are a fundamental instrument in the analysis of $\omega$-categorical groups and rings of finite dimension.

**Definition.** A bilinear quasi-form $\lambda$ from $G, H$ to $K$ abelian groups is a partial function from $G \times H$ to $K$ such that, for every $g \in G$ and $h \in H$, the partial functions

$$ \lambda_g := \lambda(g, -) : h \in \dim(\lambda(g, -)) \to \lambda(g, h) \in K $$

and

$$ \lambda'_h := \lambda(-, h) : g \in \dim(\lambda(-, h)) \to \lambda(g, h) \in K $$

are partial endomorphisms. We denote $\text{ann}_H(g) = \{h \in H : \lambda(g, h) = 0_K\}$ and similarly $\text{ann}_G(h)$ for $h \in H$.

We define the almost annihilator of a bilinear quasi-form.

**Definition.** Let $\lambda$ be a bilinear quasi-form from $G \times H$ to $K$ and let $A \le G$ and $B \le H$. We denote by $\widetilde{\text{ann}}_H(A)$ the almost annihilator of $A$ over $H$ with respect to $\lambda$ defined as

$$ \widetilde{\text{ann}}_H(A) := \{h \in H : \lambda(A, h) \text{ is finite}\} = \{h \in H : \text{ann}_G(h) \ge A\}. $$
Similarly, we define the almost annihilator of $B$ over $G$ with respect to $\lambda$ the subgroup

$$ \widetilde{\text{ann}}_G(B) := \{g \in G : \lambda(g, B) \text{ is finite}\} = \{g \in G : \text{ann}_H(g) \ge B\}. $$

If the quasi-form is clear, we call them almost annihilator of $A$ over $H$ and almost annihilator of $B$ over $G$, respectively.

It follows from Lemma 1.1 that, if $G, H, A, B$ and $\phi$ are definable in a finite-dimensional structure, the almost annihilators are definable.

**?($\langle$definability)$?** **Lemma 2.1.** Assume we are working in a finite-dimensional theory $T$. Let $\phi$ be a definable bilinear quasi-form from $G \times H$ to $K$ and $A, B$ definable subgroups of $G, H$ respectively. Then, $\widetilde{\text{ann}}_H(A)$ and $\widetilde{\text{ann}}_G(B)$ are definable.

*Proof.* From Lemma 1.1 applied the uniformly definable family of subgroups $\{\text{ann}_G(h)\}_{h \in \widetilde{\text{ann}}_H(A)}$, there exists $N \in \mathbb{N}$ such that $N > |G : \text{ann}_G(h)|$. This implies that $\widetilde{\text{ann}}_H(A) = \{h \in H : |G : \text{ann}_G(h)| \le n\}$ is definable. The conclusion for $\widetilde{\text{ann}}_G(B)$ follows similarly. $\square$

We introduce the notions of triviality, almost triviality, and virtual almost triviality.

**Definition.** A bilinear quasi-form $\lambda$ is trivial if $\lambda(g, h) = 0$ for all $g \in G, h \in H$. It is almost trivial if $\lambda(G, H) \le S$ with $S$ a finite subgroup of $K$. $\lambda$ is virtually almost trivial if there exists $G_0 \le G$ and $H_0 \le H$ subgroups of finite index such that $\lambda|_{G_0 \times H_0}$ is almost trivial.

We have a characterization of virtually almost trivial bilinear quasi-form given by [1, Corollary 5.11].

**?($\langle$almosttrivial)$?** **Lemma 2.2.** Let $\lambda : G \times H \to K$ be a definable bilinear quasi-form. Then, TFAE:

* $\lambda$ is virtually almost trivial;

* $\widetilde{\text{ann}}_G(H) \ge G$;

* $\widetilde{\text{ann}}_H(G) \ge H$.

From now on, we assume that the bilinear quasi-forms are definable in a finite-dimensional theory. We introduce the notion of reduced dimension of a bilinear quasi-form.

**Definition.** Let $\lambda : G \times H \to K$ be a definable bilinear quasi-form of finite dimension. Then, we define

* the reduced dimension of $G$, denoted as $\text{rdim}_\lambda(G)$, as $\max\{\dim(G/\text{ann}_G(\bar{h}))\}_{\bar{h} \text{ finite tuple in } H}$;

* the reduced dimension of $H$ denoted as $\text{rdim}_\lambda(H)$ as $\max\{\dim(H/\text{ann}_H(\bar{g}))\}_{\bar{g} \text{ finite tuple in } G}$;

* the reduced dimension of $\lambda$ denoted as $\text{rdim}(\lambda)$ as $\text{rdim}_\lambda(G) + \text{rdim}_\lambda(H) + \dim(K)$.

We also introduce the notion of induced bilinear quasi-form.
**Definition.** A bilinear quasi-form $\bar{\lambda}: A \times B \to D/C$ is induced from $\lambda: G \times H \to K$ if $A \le G, B \le H, C, D \le K$ and $\pi \circ \lambda: \lambda^{-1}(D) \cap A \times B \to D/C$ coincides with $\bar{\lambda}$ where $\pi$ is the projection from $D$ to $D/C$.

Observe that, in order to be $\bar{\lambda}$ well-defined, $A \times B$ must be almost contained in $\lambda^{-1}(D)$.

We verify that, given $\lambda$ a bilinear quasi-form, the reduced dimension of any induced bilinear quasi-form is less than rdim($\lambda$).

?(induced)? **Lemma 2.3.** if $\bar{\lambda}: A \times B \to D/C$ is induced from $\lambda$, then $\text{rdim}_{\bar{\lambda}}(B) \le \text{rdim}_{\lambda}(H)$ and $\text{dim}(D/C) \le \text{dim}(K)$. In particular, $\text{rdim}(\bar{\lambda}) \le \text{rdim}(\lambda)$.

*Proof.* The latter conclusion follows easily from the previous one. For the first point, it is sufficient to prove that

$$ \max\{\dim(H/\ker(\lambda_g))\}_{g \in G} \ge \max\{\dim(B/\ker(\bar{\lambda}_a))\}_{a \in A}. $$

For every $\bar{a} \in A$, denote $\lambda_a^C$ the function that sends $h \in H$ to $\lambda_a(h) + C$. Then,

$$ \dim(H/\ker(\lambda_a)) \ge \dim(H/\ker(\lambda_a^C)) \ge \dim(A/\ker(\lambda_a^C \cap A)) = \dim(A/\ker(\bar{\lambda}_a)). $$

This completes the proof. □

Finally, we recall [1, Lemma 4.6].

?(Lemm 4.6)? **Lemma 2.4.** Let $G$ be an $\omega$-categorical abelian group and $f$ a definable quasi-endomorphism of $G$. Then, there is $n < \omega$ such that $\text{im}(f^{\circ n}) \hat{\otimes} \ker(f^{\circ n}) \sim G$.

# 3 Principal indiscernible sequences

We introduce principal dimensional-generic indiscernible sequences. As proven in Lemma 1.8, any dimensional-generic is subgroup-generic. Therefore, these sequences are particular examples of principal subgroup-generic sequences introduced in [1]. On the other hand, the existence of principal indiscernibles finite-dimensional sequences will substantially simplify the proof of Theorem 0.7.

**Definition.** Let $G$ be an infinite definable group and $A$ a set of parameters. An A-indiscernible sequence $\langle (g_i, \bar{a}_i); i \in I \rangle$ with $g_i \in G$ for each $i \in I$ is principal indiscernible if, for any $i \in I$ and $A_i = A \cup \{g_j, \bar{a}_j : j \ne i\}$, whenever $C$ is an $A_i$-definable coset of some subgroup $H$ and $g_i \in C$, then $g_i \in H_{A_i}^0$. A sequence $\{g_i\}_{i \in I}$ is principal indiscernible dimensional-generic if it is principal, indiscernible, and dimensional-generic.

We modify slightly [1, Proposition 6.4].

?(Prinseq)? **Proposition 3.1.** Let $\langle (g_i, \bar{a}_i) : i \in \mathbb{Q} \cup (\mathbb{Q} + \varepsilon) \rangle$, with $\varepsilon$ an infinitesimal, be an A-indiscernible sequence. Then, the sequence $\langle (h_i, \bar{a}_i \bar{a}_{i+\varepsilon}) : i \in \mathbb{Q} \rangle$ with $h_i = g_{i+\varepsilon}^{-1}g_i$ is principal indiscernible. If, moreover, $\langle (g_i, \bar{a}_i) : i \in \mathbb{Q} \cup (\mathbb{Q} + \varepsilon) \rangle$ is dimensional-generic, then also $\langle (h_i, \bar{a}_i \bar{a}_{i+\varepsilon}) : i \in \mathbb{Q} \rangle$ is dimensional-generic. In particular, principal indiscernible dimensional-generic sequences exist.
*Proof.* The first part follows from [1, Proposition 6.4]. The second part by the fact that if $\mathrm{tp}(g_{i+\varepsilon}/A, \{g_j, \bar{a}_j : j \le i\})$ is dimensional-generic, also $\mathrm{tp}(g_{i+\varepsilon}^{-1}/A, \{g_j, \bar{a}_j : j \le i\})$ and $\mathrm{tp}(g_{i+\varepsilon}^{-1}g_i/A, \{g_j, \bar{a}_j : j \le i\})$ are dimensional-generic. From Lemma 1.7, there exists an indiscernible dimensional independent sequence. Therefore, applying the construction of the first part, we may obtain a principal indiscernible dimensional-generic sequence. This completes the proof of the Lemma. $\square$

# 4 Virtual almost triviality

We now prove Theorem 0.7. By Lemma 2.2, it is sufficient to show that one of the two almost annihilators is of finite index in $G$ or $H$, respectively.

We prove Theorem 0.7 by induction on the reduced dimension of the bilinear quasi-form $\lambda$. Since for dimension 0, the conclusion is clear, we may assume that $\mathrm{rdim}(\lambda) = N$ and that any bilinear quasi-form of reduced dimension strictly less than $N$ is virtually almost trivial.

We need some preliminary lemmas.

*?($\langle$Comparability$)$?* **Lemma 4.1.** Let $\lambda : G \times H \to K$ be a definable bilinear quasi-form, and that $G, H, K$ are $\omega$-categorical groups of finite dimension.

Assume that any induced quasi-form of strictly smaller reduced dimension is virtually almost trivial. Let $\langle y_i \rangle_{i \in \mathbb{Q}}$ be a principal indiscernible sequence in $G$ over $\emptyset$. Then, $\mathrm{im}(\lambda_{y_i})$ and $\mathrm{im}(\lambda_{y_j})$ are comparable. The same holds for the annihilators.

*Proof.* Let $C = \sum_{i=0}^n \mathrm{im}(\lambda_{y_{l_i}})$ be a sum of maximal dimension among the finite sums of images of $\lambda_y$ for $y \in \langle y_i \rangle_{i \in \mathbb{Q}}$. Moreover, we may assume that, for any $j \le n$, $\mathrm{dim}(\sum_{i=0, i \ne j}^n \mathrm{im}(\lambda_{y_{l_i}})) < \mathrm{dim}(\sum_{i=0}^n \mathrm{im}(\lambda_{y_{l_i}}))$. By maximality of the dimension, for every $y \in \langle y_i \rangle_{i \in \mathbb{Q}}$, $\mathrm{im}(\lambda_y)$ is almost contained in the previous sum.

We consider the induced bilinear quasi-form $\bar{\lambda} : G \times H \to K/C$ and denote by $A$ the almost annihilator of $\bar{\lambda}$ in $G$ and by $B$ the subgroup $\{h \in H : \lambda_h(A)$ is finite$\}$. $A, B$ are definable by Lemma 2.1, and moreover, $H$ is almost contained in $B$ by Lemma 1.4. By definition, $\lambda_a[B] \le \lambda_a[H] \le C$ for any $a \in A$ and similarly for $h \in B$. The restriction

$$ \lambda : A \times B \to C $$

is an induced $\{y_{l_i}\}_{i \in \mathbb{Q}}$ form as is

$$ \lambda_i : A \times B \to C/\mathrm{im}(\lambda_i) \cap C. $$

Each of the $\lambda_i$ has dimension strictly less than $\mathrm{dim}(\lambda)$ (by minimality of $n$) and so it must be virtually almost trivial. Therefore, the almost annihilator is a $\{y_i\}_{i \le n}$-definable subgroup of finite index in $A$ i.e. it contains $A_{\{y_{l_i} : i \le n\}}^0$. Since $y$ is principal, it must be contained in the connected component and so $\mathrm{im}(\lambda_y) \le \mathrm{im}(\lambda_{y_i})$. By principal indiscernibility, the conclusion follows.

For the annihilators, take an intersection $B := \bigcap_{i \le n} \mathrm{ann}(\lambda_{y_{l_i}})$ of minimal dimension possible and with $n$ minimal among the intersections of minimal dimension.

Let $A := \{g \in G : B \le \mathrm{ann}_H(g)\}$ and consider the following quasi bilinear form

$$ \lambda_i : A \times \mathrm{ann}_H(y_{l_i}) \to K. $$
We show that the reduced dimension of $\lambda_i$ is strictly less than $\mathrm{rdim}(\lambda)$. It is sufficient to prove that $\mathrm{rdim}_{\lambda_i}(\mathrm{ann}_H(y_{i_l})) < \mathrm{rdim}_{\lambda}(H)$. Taken $\bar{g} \in G$, then $\mathrm{ann}_H(\bar{g}) \ge B$ and so the dimension of $\mathrm{ann}_H(y_{i_l})/\mathrm{ann}_{\mathrm{ann}_H(y_{i_l})}(\bar{g})$ is less than $\mathrm{dim}(\mathrm{ann}_h(y_{i_l})/B)$. By minimality, $\mathrm{dim}(\bigcap_{j=1, j \ne i} \mathrm{ann}_H(y_j)/B) > 0$ and so

$$ \dim(\mathrm{ann}_H(y_{i_l})/B) < \dim(\mathrm{ann}_H(y_{i_l})/B) + \dim\left(\bigcap_{j=1, j \ne i} \mathrm{ann}_H(y_j)/B\right) \le \dim(H/B) $$

that is less than the reduced dimension of $H$. Therefore, $\lambda_i$ is virtually almost trivial. Consequently, the almost annihilator is a subgroup of finite index in $A$ and so it contains the connected component. This implies that $\mathrm{ann}_H(y_{i_l}) \le \mathrm{ann}_H(y_{l_j})$. $\square$

?(NilpoIsom)? **Lemma 4.2.** Suppose that $\lambda$ is definable in a finite-dimensional theory and that any induced bilinear quasi-form of strictly smaller dimension is virtually almost trivial. Let $\langle y_i \rangle_{i \in Q}$ be a principal dimensional-generic sequence. Then, for any $j \in Q$, any definable quasi-endomorphism $f$ of $H/\mathrm{ann}_H(y_j)$ is either invertible or nilpotent.

*Proof.* By Lemma 2.4, there exists $n < \omega$ such that $G \le \mathrm{im}(f^n) \oplus \ker(f^n)$. The group

$$ A := \{g \in G : \mathrm{ann}_H(y_j) \le \mathrm{ann}_H(g)\} $$

is a definable subgroup by Lemma 1.1. Denote $B_1, B_2$ the preimages in $H$ of $\mathrm{im}(f^n)$ and $\ker(f^n)$ respectively. Then $B_1 \cap B_2$ is a finite extension of $\mathrm{ann}_H(y_j)$. If $f$ is neither invertible nor nilpotent, then both the $B_i$ are infinite and the restriction of $\lambda$ to $A \times B_i$ has strictly smaller dimension. Therefore, it is virtually almost trivial and the same for the bilinear quasi-form

$$ \lambda' : A \times H \to K $$

given by the sum of the two. By Lemma 4.1, $y_i \in A$ for either $i < j$ or $j > i$. Therefore, $y_i \in \mathrm{ann}_G(H)$ and the latter must be of finite index since $y_i$ is dimensional-generic. Therefore, by Lemma 2.2, $\lambda$ is virtually almost trivial. $\square$

We start the proof of Theorem 0.7.

*Proof.* Let $\lambda$ be a counterexample of Theorem 0.7 of minimal dimension. Up to add finitely many parameters, we can assume that everything is $\emptyset$-definable. Let $\langle (x_i, x'_i) : i \in Q \cup Q + \varepsilon \rangle$ be a dimensional-generic sequence, that exists by Lemma 1.7. Define $y_i = x_i - x_{i+\varepsilon}$ and $y'_i = x'_i - x'_{i+\varepsilon}$. Then, by Lemma 3.1, $\langle (y_i, y'_i) : i \in Q \rangle$ is a principal indiscernible dimensional-generic sequence.

**Claim.** For $i < j$, $\mathrm{ann}_H(y_j) \le \mathrm{ann}_H(y_i)$ and $\mathrm{im}(\lambda_i) \le \mathrm{im}(\lambda_j)$.

*Proof.* The second is clearly a consequence of the first and of Lemma 4.1. $\widetilde{\mathrm{ann}}_H(G)$ is definable and the indexes $|G : \mathrm{ann}_G(h)|$ takes only finitely many values. Let $n$ be the maximal one and let $h$ realize a generic type for $\widetilde{\mathrm{ann}}_H(G)$. Then, taken $x_0, x_1, \dots, x_n$, there exist $i \ne j$ such that $x_i - x_j \in \mathrm{ann}_G(h)$ and so $h \in \mathrm{ann}_H(x_i - x_j)$. By dimensional-genericity of $h$, the subgroup $\mathrm{ann}_H(x_i - x_j) \cap \widetilde{\mathrm{ann}}_H(G)$ is of finite index in $\widetilde{\mathrm{ann}}_H(G)$ i.e. $x_i - x_j \in \widetilde{\mathrm{ann}}_G(\widetilde{\mathrm{ann}}_H(G))$. By indiscernibility, $\mathrm{ann}_H(y_0) \ge \widetilde{\mathrm{ann}}_H(G)$. Suppose $\mathrm{ann}_H(y_0) \le \mathrm{ann}_H(y_1)$. Then $y_1 \in \widetilde{\mathrm{ann}}_G(\mathrm{ann}_G(y_0))$. By dimensional-genericity,
$\overline{\text{ann}}_G(\text{ann}_G(y_0))$ should be of finite index. Consequently, $\overline{\text{ann}}_H(G) \ge \overline{\text{ann}}_H(y_0)$ and the same must hold for every $y_i$ for $i \in \mathbb{Q}$ by indiscernibility. This implies that $\text{ann}_H(y_0) \sim \overline{\text{ann}}_H(G) \sim \text{ann}_H(y_1)$. If $\text{ann}_H(y_1) \le \text{ann}_H(y_0)$, the conclusion follows. $\square$

We define $\lambda_{y_1,y_j} = \lambda_{y_j}^{-1} \circ \lambda_{y_i}$ to be the definable additive relation in $H \times H$ given by

$$ \{(h, h') : \lambda(y_i, h) = \lambda(y_j, h')\}. $$

Clearly, the kernel of this relation is $\text{ann}_H(y_i)$ and $\text{kat}(\lambda_{y_i,y_j}) = \text{ann}_H(y_j)$. By Claim 1, this defines a quasi-homomorphism $\bar{\lambda}_{y_i,y_j}$ of $H/\text{ann}_H(y_j)$. For the definition of additive relation and quasi-homomorphism see section 3 of [1].

**Claim.** For $i \ne j$, $\text{ann}_H(y_i) \not\subset \text{ann}_H(y_j)$.

*Proof.* Suppose otherwise and put $\overline{H} = H/\text{ann}_H(y_0)$. If $\text{ann}_H(y_0)$ is of finite index, $y_0 \in \overline{\text{ann}}_G(H)$ that, being $y_0$ dimensional-generic, should imply that $\overline{\text{ann}}_G(H)$ is of finite index in $G$. Consequently, $\lambda$ is virtually almost trivial.

Assume, for a contradiction, that $\text{ann}_H(y_0) \sim \text{ann}_H(y_i)$. Then $\text{ann}_H(y_0) \sim \text{ann}_H(y_i)$ for every $i$ by indiscernibility. This implies that $\text{im}(\lambda_{y_i}) \sim \text{im}(\lambda_{y_j})$ since $\text{im}(\lambda_0) \le \text{im}(\lambda_i)$ and

$$ \dim(\text{im}(\lambda_0)) = \dim(H) - \dim(\text{ann}_H(y_0)) = \dim(H) - \dim(\text{ann}_H(y_i)) = \dim(\text{im}(\lambda_i)). $$

Let $R$ be the ring of quasi-endomorphisms of $\overline{H}$ modulo equivalence. Then, every element of $R$ is nilpotent or invertible by Lemma 4.2. The subset $I = \{r \in R : r \text{ is nilpotent}\}$ is an ideal in $R$. Indeed, it is clearly invariant for left and right multiplication. On the other hand, if $f, g \in I$, then $f + g$ is nilpotent. If not, there is $h \in R$ such that $h(f + g) = 1$. But $(id - hg)(id + hg + (hg)^2 + \dots) = id$ (this sum is finite since $hg$ is nilpotent) implies that $id - hg = hf$ is invertible, a contradiction. Therefore, $R/I$ is a division ring. Let $\bar{r}$ be a finite tuple in $R/I$ definable over a finite set $A$. Then, by $\omega$-categoricity, $\bar{r}$ generates a finite subring of $R$. Therefore, by little Weddleburn's theorem, this is a locally finite field. Any quasi-endomorphism $\lambda_{y_i,y_j}$ is a quasi-automorphism on $\overline{H}$. Indeed, $\text{ker}(\lambda_y) = \text{ker}(\lambda_{y'})$ and therefore $\lambda_y$ is a quasi-isomorphism from $\overline{H}$ to $\text{im}(\lambda_0)$. This implies that the composition of $\lambda_y$ and $\lambda_{y'}^{-1}$ is a quasi-automorphism. In particular, it is not nilpotent. By locally finiteness and indiscernibility, $\lambda_{y,y'}$ in $R/I$ has finite order equal to $N$. Since there are only finitely many $N$-roots of the unity, there are finitely many couples $\{(y_i, y_j)\}_{(i,j) \in \mathbb{Q}^2}$ such that, for any $y, y' \in \{y_i\}_{i \in \mathbb{Q}}$, there exists a couple $(y_i, y_j)$ such that $\lambda_{y_i, y_j} \sim \lambda_{y, y'}$. Therefore, by indescernibility, we may assume that all $\lambda_{y_i, y_j}$ are equal to the same $\eta + I$. Then, $\eta^2 + I = \lambda_{y_j, y_k} \lambda_{y_i, y_j} + I = \eta + I$ and so $\eta = 1$ (since $\eta \ne 0$).

Since $\lambda_{x_1-x_3, x_2-x_3} - id_H \in I$ then $B := \text{im}(\lambda_{x_1-x_3, x_2-x_3} - id_H)$ is of infinite index in $H$ almost containing $\text{ann}_H(y_0)$.

Consequently, for any $h \in H_{x_1, x_2, x_3}^{\circ} \le \text{dom}(\lambda_{x_1-x_3, x_2-x_3} - id_H)$, there exists $b \in B$ such that $h + b \in \lambda_{x_1-x_3, x_2-x_3}(h)$. Hence,

$$ \lambda(x_1 - x_3, h) = \lambda(x_2 - x_3, h + b) = \lambda(x_2 - x_3, h) + \lambda(x_2 - x_3, b) $$
whence

$$ \lambda(x_1 - x_2, h) = \lambda(x_1 - x_3, h) - \lambda(x_2 - x_3, h) = \lambda(x_2 - x_3, b). $$

This means that $\mathrm{im}(\lambda_{x_2-x_3}) \simeq \mathrm{im}(\lambda_{x_1-x_2})$ is almost contained in $\lambda_{x_2-x_3}[B]$. But $\mathrm{ker}(\lambda_{x_2-x_3}) \le B$ so

$$ \dim(\lambda_{x_2-x_3}[B]) = \dim(B) + \dim(\mathrm{ker}(\lambda_{x_2-x_3})) < \dim(\lambda_{x_2-x_3}). $$

This is clearly a contradiction.

□

We now complete the proof.

Observe that $\mathrm{im}(\lambda_{y_i})$ is of infinite index in $K$ for $i \in \mathbb{Q}$. If not, there exists $n < \omega$ such that the following formula is satisfied by $(x_i, x_{i+\epsilon})$:

$$ \exists k_1, \dots, k_n \in K \forall k \in K \exists h \in H \bigvee_{i=1}^{n} k_i \lambda(x_i - x_{i+\epsilon}, h) = k. $$

By indiscernibility, every $y_j$ realizes this formula. Therefore, the images of $\lambda_{y_j}$ are all of finite index in $K$, implying that $\mathrm{ann}_H(y_j) \sim \mathrm{ann}_H(y_i)$ for all $i, j \in \mathbb{Q}$, a contradiction by the previous claim.
Take $0 < k \in \mathbb{Q}$ and put

$$ A := \{g \in G : \mathrm{im}(\lambda_g) \le \mathrm{im}(\lambda_{y_k})\} $$

and

$$ H' := \{h \in H : \lambda'_h[A] \le \mathrm{im}(\lambda_{y_k})\}. $$

We verify that $\dim(H') = \dim(H)$ and so $H' \sim H$. Let

$$ X := \{(a, h) \in A \times H : \lambda(a, h) \in \mathrm{im}(\lambda_{y_k})\}. $$

$X$ is clearly definable. We prove that $\dim(X) = \dim(A \times H)$. Taken the first projection $\pi_1$, this is a surjective map from $X$ to $A$ with fibers $\pi_1^{-1}(a) = \{h \in H : \lambda(a, h) \in \mathrm{im}(\lambda_{y_k})\}$ that is of dimension equal to $\dim(H)$ by the assumption on $A$. By fibration, $\dim(X) = \dim(H) + \dim(A)$. Taken the second projection $\pi_2$, the image is $H$ and the fibers are $\pi_2^{-1}(h) = \{a \in A : \lambda(a, h) \in \mathrm{im}(\lambda_{y_k})\}$. $H'$ is clearly the set (definable by Lemma 1.1) of $h \in H$ with fibers of maximal dimension. By union, $\dim(X) \le \max\{\dim(H') + \dim(A), \dim(H - H') + \dim(A) - 1\}$. $\dim(X) = \dim(A) + \dim(H)$ and necessarily $\dim(H') = \dim(H)$.

We now study the bilinear quasi-form

$$ \tilde{\lambda} : A \times H' \to \mathrm{im}(\lambda_{y_k}). $$

This has reduced dimension strictly less than $\mathrm{rdim}(\lambda)$. Indeed, $\mathrm{rdim}_{\tilde{\lambda}}(A) \le \mathrm{rdim}_{\lambda}(G)$ by Lemma 2.3. The same holds for $\mathrm{rdim}_{\tilde{\lambda}}(H')$. Finally, $\dim(\lambda_{y_i}) < \dim(K)$ by previous proof. By assumption on the minimality of $\dim(\lambda)$, $\tilde{\lambda}$ is virtually almost trivial and so $\overline{\mathrm{ann}}_A(H')$ is a $(x_k, x_{k+\epsilon})$-definable subgroup of finite index in $A$ such that, for all $g \in \overline{\mathrm{ann}}_A(H')$, $\lambda(g, \overline{\mathrm{ann}}_{H'}(A))$ is finite. Since $\overline{\mathrm{ann}}_{H'}(A)$ is of finite index in $H'$ and so also in $H$, $\overline{\mathrm{ann}}_A(H') \le \overline{\mathrm{ann}}_G(H)$. Taken $y_l$ for $l < k$, $\dim(y_l) \le \dim(y_k)$ by Claim 1 and so $y_l \in A$. By principality, $y_l \in A_{y_k}^o$. This implies that $y_l \in \overline{\mathrm{ann}}_G(H)$. By indiscernibility, this holds for every $y_i$. Since $y_0$ is dimensional-generic, this implies that $\overline{\mathrm{ann}}_G(H)$ is of finite index in $G$. Finally, the proof follows from Lemma 2.2. □
# 5 Groups and rings

We show that finite-dimensional $\omega$-categorical groups are finite-by-abelian-by-finite and that finite-dimensional $\omega$-categorical rings are virtually finite-by-null.

We start by verifying that the case (2) in Theorem 0.5 is impossible for finite-dimensional groups. In other words, we need to prove that a definable characteristically simple group $G$ of finite dimension cannot be isomorphic to $B(F)$ or $B^{-}(F)$. This is a consequence of the following Lemma.

*(B(F))?** **Lemma 5.1.** The groups $B(F)$ and $B^{-}(F)$ are not of finite dimension i.e. there is no finite-dimensional theory $T$ in which $B(F)$ and $B^{-}(F)$ can be defined.

*Proof.* It is sufficient to observe that there exists an infinite chain of definable subgroups $\{C_i\}_{i<\omega}$ such that $|C_i/C_{i+1}|$ is infinite, contradicting $\omega$-DCC.

Let $\{A_i\}_{i<\omega}$ be a strictly ascending chain of clopen subsets in the Cantor set (this clearly exists). We take, for $f$ a non central element in $F$, the function $g_1$ such that $g_1(A_1) = \{f\}$ and $g_1(C - A_1) = \{0\}$. Let $C_1$ be the centraliser in $B(F)$ of $g_1$. This contains all the functions $g$ such that $g(A_1) \le C_F(f)$ and $g$ assumes arbitrary values outside $A_1$. Let $g_2$ be a function with $g_2(A_2) = f$ and let $C_2$ be the centraliser of $g_2$ in $B(F)$. We verify that $|C_2 : C_1|$ is infinite. Indeed, there exist pairwise disjoint clopen subsets $\{D_j\}_{j<\omega}$ of $C$ such that $D_j \subseteq A_2$ and $D_j \cap A_1 = \emptyset$. Take the functions $s_j \in B(F)$ such that $s_i(A_1) = f$ and $s_i(D_i) = d_i \in F - C_F(f)$. Then, $s_j^{-1}s_i \in C_1$ but not in $C_2$. This verifies that the index of $C_2$ in $C_1$ is infinite. Iterating the proof for any $i$, we obtain the contradiction. The proof for $B^{-}(F)$ is similar. □

The second step is to show that a $\omega$-categorical finite-dimensional definable group is virtually nilpotent. To do this, we can proceed in two ways: either as in [1] or using the two following results.

**Lemma 5.2.** A locally nilpotent definable group of finite dimension is virtually nilpotent.

*Proof.* It is sufficient to prove that, given $G$ a locally nilpotent definable group of finite dimension, $\tilde{Z}(G)$ is infinite. Iterating the proof, we conclude that $G$ is almost nilpotent. Consequently, $G$ is virtually nilpotent from [2, Proposition 3.29].

Assume, by contradiction, that the almost center is finite. Then $G/\tilde{Z}(G)$ is a definable locally nilpotent group of finite dimension with trivial almost center by Lemma 1.3. We verify that this is impossible. Let $C$ be a centraliser of a tuple of minimal dimension between all the centralisers. If $\dim(C) = 0$, take $C_1$ a minimal finite centraliser of a finite tuple. Then $C_1 \cap C_G(g)$ is not empty by locally nilpotency. By minimality, $C_1 \le C_G(g)$ for all $g \in G$ i.e. $C_1 \le Z(G) = \{0\}$, a contradiction. Assume that $C$ is infinite. Then, $C \cap C_G(g)$ has same dimension as $C$ and so $C \le C_G(g)$ for all $g \in G$ i.e. $G \le \tilde{C}_G(C)$. By Lemma 1.4, $C \le \tilde{Z}(G)$. This is again contradictory. □

Since an $\omega$-categorical group of exponent $p$ is locally nilpotent, the third case of Lemma 0.5 is impossible. Since the second case is excluded by Lemma 5.1, we have the following conclusion.
?(AbeChaSim)? **Lemma 5.3.** Let $G$ be an $\omega$-categorical finite-dimensional characteristically simple group. Then, $G$ is an elementary abelian $p$-group.

By Lemma 0.4 and Lemma 5.3, for any $\omega$-categorical group $G$ of finite dimension, there exists a series of definable characteristic subgroups $\{G_i\}_{i \le n}$ such that:

* $G_i/G_{i+1}$ is characteristically simple;

* $G_i/G_{i+1}$ is an elementary abelian $p$-group.

Therefore, we have the following conclusion.

**Lemma 5.4.** Let $G$ be a soluble finite-dimensional group. Then, $G$ is soluble and of finite exponent.

Finally, the virtual nilpotency follows from the next Lemma.

**Lemma 5.5.** Let $G$ be a finite-dimensional $\omega$-categorical group. Then, $G$ is virtually nilpotent.

*Proof.* By [2, Proposition 3.29], it is sufficient to prove that $G$ is almost nilpotent. We prove the Lemma by induction on the dimension. Clearly, if $G$ has infinite almost center, the conclusion follows by induction hypothesis. Let $\{G_i\}_{i \le n}$ be a series of definable subgroups of $G$ such that $G_i/G_{i+1}$ is infinite and almost abelian. This exists by [2, Proposition 3.21]. Clearly, the almost abelian group $G_{n-2}/G_{n-1}$ acts on the abelian group $G_{n-1}/G'_{n-1}$. Let $A$ be a definable subgroup of $G_{n-1}/G'_{n-1}$ of minimal dimension between all the subgroup that are minimal for a definable subgroup of finite index in $G_{n-2}/G_{n-1}$. Let $H \le G_{n-2}/G_{n-1}$ be a definable subgroup of finite index such that $A$ is $H$-invariant. Then, by construction, $A$ is absolutely $H$-minimal. By Linearization Theorem 1.6, if the action is not almost trivial, there exists a finite subgroup $F_1$ in $A$ and a definable subgroup $F_2$ not of finite index in $H$ such that $A/F_1 \simeq R^+$ and $H/F_2$ embeds in $R^\times$. The latter contradicts the fact that $H$ is of finite exponent. Therefore, the action is almost trivial. This implies that $\tilde{C}_{G_{n-1}}(G_{n-2}) =: C$ is a definable infinite $G$-normal subgroup in $G_{n-1}$. By Lemma 1.4, $\tilde{C}_{G_{n-2}}(C)$ is of finite index in $G_{n-2}$. Moreover, by Lemma 1.5, $[C, \tilde{C}_{G_{n-2}}(G)] =: C_1$ is a finite $G$-normal subgroup. $G_{n-3}/\tilde{C}_{G_{n-2}}$ is almost abelian and acts on $C/C_1$. Iterating the proof for any $G_i$, we obtain that $\tilde{Z}(G)$ is infinite. $\square$

We can also verify that $G$ is virtually nilpotent-by-finite with a proof similar to [1]. We need the following lemma.

?(BooAlg)? **Lemma 5.6.** An atomless Boolean algebra cannot have finite dimension (i.e. it cannot be defined in a finite-dimensional theory).

*Proof.* Assume, for a contradiction, that there exists a finite-dimensional theory $T$ and $B$ an atomless Boolean algebra definable in $T$. Assume $B$ of minimal dimension $n$ among all the infinite atomless Boolean algebras definable in $T$. Let $a \in B$ be an element not equal to 0 or 1 and define the function

$$f_a : (B, \vee) \rightarrow (B_a, \vee)$$
that sends $b$ in $f_a(b) = a \wedge b$ where $B_a = \{b \in B : b \le a\} = \{b \in B : \exists c a \wedge c = b\}$. This map is clearly surjective. We verify that $B_a$ is an atomless Boolean algebra. Since $B_a$ is clearly definable, it is either finite or its dimension is equal to $n$. It is closed for $\wedge$ and $\vee$ since $a \wedge (b \vee c) = a \wedge b \vee a \wedge c$ and $a \wedge (a \wedge b) = a \wedge b$ (since $a \wedge 1 = a$, this also proves that $f_a$ is an homorphism of monoids). Therefore, $B_a$ is closed for $\vee$. It is closed also for $\wedge$ since $a \wedge b \wedge a \wedge c = a \wedge (b \wedge c)$. The other properties of Boolean algebra follow easily (assuming $1_{B_a} = a$). Finally, we prove that it is atomless. It is sufficient to show that, for any $a \wedge c \in B_a$, there exists $d \in B_a$ such that $d \ne 0$, $a \wedge c$ and $d < a \wedge c$. Since $B$ is an atomless Boolean algebra and $a \wedge c \in B$, there exists $d \in B$ such that $0 < d < a \wedge c \le a$ and so $d \in B_a$. Therefore, $n = \dim(B) \ge \dim(B_a) \ge n$ so $\dim(B_a) = n$. If we verify that $f_a^{-1}(c)$ is infinite for every $c \in B_a$, it implies that $n = \dim(B) \ge 1 + \dim(B_a) = 1 + n$ by fibration, a contradiction. We prove that $f_a^{-1}(0) = \{b \in B : b \le \neg a\}$. The latter set is infinite since $B$ is atomless and $\neg a = 0$ iff $a = 1$. Assume $b \le \neg a$. Then, $a \wedge b = a \wedge (\neg a \wedge b) = (a \wedge \neg a) \wedge b = 0 \wedge b = 0$. On the other hand, if $a \wedge b = 0$, then $b = 1 \wedge b = (a \vee \neg a) \wedge b = (a \wedge b) \vee (\neg a \wedge b) = \neg a \wedge b$ and so $b \le \neg a$.

Given $b \in f_a^{-1}(0)$ and $d \in f_a^{-1}(c)$, then $a \wedge (b \vee d) = (a \wedge b) \vee (a \wedge d) = a \wedge d = c$. Clearly, $f_a(d) = f_a(a \wedge d)$. Therefore, any element $(a \wedge d) \vee b$ with $b \in f_a^{-1}(0)$ is in $f_a^{-1}(c)$. Assume that, for $b \ne b_1 \le \neg a$, $(a \wedge d) \vee b = (a \wedge d) \vee b_1$. Then, $\neg a \wedge ((a \wedge d) \vee b) = \neg a \wedge ((a \wedge d) \vee b_1)$. But the first is equal to $\neg a \wedge b$ and the second to $\neg a \wedge b_1$. Since $b, b_1 \le \neg a$, these are equal to $b$ and $b_1$, contradicting the assumption that $b \ne b_1$. Therefore, each fiber is infinite, and we reach a contradiction. $\square$

Since any $\omega$-categorical group can be written as a chain of characteristic subgroups with characteristically simple factors, we may assume that every $\omega$-categorical finite-dimensional subgroup is soluble. If it is not nilpotent-by-finite, an atomless Boolean algebra can be defined on it by [6], contradicting Lemma 5.6.

Therefore, we may assume that every $\omega$-categorical group of finite dimension is nilpotent-by-finite. From [2, Proposition 3.29], any $\omega$-categorical finite-dimensional subgroup has a characteristic nilpotent subgroup of finite index. Finally, we prove that any nilpotent $\omega$-categorical subgroup of finite dimension is finite-by-abelian-by-finite. This verifies the meta-conjecture for the group case.

**Lemma 5.7.** A nilpotent $\omega$-categorical group of finite dimension is finite-by-abelian-by-finite.

*Proof.* We may take a counterexample of minimal nilpotent class possible.

By minimality, $G/Z(G)$ is finite-by-abelian-by-finite. Since $G$ is not finite-by-abelian-by-finite, $Z(G)$ cannot be finite. The pre-image of $\tilde{Z}(G/Z(G))$ is a characteristic subgroup of finite index in $G$, and so we may replace $G$ with it. By [2, Proposition 3.27], $F := G'Z(G)$ is a finite extension of $Z(G)$. Again by [2, Proposition 3.27], $F'$ is finite and characteristic. Therefore, the quotient $G/F'$ is again $\omega$-categorical and we can assume $F' = \{0\}$. As $F/Z(G)$ is characteristic, $C_G(F/Z(G))$ is a characteristic definable subgroup of finite index. Up to take $G = C_G(F/Z(G))$, we may assume $[G, F] \le Z(G)$ i.e. $F \le Z^2(G)$. Consequently, the map

$$[\cdot, \cdot] : G \to F/Z(G)$$
that sends $g'$ in $[g, g'] + Z(G)$ is a definable homomorphism. The kernel of $H_g$ has finite index in $G$ for any $g$. Consequently, the map

$$ [\_, g] : h \in H_g \to [h, g] \in Z(G) $$

is a homomorphism with kernel containing $H'_g Z(G)$, since the image is abelian. Let $G_0$ be a definable subgroup in $G$ of finite index such that $|G'_0 : G'_0 \cap Z(G)|$ is minimal. Then, for any definable subgroup $H$ of finite index, $G'_0 Z(G)$ and $(H \cap G_0)' Z(G)$ are finite extensions of $Z(G)$ with $G'_0 Z(G) \ge (G_0 \cap H)' Z(G)$. Moreover, $|G'_0 Z(G)/Z(G)| = |G'_0/Z(G) \cap G'_0|$ and $|(G_0 \cap H)' Z(G) : Z(G)| = |(G_0 \cap H)'/Z(G) \cap (G_0 \cap H)'|$. By minimality, the latter must be equal to the first. Therefore, $G'_0 Z(G) = (H_g \cap G_0)' Z(G) \le H'_g Z(G) \le C_G(g)$. This implies that $G'_0 \le Z(G)$ and so $Z_2(G)$ is of finite index in $G$. Since $Z_2(G)$ is a characteristic subgroup, we may assume that $G$ is nilpotent of class 2. Then the function

$$ [\_, \_] : (g_1, g_2) \in G \times G \to [g_1, g_2] \in Z(G) $$

is a bilinear quasi-form. By Theorem 0.7, it should be virtually almost trivial, and this is equivalent to saying that $G$ is finite-by-abelian-by-finite. $\square$

In conclusion, we have proven the following theorem.

**Theorem 5.8.** Let $G$ be an $\omega$-categorical group of finite dimension. Then $G$ is finite-by-abelian-by-finite.

For rings, the proof is a straightforward corollary of Theorem 0.7.

**Lemma 5.9.** Let $R$ be an $\omega$-categorical ring of finite dimension. Then, $R$ is virtually finite-by-null.

*Proof.* It is sufficient to observe that the function

$$ \_\cdot\_ : R \times R \to R $$

that sends $(r, r')$ in $r \cdot r'$ is a bilinear form. By Theorem 0.7, it must be virtually almost trivial. Therefore, $R$ is virtually finite-by-null. $\square$

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