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Below all languages are finite; if preferred, it's enough to work in the language consisting of a single binary relation.

By a simple counting argument, there is some $\mathcal{L}_{\omega_1,\omega}$-sentence which is not equivalent to any sentence in (finitary) second-order logic (with full semantics). However, this isn't constructive at all. Moreover, the set-theoretic nastiness of second-order logic means that lots of basic questions about it can be highly model-dependent.

My question is the following:

Is there a concrete example of an $\mathcal{L}_{\omega_1,\omega}$-sentence not equivalent to (= has the same models as) any second-order sentence?

This is a bit slippery; I'm ultimately interested in any natural precisiation or progress, but here are a couple candidate rephrasings:

  • Is there some transitive model $M$ of ZFC and $\varphi\in\mathcal{L}_{\omega_1,\omega}^M$ such that for every outer model $N$ of $M$ there is no second-order sentence $\theta$ equivalent to $\varphi$ in $N$?

  • What can we say about the descriptive set theoretic complexity of the set $B$ of codes for $\mathcal{L}_{\omega_1,\omega}$-sentences not equivalent to any second-order sentence, or the set $B_0$ of codes for $\mathcal{L}_{\omega_1,\omega}$-sentences not equivalent on countable models to any second-order sentence?

Note that the same counting argument shows that $B_0\not=\emptyset$, and $B_0$ isn't too complicated (it's a priori $\Pi^1_\omega$ or $\Pi^1_{\omega+1}$ depending on how we define limit stages of the extended projective hierarchy). Meanwhile, $B$ is worse: a quick glance merely gives a bound in the Levy hierarchy of $\Pi_2$.

But I'm interested in any progress on any natural precisiation of the question above.

Incidentally, note that the converse is easy: the set of structures of successor cardinality is second-order definable but not $\mathcal{L}_{\omega_1,\omega}$-definable (see here).

Noah Schweber
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1 Answers1

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Here's a partial positive answer:

It's easy to show that for $X\subseteq\omega$ the (isomorphism class of) structure $$Set_X:=(\omega; <,X)$$ is characterizable by a single second-order sentence iff $X$ is second-order-definable in $(\omega;<)$ - that is, iff $X$ is a lightface projective real. However, we obviously have that $Set_X$ is characterizable by a single $\mathcal{L}_{\omega_1,\omega}$-sentence. So, for example, letting $\sigma$ be the Scott sentence of $Set_{Th_2(\omega;<)}$ we have that no second-order sentence is equivalent to $\sigma$ (even on countable structures).

However, this isn't entirely satisfying: this property of $\sigma$ may not be absolute upwards since $Th_2(\omega;<)$ is not absolute upwards in general. Specifically, while large cardinals do yield projective absoluteness, this breaks down quite badly if we work over $L$ since $Th_2(\omega;<)^L$ is second-order definable over $(\omega;<)$ in the sense of $L^G$ when $G$ is $Col(\omega,\omega_1^L)$-generic over $L$.

  • The point is that - regardless of $V$ - if $\theta$ is a second-order sentence then $L\models((\omega;<)\models\theta)$ iff $L_{\omega_1^L}\models\hat{\theta}$ for an appropriate first-order sentence $\hat{\theta}$ in the language of set theory. If $\omega_1^L$ is countable, then $L_{\omega_1^L}$ is characterizable up to isomorphism as a countable well-founded structure satisfying the obvious fragment of $ZFC+V=L$ and such that there is no larger countable well-founded model of that same theory which is locally countable. For each second-order sentence $\sigma$, the sentence $\sigma' \equiv$ "every such structure thinks $\sigma$ is true" is then a second-order sentence over $(\omega;<)$. (And the maps $\theta\mapsto\hat{\theta},\sigma\mapsto\sigma'$ are simple enough that they don't cause issues.)

Indeed, it's not hard to show that there is a parameter-freely-definable set forcing in $L$ such that for every generic $G$, all constructible reals are second-order definable over $(\omega;<)$ in the sense of $L[G]$. So this solution isn't "persistent to outer models," even if we restrict attention to pretty mild constructions.

And in fact Fedor Pakhomov found a negative result preventing any too-natural improvement on the situation. So I think this is ultimately a question whose answer is "morally" negative, even if the original phrasing happened to yield a positive answer.

Noah Schweber
  • 245,398