Stably embedded clone

Question

Let $M$ be a first-order structure, considered as an element of the ambient set-theoretic universe $V$.

Clearly, for any $L_{\infty,\infty}$-formula $\varphi(\bar x)$ (with $\bar x$ finite, say) in the language of $M$, without parameters, we can find a formula $\varphi'(\bar x,y)$ in the language of set theory such that for any tuple $\bar a\in M$ we have $V\models \varphi'(\bar a,M)\iff M\models\varphi(\bar a)$.

Let us say that $M$ is stably embedded (for $L_{\infty,\infty}$) if the converse is also true (i.e. for every $L_{\infty,\infty}$-formula $\varphi'(\bar x,y)$, we can find a corresponding $\varphi(\bar x)$)

Clearly, not every structure is stably embedded. For example, $M=\omega$ with the trivial structure is not stably embedded (even if we allow finitely many extra parameters). On the other hand, if $M$ has full structure (i.e. we have predicates for all subsets of all powers), then $M$ is stably embedded.

I guess that in $L$, "most" structures are not stably embedded, since we can do weird stuff with the definable well-ordering.

On the other hand, it seems conceivable that even if $M$ is not stably embedded, there is an $M'\cong M$ which is stably embedded, let us call this a stably embedded clone of $M$.

Now, I have the following, closely related questions:

1. Is it consistent that every $M$ has a stably embedded clone?
2. Is it true that given $M$, there is a forcing extension $V[G]$ which has a stably embedded clone of $M$?

I suspect the answer might be no because an $L_{\infty,\infty}$ formula can be used to code too much. If this the case, what if we consider instead just $L_{\infty,\omega}$, or even first-order formulas?

(This question is motivated by this math.se question and my answer to it.)

Answer 1

As was pointed out in the comments, the question as stated is a little bit ambiguous (specifically is the formula in the language of set theory allowed to be an $L_{\infty,\infty}$-formula or just an ordinary first-order formula). Regardless, $L_{\infty,\infty}$ is so strong that either version of the question isn't going to have a terribly interesting answer.

The following proposition was originally shown by McGee, but McGee's formalism is a little bit unfamiliar, so I will reproduce the proof here.

Proposition. Given a structure $M$ in a language $L$ and an ordinal $\alpha$, a set $X \subseteq M^\alpha$ is $L_{\infty,\infty}$-definable without parameters if and only if it is invariant under all automorphisms of $M$.

Proof. First note that $L_{\infty,\infty}$-definable sets are obviously automorphism-invariant, so we just need to show that any automorphism-invariant set is $L_{\infty,\infty}$-definable.

Fix an automorphism-invariant set $X \subseteq M^\alpha$. Let $\kappa = |M|$. Fix an enumeration $(a_j)_{j<\kappa}$ of the elements of $M$. Let $X^\ast$ be the set of functions $f: \alpha \to \kappa$ satisfying that $(a_{f(i)})_{i < \alpha} \in X$. Let $\Delta(y_0,\dots,y_{j < \kappa},\dots)$ be the atomic diagram of $M$ regarded as an $L_{\infty,\infty}$-formula. (In other words, $\Delta$ is the conjunction of all atomic or negated atomic formulas $\psi$ with variables among $(y_j)_{j<\kappa}$ such that $M$ satisfies $\psi$ with the variable assignment $y_j \mapsto a_j$.) Consider the following $L_{\infty,\infty}$-formula: $$\varphi_X(x_0,\dots,x_{i < \alpha},\dots) = \exists(y_0,\dots,y_{j < \kappa},\dots)\left[\Delta(\bar{y}) \wedge \forall w \bigvee_{j < \kappa}w=y_j \wedge \bigvee_{f \in X^\ast} \bigwedge_{i<\alpha} x_i = y_{f(i)}\right].$$ What $\varphi_X$ means is that there is an enumeration $\bar{y}$ of $M$ such that $\bar{x}$ is in $X$ up to the automorphism corresponding to the enumeration $\bar{y}$, but this is true if and only if $\bar{x}$ is actually an element of $X$, since $X$ is automorphism-invariant. $\square$

Assuming that your question is meant to be interpreted with $L_{\infty,\infty}$ set-theoretic formulas, we get a precise characterization: A structure is stably embedded for $L_{\infty,\infty}$ if and only if it is rigid (i.e., has no non-trivial automorphisms). This is because every set can be uniquely defined by an $L_{\infty,\infty}$-formula in the language of set theory (as demonstrated in this answer by Joel David Hamkins). In particular, this gives negative answers for your questions 1, since not every structure is rigid, and 2, since rigidity is downwards-absolute.

The versions of the your question with weaker logics are interesting but seem a lot harder to me.