This question is inspired by the excellent question by Douglas Ulrich [When is $L$-Rank definable in inner models of $V=L$?](https://mathoverflow.net/questions/119882/when-is-mathbbl-rank-definable-in-inner-models-of-mathbbv-mathbbl)


Suppose $M \in L$ is a countable model of $ZFC$, and furthermore suppose $M \vDash V \neq L$. These models have the funny property that, although every set in them is constructible, the model does not recognize this fact -- the "nonconstructible" sets simply arise at a level of the $L$-hierarchy that is greater than the ordinals of $M$. Provided any countable $ZFC$ model exists, then such models abound -- for example, we can easily build forcing extensions of countable models by building an $M$-generic filter directly, and the forcing extension will not recognize that the generic is constructible.


In the answer to the question linked above it is shown that such a model $M$ cannot define the $L$-order on all of its members, at least not with the usual definition -- it can only define $<\_L$ for those elements that it recognizes as constructible (that is, for members of $L^M$). However, I am interested whether this limitation is inherent, or simply a limitation of definability over $M$. In particular, is it consistent for us to add the $L$-order on all of $M$, as a class, without destroying $ZFC$? I will phrase the question in terms of $GBC$ models, to make the use of classes explicit.



> 
> Is it consistent that $M \in L$ is a countable model of $GBC$ with $M \vDash V \neq L$, and there is class $U \in M$ such that $U$ gives the $L$-order on $M$? (That is, $\langle x,y \rangle \in U$ if and only if $x <\_L y$, for all sets $x, y \in M$).
> 
> 
> 


Note: In the context of $GBC$ the statement $V \neq L$ may be ambiguous -- I intend it to refer only to sets, and not to classes, e.g. $V \neq L$ means "there exists a nonconstructible set."


Such a $U$ would give us a proper class well order with order type larger than $ORD^M$, but this does not seem immediately problematic - many such well orders are definable over any model of $ZFC$. One way in which such a $U$ might be inconsistent is if, from $U$, we could show how to "construct" every set in $M$, but these constructions would be of more-than-$ORD^M$ length, and so this might not directly contradict $V \neq L$.


If such a $U$ is inconsistent, it would be nice to see why this limitation exists. On the other hand, if we can have such a $U$, is it universally possible?



> 
> If such a $U$ is consistent, are there any restrictions on the models $M$ that can have them?
> 
> 
> 


For example, it might be consistent only if every member of $M$ has $L$-rank "not too much greater" than $ORD^M$.


EDIT: changed a stray occurence of "GBC" to "ZFC" in paragraph 1, for clarity.



Can a model of $V\neq L$ contain a class giving the $L$-ordering on all its sets?
This answer just considers the version of the question for transitive models $M$.


Under a reasonable interpretation of the order $<\_L$ of constructibility, there is no such transitive model $M$. However, that interpretation is fine structurally motivated, and with a more obvious interpretation, I’m not sure how to prove it in general. There are many cases, however, where it can be ruled out with the more obvious one, and I’ll start with an example of that.


So let us start with the *non*-fine structurally motivated definition of $<\_L$:
We order elements $x\in L$ first on stage $L\_{\alpha+1}$ of construction, then on complexity $\Sigma\_n$ of formula used to define $x$ over $L\_\alpha$, then on finite set $s$ of ordinals used as parameters in such a definition, and finally on the specific $\Sigma\_n$ formula used. Here we order $[\mathrm{OR}]^{<\omega}$ (finite sets of ordinals) top-down lexicographically, so $s<t$ iff $s\neq t$ and $\max((s\backslash t)\cup(t\backslash s))\in t$. A little more precisely, fix a recursive enumeration $\varphi\_0,\varphi\_1,\ldots$ of all formulas in the language of set theory. Now given $x\in L$, the rank of $x$ is the lexicographic rank of the lexicographically least tuple $(\alpha,n,s,k)$ where $x\in L\_{\alpha+1}$ and $n,k<\omega$ and $\varphi\_k$ is $\Sigma\_n$ and $s=\{s\_0,\ldots,s\_{\ell-1}\}\in [\alpha]^{<\omega}$ and $x=\{y\in L\_\alpha\bigm| L\_\alpha\models\varphi\_k(s\_0,\ldots,s\_{\ell-1},y)\}$.


Now let’s consider what is presumably the most obvious example of a candidate model $M$: let $\alpha$ be least such that $L\_\alpha\models\mathrm{ZFC}$ and let $c$ be the $<\_L$-least Cohen generic over $L\_\alpha$, and $M=L\_\alpha[c]$. Then $<\_L\upharpoonright M$ cannot be added to $M$.


(Remark:  This should be contrasted to when we restrict attention to the prewellorder $\leq^\*$ of $L$
given just by stage of construction ($x\leq^\* y$ iff for all $\beta$, if $y\in L\_\beta$ then $x\in L\_\beta$). For this example, $\leq^\*\upharpoonright M$ is definable from parameter $c$; see [When is $L$-Rank definable in inner models of $V=L$?](https://mathoverflow.net/questions/119882/when-is-mathbbl-rank-definable-in-inner-models-of-mathbbv-mathbbl))


Now $L\_\alpha$ is pointwise definable, and this yields a surjection $\pi:\omega\to L\_\alpha$ which is $\Sigma\_1$-definable over $L\_{\alpha+1}$ from the parameter $\alpha$. Therefore there is a Cohen generic over $L\_\alpha$ in $L\_{\alpha+2}$,
and in fact one that is $\Sigma\_1$-definable over $L\_{\alpha+1}$ from the parameter $\alpha$. This is optimal for such a Cohen generic: Because $L\_\alpha\models\mathrm{ZFC}$, we have $c\notin L\_{\alpha+1}$ and there is no $\Sigma\_0$ definition of $c$ over $L\_{\alpha+1}$ from parameters. It also implies that $L\_\alpha\preceq\_{\Sigma\_1} L\_{\alpha+1}$, and hence there is no $\Sigma\_1$ definition of $c$ over $L\_{\alpha+1}$ from any $s\in[\mathrm{OR}]^{<\omega}$ with $s<\_{\mathrm{lex}}\{\alpha\}$.


So in terms of the 4-tuples defining $<\_L$, the $<\_L$-least Cohen generic $c$ has tuple $(\alpha+1,1,\{\alpha\},k)$ for some $k<\omega$.


Assuming that $<\_L\upharpoonright M$ is $M$-compatible (meaning that $(M,{<\_L\upharpoonright M})$ is a model of ZFC, in the expanded language with the predicate $<\_L\upharpoonright M$), we want to deduce that $M$ can understand enough about $\pi$ for a contradiction.
That is, for $\beta\in\alpha\backslash\omega$ let $c\_\beta=c\cup\{\beta\}$. Then note that $c\_\beta$ has tuple $(\alpha+1,1,\{\alpha\},k\_\beta)$ for some $k\_\beta<\omega$ (using here the surjection $\pi:\omega\to L\_\alpha$ mentioned above). But the class $C=\{c\_\beta\bigm|\beta\in[\omega,\alpha)\}\subseteq M$ is definable over $M$, so if $<\_L\upharpoonright M$ is $M$-compatible, then $<\_L\upharpoonright C$ is too. But this is an ordering of $\mathrm{OR}^M\backslash\omega$ in ordertype $\omega$, which is impossible.


Now it seems one would like to generalize this to show that there is no such transitive $M$ at all. But we used some particular fine structural facts which don't easily (seem to) generalize. In particular, the use of the standard $\Sigma\_n$ hierarchy becomes inconvenient when $n>1$. This is modified in fine structure theory, and replaced with a slightly different definability hierarchy (e.g. $\mathrm{r}\Sigma\_n$ instead of $\Sigma\_n$). One also uses the $\mathcal{J}$-hierarchy $\mathcal{J}\_\alpha$, and definability over $\mathcal{J}\_\alpha$, instead of the $L$-hierarchy $L\_\alpha$ (though I think the more crucial thing is the definability hierarchy). If one defines $<\_L$ in a natural way using these things, also incorporating the fine structural standard parameters in a natural way, then there can be no such transitive $M$ for which $<\_L\upharpoonright M$ is $M$-compatible.


That is (now assuming fine structure), define $<\_L$ as follows: Given $x\in L$, let the $<\_L$-rank of $x$ be the lexicographic rank of the lexicographically least tuple $(\alpha,n,s,k)$  such that $x\in\mathcal{J}\_{\alpha+1}$, $\varphi\_k$ is $\mathrm{r}\Sigma\_{n}$, $s\in[\mathrm{OR}\cap\mathcal{J}\_\alpha]^{<\omega}$,
and $x=\{y\in\mathcal{J}\_\alpha\bigm|\mathcal{J}\_\alpha\models\varphi\_k(\vec{p}\_{n-1},s,y)\}$, where $\vec{p}\_{n-1}=(p\_1,p\_2,\ldots,p\_{n-1})^{\mathcal{J}\_\alpha}$, the first $n-1$ standard parameters of $\mathcal{J}\_\alpha$ (and $\vec{p}\_{-1}=\emptyset$). Then there is no transitive $M\models\mathrm{ZFC}+V\neq L$ such that $<\_L\upharpoonright M$ is $M$-compatible.


Proof: Suppose $M$ is otherwise. Let $\alpha=\mathrm{OR}^M$. Let $(\gamma,n',s,k)$
be a minimal tuple determining a set $x\in M\backslash \mathcal{J}\_\alpha$. Note that $n'=n+1>0$, by rank considerations. So $x\in \mathcal{J}\_{\gamma+1}\backslash \mathcal{J}\_\gamma$ and $x$ is $\mathrm{r}\Sigma\_{n+1}^{\mathcal{J}\_\gamma}(\{\vec{p}\_n^{\mathcal{J}\_\alpha},s\})$. Let $\xi=\sup(x)<\alpha$. Now observe that $\rho\_{n+1}^{\mathcal{J}\_\gamma}\leq\xi$ and $s\backslash\xi=p\_{n+1}^{\mathcal{J}\_\gamma}\backslash\xi$. (If $s\backslash\xi<\_{\mathrm{lex}}p\_{n+1}^{\mathcal{J}\_\gamma}\backslash\xi$ then use an $(n+1)$-solidity witness, which is in $\mathcal{J}\_\gamma$,  to compute $x\in \mathcal{J}\_\gamma$, contradicting the minimality of $\gamma$. The other direction is by $(n+1)$-soundness.) Now for $\beta\in[\xi,\alpha)$ define $x\_\beta=x\cup\{\beta\}$, and note that the tuple for $x\_\beta$ is $(\gamma,n+1,s\_\beta,k\_\beta)$, where $s\_\beta\backslash\xi=s\backslash\xi$ and $k\_\beta<\omega$. (If $s\_\beta\backslash\xi<\_{\mathrm{lex}} s\backslash\xi$, note that $x\_\beta$ violates the minimality of the tuple for $x$ for giving something in $M\backslash L\_\alpha$.) But then if ${<\_L}\upharpoonright M$ is $M$-compatible, then $(M,{<\_L\upharpoonright M})$ can define an ordering of its ordinals in ordertype $\leq$ that of the lexicographic ordering on $[\xi]^{<\omega}\times\omega$, which is impossible.


It would be very interesting to know what the answer is in general for the first (non-fine structurally motivated) definition of $<\_L$.


