Generalised Vector Spaces and Finitary Matroids

I first learnt about the notion of a generalised vector space from a video by alg-d. ^{Footnote¹I learnt about it from alg-d's video [Alg]. The corresponding article and PDF version are available as [Algb; Alg14]. The video description and the article cite Bleicher [Ble64] as a reference.1} When I first looked at the definition, my impression was that it was quite close to the closure-operator axioms for matroids.

Indeed, the relation $X \Subset Y$ appearing in the definition of a generalised vector space can be read as saying that “$X$ is generated by $Y$”, or that “$X$ is spanned by $Y$”. For an ordinary vector space, one should think of this as

and that is the right guiding picture.

On the other hand, matroids also have a definition in terms of a closure operator

The assertion $x \in \cl(A)$ may be interpreted as saying that “$x$ is generated by $A$”. In particular, for a linear matroid this is exactly the condition that the vector corresponding to $x$ lies in the subspace spanned by the vectors corresponding to the elements of $A$.

In this note, I rewrite generalised vector spaces in the language of matroid closure operators. The conclusion, stated first, is that a generalised vector space is essentially the same thing as a finitary matroid defined by a closure operator, or equivalently a pregeometry. However, if the term “infinite matroid” is used in its modern sense, then generalised vector spaces do not cover all infinite matroids. They correspond to the finitary subclass, the one with no infinite circuits.

Since the axiom of choice itself is part of the story, I do not include the existence of bases as part of the definition in this note. Thus, by a finitary matroid here I mean a structure whose closure operator satisfies the matroid closure axioms. Under the axiom of choice, such a closure operator does give bases in the usual sense. Without the axiom of choice, however, the existence of bases is itself a non-trivial assertion.

Let us first write the definition of a generalised vector space in the form used in this note.

This definition becomes easier to parse if one reads $X \Subset Y$ as “$X$ is generated by $Y$”. Then (GV1)If $Y \subseteq X$, then $Y \Subset X$.(GV1) says that “a subset of $X$ is, of course, generated by $X$”, (GV2)If $X_{\lambda} \Subset Y$ for every $\lambda \in \Lambda$, then $\bigcup_{\lambda \in \Lambda} X_{\lambda} \Subset Y$.(GV2) says that “if each $X_{\lambda}$ is generated by $Y$, then collecting all of them together is still generated by $Y$”, and (GV3)If $X \Subset Y$ and $Y \Subset Z$, then $X \Subset Z$.(GV3) says that “if $X$ is generated by $Y$, and $Y$ is generated by $Z$, then $X$ is generated by $Z$”.

The especially important conditions are (GV4)If $\{x\} \Subset X \cup \{y\}$ and $\{x\} \not\Subset X$, then $\{y\} \Subset X \cup \{x\}$.(GV4) and (GV5)If $\{x\} \Subset X$, then there is a finite subset $Y \subseteq X$ such that $\{x\} \Subset Y$.(GV5). Condition (GV4)If $\{x\} \Subset X \cup \{y\}$ and $\{x\} \not\Subset X$, then $\{y\} \Subset X \cup \{x\}$.(GV4) is the Mac Lane–Steinitz exchange property. It abstracts the familiar phenomenon from linear algebra: if adding $y$ is what first makes it possible to generate $x$, then, conversely, adding $x$ makes it possible to generate $y$. Condition (GV5)If $\{x\} \Subset X$, then there is a finite subset $Y \subseteq X$ such that $\{x\} \Subset Y$.(GV5) is a finiteness condition: if one point is generated by all of $X$, then in fact it is generated already by a finite subset of $X$. Since linear combinations are finite sums, this condition holds naturally in ordinary vector spaces.

Independent sets and bases in a generalised vector space are defined as follows.

In matroid language, this says that an independent set is a set in which no element is closed by the others, and a basis is an independent set which spans the whole set. Below, we translate this literally into closure-operator language.

Let $(V,\Subset)$ be a generalised vector space. For $A \subseteq V$, define

Thus $\cl(A)$ is the set of all points generated by $A$.

With this definition, the original relation $\Subset$ can be recovered completely.

Proof

First suppose that $X \Subset A$. Take any $x \in X$. Since $\{x\} \subseteq X$, condition (GV1)If $Y \subseteq X$, then $Y \Subset X$.(GV1) gives $\{x\} \Subset X$. Since $X \Subset A$, condition (GV3)If $X \Subset Y$ and $Y \Subset Z$, then $X \Subset Z$.(GV3) then gives $\{x\} \Subset A$. Hence $x \in \cl(A)$, and so $X \subseteq \cl(A)$.

Conversely, suppose that $X \subseteq \cl(A)$. This means that $\{x\} \Subset A$ for every $x \in X$. If $X = \emptyset$, then $\emptyset \subseteq A$, so condition (GV1)If $Y \subseteq X$, then $Y \Subset X$.(GV1) gives $\emptyset \Subset A$. If $X \neq \emptyset$, then applying (GV2)If $X_{\lambda} \Subset Y$ for every $\lambda \in \Lambda$, then $\bigcup_{\lambda \in \Lambda} X_{\lambda} \Subset Y$.(GV2) to the family $\{\{x\}\}_{x \in X}$ gives

$$X = \bigcup_{x \in X} \{x\} \Subset A.$$

Therefore $X \Subset A$.

End of proof□

This lemma shows that $\Subset$ was in fact just the relation of being contained in a closure. Thus the information in a generalised vector space is completely described by the closure operator $\cl$.

Next, in a form that also works over infinite sets, we define finitary matroids by closure operators. In this note, I will call the following closure structure over ZF a “pregeometry with finite character” or a “finitary matroid closure structure”, without building the existence of bases into the definition. Under ZFC, this gives finitary matroids in the usual sense. One can think of a finitary matroid as a matroid closure operator on a finite set, but with an additional finite-character condition added for infinite ground sets.

If $E$ is finite, then (FCL5)For every $A \subseteq E$ and $x \in E$, if $x \in \cl(A)$, then there is a finite subset $A_{0} \subseteq A$ such that $x \in \cl(A_{0})$ (finite character).(FCL5) holds automatically. Thus, when one only works with finite matroids, (FCL1)For every $A \subseteq E$, one has $A \subseteq \cl(A)$ (extensibility).(FCL1)–(FCL4)For every $A \subseteq E$ and $x,y \in E$, if $x \in \cl(A \cup \{ y \}) \setminus \cl(A)$, then $y \in \cl(A \cup \{ x \})$ (Mac Lane–Steinitz exchange property).(FCL4) are the usual closure-operator axioms for matroids.

When $E$ is infinite, however, (FCL5)For every $A \subseteq E$ and $x \in E$, if $x \in \cl(A)$, then there is a finite subset $A_{0} \subseteq A$ such that $x \in \cl(A_{0})$ (finite character).(FCL5) is an essential condition. It rules out the phenomenon that dependence appears only after genuinely using infinitely many elements. In the usual matroid setting, where the existence of bases is also included, this finite-character condition corresponds to all circuits being finite.

We can now state the correspondence between generalised vector spaces and finitary matroid closure operators.

Proof

Constructing a finitary matroid closure operator from a generalised vector space.

Let $(V,\Subset)$ be a generalised vector space, and define $\cl(A) \coloneqq \{ x \in V \mathrel{:} \{ x \} \Subset A \}$. We verify (FCL1)For every $A \subseteq E$, one has $A \subseteq \cl(A)$ (extensibility).(FCL1)–(FCL5)For every $A \subseteq E$ and $x \in E$, if $x \in \cl(A)$, then there is a finite subset $A_{0} \subseteq A$ such that $x \in \cl(A_{0})$ (finite character).(FCL5) in turn.

Extensibility (FCL1)For every $A \subseteq E$, one has $A \subseteq \cl(A)$ (extensibility).(FCL1)

Let $A \subseteq V$ and $x \in A$. Since $\{ x \} \subseteq A$, condition (GV1)If $Y \subseteq X$, then $Y \Subset X$.(GV1) gives $\{x\}\Subset A$. Hence $x \in \cl(A)$, and therefore $A \subseteq \cl(A)$.

Monotonicity (FCL2)For every $A \subseteq B \subseteq E$, one has $\cl(A) \subseteq \cl(B)$ (monotonicity).(FCL2)

Suppose $A \subseteq B \subseteq V$. If $x \in \cl(A)$, then $\{ x \} \Subset A$. Since $A \subseteq B$, condition (GV1)If $Y \subseteq X$, then $Y \Subset X$.(GV1) gives $A \Subset B$. Hence, by (GV3)If $X \Subset Y$ and $Y \Subset Z$, then $X \Subset Z$.(GV3), we have $\{ x \} \Subset B$, and so $x \in \cl(B)$. Therefore $\cl(A) \subseteq \cl(B)$.

Idempotency (FCL3)For every $A \subseteq E$, one has $\cl(\cl(A))=\cl(A)$ (idempotency).(FCL3)

Extensibility and monotonicity imply $\cl(A) \subseteq \cl(\cl(A))$. We prove the reverse inclusion.

Let $x \in \cl(\cl(A))$. This means that $\{ x \} \Subset \cl(A)$. On the other hand, for every $y \in \cl(A)$, the definition gives $\{ y \} \Subset A$. Hence, by (GV2)If $X_{\lambda} \Subset Y$ for every $\lambda \in \Lambda$, then $\bigcup_{\lambda \in \Lambda} X_{\lambda} \Subset Y$.(GV2),

$$\cl(A) = \bigcup_{y \in \cl(A)} \{ y \} \Subset A$$

holds. By (GV3)If $X \Subset Y$ and $Y \Subset Z$, then $X \Subset Z$.(GV3), we get $\{ x \} \Subset A$. Thus $x \in \cl(A)$, and so $\cl(\cl(A)) \subseteq \cl(A)$.

Exchange (FCL4)For every $A \subseteq E$ and $x,y \in E$, if $x \in \cl(A \cup \{ y \}) \setminus \cl(A)$, then $y \in \cl(A \cup \{ x \})$ (Mac Lane–Steinitz exchange property).(FCL4)

Suppose $x \in \cl(A \cup \{ y \}) \setminus \cl(A)$. This means precisely that

$$\{ x \} \Subset A \cup \{ y \} \quad\text{and}\quad \{ x \} \not\Subset A.$$

By (GV4)If $\{x\} \Subset X \cup \{y\}$ and $\{x\} \not\Subset X$, then $\{y\} \Subset X \cup \{x\}$.(GV4), it follows that

$$\{ y \}\Subset A \cup \{ x \}.$$

Therefore $y \in \cl(A \cup \{ x \})$.

Finite character (FCL5)For every $A \subseteq E$ and $x \in E$, if $x \in \cl(A)$, then there is a finite subset $A_{0} \subseteq A$ such that $x \in \cl(A_{0})$ (finite character).(FCL5)

Suppose $x \in \cl(A)$. This means that $\{ x \} \Subset A$. By (GV5)If $\{x\} \Subset X$, then there is a finite subset $Y \subseteq X$ such that $\{x\} \Subset Y$.(GV5), there is a finite subset $A_{0} \subseteq A$ such that $\{ x \} \Subset A_{0}$. Hence $x \in \cl(A_{0})$.

Thus $(V, \cl)$ is a finitary matroid.

Constructing a generalised vector space from a finitary matroid.

Let $(E,\cl)$ be a finitary matroid, and define

$$X \Subset Y \quad \Longleftrightarrow \quad X \subseteq \cl(Y).$$

We verify (GV1)If $Y \subseteq X$, then $Y \Subset X$.(GV1)–(GV5)If $\{x\} \Subset X$, then there is a finite subset $Y \subseteq X$ such that $\{x\} \Subset Y$.(GV5).

(GV1)If $Y \subseteq X$, then $Y \Subset X$.(GV1)

Suppose $Y \subseteq X$. By extensibility, $X \subseteq \cl(X)$, so $Y \subseteq \cl(X)$. Hence $Y \Subset X$.

(GV2)If $X_{\lambda} \Subset Y$ for every $\lambda \in \Lambda$, then $\bigcup_{\lambda \in \Lambda} X_{\lambda} \Subset Y$.(GV2)

Suppose $X_{\lambda}\Subset Y$ for every $\lambda \in \Lambda$. This means that $X_{\lambda}\subseteq \cl(Y)$ for every $\lambda$. Therefore

$$\bigcup_{\lambda\in\Lambda}X_{\lambda} \subseteq \cl(Y),$$

and so $\bigcup_{\lambda \in \Lambda} X_{\lambda} \Subset Y$.

(GV3)If $X \Subset Y$ and $Y \Subset Z$, then $X \Subset Z$.(GV3)

Suppose $X \Subset Y$ and $Y \Subset Z$. This means that $X \subseteq \cl(Y)$ and $Y \subseteq \cl(Z)$. By monotonicity,

$$\cl(Y) \subseteq \cl(\cl(Z)).$$

By idempotency, $\cl(\cl(Z))=\cl(Z)$. Hence $X \subseteq \cl(Z)$, and so $X \Subset Z$.

(GV4)If $\{x\} \Subset X \cup \{y\}$ and $\{x\} \not\Subset X$, then $\{y\} \Subset X \cup \{x\}$.(GV4)

Suppose $\{ x \} \Subset X \cup \{ y \}$ and $\{ x \} \not\Subset X$. This means that

$$x \in \cl(X \cup \{ y \}) \setminus \cl(X).$$

By exchange,

$$y \in \cl(X \cup \{ x \}),$$

and therefore $\{ y \} \Subset X \cup\{ x \}$.

(GV5)If $\{x\} \Subset X$, then there is a finite subset $Y \subseteq X$ such that $\{x\} \Subset Y$.(GV5)

Suppose $\{ x \} \Subset X$. This means that $x \in \cl(X)$. By finite character, there is a finite subset $Y \subseteq X$ such that $x \in \cl(Y)$. Hence $\{ x \} \Subset Y$.

Thus $(E, \Subset)$ is a generalised vector space.

Finally, let us check that the two constructions are inverse to one another. If we start from a generalised vector space, then by Lemma Lemma 1, the original relation $\Subset$ is equivalent to $X \subseteq \cl(Y)$, and is therefore recovered. Conversely, if we start from a finitary matroid closure operator, the newly obtained closure is

$$\{ x \in E \mathrel{:} \{ x \} \Subset A \} = \{ x \in E \mathrel{:} x \in \cl(A) \} = \cl(A).$$

Hence the two constructions are inverse to one another.

End of proof□

This theorem says that a generalised vector space can be regarded as another presentation of a finitary matroid. In particular, instead of using the binary relation $\Subset$, one may use just the closure operator

without losing any information.

Under the correspondence above, independent sets and bases in a generalised vector space correspond exactly to independent sets and bases in the associated matroid.

Proof

The first statement is just a rephrasing of the definition. Indeed,

$$\{ x \} \Subset I \setminus \{ x \} \quad \Longleftrightarrow \quad x \in \cl(I \setminus \{ x \}).$$

For the second statement, the definition of a basis in a generalised vector space is: “independent and satisfying $V \Subset B$”. By Lemma Lemma 1,

$$V \Subset B \quad \Longleftrightarrow \quad V \subseteq \cl(B).$$

Since always $\cl(B) \subseteq V$, this is equivalent to $\cl(B) = V$.

End of proof□

Therefore the statement in alg-d's article that “every generalised vector space has a basis”, which is equivalent to the axiom of choice, can be read in matroid language as “every finitary matroid closure operator has a basis”. Under this assumption, in a finitary matroid the usual exchange property (FCL4)For every $A \subseteq E$ and $x,y \in E$, if $x \in \cl(A \cup \{ y \}) \setminus \cl(A)$, then $y \in \cl(A \cup \{ x \})$ (Mac Lane–Steinitz exchange property).(FCL4) shows that any two bases of the same closed set have the same cardinality. We then define the dimension $\dim F$ of a closed set $F$ to be the cardinality of a basis of $F$. For a general subset $A \subseteq E$, define

The reason why the basis-existence theorem for generalised vector spaces is equivalent to the axiom of choice is also quite natural in the language of pregeometries.

In the direction from the axiom of choice to the existence of bases, one applies Tukey's lemma or Zorn's lemma to the collection of all independent sets, and obtains a maximal independent set. Because of the finite-character condition (FCL5)For every $A \subseteq E$ and $x \in E$, if $x \in \cl(A)$, then there is a finite subset $A_{0} \subseteq A$ such that $x \in \cl(A_{0})$ (finite character).(FCL5), independence can be tested on finite subsets. This is exactly what makes the standard argument for maximal independent sets work.

More concretely, order the collection of independent sets by inclusion. For a chain of independent sets $\{ I_{\alpha} \}_{\alpha \geq 1}$, put $\displaystyle I \coloneqq \bigcup_{\alpha \geq 1} I_{\alpha}$. Then $I$ is independent. Indeed, if some $x \in I$ lies in the closure of $I \setminus \{ x \}$, then by finite character there is a finite set $F \subseteq I \setminus \{ x \}$ such that $x \in \cl(F)$. By the chain condition, $F \cup \{ x \}$ is contained in some $I_{\alpha}$, contradicting the independence of $I_{\alpha}$.

Let us also see why a maximal independent set spans the whole set. Argue by contradiction, and suppose that a maximal independent set $B$ does not span the whole set, that is, $\cl(B)\ne E$. Choose $x \in E \setminus \cl(B)$. Then $B \cup \{ x \}$ is independent. Indeed, if $B \cup \{ x \}$ were dependent, then some $y \in B \cup \{ x \}$ would lie in the closure of the other elements. The case $y = x$ contradicts $x \notin \cl(B)$, and the case $y \in B$ implies $x \in \cl(B)$ by exchange. Thus adding any $x \in E \setminus \cl(B)$ would not destroy independence, contradicting the maximality of $B$. Hence $B$ is a basis.

For the reverse direction, deriving the axiom of choice from the existence of bases, it is helpful to look at partition matroids. For a family of non-empty sets which are not necessarily disjoint, one may replace $X_{\lambda}$ by $\{ \lambda \} \times X_{\lambda}$, so it is enough to consider a pairwise disjoint family of non-empty sets $\{ X_{\lambda} \}_{\lambda \in \Lambda}$. Put

For $A\subseteq E$, define

This is a finitary matroid closure operator. Intuitively, once one chooses even a single element from a block $X_{\lambda}$, the whole block enters the closure.

In this matroid, an independent set is exactly a set which contains at most one element from each block $X_{\lambda}$. To be a basis means to be independent and to span every block. Therefore a basis $B$ satisfies, for every $\lambda$,

In other words, $B$ is precisely a choice set for the family $\{X_{\lambda}\}_{\lambda\in\Lambda}$.

Seen this way, it is quite direct that the existence of bases for generalised vector spaces contains the axiom of choice. If every finitary matroid has a basis, then in particular the partition matroid above has a basis, and that basis gives a choice function.

The discussion so far shows that generalised vector spaces are the same thing as finitary matroids. But are they the same thing as “infinite matroids”?

The answer depends on how the words are being used. If an infinite matroid simply means a matroid-like closure structure whose ground set may be infinite, then a generalised vector space is certainly one kind of such object. However, if one means infinite matroids in the modern sense developed after Bruhn–Diestel–Kriesell–Pendavingh–Wollan [BDKP+13], then generalised vector spaces do not cover all of them.

The difference lies in the finite-character condition. In a generalised vector space, condition (GV5)If $\{x\} \Subset X$, then there is a finite subset $Y \subseteq X$ such that $\{x\} \Subset Y$.(GV5) in Definition Definition gives

Thus dependence is always detected by a finite subset. Consequently, infinite circuits do not appear.

Modern infinite matroids, on the other hand, allow examples with infinite circuits. Instead, they impose stronger maximality conditions on independent sets, bases, circuits, closure, and related notions, so that bases and duality behave as they do for finite matroids. This theory can handle, for instance, duals of finitary matroids, which need not be finitary in general.

As this example shows, generalised vector spaces are not all modern infinite matroids. They correspond precisely to the finitary ones among them, that is, those whose circuits are all finite.

We have seen that generalised vector spaces have the same content as finitary matroids, or pregeometries. What, then, are such structures good for outside ordinary vector spaces?

In short, matroids and pregeometries are tools for carrying the linear-algebraic vocabulary \begin{quote} “generation”, “dependence”, “independence”, “basis”, and “dimension” \end{quote} into structures which are not vector spaces. In a vector space, saying that a vector is generated by other vectors means that it can be written as a linear combination of them. In mathematics, however, many other kinds of dependence relation occur. Examples include algebraic dependence given by polynomial equations, and algebraic dependence in model theory given by logical formulae.

As a first standard example away from vector spaces, consider algebraic dependence in a field extension. Let $K \subseteq L$ be a field extension. For $A \subseteq L$, define

Here $K(A)$ is the subfield obtained by adjoining all elements of $A$ to $K$. The exchange property corresponds to Steinitz's exchange lemma in field theory, or equivalently to the exchange property for algebraic independence.

This closure expresses generation not by linear combinations, but by polynomial equations. That is, $x \in \cl(A)$ means that $x$ is determined as a solution of some algebraic equation whose coefficients may use elements of $A$.

In this case, independent sets are algebraically independent sets. Thus $A \subseteq L$ is independent precisely when there is no non-trivial polynomial relation with coefficients in $K$ among the elements of $A$. Moreover, bases are transcendence bases of $L/K$. Consequently, the dimension of the pregeometry agrees with the transcendence degree of the field extension. The general existence proof for transcendence bases also uses Zorn's lemma, and hence the axiom of choice. For transcendence bases and transcendence degree, see for example [Aut, Section 9.26].

The dependence relation in this example is not linear. For instance, the relation $y = x^{2}$ is not a linear relation, but $y$ is algebraically determined by $x$. In the language of pregeometries, this non-linear dependence can still be written as $y \in \cl(\{ x \})$.

Thus matroids and pregeometries can be viewed not merely as “generalisations of vector spaces”, but as a general theory of dependence relations and dimension.

One area where pregeometries appear especially naturally is model theory. Model theory treats groups, rings, fields, ordered sets, and other objects as structures for first-order logic, and studies the sets definable inside those structures. For the discussion of model theory, I have used Masanori Itai's Introduction to Geometric Model Theory, revised edition [Ita20], as a reference. Please see that book for a fuller account.

Roughly speaking, an infinite structure $M$ is minimal if every definable subset of $M$ in one variable is either finite or cofinite. If the same property holds in every structure elementarily equivalent to $M$, then $M$ is called strongly minimal. More precisely, a structure $M$ is strongly minimal if, in every structure $N$ elementarily equivalent to $M$, every one-variable subset of $N$ definable with parameters is finite or cofinite.

For a strongly minimal structure $M$, define the algebraic closure $\acl(A)$ of $A \subseteq M$ as follows. ^{Footnote²Strictly speaking, this is usually defined inside a sufficiently saturated monster model. For the purposes of this explanatory note, I define it inside the structure $M$ in order to keep the intuition visible.2} An element $a \in M$ belongs to $\acl(A)$ if there is a formula $\varphi(x, \bar{a})$ with a finite tuple of parameters $\bar{a} \in A^{<\omega}$ such that $\varphi(M)$ is finite and

In other words, $a$ lies in a finite set defined using parameters from $A$. It is well known that, in a strongly minimal set, $\acl$ defines a pregeometry [Mar98; Ita20].

This is a logical generalisation of algebraic closure in field theory. For an algebraically closed field, a formula with only finitely many solutions can be thought of, intuitively, as describing the finitely many solutions of a polynomial equation. Thus model-theoretic $\acl$ corresponds, in algebraically closed fields, to field-theoretic algebraic closure.

In a strongly minimal structure, this $\acl$ defines a pregeometry. Therefore one can define independence, bases, and dimension inside strongly minimal structures. This dimension plays the same kind of role as dimension in vector spaces or transcendence degree in field extensions.

From this perspective, a pregeometry is rather like the combinatorial shadow of a strongly minimal structure. The original structure may be a complicated object described by logical formulae, but when one extracts $\acl$, a matroid-like geometry appears. The properties of that geometry can then be used to study the original structure.

Among the properties of pregeometries, triviality, modularity, and local modularity are especially important in model theory.

A pregeometry $(M,\cl)$ is trivial if, for every $A\subseteq M$,

holds. This means that combining several elements produces no new dependence relation. If something is generated by $A$, then in fact it is generated by just one element of $A$.

On the other hand, a modular pregeometry is one in which, for finite-dimensional closed sets $A$ and $B$,

holds. This is analogous to the dimension formula for subspaces of a vector space. For this reason, modular pregeometries can be thought of as having “vector-space-like” or “projective-geometric” behaviour. Local modularity means, roughly, that after quotienting out $\cl(\emptyset)$ and parallel elements, or after localising at a finite set, the modular dimension formula holds. For the present intuitive discussion, it is enough to think of it as the property that once a base point is fixed, the geometry satisfies a vector-space-like dimension formula.

The pregeometry arising from an algebraically closed field is not modular in general. In an algebraically closed field, dependence is not merely linear dependence; it is algebraic dependence given by polynomial equations. For example, if $a$, $b$, and $x$ are independent and we set $y=ax+b$, then $a$ and $b$ determine a line, while $x$ and $y$ determine a point on that line. This creates a geometric interaction. More concretely, over $\acl(\emptyset)$, take algebraically independent elements $a$, $b$, and $x$, and put $y = ax + b$. Consider the closed sets $A = \acl(a,b)$ and $B = \acl(x,y)$. Then $\dim A = \dim B = 2$, and since $A \cap B = \acl(\emptyset)$, we have $\dim(A \cap B) = 0$. On the other hand,

holds. This disagrees with the value $4$ demanded by the modular dimension formula. The interaction between a pair of parameters and a pair of points on the line is not captured by a purely vector-space-like dimension formula, and appears as non-modularity.

Thus, by studying the pregeometry attached to a strongly minimal structure, one can distinguish whether the structure is

• a trivial structure with almost no interaction,

• a vector-space-like or projective-geometric structure, or

• a non-linear algebraic-geometric structure, such as an algebraically closed field.

This point of view lies behind Zilber's trichotomy. Very roughly, it is the idea that the geometry of a strongly minimal structure should fall into one of three types: trivial, vector-space-like, or field-like. Hrushovski's construction [Hru93] gives counterexamples to the naive form of Zilber's trichotomy, but it also illustrates how central pregeometries are in the classification of strongly minimal structures.

Another related class of examples is algebraic matroids. These record algebraic dependence relations among finitely many quantities as a matroid.

For instance, suppose we have an object defined by a system of polynomial equations, and on it there are coordinates or observables

More precisely, one records algebraic dependence among coordinate functions $x_{1}, \dots, x_{n}$ in the function field of an irreducible algebraic variety, or among elements $x_{1}, \dots, x_{n}$ of a field extension. A subset $S\subseteq \{ 1, \dots, n \}$ is independent if the corresponding coordinates satisfy no non-trivial polynomial relation. Conversely, a minimal algebraically dependent set is a circuit. Thus a circuit is a minimal set of variables on which a non-trivial algebraic relation first appears.

In this way, algebraic matroids are combinatorial tools for describing which quantities are algebraically determined by which other quantities. Linear matroids deal with linear relations, whereas algebraic matroids deal with polynomial relations. They therefore arise naturally in settings with non-linear algebraic constraints, such as algebraic statistics, low-rank matrix completion, and rigidity theory.

From this viewpoint as well, generalised vector spaces and pregeometries are not merely analogues of vector spaces. They are a common language for abstracting dependence relations and measuring dimension.

For an introductory account of algebraic matroids, see Rosen–Sidman–Theran [RST20].

The main point of this note is that generalised vector spaces and finitary matroid closure operators are the same concept. The correspondence is very simple. Starting from a generalised vector space $(V,\Subset)$, define

This gives a finitary matroid closure operator. Conversely, starting from a finitary matroid closure operator $\cl$, define

This gives a generalised vector space.

Under this correspondence, independent sets and bases also agree. Thus the statement appearing in alg-d's article, “every generalised vector space has a basis”, can be read in matroid language as “every finitary matroid closure operator has a basis”. By considering partition matroids, one can also see naturally why this basis-existence statement contains the axiom of choice.

However, modern infinite matroids form a broader class than finitary matroids. They allow matroids with infinite circuits. Therefore it is most accurate to think of a generalised vector space not as “an arbitrary infinite matroid”, but as a finitary infinite matroid, or a pregeometry.

Footnotes

1. I learnt about it from alg-d's video [Alg]. The corresponding article and PDF version are available as [Algb; Alg14]. The video description and the article cite Bleicher [Ble64] as a reference. ↩
2. Strictly speaking, this is usually defined inside a sufficiently saturated monster model. For the purposes of this explanatory note, I define it inside the structure $M$ in order to keep the intuition visible. ↩

References

1. [BDKP+13] Henning Bruhn, Reinhard Diestel, Matthias Kriesell, Rudi Pendavingh, and Paul Wollan. Axioms for infinite matroids. Adv. Math., vol. 239, pp. 18–46, 2013. doi:10.1016/j.aim.2013.01.011 Citation contextThe answer depends on how the words are being used. If an infinite matroid simply means a matroid-like closure structure whose ground set may be infinite, then a generalised vector space is certainly one kind of such object. However, if one means infinite matroids in the modern sense developed after Bruhn–Diestel–Kriesell–Pendavingh–Wollan [BDKP+13], then generalised vector spaces do not cover all of them.↩
2. [Aut] The Stacks Project Authors. The Stacks Project, Section 9.26: Transcendence Citation contextIn this case, independent sets are algebraically independent sets. Thus $A \subseteq L$ is independent precisely when there is no non-trivial polynomial relation with coefficients in $K$ among the elements of $A$. Moreover, bases are transcendence bases of $L/K$. Consequently, the dimension of the pregeometry agrees with the transcendence degree of the field extension. The general existence proof for transcendence bases also uses Zorn's lemma, and hence the axiom of choice. For transcendence bases and transcendence degree, see for example [Aut, Section 9.26].↩
3. [Ita20] 板井 昌典 (Masanori Itai). 幾何的モデル理論入門［改訂版］ (Introduction to Geometric Model Theory, revised edition). 日本評論社 (Nippon Hyoron Sha), 2020 Citation contextOne area where pregeometries appear especially naturally is model theory. Model theory treats groups, rings, fields, ordered sets, and other objects as structures for first-order logic, and studies the sets definable inside those structures. For the discussion of model theory, I have used Masanori Itai's Introduction to Geometric Model Theory, revised edition [Ita20], as a reference. Please see that book for a fuller account.↩1 Citation contextIn other words, $a$ lies in a finite set defined using parameters from $A$. It is well known that, in a strongly minimal set, $\acl$ defines a pregeometry [Mar98; Ita20].↩2
4. [Mar98] David Marker. Strongly minimal sets and geometry. Logic Colloquium '95 (Haifa), vol. 11, pp. 191–213, 1998. doi:10.1007/978-3-662-22108-2_12 Citation contextIn other words, $a$ lies in a finite set defined using parameters from $A$. It is well known that, in a strongly minimal set, $\acl$ defines a pregeometry [Mar98; Ita20].↩
5. [Hru93] Ehud Hrushovski. A new strongly minimal set. Ann. Pure Appl. Logic, vol. 62, no. 2, pp. 147–166, 1993. doi:10.1016/0168-0072(93)90171-9 Citation contextThis point of view lies behind Zilber's trichotomy. Very roughly, it is the idea that the geometry of a strongly minimal structure should fall into one of three types: trivial, vector-space-like, or field-like. Hrushovski's construction [Hru93] gives counterexamples to the naive form of Zilber's trichotomy, but it also illustrates how central pregeometries are in the classification of strongly minimal structures.↩
6. [RST20] Zvi Rosen, Jessica Sidman, and Louis Theran. Algebraic matroids in action. Amer. Math. Monthly, vol. 127, no. 3, pp. 199–216, 2020. doi:10.1080/00029890.2020.1689781 Citation contextFor an introductory account of algebraic matroids, see Rosen–Sidman–Theran [RST20].↩
7. [Alg] alg-d. Generalized vector space [Axiom of Choice] Citation contextI learnt about it from alg-d's video [Alg]. The corresponding article and PDF version are available as [Algb; Alg14]. The video description and the article cite Bleicher [Ble64] as a reference.↩
8. [Algb] alg-d. 線型空間の基底の存在 : 選択公理 Citation contextI learnt about it from alg-d's video [Alg]. The corresponding article and PDF version are available as [Algb; Alg14]. The video description and the article cite Bleicher [Ble64] as a reference.↩
9. [Alg14] alg-d. 線型空間の基底の存在と選択公理. 2014 Citation contextI learnt about it from alg-d's video [Alg]. The corresponding article and PDF version are available as [Algb; Alg14]. The video description and the article cite Bleicher [Ble64] as a reference.↩
10. [Ble64] M. Bleicher. Some theorems on vector spaces and the axiom of choice. Fundamenta Mathematicae, vol. 54, no. 1, pp. 95–107, 1964. doi:10.4064/fm-54-1-95-107 Citation contextI learnt about it from alg-d's video [Alg]. The corresponding article and PDF version are available as [Algb; Alg14]. The video description and the article cite Bleicher [Ble64] as a reference.↩