End-extension in Gödel's constructible universe

Question

Given two ordinals $\alpha < \beta$, considering the subsets of Gödel's constructible universe, one say that $L_\beta$ is a $\Sigma_n$ end-extension of $L_\alpha$ (and $L_\alpha$ is an $\Sigma_n$ elementary submodel of $L_\beta$), written $L_\alpha \prec_{\Sigma_n} L_\beta$, when for all $\Sigma_n$ formulas $\varphi$, \begin{align} L_\beta \models \varphi \iff L_\alpha \models \varphi \end{align}

Given such sets $L_\beta$ there exists results, showed by Jensen and others using Skolem functions, that helps us constructing elementary submodels $L_\alpha \subset L_\beta$. However I have had a hard time finding results that lets us do the converse, that is constructing end-extensions $L_\delta \supset L_\beta$.

Does there exists such general results about end-extension of Gödel's constructible sets ?

And more particulary, let ($\zeta$, $\Sigma$) be the least ordered-pair such that $L_\zeta \prec_{\Sigma_2} L_\Sigma$, as characterized by Welch using infinite time Turing machines or by Harrington using $\Sigma_2$ truth sets :

Can we find an ordinal $\kappa > \Sigma$ such that $L_\Sigma \prec_{\Sigma_2} L_\kappa$ ? And if so, can we continue this sequence of end-extensions $\zeta, \Sigma, \kappa \ldots$ ?

Answer 1

This is in the same vein as Monroe Eskew's comments. For $n\geq 1$, given that $\alpha<\beta$ and $L_\alpha\prec_{\Sigma_n}L_\beta$, it is not always possible to produce even a $\Sigma_1$ end-extension of $L_\beta$. In particular,

Theorem 1: Let $n\geq 1$ be a natural number, and let $\beta$ be least such that there is an $\alpha<\beta$ where $L_\alpha\prec_{\Sigma_n}L_\beta$. Then $L_\beta$ has no $\Sigma_1$ end-extension.

Proof: There is a $\Pi_n$ formula $\varphi_n$ such that for any ordinals $\gamma<\delta$, $L_\delta\vDash\varphi_n(\gamma)$ iff $L_\gamma\prec_{\Sigma_n}L_\delta$. (This is theorem 1.8 in Kranakis's "Reflection and partition properties of admissible ordinals", Annals of Mathematical Logic, vol. 22, issue 3.) Assume for the sake of a contradiction that there is an ordinal $\kappa>\beta$ such that $L_\beta\prec_{\Sigma_1}L_\kappa$. Then $L_\kappa\vDash\exists\zeta,\sigma(\zeta<\sigma\land L_\sigma\vDash\varphi_n(\zeta))$. As $L_\kappa$ is admissible and correct about levels of $L$, the function $\xi\mapsto L_\xi$ is $\Sigma_1$ on $L_\kappa$ (Barwise, Admissible Sets and Structures, p.58), and as Monroe Eskew mentioned in the comments, satisfaction is $\Delta_1$, so this is a $\Sigma_1$ sentence. Then by $\Sigma_1$-elementarity, $L_\beta$ satisfies this sentence as well, so there must be ordinals $\zeta,\sigma$ in $L_\beta$ such that $\zeta<\sigma$ and $L_\zeta\prec_{\Sigma_n}L_\sigma$. However, $\beta$ was assumed to be least such that $L_\beta$ has an $L_\alpha$ that's a $\Sigma_n$-elementary substructure, which is a contradiction. $\square$

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In fact you have to travel quite a bit past the least such $\alpha,\beta$ in order to find ordinals $\alpha<\beta$ where $L_\alpha\prec_{\Sigma_n}L_\beta$ and $L_\beta$ has a $\Sigma_1$ end-extension.

Theorem 2: Let $n\geq 1$ be a natural number, and let $\beta$ be least such that there is an $\alpha<\beta$ where $L_\alpha\prec_{\Sigma_n}L_\beta$ and $L_\beta$ has a $\Sigma_1$ end-extension $L_\kappa$. Then $\alpha$ is a limit of ordinals $\zeta$ such that $L_\zeta$ has a $\Sigma_n$ end-extension.

Proof: Let $\varphi_n$ be as before. Choose an arbitrary ordinal $\xi<\alpha$. The end-extension $L_\kappa$ satisfies $\exists\zeta,\sigma(\zeta>\xi\land L_\sigma\vDash\varphi_n(\zeta))$, with $\alpha$ and $\beta$ as witnesses respectively. This formula has parameter $\xi\in L_\alpha$ and is $\Sigma_1$ by similar reasoning to the kind appearing in the proof of theorem 1. As $L_\alpha\prec_{\Sigma_n}L_\beta$, $L_\beta\prec_{\Sigma_1}L_\kappa$, and $n\geq 1$, it follows that $L_\alpha\prec_{\Sigma_1}L_\kappa$. By $\Sigma_1$-elementarity, $L_\alpha$ also satisfies $\exists\zeta,\sigma(\zeta>\xi\land L_\sigma\vDash\varphi_n(\zeta))$. As $\xi<\alpha$ was arbitrary, this produces a pair of ordinals $(\zeta,\sigma)$ arbitrarily high below $\alpha$ such that $L_\zeta\prec_{\Sigma_n}L_\sigma$. $\square$

This is stronger than theorem 1, as it shows that the least such $\alpha$ is not only larger than the least $\zeta$ such that $L_{\zeta}$ has a $\Sigma_n$ end-extension, but it is also larger than the $n$th such $\zeta$ for any finite $n$. Maybe it is possible to strengthen this result to consider limits of limits of ... of such $\zeta$, using an approach similar to lemma 4.15 of Marek and Srebrny's "Gaps in the Constructible Universe" (Annals of Mathematical Logic vol. 6, 1974).