MathOverflow Asked on January 3, 2022
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 $mathcal{L}_{kappakappa}$, which yields Chang model. The inner model given by $mathcal{L}_{kappakappa}$ 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 $L_{kappakappa}$.) 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.
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 $Ain M$ as Hamkins defined: to repeat the definition, $$psi_A(x):= bigvee_{ain A} (forall v : vin uleftrightarrow psi_a(u)).$$ Then $psi_A(x)$ is a member of $M$ by induction on $Ain M$.
We can see that if $Ain M$, $Asubseteq V_alpha^M$ then $$A={uin V^M_alpha mid V^M_alphamodels psi_A(u)}.$$
Hence the $alpha$th hierarchy $L_alpha^mathcal{L}$ contains $V^M_alpha$ (It can be shown by induction on $alpha$.) Therefore $Msubseteq L^mathcal{L}$. 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 $L^mathcal{L}subseteq M$.
Answered by Hanul Jeon on January 3, 2022
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