This is a question related to ideas raised in http://arxiv.org/abs/1410.1224 and http://arxiv.org/pdf/1405.7456.pdf. Basically, the idea is the following:

Suppose I have a first-order theory $T$. Under what conditions are there "few" models of $T$ across all possible forcing extensions of the universe?

Maybe "few" is the wrong word - what I really mean is, when does the collection of "models of $T$ you can get by forcing" have some nontrivial structure?

There are a bunch of aspects of this question; the one I'm interested in right now is the following:

Say that a theory $T$ is generically embeddable if - whenever $\mathbb{P}$ is a forcing notion and $\nu$ is a $\mathbb{P}$-name such that $\Vdash_\mathbb{P}``\nu\models T"$ - there is some forcing notion $\mathbb{Q}$ and $\mathbb{Q}$-name $\mu$ such that $$\Vdash_{\mathbb{P}\times\mathbb{Q}} ``\mbox{$(\mu[G_1]\models T)$ and there is a homomorphism $h\colon \nu[G_0]\rightarrow\mu[G_1]$}."$$ (Note that this homomorphism of course exists in $V[G_0\times G_1]$.)

To get a sense of what this means, let's look at a counterexample. Take $T$ to be the true theory of second-order arithmetic - that is, $T=Th(\omega\sqcup 2^\omega; +, \times, \in)$. Then models of $T$ code reals, and this is bad. Specifically, let $\mathbb{P}$ be Cohen forcing, and let $\nu$ be the $\mathbb{P}$-name picking out the "true" model $(\omega\sqcup 2^\omega; +, \times, \in)^{V[G]}$. Then if $G_0\times G_1$ is $\mathbb{P}\times\mathbb{Q}$ generic - for any $\mathbb{Q}$! - since the real $G_0$ won't be in $V[G_1]$, no model $N$ of $T$ in $V[G_1]$ will embed $\nu[G_0]$, since such a homomorphism would have to biject $\nu[G_0]$ with the well-founded segment of $N$, and from this we could recover $G_0$ in $V[G_1]$.

More generally, theories which let you code sets are bad.

My question is: When is a theory generically embeddable? Specifically, are there model-theoretic niceness properties which guarantee this? For instance, I can't come up with a stable example of a non-generically embeddable theory $T$, but I also can't prove there isn't one. I can prove that if $T$ is totally categorical, then $T$ is generically embeddable, but this isn't very interesting (in fact, it's hard not to prove this).

EDIT: Note that there are many natural strengthenings of this: e.g.

  • We may demand that $\mathbb{Q}=\mathbb{P}$, or even $\mathbb{Q}=\mathbb{P}$ and $\mu=\nu$!

  • We may ask for the homomorphism $h$ to be an elementary embedding, or satisfy some other strong property (e.g. the papers linked above looked at isomorphisms, not homomorphisms).

  • We may restrict attention to certain classes of forcing notions, or certain names (e.g. the papers linked above looked only at $\nu$ such that $\Vdash_\mathbb{P}$``$\nu[G]$ is countable").

For now, though, I'm interested in the question as it stands.

  • $\begingroup$ When you take a trivial forcing, then it's not very wild. :-P $\endgroup$ – Asaf Karagila Jun 1 '16 at 20:16
  • $\begingroup$ @AsafKaragila Hehe. Note the quantification over $\mathbb{P}$, though. $\endgroup$ – Noah Schweber Jun 1 '16 at 20:48

For an example of a super-stable theory failing the property, you can take infinitely many unary predicates (so $Th(2^\omega, U_n: n \in \omega)$ where $U_n(\eta)$ holds iff $\eta(n) = 1$). Then if $\nu$ and $\mu$ satisfy that if $\nu[G_0]$ embeds into $\mu[G_1]$ in $\mathbb{V}[G_0 \times G_1]$ then every quantifier-free type realized in $\nu[G_0]$ is also realized in $\mu[G_1]$, hence is in $\mathbb{V}[G_0] \cap \mathbb{V}[G_1] = \mathbb{V}$. So let $\mathbb{P}$ adjoin a Cohen real $x$ and let $\nu$ be a name for the model of $T$ corresponding to $\{\eta: \mbox{supp}(\eta) \mbox{ is finite}\} \cup \{x\}$.

I can give an exact characterization if we replace homomorphism with elementary embedding (which isn't much of a loss, since we can just Morleyize) and require $T$ to be complete, in a countable language:

Say that $T$ is generically elementarily embeddable (g.e.e.) if for all $\mathbb{P}$-names $\nu$ for a model of $T$ there is a $\mathbb{Q}$-name $\mu$ for a model of $T$ such that $\mathbb{P} \times \mathbb{Q} \Vdash \nu[G_0]$ elementarily embeds into $\mu[G_1]$.

Claim. If $T$ is a complete theory in a countable language, then $T$ is g.e.e. iff $T$ is small.

Proof. It follows from the previous example essentially that g.e.e. implies $T$ is small. (If $T$ is not small we can arrange $\nu \models T$ realizes a type not in $\mathbb{V}$, which is enough.)

For the reverse direction, let $\mathbb{P}, \nu$ be given. Let $\mathbb{Q}$ force that $\nu$ (i.e., the name in $\mathbb{V}$) becomes countable and let $\mu$ be a name for a countable saturated model of $T$. Then in $\mathbb{V}[G_0 \times G_1]$, $\nu[G_0]$ is countable, $\mu[G_1]$ is countable and saturated (since being $\aleph_0$-saturated is absolute for small theories) and so $\nu[G_0]$ elementarily embeds into $\mu[G_1]$.

P.S. Chris Laskowski, Richard Rast and myself independently investigated several similar things to the papers you quoted, see A New Notion of Cardinality for Countable First Order Theories

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  • $\begingroup$ What does "small" mean in this context? $\endgroup$ – Noah Schweber Jun 1 '16 at 21:00
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    $\begingroup$ Equivalently: only countably many complete types over the empty set; or only countably many complete types over finite sets; or has a countable saturated model. $\endgroup$ – Douglas Ulrich Jun 1 '16 at 21:05
  • $\begingroup$ Very nice! Let me just comment that the restriction to complete theories is actually significant, at least on the face of things - since new completions can be added, it's not clear to me that an incomplete $T$ is generically embeddable even if $T$ is Morleyized and every completion of $T$ is small (note that this is absolute, assuming some large cardinals). Naively I'd take $\mathbb{Q}$ to be the forcing collapsing "enough" to $\aleph_0$, and let $\mu$ be the name for the ultrapower of a bunch of saturated models of completions of $T$, but it's not clear to me that this works - thoughts? $\endgroup$ – Noah Schweber Jun 1 '16 at 21:55
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    $\begingroup$ Trivial observation: any $T$ with a perfect set of completions fails to be g.e.e. (hence the "g.e." in my above comment, instead of "g.e.e."). $\endgroup$ – Noah Schweber Jun 1 '16 at 21:57
  • $\begingroup$ Your last comment is basically the only thing that can go wrong: if $T$ is Morleyized and countable, then $T$ is g.e.e. iff $T$ has only countably many complete extensions, each of which is small. $\endgroup$ – Douglas Ulrich Jun 1 '16 at 22:03

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