This is a "forcing-absolute" followup to [this question](https://math.stackexchange.com/questions/3568276/is-there-a-specific-infinitary-sentence-second-order-logic-cant-capture), 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 $\mathsf{ZFC}$.



Is there an infinitary sentence which is absolutely not second-order expressible?
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:
$$\exists \in,\in\_u,\alpha,\psi(\mathsf{KPUL}\_2(\in,\in')\land \mathsf{MSt}(\alpha,\in,\in')\land \mathsf{LC}(\alpha,\psi,\in,\in')\land \mathsf{Tr}(\psi,\in,\in')).$$


