Non-Standard Axioms for Various Math Structures 3

In this post, we continue the discussion from the previous in the series.

We make some additions. First, it is clear that if an equational identity is true for a group $G$ , then it must be true for a subgroup $H$ of $G$ — by definition, the identity must be of the form $\forall x_{1}\ldots\forall x_{n}(\ldots)$ , and if this holds with any combination of the $x_{i}$ taken from $G$ then it certainly holds with any combination of the $x_{i}$ taken from a subset of $G$ , including a subgroup. This is generally true: for any algebra $A$ , if an equation $\lambda$ is true of $A$ , then $\lambda$ is true of any subset $A^{'}$ of $A$ .

This actually shows that: the class of all groups is the equational wide closure of the class of all symmetry groups. In other words, we can alternatively define a group as follows:

Definition. For any set $S$ , let $P(S)$ denote the set of permutations of $S$ with the operations of composition, identity, and inverse. Let $E$ be the set of all equations that hold for all such $P(S)$ over all sets $S$ . Then, a group can be defined as any model of $E$ .

In particular, the three equations consisting of the associative property, existence of identity, and existence of inverse are enough to generate all of $E$ — in the language used in the Wikipedia article on abstract Boolean algebras, these three equations are a basis. This inspires similar questions: given our kinds of characterizations of structures that result from the concepts of narrow and wide closures, we can ask about finite or even numerically “small” bases for the relevant sets of equations or first-order statements.

Note that our logic doesn’t show that a group can be defined as a model of $E$ generated from the $P(S)$ that only include the operation of composition, excluding identity and inverse, since we haven’t seen that a group could be defined with purely equations involving composition (multiplication.) Actually, is that really true though? Let’s step through our logic a bit more slowly. Let’s ignore the associativity, existence of identity, and existence of inverse properties completely. (By the way, whatever we do here can shed insight on our question on dependence in general on which operations are included in the signature.)

What does Cayley’s Theorem say? Let’s say our first-order language is just multiplication, $L = \left\{ \times \right\}$ . Let $C$ be the class of subgroups of symmetry groups, as models of $L$ (so including just multiplication.) Then, Cayley’s Theorem says that any group $G$ must be isomorphic to a member of $C$ — with isomorphism defined with respect to just multiplication. However, we know that $f(ab) = f(a)f(b)$ for a group implies $f(1) = 1$ and $f\left( a^{- 1} \right) = f(a)^{- 1}$ . So this implication is what will make the translation into a different signature possible. For example, for rings this implication doesn’t hold (it is possible for $f(a + b) = f(a) + f(b)$ and $f(ab) = f(a)f(b)$ but $f(1) \neq 1$ . As an example, take $f\mathbb{:Z \rightarrow Z}$ with $f(x) = 0$ .) Going back to groups, the isomorphism for just multiplication implies that any first-order statement in $\left\{ \times \right\}$ will be true of one group if and only if it is true of the other (when both are isomorphic.) But the fact that a just-multiplication isomorphism also preserves identity and inverse then implies that any first-order statement in $\left\{ \times ,e,a \rightarrow a^{- 1} \right\}$ will be true of one group if and only if it is true of the other. The set of first-order statements in the second statement is a superset of that in the first statement, so this is a great improvement!

So then let $C$ now stand for the class of all symmetry groups, as models of $\left\{ \times ,e,a \rightarrow a^{- 1} \right\}$ (all three operations.) Cayley’s Theorem implies isomorphism with respect to this expanded language even though it just states isomorphism with respect to the smaller language $\left\{ \times \right\}$ , and the isomorphism for the expanded language shows that any equation in the expanded language true for all symmetry groups is also true for all groups. Furthermore, some equations in the expanded language true for all symmetry groups include the three properties of associativity, existence of identity, and existence of inverse, so we have that the class of groups is the class of all structures that satisfy every equation in the expanded language true for all symmetry groups.

This last part is what can break down with an analogous statement for just multiplication. Clearly, we have that isomorphism with respect to $\left\{ \times ,e,a \rightarrow a^{- 1} \right\}$ implies isomorphism with respect to just $\left\{ \times \right\}$ . Thus, any equation in just $\left\{ \times \right\}$ true for all symmetry groups is also true for all groups. This implies that the class of all groups is a subclass of the wide closure of the class of all symmetry groups with respect to $\left\{ \times \right\}$ . But for it to be equal, our logic would need an equational axiomatization of a group with just multiplication. If such an axiomatization didn’t exist, then it’s possible that there could be structures that are not groups but that satisfy all equations in $\left\{ \times \right\}$ true for all symmetry groups — intuitively, this set of equations could be smaller, so that the class of structures satisfying them could be larger.

So this really boils down to equivalent axiomatization, while taking care to respect different kinds of statements (different logics.) Let’s state this more precisely in general:

Statement. Let $\mathcal{L}$ be a logic that is a subset of first-order logic. (So $\mathcal{L}$ could be first-order logic itself, or equational logic, or something else.) Let $L$ be a first-order language (so that $L$ in particular applies to $\mathcal{L}$ .) Let $C$ be a class of first-order models of $L$ . Then, let $W$ be the $\mathcal{L}$ -wide closure of $C$ with respect to $L$ . If $C$ can be equivalently axiomatized in first-order logic by a different first-order language $L^{'}$ , but not in the logic of $\mathcal{L}$ , then the $\mathcal{L}$ -wide closure of $C$ with respect to $L^{'}$ may be different from $W$ (at least, our reasoning doesn’t show that they’re the same.)

Let’s further try a counterexample to hammer home this point. Let’s use rings. But first actually, let’s update our notation: we see that narrow and wide closures are dependent on both the first-order language as well as the logic. So especially in the presence of possible confusion, we should introduce terminology that adequately clarifies both. (I’m not going to do anything now with logics that aren’t subsets of first-order logic.) Let’s say that given a first-order language $L$ and a logic $\mathcal{L}$ , we can speak of the “ $L$ -language $\mathcal{L}$ -logic narrow/wide closure.” Thus, with this terminology, we can say definitively that: the class of all groups is the $\left\{ \times ,e,a \rightarrow a^{- 1} \right\}$ -language equational-logic wide closure of the class of all symmetry groups. (Intuitively, this views groups as their underlying sets, where we then impose the language on them, as opposed to groups containing data about their language themselves.)

So, for our counterexample. Let $R$ be the class of all rings. We want to exhibit a subclass $R^{'}$ such that $R$ is the equational-logic $\left\{ + , \times \right\}$ -language wide closure of $R^{'}$ , but not the equational-logic $\left\{ + , \times ,1 \right\}$ -language wide closure of $R^{'}$ . Actually, this is just the same as what we said for groups: our reasoning would break down if we couldn’t find an equational axiomatization for rings without the multiplicative identity operation.

We’ve stated that “our reasoning can’t conclude X”; can we go further and say that without such an equivalent axiomatization, “X can’t be true”? In other words, can we show that: if there doesn’t exist an equational axiomatization for groups in just multiplication, then the equational wide closure in just multiplication must be strictly larger than the equational wide closure in the expanded language? Well, if these two wide closures were equal, then every equation $E$ true of all members of the “representative” $C$ would be true of all groups, in just multiplication. But then the set of such equations $E$ is literally the equational axiomatization of groups in just multiplication! So we’ve shown the contrapositive, which yields our desired result. We can generalize this reasoning:

Result. If $C$ is a class of models of $L$ and also a class of models of $L^{'}$ , then the following are equivalent:

The $\mathcal{L}$ -logic wide closures of $C$ in the languages $L$ and $L^{'}$ are equal;
The $\mathcal{L}$ -logic wide closure of $C$ in the language $L$ can be equivalently axiomatized in the logic $\mathcal{L}$ and language $L^{'}$ , and vice versa.

So determining whether the class of groups is the wide closure of the class of symmetry groups in just multiplication is the same as determining whether groups are an equational class under just multiplication. Can we axiomatize groups via just equations in multiplication?

From Working On It (unm.edu) (which I referenced in the first post of the series), it looks like we can’t just do multiplication for equational axioms, we need inverse too. Of course, this is single equational axioms, but if the axiomatization doesn’t exist for just multiplication then we’d have an answer to our question.

To be fair, we could also ask about finite equational axiomatizations; in the reasoning above, the set of all equations $E$ true of the representative could very well be infinite.

Sidenote: the fact that we have the terminology of narrow and wide closures has allowed us to sunset the “representative” term, which is great; I like “narrow closure” and “wide closure” a lot more. Or maybe better terms could be “weaker closure” or “stronger closure.” (Although the appearance of “weak” and “strong” seem rampant throughout math, and therefore prone to ambiguity.)

Actually, is this result fully true? Did we miss something around the isomorphism implication that was necessary earlier?

Let’s step through our reasoning slowly to be sure. Let’s say that we have a logic $\mathcal{L}$ , and we have a class $C$ of models of a language $L$ . Let $W$ be the $\mathcal{L}$ -logic $L$ -language wide closure of $C$ (we explicitly say in the language of $L$ , although the fact that we define $C$ to consist of models of $L$ makes this redundant here.) Assume that we can also express all of $C$ as models of another language $L^{'}$ .

First, let’s show (2) implies (1). Let’s say that we can find a set $S$ of $\mathcal{L}$ -logic statements in $L$ that axiomatize $W$ , or in other words that $W$ is the class of all models of $L$ that satisfy $S$ . Let’s also say that there exists an alternative axiomatization of $W$ in the language of $L^{'}$ , or more formally, there exists a set $S^{'}$ of $\mathcal{L}$ -logic statements in $L^{'}$ such that $W$ is the class of all models of $L^{'}$ that satisfy $S^{'}$ . (When we say “is,” we mean equality of underlying universal sets.) Then, let $W^{'}$ be the $L^{'}$ -language ( $\mathcal{L}$ -logic) wide closure of $C$ . We want to show that $W = W^{'}$ . It remains to continue this.

We continue this discussion in a follow-up post.

Share and like (need WordPress account to like):

Leave a comment Cancel reply