Is there an infinitary sentence which is absolutely not second-order expressible?

Question

This is a "forcing-absolute" followup to this question, whose answer was largely unsatisfying. The question is:

Suppose $V=L$. Is there an $\mathcal{L}_{\infty,\omega}$-sentence $\varphi$ such that in no forcing extension is $\varphi$ equivalent to a second-order sentence?

Throughout, by "forcing" I mean "set forcing," although the (tame) class forcing version also seems potentially interesting.

EDIT: I forgot to add that I'm restricting attention to infinite structures here (which is key to my comment below that every projectively-definable infinitary sentence is second-order expressible). As Fedor Pakhomov commented below, without this restriction the problem is trivial since second-order theories of finite structures can't be changed by forcing. I do not, however, want to restrict attention to countable structures.

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I've decided to focus on models of $\mathsf{V=L}$ since that hypothesis seems to add interesting flavor to the question in a few ways:

• It implies that such a $\varphi$ must not be in $\mathcal{L}_{\omega_1,\omega}$. This is because $(1)$ every constructible real is projective in some forcing extension and $(2)$ every projectively definable $\mathcal{L}_{\omega_1,\omega}$-sentence is equivalent to a second-order sentence. So a positive answer would have to crucially rely on uncountable Boolean combinations - which seems a bit odd, because each specific $\mathcal{L}_{\infty,\omega}$-sentence is equivalent to some $(\mathcal{L}_{\omega_1,\omega})^{V[G]}$-sentence in an appropriate forcing extension $V[G]$ (just collapse the size of the sentence), but isn't an obvious contradiction since the "potential projectivity" fact about $L$ doesn't seem to lift to arbitrary forcing extensions of $L$.

• It rules out the "silly" solution provided by large cardinals. If large cardinals exist - specifically, enough to guarantee projective absoluteness - then any infinitary sentence which is not equivalent to a second-order sentence in $V$ remains so in all forcing extensions. (Note that this would give us an example in $\mathcal{L}_{\omega_1,\omega}$ for that matter.) But $\mathsf{V=L}$ breaks this "hammer," so that we seem to be forced to do some actual work.

• If the answer is yes, there is in fact a definable example, namely the least such sentence with respect to the $L$-ordering. Of course this is silly, but it suggests that there might be canonical examples in a more interesting sense. By contrast I could imagine models of $\mathsf{ZFC+V\not=L}$ where the existence of such a sentence is guaranteed nonconstructively (e.g. by a more intricate counting argument), and so no canonical example need exist.

That said, since it seems plausible that the $\mathsf{V=L}$-situation is more difficult to attack than I'd hoped, I'm also interested in results for other extensions of $\mathsf{ZFC}$.

Answer 1

For any given finite signature $\Omega$ there is a second-order sentence $\varphi$ of the signature $\Omega$ such that $\mathsf{ZFC}+V=L$ proves that for any $\mathcal{L}_{\infty,\omega}$-formula $\psi$ of the signature $\Omega$ there is a poset $P$ for which it is $\Vdash_P$-forced that for any infinite model $\mathfrak{M}$ we have $\mathfrak{M}\models \varphi$ iff $\mathfrak{M}\models \psi$.

$\varphi$ should be a second-order sentence expressing in all infinite models $\mathfrak{M}$ that the following holds (see below for a more explicit construction of $\varphi$):

1. there exists the greatest ordinal $\alpha$ such that the value $\aleph_\alpha^{L_\beta}$ is the same for all large enough countable limit ordinals $\beta$;
2. $\alpha$ is the position in $<_L$ of some $\mathcal{L}_{\omega_1,\omega}$-sentence $\psi$
3. $\psi$ is true in $\mathfrak{M}$.

Given $\mathcal{L}_{\infty,\omega}$-formula $\psi$ we consider $\alpha_0$ that is the position of $\psi$ in $<_L$. Let $P$ be the Levy collapse of $\aleph_{\alpha_0}$ onto $\omega$. Let us check that for this $\varphi,\psi$, and $P$ we have the desired equivalence.

Indeed let us reason inside $\Vdash_P$. Clearly, $\psi$ became an $\mathcal{L}_{\omega_1,\omega}$-formula.

Observe that $\alpha$ from 1. will always be equal to $\alpha_0$. Indeed, let us choose countable limit $\beta_0>\alpha_0$ such that all ordinals $\gamma<\alpha_0$ that aren't $L$-cardinals aren't cardinals in $L_{\beta_0}$. Clearly, for any limit $\beta>\beta_0$, we have $\aleph_{\alpha}^{L_\beta}=\aleph_\alpha^L$. For any $\alpha'>\alpha_0$ since $\aleph_{\alpha'}^L$ isn't countable and any countable $\gamma$ we could find limit countable $\beta'$ such that $\aleph_{\alpha'}^{L_{\beta'}}$ is either undefined or has the value $>\gamma$. Hence $\alpha_0$ is the ordinal $\alpha$ from 1.

Using this the proof of semantical equivalence of $\varphi$ and $\psi$ in infinite models is trivial.

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Now let me give more detailed description of $\varphi$. Let $\mathsf{KPUL}_2(\in,\in_u)$ be a second-order formula depending on element variable $\alpha$ and binary predicate symbols $\in,\in_u$ asserting that we have $(\mathsf{KPU}-\mathsf{Foundation})+\textsf{Second-Order-Foundation}+L[\mathfrak{M}]=V$ for the structure where each element of the underlying domain simultaniously represent itself (as an urelement) and some set, $\in$ gives membership of sets in sets, $\in_u$ gives membership of urelements in sets. Formula $\mathsf{Emb}(R,\in,\in_u,\in',\in_u')$ expresses that $\mathsf{KPUL}_2(\in,\in_u)\land \mathsf{KPUL}_2(\in',\in_u')$ and the unary function $f$ gives an end embedding of the admissible set given by $(\in,\in_u)$ into the admissible set $(\in',\in_u')$. Formula $\mathsf{St}(\alpha,\beta,\kappa,\in,\in_u)$ expresses that $\mathsf{KPUL}_2(\in,\in_u)$ and in the corresponding admissible set $\alpha$ is a countable ordinal, $\beta$ is a countable limit ordinal, $\kappa$ is the ordinal that is $\alpha$-th cardinal according to $L_\beta$, and for any $f,\in',\in_u',\beta'$ if $\mathsf{Emb}(f,\in,\in_u,\in',\in_u')$ and $\beta'>f(\beta)$ is a countable limit ordinal according to the admissible set given by $(\in',\in_u')$, then there is $f(\alpha)$-th cardinal according to $L_{\beta'}$ and it is equal to $f(\kappa)$. Formula $\mathsf{MSt}(\alpha,\in,\in_u)$ expresses that $\mathsf{St}(\alpha,\in,\in_u)$ and we don't have $\mathsf{St}(\alpha+1,\in,\in_u)$. Formula $\mathsf{LC}(\alpha,\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\alpha$ is an ordinal, $\psi$ is a constructible $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$, and its place in $<_L$-order is $\alpha$. Formula $\mathsf{Tr}(\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\psi$ is an $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$ that is true in the underlying model $\mathfrak{M}$. We put $\varphi$ to be: $$\exists \in,\in_u,\alpha,\psi(\mathsf{KPUL}_2(\in,\in')\land \mathsf{MSt}(\alpha,\in,\in')\land \mathsf{LC}(\alpha,\psi,\in,\in')\land \mathsf{Tr}(\psi,\in,\in')).$$