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It is a known result by Scott and Myhill that the second-order version of $L$ yields $\mathrm{HOD}$. Recently, Kennedy, Magidor, and Väänänen (Inner models from extended logics: Part I and II) investigated inner models given by logics with generalized quantifiers, which yields a logic intermediate between first-order and second-order logic. It motivates the following question:

Is there a logic that produces the mantle?

(Here the choice of the mantle is somewhat arbitrary; we may replace it by 'generic mantle', 'symmetric mantle' or whatever. I will focus on the mantle in this question, but I welcome discussing other cases.)

Of course, the answer is trivial if we assume, like $V=L$ or $V=L[G]$ for some $L$-generic $G$. I want to ask the existence of logic which defines the mantle uniform to models of ZFC.

Is there a (ZFC-definable) abstract logic $\mathcal{L}$ such that the inner model given by $\mathcal{L}$ is (ZFC-provably) the mantle?

(Under model-theoretic terms, is there $\mathcal{L}$ such that for any model $M$ of $\mathsf{ZFC}$, the inner model given by $\mathcal{L}$ is the mantle of $M$?)

Here are some of my rough thoughts:

  • Sublogics of higher-order logics are not the candidate for $\mathcal{L}$: the corresponding inner models of higher-order logics are $\mathrm{HOD}$ (if my reasoning is correct), so the sublogics yield a submodel of $\mathrm{HOD}$. However, $\mathrm{HOD}$ need not be the mantle. (Theorem 70 of Fuchs, Hamkins, and Reitz (Set-theoretic geology).)

  • We can rule out $\mathcal{L}_{\kappa\kappa}$, which yields Chang model. The inner model given by $\mathcal{L}_{\kappa\kappa}$ is the least transitive model of ZF that contains all ordinals and is closed under $<\kappa$-sequences (Theorem II of Chang's Sets constructible using $L_{\kappa\kappa}$.) However, the mantle need not be closed under $<\kappa$-sequences. (A generic extension of $L$ would be an example.)

I would appreciate any comments or answers.

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    $\begingroup$ An observation: if the mantle has the form $L[A]$, where $A$ is a class of ordinals, then we may define an artificial $A$-recovering quantifier $Q^A$ as follows: $N\vDash (Q^A xy)\varphi(x,y,\vec{a})$ iff $\{(x,y)\in N^2\mid N\vDash \varphi(x,y,\vec{a})\}$ is a linear order of ordertype in $A$. Then the resulting model is just $L[A]$. $\endgroup$
    – Jason Zesheng Chen
    Commented Jul 25, 2020 at 0:04
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    $\begingroup$ Can you be more specific about what you mean by an abstract logic? It is hard to find the definition online, or more accurately, it is easy to find a number of inequivalent definitions. It seems to me it might be possible to realize any inner model $M$ as $C(\mathcal L)$ for some proper class abstract logic $\mathcal L$ which is ad hoc yet definable from the predicate $M$. $\endgroup$
    – Gabe Goldberg
    Commented Jul 26, 2020 at 22:09
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    $\begingroup$ @GabeGolsberg Thank you for your comment. I did not know that it has an ambiguous definition (I initially assumed the definition in Model Theory by Chang and Keisler.) $\endgroup$
    – Hanul Jeon
    Commented Jul 27, 2020 at 2:21
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    $\begingroup$ What if $\mathcal L$ has formulas $\varphi_x$ for every $x\in M$ such that $\varphi_x$ is false unless $\mathfrak{A}$ is a structure with one relation symbol and for some transitive $y\in M$, there is an isomorphism $f : \mathfrak{A}\to (y,\in)$ such that $f(x)\in y$. In other words, $\mathcal L$ has formulas coding all elements of $M$. It seems like by induction, the $\alpha$-th level of the $C(\mathcal L)$-hierarchy is $(V_\alpha)^M$. I worry that I am misunderstanding the definition of an abstract logic. $\endgroup$
    – Gabe Goldberg
    Commented Jul 27, 2020 at 4:28
  • $\begingroup$ @GabeGoldberg Your argument seems relevant to Hamkins' previous answer. $\endgroup$
    – Hanul Jeon
    Commented Jul 27, 2020 at 9:12

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Combining Goldberg's comment and Hamkins' answer seems to work. Especially, for any inner model $M$ of ZF, we have an abstract logic $\mathcal{L}$ whose corresponding inner model $L^\mathcal{L}$ is $M$.

Consider the sublogic of $\mathcal{L}_{\infty,\omega}$ such that infinite conjunction and disjunctions are only allowed to set of formulas in $M$. In fact, $\mathcal{L}=\mathcal{L}_{\infty,\omega}^M$.

Define $\psi_A$ for $A\in M$ as Hamkins defined: to repeat the definition, $$\psi_A(x):= \bigvee_{a\in A} (\forall v : v\in u\leftrightarrow \psi_a(u)).$$ Then $\psi_A(x)$ is a member of $M$ by induction on $A\in M$.

We can see that if $A\in M$, $A\subseteq V_\alpha^M$ then $$A=\{u\in V^M_\alpha \mid V^M_\alpha\models \psi_A(u)\}.$$

Hence the $\alpha$th hierarchy $L_\alpha^\mathcal{L}$ contains $V^M_\alpha$ (It can be shown by induction on $\alpha$.) Therefore $M\subseteq L^\mathcal{L}$. On the other hand, an inductive argument shows that the $\alpha$th hierarchy $L^\mathcal{L}_\alpha$ is a member of $M$ (we need the absoluteness of the satisfaction relation for $\mathcal{L}$ between $M$ and $V$), so $L^\mathcal{L}\subseteq M$.

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    $\begingroup$ Of course, this is a proper-class-sized logic - it would be quite reasonable to ask for a set-sized logic, that is, one whose class of $\Sigma$-formulas is a set for each language $\Sigma$. And then I see no similar trick. $\endgroup$
    – Noah Schweber
    Commented Jul 27, 2020 at 23:46
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    $\begingroup$ In that case, it seems like the $C(\mathcal L)$ must be contained in $\text{HOD}_{\text{Ord}^\kappa,S}$ for some set $S$ (corresponding to the set of sentences) and some cardinal $\kappa$ (corresponding to the number of free variables allowed in a single sentence sentence). I bet you can get a model where the mantle is not contained in $\text{HOD}_{\text{Ord}^\kappa,S}$ for any $\kappa$ and $S$. It may be interesting to look instead at logics that have a Lowenheim number. $\endgroup$
    – Gabe Goldberg
    Commented Jul 28, 2020 at 0:06

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