The question is exactly that of the title. Suppose $\varphi\in V$ is an $\mathcal{L}\_{\infty\omega}$-sentence, and $W$ is an inner model of $V$ such that $\varphi\in W$. Is the statement 



> 
> $\varphi$ has a model
> 
> 
> 


absolute between $V$ and $W$? It is clearly upwards absolute, by induction on rank, so I'm really asking about the downwards direction.


The main obstacle that I see right now is the lack of a Lowenheim-Skolem theorem, which prevents us from using Shoenfield absoluteness. As a specific example, let $A$ be $\omega\_1^W$ as a linear order, and suppose $V$ is a forcing extension of $W$ in which $\omega\_1$ is collapsed. In $V$, there is a sentence $\varphi\in\mathcal{L}\_{\omega\_1\omega}\subset\mathcal{L}\_{\infty\omega}$ whose only countable model up to isomoprhism is $A$: the Scott sentence of $A$. Now there is a formula $\psi\in \mathcal{L}\_{\infty\omega}\cap W$ such that, in $V$, $\psi$ and $\varphi$ are equivalent; this is a roundabout way of saying "$\varphi\in W$," which is not exactly true. But then $\psi$ has no countable models in $W$, since such a model would have to be isomorphic to $A$ in $V$.


I assume this is well-known, but I haven't been able to find an answer myself.



Is the consistency of $\mathcal{L}_{\infty\omega}$-sentences absolute?
Here is a somewhat easier counterexample. 


Let $I$ be a countable set, which is uncountable in $W$. Let $c\_n$ and $d\_\alpha$ be constant symbols, for $n\in\mathbb{N}$ and $\alpha\in I$. Consider the formula $\varphi$ that asserts that all the $d\_\alpha$'s are different, but that every $d\_\alpha$ is equal to some $c\_n$. That is, 
$$\varphi=\bigl(\bigwedge\_{\alpha\neq\beta\in I}d\_\alpha\neq d\_\beta\bigr)\wedge\bigl(\bigwedge\_\alpha\bigvee\_n d\_\alpha=c\_n\bigr).$$ A model of this sentence is essentially providing an injective function from $I$ to $\mathbb{N}$, mapping $\alpha\mapsto n$ when $n$ is least for which $d\_\alpha=c\_n$. Thus, the sentence is not satisfiable in $W$, since $I$ is uncountable there, but it is satisfiable in $V$, since $I$ is countable in $V$.


Using the same idea, one can make a counterexample in a finite signature language as follows. Suppose that $P(\mathbb{N})^W$ is countable in $V$. First, we can make a sentence that asserts that we have a copy of $\langle\mathbb{N},S\rangle$, by asserting that everything in that sort is a finite successor of $0$. Next, we can have another sort that will be a copy of $P(\mathbb{N})^W$, each subset represented by a vertex pointing exactly at its elements. In a single formula (of size continuum in $W$), we can say that every subset of $\mathbb{N}$ that it is represented by some vertex. Finally, with one more binary relation, we can say that every subset is associated with a distinct natural number. The point now is that the overall assertion is not satisfiable in $W$, since $P(\mathbb{N})^W$ is uncountable in $W$, but it will be satisfiable in $V$, since the set is countable there. 


**Update.** Meanwhile, the satisfiability of sentences in $\cal{L}\_{\omega\_1,\omega}$ logic, that is, using only countable meets and joins, is a $\Sigma^1\_1$ assertion and hence absolute between models with the same countable ordinals.


