6
$\begingroup$

Short version: what are some interesting hyperdoctrines for classical (not intuitionistic) first-order logic, that are not models in the traditional sense? (Where the terminal and initial hyperdoctrines are "uninteresting".)

Long version:

The categorical semantics of first-order logic are given by hyperdoctrines. This is in contrast with the traditional semantics in terms of model theory.

In brief, in the traditional picture, we think of the semantics of a first-order theory as being some set $U$ ("the universe") plus an interpretation of every constant symbol / function symbol / predicate symbol in that universe (so, eg, for the unary function symbol $f$, the model provides $[\![f]\!] : U \to U$, and so on and so forth.) We can then interpret every proposition in the theory, interpreting (for example) $f x = x$ as the subset of $U$ where $[\![f]\!]$ is fixed.

By contrast, in the categorical style, we think of the semantics of a first-order theory T as being a hyperdoctrine plus some interpretations for the symbols and etc, which I'll call a "hyperdoctrine for T". This is more-or-less a stream of lattices, equipped with some strucures relating the lattices to one another, plus interpretations for the various symbols in terms of the lattices. Roughly speaking, the nth lattice is thought of as the possible interpretations of a proposition with n variables free, and the structures relating the lattices are about substitution and quantification.

The latter framework is more general. For instance, we can turn a traditional model for a theory T into a hyperdoctrine for T by letting the nth lattice be the lattice of all subsets of $U^n$. But we also have new hyperdoctrines: most notably a terminal one, corresponding to the choice where every lattice in the stream is the trivial (one-object) lattice; and the initial one, corresponding to the syntax.

And presumably, the categorical semantics also add a whole host of more interesting "new models". Like, presumably there are hyperdoctrines (for, say, 1st order arithmetic) that assert some combination of sentences, that no traditional model asserts. (Such combinations must necessarily be infinite, on account of the traditional completeness theorem, but still. (ETA: Not quite; see the comments below.)) And, like, yes, the terminal and initial hyperdoctrines show us some boring ways that this is true, but surely this newfound generality does more than just bolt a new "initial" and "terminal" model onto the traditional models. So, what are some of these new models?

(Ideally in the classical setting; I know we can get topological models and stuff if we consider intuitionistic logic, but it still seems to me that even classically we must have additional interesting hyperdoctrines, and I'd like to know what they are.)

(Ideally I'm looking for hyperdoctrines that feel motivated in their own right, more like "the lattices contain only the propositions that satisfy the following natural property" than "well given any hyperdoctrine we can generate a new one by bolting on a spandrel; just do that to the initial model". My apologies for the vagueness of this constraint. What I'm really after here are intuitions about how hyperdoctrines expand the space of models.)

(If I'm wrong in my assumption that there are interesting hyperdoctrines aside from the initial and terminal one, I'd also be happy to hear about that.)

$\endgroup$
5
  • 1
    $\begingroup$ "presumably there are hyperdoctrines (for, say, 1st order arithmetic) that assert some combination of sentences, that no traditional model asserts. (Such combinations must necessarily be infinite, on account of the traditional completeness theorem, but still.)" It seems to me the opposite is true: unless I'm wildly misunderstanding the set-up, by the compactness theorem a hyperdoctrine $H$ does this iff $H$ asserts some specific first-order sentence which is unsatisfiable. $\endgroup$ Commented Oct 1, 2021 at 17:50
  • $\begingroup$ Ah, right, I had forgotten about compactness. So... consider the initial hyperdoctrine for 1st order arithmetic. This asserts all and only the set of provable (1st order arithmetical) sentences. Clearly, every finite subset has a model (take the standard model). So the initial hyperdoctrine is, apparently, a traditional model? And the terminal one is not, as it asserts a falsehood. New hypothesis: every "new" hyperdoctrine asserts a falsehood, so in the classical setting, there aren't any interesting new models. (And the Q is incorrect in asserting the initial hyperdoctrine is new.) $\endgroup$
    – So8res
    Commented Oct 1, 2021 at 18:05
  • $\begingroup$ I'm not entirely sure that the above analysis is correct, but if it is and you or someone else tighten it up and post it as an answer, I'd happily accept it :-). Thanks either way! $\endgroup$
    – So8res
    Commented Oct 1, 2021 at 18:07
  • $\begingroup$ (I'm assuming "first order arithmetic" means first-order $\mathsf{PA}$ here, which is incomplete.) "So the initial hyperdoctrine is, apparently, a traditional model?" No, but my point is that in fact the opposite problem happens - it doesn't assert enough. The theory of a (traditional) model is always complete. So the initial hyperdoctrine, if I understand correctly, is really more appropriately identified with the class of all traditional models of $\mathsf{PA}$ than with any particular traditional model. Maybe in general the right connection is between hyperdoctrines and classes of models? $\endgroup$ Commented Oct 1, 2021 at 18:08
  • $\begingroup$ Ah, yeah, I think I see what you're saying -- it's not so much that a hyperdoctrine can assert a new infinite combo, as that a hyperdoctrine can assert a bunch of sentences while failing to either assert or deny a traditional consequence thereof. (And, the flaw in my argument above is that the traditional model generated by compactness is allowed to satisfy additional sentences, so the initial hyperdoctrine is not necessarily a traditional model.) (And yeah, I mean PA, thanks.) That feels like it's at least a hint. $\endgroup$
    – So8res
    Commented Oct 1, 2021 at 18:27

3 Answers 3

5
$\begingroup$

Maybe I am wrong, but it seems to me that the other answers are misunderstanding the question. The emphasis on syntactic hyperdoctrines seems to me beside the point.

A (classical, first-order) hyperdoctrine is a categorical (semantic) structure, consisting of a category $C$ and a functor $P: C \to \rm BoolAlg$ together with adjoints and a Beck-Chevalley condition. It seems to me that the OP wants to consider a particular construction of such a hyperdoctrine as follows: given a set $U$, let $C$ be the full subcategory of $\rm Set$ determined by the objects $U^n$, and let $P(U^n)$ be the powerset of $U^n$. Let's call this hyperdoctrine ${\rm Set}|_U$.

For a particular theory $T$ (which, in general, should be multi-sorted) and hyperdoctrine $C$, one can then consider the notion of a "model of $T$ in $C$". This consists of an interpretation function assigning an object of $C$ to each sort of $T$, a morphism of $C$ to each function symbol of $T$, and a predicate in some $P(X)$ to each relation symbol of $T$, such that the axioms of $T$ are "satisfied". I think this is what the OP means by a "hyperdoctrine for $T$". One can rephrase this in a more highbrow way by building a "syntactic hyperdoctrine" $S_T$ out of $T$ and saying that a model of $T$ in $C$ is a morphism of hyperdoctrines $S_T \to C$, but that isn't necessary.

It seems to me that the OP is asking whether there are any interesting models of $T$ in hyperdoctrines $C$ not of the form ${\rm Set}|_U$ (and also where $C$ is not terminal and not the initial $T$-hyperdoctrine $S_T$).

There are some fairly trivial answers to that question. One, which I think appears in the other answers, is that for any theory $T'$ extending $T$, there is a canonical model of $T$ in $S_{T'}$. Another is that instead of models in ${\rm Set}|_U$, we can consider models in $\rm Set$ itself; if $T$ has only one (base) sort then any such model factors uniquely through some ${\rm Set}|_U$, so it is not very different (but if $T$ has more than one sort, then this more general kind of model is the "correct" one to think about).

However, I think the most satisfying answer is that

You can replace $\rm Set$ by any (Boolean) category.

In other words, given any category $C$, you can define a hyperdoctrine over $C$ where $P(X)$ is the poset of subobjects of $X\in C$. If $C$ is a Boolean category, this will be a classical first-order hyperdoctrine, call it ${\rm Sub}(C)$. Then you can consider models of any theory $T$ in ${\rm Sub}(C)$, and they will be new, different, and interesting, and can satisfy principles that don't hold in $\rm Set$.

You get many more interesting models if you generalize to intuitionistic logic, in which case you use Heyting categories instead of Boolean categories and every elementary topos is an example. This leads to the whole field of topos theory and the categorical semantics of internal languages. But there are also interesting Boolean categories other than $\rm Set$, such as ${\rm Set}^X$ for any set $X$, or even for any groupoid $X$.

$\endgroup$
8
  • $\begingroup$ Thanks! (And, I endorse your rephrasing of my question, and greatly appreciate your help in making it clear.) "Use a Boolean category other than Set" is much closer to an answer to my question. "Use $\mathrm{Set}^2$" is more concrete and therefore closer still. IIUC this is saying that some hyperdoctrines interpret formulas not as a set, but as a pair of sets (or, more generally, as an $X$-indexed set-family for $X$ an arbitrary groupoid)? And that's cool, but even cooler would be a specific interpretation of formulas of (say) PA as a pair of sets and which has intuitive pull in its own right. $\endgroup$
    – So8res
    Commented Oct 4, 2021 at 21:41
  • $\begingroup$ Attempting to work out a concrete example using this idea... Letting $X = 2$, I could pick any two traditional models of PA, and sew together a hyperdoctrine that interprets each formula into the corresponding pair of sets. But that doesn't quite feel "new and interesting" to me. On the other extreme, letting $X = \mathrm{Set}$ (considered as a groupoid) suggests hyperdoctrines that interpret each formula as an endofunctor(?) of $\mathrm{Set}$ (considered as a groupoid), which sounds like the type of thing that could feel new and interesting, though no specific instance springs to my own mind. $\endgroup$
    – So8res
    Commented Oct 4, 2021 at 21:42
  • $\begingroup$ So what does the hyperdoctrine of subobjects in Set^Set look like? We could interpret a formula w/ n free variables as a subobject of (const U^n ∈ Set^Set). Which is essentially a groupoid homomorphism (F : Set^Set) paired with an injective transformation to (const U^n). The components of the transformation are like, for every set A, an injection from F(A) to U^n. So an interpretation for a formula is a sort of natural family of subobjects. The remaining Q is to find an actual interpretation of some theory (eg, PA) into Sub(Set^Set), that isn't just the boring one (where F is always constant). $\endgroup$
    – So8res
    Commented Oct 5, 2021 at 15:37
  • 1
    $\begingroup$ You get a lot more interesting examples if you generalize to constructive logic. Then you can use, for instance, the category of sheaves on any topological space, or the category of presheaves on any small category. For classical logic, I agree that set-indexed powers of sets are not all that interesting. I think $\rm Set^{core(Set)}$ is a bit too big to be very interesting, but ${\rm Set}^G$ for a group $G$ is interesting: the objects are sets with a $G$-action, the subsets are sub-actions. So you are doing "equivariant mathematics" for any group of equivariance. $\endgroup$ Commented Oct 5, 2021 at 19:25
  • 1
    $\begingroup$ Another big class of Boolean examples is double-negation sheaves on forcing posets. This yields, in particular, forcing models of ZFC in the classical sense. So in particular this way you get a hyperdoctrine for ZFC that violates the continuum hypothesis. $\endgroup$ Commented Oct 5, 2021 at 19:26
5
$\begingroup$

Given any theory $T$, we have a hyperdoctrine $H(T)$ (the "syntactic hyperdoctrine for $T$"), where (in the single sorted case) the $n^{\text{th}}$ lattice is the lattice of formulas in $n$ free variables, up to equivalence modulo $T$.

As Zhen Lin's answer shows, in fact every hyperdoctrine is the syntactic hyperdoctrine for some theory (the "full theory" of the hyperdoctrine).

Now what is a morphism of hyperdoctrines? It's an interpretation of theories. That is, a morphism of hyperdoctrines $H(T)\to H(T')$ gives a translation of formulas relative to $T$ into formulas relative to $T'$, which preserves the logical structure, i.e., an interpretation of $T$ in $T'$. More generally, any morphism of hyperdoctrines $H\to H'$ can be viewed as an interpretation of the full theory of $H$ in the full theory of $H'$.

When we view a traditional model $M$ of $T$ as a hyperdoctrine under $H(T)$, we form the hyperdoctrine $H(M)$ where (in the single sorted case) the $n^{\text{th}}$ lattice is $\mathcal{P}(M^n)$. And we get a morphism of hyperdoctrines $H(T)\to H(M)$ mapping a formulas relative to $T$ to its evaluation in $M$ (the set of tuples in $M$ satisfying the formula). Note that we can also view this as an interpretation of $T$ in the full theory of $M$: This theory has relation symbols for every subset of $M^n$ for all $n$, and the interpretation sends a formula to the relation symbol naming its evaluation in $M$. This is one of the features of categorical logic: it puts the notion of "model of a theory" and "interpretation of theories" on the same footing.

Ok, so addressing your question: if a "hyperdoctrine for $\mathsf{PA}$" is a hyperdoctrine $H$ equipped with a morphism $H(\mathsf{PA})\to H$, then we have initial hyperdoctrine for $\mathsf{PA}$ (which is the syntactic hyperdoctrine $H(\mathsf{PA})$ itself) and all the standard models of $\mathsf{PA}$ in the form $H(M)$. What else? Well, we have a "hyperdoctrine for $\mathsf{PA}$" for every interpretation of $\mathsf{PA}$ into a theory $T$. For example, the standard interpretation of $\mathsf{PA}$ in $\mathsf{ZFC}$ gives a morphism of hyperdoctrines $H(\mathsf{PA})\to H(\mathsf{ZFC})$. And in a precise sense, every hyperdoctrine for $\mathsf{PA}$ has this form, since every morphism of hyperdoctrines can be viewed as an interpretation of theories. The terminal hyperdoctrine for $\mathsf{PA}$ is the interpretation of $\mathsf{PA}$ in the inconsistent theory.

So typical hyperdoctrines for $\mathsf{PA}$ include things like $H(T)$ where $T$ is a stronger theory than $\mathsf{PA}$ (obtained by adding axioms) or $H(T)$ where $T$ is an expansion of $\mathsf{PA}$ to include extra structure. In the comments, you wrote that you were hoping for something like "there's a hyperdoctrine for $\mathsf{PA}$ of only the decidable predicates". This doesn't make sense, since any hyperdoctrine for $\mathsf{PA}$ must interpret all the predicates definable in $\mathsf{PA}$, not just the decidable ones.

I've been intentionally vague about some details here (like what counts as a morphism of hyperdoctrines), hoping that a more high level view will help get your thinking on the right track.

$\endgroup$
5
  • $\begingroup$ So, is the sub-hyperdoctrine of $H(\textsf{PA})$ generated by decidable predicates (if such thing exists) the whole of $H(\textsf{PA})$? $\endgroup$ Commented Oct 4, 2021 at 16:46
  • $\begingroup$ @მამუკაჯიბლაძე Should be, since all formulas are generated (under the logical connectives) by the symbols in the language, which are decidable. But note also that we're talking about hyperdoctrines admitting a morphism from the syntactic hyperdoctrine of PA. There's no reason a sub-hyperdoctrine should admit such a morphism, unless it is the whole syntactic hyperdoctrine. $\endgroup$ Commented Oct 4, 2021 at 16:50
  • $\begingroup$ Thanks! $H(\text{ZFC})$ is a lot closer to the sort of concrete example I'm looking for. And while I buy the argument that every hyperdoctrine is "syntactic" for some (not necessarily enumerable) theory, I'm still interested in hyperdoctrines for PA that aren't enumerable, and that in a sense "feel like" they're taking advantage of the hyperdoctrine-ability to interpret formulas using non-powerset algebras, in some manner other than "quotient sentences by provability in some enumerable extension of PA". $\endgroup$
    – So8res
    Commented Oct 4, 2021 at 16:58
  • $\begingroup$ With regards to a hyperdoctrine-for-PA of decidable predicates, I'm aware this doesn't work as stated (I tried it before asking my Q :-p), but I don't see your argument as working against plausible repairs thereof. Like, is there a hyperdoctrine-for-PA where the nth BoolAlg is the decidable predicates (on n numbers) with an extra "undecidable" point adjoined? Probably not (I looked briefly at that too), but I'm still holding out for there being some sort of cool computational / topological / domain-theoretic / ??? thing you can do w/ hyperdoctrines that you can't do with models. $\endgroup$
    – So8res
    Commented Oct 4, 2021 at 16:59
  • $\begingroup$ Like, hyperdoctrines are cool, and this whole idea of putting models and interpretations on the same footing is cool, and it feels to me like I should be able to go back to the 1920s and tell Skolem about this cool and more general tool for doing semantics. And he could rightly ask "if this is so cool, what interpretations do you have that I don't?". And I could say "I can interpret PA using the syntax of PA, or of PA + Con(PA), or of ZFC, using exactly the same machinery I use to interpret PA using the nats". And that is indeed cool! But are there any other new interpretations I can show off? $\endgroup$
    – So8res
    Commented Oct 4, 2021 at 17:06
3
$\begingroup$

Every hyperdoctrine is "syntactic", in the sense that given any hyperdoctrine you can construct a theory whose syntactic hyperdoctrine is equivalent to the one you start with. Thus, hyperdoctrines correspond not to models (i.e. complete theories) but rather (deductively closed) theories.

Suppose we have a hyperdoctrine, that is, a cartesian monoidal category $\mathcal{S}$ and a contravariant functor $\Omega$ from $\mathcal{S}$ to the category of boolean algebras satisfying certain axioms (corresponding to logical axioms for $\exists$ and $=$). Consider the following (multisorted) theory $T$.

  • The sorts are the objects in $\mathcal{S}$.
  • For each tuple $(X_1, \ldots, X_n, Y)$ (where $n \ge 0$) of objects in $\mathcal{S}$ and each morphism $X_1 \times \cdots \times X_n \to Y$ in $\mathcal{S}$ we have an $n$-ary function symbol of the same name and type.
  • For each tuple $(X_1, \ldots, X_n)$ of objects in $\mathcal{S}$ and each element of $\Omega (X_1 \times \cdots \times X_n)$ we have an $n$-ary relation symbol of the same name and type.
  • We introduce as axioms equations encoding projections and composition in $\mathcal{S}$.
  • For every morphism $\langle f_1, \ldots, f_n \rangle : X_1 \times \cdots \times X_m \to Y_1 \times \cdots \times Y_n$ in $\mathcal{S}$, every tuple $(R_1, \ldots, R_p)$ of elements in $\Omega (X_1 \times \cdots \times X_m)$, and every $S \in \Omega (Y_1 \times \cdots \times Y_n)$, if $R_1 \wedge \cdots \wedge R_p \le \langle f_1, \ldots, f_n \rangle^* S$ then we introduce the axiom $$R_1 (x_1, \ldots, x_m) \land \cdots \land R_p (x_1, \ldots, x_m) \Rightarrow S (f_1 (x_1, \ldots, x_m), \ldots, f_n (x_1, \ldots, x_m))$$ Similarly, if $\langle f_1, \ldots, f_n \rangle^* S \le R_1 \vee \cdots \vee R_p$ then we introduce the axiom $$S (f_1 (x_1, \ldots, x_m), \ldots, f_n (x_1, \ldots, x_m)) \Rightarrow R_1 (x_1, \ldots, x_m) \lor \cdots \lor R_p (x_1, \ldots, x_m)$$
  • For every object $X$ in $\mathcal{S}$, if $R \in \Omega (X \times X)$ is the hyperdoctrinal equality then we introduce the axiom $R (x_1, x_2) \Rightarrow (x_1 = x_2)$.

And so on. I think it is clear that there is a "tautological" morphism from the syntactic hyperdoctrine of $T$ to $(\mathcal{S}, \Omega)$ defined by sending the generators back to the objects/morphisms/elements of the same name, and that this is an equivalence of hyperdoctrines.

$\endgroup$
20
  • $\begingroup$ I buy that this construction works (thanks!), but insofar as you're trying to argue that all hyperdoctrines are "non-interesting" (in the sense of the question), I don't yet see the argument. The sort of thing I'm looking for is more like "there's a hyperdoctrine where the nth lattices is only the computable properties of n numbers" or something (I don't think that's true as stated). While I agree all hyperdoctrines admit your construction here, I don't know how to use it to produce a specific "new" hyperdoctrinal model, nor how to see it as essentially ruling all interesting ones out. $\endgroup$
    – So8res
    Commented Oct 2, 2021 at 1:47
  • 2
    $\begingroup$ Hyperdoctrines are not models, as I wrote in the first paragraph. If you insist, a model is a morphism of hyperdoctrines – the domain being the theory being interpreted and the codomain being the universe of interpretation. "Interesting" is in the eye of the beholder. For example, you could take the syntactic hyperdoctrine of the first order theory of an infinite structure over a countable signature, then regard it as a model of some subtheory. This will be a non-classical model because its lattices are countably infinite. $\endgroup$
    – Zhen Lin
    Commented Oct 2, 2021 at 4:59
  • $\begingroup$ Yeah, I agree that hyperdoctrines are not (in general) traditional models, sorry for my confusing wording. In an answer, I'm looking for concrete examples of hyperdoctrines that aren't traditional models. Perhaps your example can be used to generate one (thanks!), though my current guess is that it's going to look more like the negative example in the penultimate paragraph of my original question, than the positive example. I'm still hoping for something more like "there's a hyperdoctrine for PA of only the decidable predicates", or something similarly specific and concrete. $\endgroup$
    – So8res
    Commented Oct 2, 2021 at 16:46
  • $\begingroup$ @So8res Zhen Lin's point is that hyperdoctrines are not just not traditional models in general, they are not traditional models in any sense. That is, every hyperdoctrine is an example of a hyperdoctrine which is not a traditional model. Your question is like saying "I'm looking for concrete examples of theories in the language of groups which are not groups in the traditional sense." It's a type error. $\endgroup$ Commented Oct 2, 2021 at 17:10
  • 1
    $\begingroup$ @So8res Two clarifying questions: (1) What, precisely, is your definition of "hyperdoctrine for PA"? (2) Let's say we have a traditional model, say the standard model $\mathbb{N}$ for PA. Now you form the hyperdoctrine where the lattices are the powerset algebras of cartesian powers of $\mathbb{N}$. How does this construction encode the fact that we're thinking of $\mathbb{N}$ as a model of PA, not just as a pure set? $\endgroup$ Commented Oct 3, 2021 at 11:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .