Groups with Locally Modular Homogeneous Pregeometries are Commutative

TL;DR

This paper proves that groups with locally modular homogeneous pregeometries are necessarily commutative, extending the known result for strongly minimal groups to a broader class.

Contribution

It establishes that groups with locally modular homogeneous pregeometries are commutative, generalizing the classical result for strongly minimal groups.

Findings

Groups with locally modular homogeneous pregeometries are commutative.

Extends the known commutativity result from strongly minimal groups to a wider class.

Supports the conjecture that certain geometric conditions imply group commutativity.

Abstract

It is well known that strongly minimal groups are commutative. Whether this is true for various generalisations of strong minimality has been asked in several different settings (see Hyttinen [2002], Maesono [2007], Pillay and Tanovi\'c [2011]). In this note we show that the answer is positive for groups with locally modular homogeneous pregeometries.

Full text

1 Introduction

The basic setting of geometric stability theory is a strongly minimal theory where algebraic closure induces a pregeometry. In the literature there are several approaches in generalising this basic setting.

One approach is to ask for every definable set to be countable or cocountable (instead of finite or cofinite in strongly minimal structures). An uncountable structure satisfying this condition is called quasiminimal.

Another option is to substitute the finite/cofinite dichotomy by large/small dichotomy with respect to some ultrafilter on the boolean algebra of definable sets (i.e. a type). So given a type $p\in\mathrm{S}(M)$ on a structure $M$, one can look at the operator associating to $A\subseteq M$ the union of all small $A$-definable sets (i.e. those that are not in $p$). The type $p$ is called strongly regular if this operator is a closure operator (i.e. a pregeometry possibly without exchange).

The third possibility is to simply require the existence of a homogeneous pregeometry on a structure $M$. The pregeometry $\operatorname{cl}:\mathcal{P}(M)\to\mathcal{P}(M)$ is called homogeneous if $M$ is infinite dimensional with respect to $\operatorname{cl}$ and for every finite $A\subset M$ and $b,c\in M\setminus\operatorname{cl}(A)$ there is an automorphism $f$ of $M$ (and of $\operatorname{cl}$) fixing $A$ pointwise with $f(b)=c$.

By a well known result of Reineke (1975) all strongly minimal groups are commutative. It has been asked whether the same is true in each of the above three settings (in Maesono (2007) for quasiminimality, in Pillay and Tanović (2011) for regular types and in Hyttinen (2002); Hyttinen et al. (2005) for groups with homogeneous pregeometries). Reineke’s argument has been generalised to show that if a noncommutative group has one of the above properties, then it must have the strict order property, so in particular is unstable. (See also Gogacz and Krupiński (2014) which, as a step towards a potential counterexample, constructs an uncountable group were all nontrivial elements are conjugate and have countable centralisers.) However not much is known beyond this. The main result of this paper is the following

Theorem 1**.**

A group with a locally modular homogeneous pregeometry is commutative.

Using the connections between quasiminimality, strong regularity and homogeneous pregeometries (see Pillay and Tanović (2011)), we have an analogues statement for quasiminimal groups of cardinality at least $\aleph_{2}$ and groups with symmetric strongly regular types, whose respective pregeometries are locally modular.

The initial idea of the proof was suggested to me by Alfonso Ruiz Guido. It used the representation theorem of projective geometries and required a stronger homogeneity hypothesis. Ivan Tomasic suggested to eliminate the use of the representation theorem by finding the contradiction explicitly. Doing that helped relax the homogeneity hypothesis. I express my gratitude to them and to Rahim Moosa for encouraging me to write this up.

2 Background on Pregeometries

Although the motivation for Theorem 1 comes from model theory, its statement and proof involve none. In order to make the presentation accessible to a wider audience, we provide a brief introduction to pregeometries.

Definition 2**.**

A pregeometry (also called matroid) on a set $X$ is an operator $\operatorname{cl}\colon\mathcal{P}(X)\to\mathcal{P}(X)$ satisfying the following conditions for all $A\subseteq X$ and $a,b\in X$.

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Reflexivity: $A\subseteq\operatorname{cl}(A)$.

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Transitivity: $\operatorname{cl}(\operatorname{cl}(A))=\operatorname{cl}(A)$.

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Finite character: $a\in\operatorname{cl}(A)$ if and only if $a\in\operatorname{cl}(A_{0})$ for some finite $A_{0}\subseteq A$.

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Exchange: $a\in\operatorname{cl}(A,b)\setminus\operatorname{cl}(A)$ implies $b\in\operatorname{cl}(A,a)$.

(Here and throughout we may use comma for union and omit the brackets for singletons, so that e.g. $\operatorname{cl}(A,b)$ means $\operatorname{cl}(A\cup\{b\})$.)

On any set $X$ the operator $\operatorname{cl}$ defined by $\operatorname{cl}(A)=A$ is a pregeometry. Other examples of pregeometries include a vector space $V$ where $\operatorname{cl}(A)$ is the linear span of $A$ and an algebraically closed field $F$ where $\operatorname{cl}(A)$ is the algebraic closure of the field generated by $A$.

A subset $A\subseteq X$ is called independent if $a\not\in\operatorname{cl}(A\setminus{a})$ for every $a\in A$. It is called generating if $\operatorname{cl}(A)=X$ and a basis if it is both independent and generating. The exchange axiom can be used to show that all bases have the same cardinality which is then called the dimension of $X$ and denoted by $\dim(X)$.

Given $Y\subseteq X$ we can define the restriction $\operatorname{cl}^{Y}$ of $\operatorname{cl}$ onto $Y$. This is a pregeometry on $Y$ defined by $\operatorname{cl}^{Y}(A)=\operatorname{cl}(A)\cap Y$. We also define the localisation $\operatorname{cl}_{Y}$ of $\operatorname{cl}$ on $Y$. The localisation is a pregeometry on $X$ defined by $\operatorname{cl}_{Y}(A)=\operatorname{cl}(A\cup Y)$. By $\dim(Y)$ we mean the dimension of $\operatorname{cl}^{Y}$ and by $\dim(Y/Z)$ we mean the dimension of $Y$ in $\operatorname{cl}_{Z}$.

A pregeometry $\operatorname{cl}$ is called a geometry if $\operatorname{cl}(\emptyset)=\emptyset$ and $\operatorname{cl}(a)=\{a\}$ for every $a\in A$. The idea is that closed subsets of dimension $1$, $2$, $3$, etc correspond to points, lines, planes, etc. (Note that what we call the dimension is $1$ more than the geometric dimension.) In such a geometry one can only talk about incidence however it turns out to be surprisingly rich.

There is a canonical way of associating a geometry to the pregeometry $\operatorname{cl}$ on $X$. We define an equivalence relation $\sim$ on $\hat{X}=X\setminus\operatorname{cl}(\emptyset)$ by $a\sim b$ iff $\operatorname{cl}(a)=\operatorname{cl}(b)$. Then the operator $\hat{\operatorname{cl}}$ defined as $\hat{\operatorname{cl}}(A/\sim)=\{b/\sim:b\in\operatorname{cl}(A)\}$ is a geometry on $\hat{X}$.

Let $\operatorname{cl}$ be a pregeometry on $X$. We distinguish the following types of pregeometries.

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The pregeometry is called trivial if $\operatorname{cl}(A)=\bigcup_{a\in A}\operatorname{cl}(a)$. In the associated geometry this means that every set is closed.

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It is called modular if $\dim(A\cup B)+\dim(A\cap B)=\dim(A)+\dim(B)$ for all finite dimensional closed $A$ and $B$. The pregeometry associated with the linear closure in a vector space is modular.

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The pregeometry is called locally modular if the above modularity equation holds provided $\dim(A\cap B)>0$. This is equivalent to $\operatorname{cl}_{a}$ being modular for every $a\in X\setminus\operatorname{cl}(\emptyset)$. An example of locally modular pregeometry that is not modular is the affine closure in a vector space. (The closed sets are translates of linear subspaces.)

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A geometry that is modular but not trivial is called projective. A classical result in projective geometry is that every projective geometry of dimension at least $4$ comes from a vector space over a division ring.

A pregeometry on a structure $M$ is a pregeometry $\operatorname{cl}$ on the base set of $M$ such that every automorphism of $M$ is an automorphism of $\operatorname{cl}$ (i.e. preserves closed sets). The pregeometry $\operatorname{cl}$ on $M$ is called homogeneous if $\dim(M)$ is infinite and for every finite $A\subset M$ and $b,c\in M\setminus\operatorname{cl}(A)$ there is an automorphism of $M$ fixing $A$ pointwise and taking $b$ to $c$.

3 Main Result

From now on let $G$ be a group and $\operatorname{cl}:\mathcal{P}(G)\to\mathcal{P}(G)$ be a homogeneous pregeometry in the above sense. We call a tuple $\bar{a}\in G^{n}$ generic over $A\subset G$ if $\dim(\bar{a}/A)=n$. For $A,B\subseteq G$ we say that $B$ is $A$-invariant if $B$ is fixed setwise by all automorphism of $G$ that fix $A$ pointwise.

We start with some simple observations.

Proposition 3**.**

If $a$ is $A$-invariant for some finite $A\subset G$, then $a\in\operatorname{cl}(A)$.

Proof.

Indeed, pick $c\in G$ generic over $A\cup a$. This is possible since $G$ is infinite dimensional. If $a\not\in\operatorname{cl}(A)$, then there is an automorphism fixing $A$ and taking $a$ to $c$. ∎

In particular if $a$ is first-order definable from $A$, then $a\in\operatorname{cl}(A)$. We will mostly use the Proposition in this form.

Corollary 4**.**

Let $A\subset G$ be a finite subset. If $a$ is generic over $A\cup b$, then $a\cdot b$ is also generic over $A\cup b$.

Proof.

Indeed, $a$ is definable from $b$ and $a\cdot b$. Hence $a\in\operatorname{cl}(A,b,a\cdot b)$. So in particular $a\cdot b\not\in\operatorname{cl}(A,b)$. ∎

As another corollary we can exclude one type of pregeometry.

Corollary 5**.**

The pregeometry $(G,\operatorname{cl})$ is not trivial.

Proof.

Pick a generic pair $a,b$ (i.e. $\dim(a,b)=2$) and consider $a\cdot b$. By the above $a\cdot b\not\in\operatorname{cl}(a)\cup\operatorname{cl}(b)$. However $a\cdot b\in\operatorname{cl}(a,b)$ demonstrating that $\operatorname{cl}$ is not trivial. ∎

Another consequence is that proper invariant (in particular definable) subgroups are small.

Proposition 6**.**

Let $A$ be a finite subset and $H\leq G$ an $A$-invariant subgroup. If $H$ contains an element generic over $A$, then $H=G$.

Proof.

Let $b\in G$ be an arbitrary element. Pick $a$ generic over $A\cup b$. Since $H$ is $A$-invariant and contains a generic element, we have that $a\in H$. (Recall that there is an automorphism over $A$ mapping one generic element to another.) By Proposition 3, $a\cdot b$ is also generic over $A$. Hence as above $a\cdot b\in H$. Finally since $H$ is a subgroup, we have that $b\in H$. ∎

As a corollary we get the following criterion for a group with homogeneous pregeometry to be commutative.

Proposition 7**.**

If a generic pair commutes in a group with a homogeneous pregeometry, then the group is commutative.

Proof.

Let $(a,b)$ be a generic pair and assume that $a$ and $b$ commute. Consider the centraliser $C_{G}(a)$ of $a$. It is definable over $a$ and contains a generic element $b$. So by the previous proposition, we have $C_{G}(a)=G$. By homogeneity the same is true for every generic element. Now given $c\in\operatorname{cl}(\emptyset)$, by the above it commutes with $a$. Hence $C_{G}(c)$ is $c$-definable and contains a generic element and therefore must be the whole group $G$. ∎

We will apply the following improved criterion.

Proposition 8**.**

Let $(a,b)$ be a generic pair over some finite $A$. If $b\cdot a\in\operatorname{cl}(A,a\cdot b)$, then $G$ is commutative.

Proof.

Assume that $b\cdot a\in\operatorname{cl}(A,a\cdot b)$. By homogeneity it is true for every generic pair over $A$. Now $(b,b^{-1}\cdot a)$ is a generic pair over $A$. And so $b^{-1}ab\in\operatorname{cl}(A,a)$. Now consider the set $\{x\in G:x^{-1}ax=b^{-1}ab\}$. It is definable over $A,a,b^{-1}ab$ and contains a generic element $b$. Hence it contains all generic elements over $A,a$. Pick $c$ generic over $A,a,b$. Then $c^{-1}ac=b^{-1}ab$ and therefore $(bc^{-1})a=a(bc^{-1})$. Since $(a,bc^{-1})$ is a generic pair, by the above $G$ is commutative. ∎

So far we have not used local modularity of the pregeometry. So everyting said so far also applies to groups with arbitrary homogeneous pregeometry. We now use the full strength of our assumptions for the main result.

Theorem 1.

A group with locally modular homogeneous pregeometry is commutative.

Proof.

Assume that $\operatorname{cl}$ is a locally modular homogeneous pregeometry on a group $G$. Pick a finite $A\not\subseteq\operatorname{cl}(\emptyset)$. Then the localisation $\operatorname{cl}_{A}$ of $\operatorname{cl}$ at $A$ is modular. (Recall that $\operatorname{cl}_{A}$ is defined as $\operatorname{cl}_{A}(X)=\operatorname{cl}(A\cup X)$.) Consider a generic pair $(a,b)$ over $A$. By the previous proposition $b\cdot a\in\operatorname{cl}_{A}(a\cdot b)$ implies that $G$ is commutative. So assume that $b\cdot a\not\in\operatorname{cl}_{A}(a\cdot b)$. Pick an element $c\not\in\operatorname{cl}_{A}(a,b)$. Similarly we assume that $c\cdot a\not\in\operatorname{cl}_{A}(a\cdot c)$. Consider the geometry of the operator $\operatorname{cl}_{A}$ on $\operatorname{cl}_{A}(a,b,c)$. Since the geometry is modular but not trivial (Corollary 5), it is a projective plane. (This essentially means that any two lines intersect.)

We claim that the lines $(b,c)$, $(ab,ac)$ and $(ba,ca)$ meet in a single point. Assume the opposite. Then these three lines meet in three different points $d_{1},d_{2}$ and $d_{3}$ as shown in the diagram below.

a$$b$$ab$$ba$$c$$ac$$ca$$d_{1}$$d_{2}$$d_{3}$$e_{1}$$e_{2}

Now consider the line $(a,d_{1})$ and assume that it meets the lines $(ba,ca)$ and $(ab,d_{2})$ in points $e_{1}$ and $e_{2}$ respectively. By the homogeneity assumption there is an automorphism $f$ of $G$ that fixes $a,e_{1},e_{2},d_{1}$ and takes $b$ to $c$. (Note that $e_{1},e_{2},d_{1}$ are points in the geometry so are equivalence classes of elements of $G$. What we mean by $f$ fixing them is $f$ fixing a representative.) Then $f$ induces a collineation of the projective plain (which we again denote by $f$). Since $f$ is an automorphism of $G$ it takes $ab$ to $ac$ and $ba$ to $ca$ respectively. Then $f$ fixes the lines $(b,c)$ and $(ba,ca)$ (since it also fixes two points $d_{1}$ and $e_{1}$ on them). Thus $f$ fixes their intersection point $d_{2}$. But this implies that $f$ fixes the line $(d_{2},e_{2})$. However $ab$ is on that line whereas $ac$ is not. This contradiction proves the claim.

But now note that $c^{-1}b\in\operatorname{cl}_{A}(b,c)\cap\operatorname{cl}_{A}(ab,ac)$ and $bc^{-1}\in\operatorname{cl}_{A}(b,c)\cap\operatorname{cl}_{A}(ba,ca)$. Thus $c^{-1}b\in\operatorname{cl}_{A}(bc^{-1})$. Since $(b,c^{-1})$ is a generic pair over $A$, by Proposition 8 the group $G$ is commutative. ∎

As a final remark, note that under the assumption of local modularity the geometry of $\operatorname{cl}$ on $G$ (after localisation) is isomorphic to the projective geometry of some vector space over a division ring. The group structure of $G$ however need not come from this vector space. Examples of locally modular strongly minimal groups that are not vector spaces can be found in Pillay (1996).