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 extensionis $\varphi$ 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.

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 $\mathsf{ZFC}$.