**Notations:** Let $L\_\alpha$ stand for the Gödel constructible hierarchy ($L\_0=\varnothing$ and $L\_{\alpha+1} = \mathrm{def}(L\_\alpha)$ is the set of definable subsets of $L\_\alpha$ and $L\_\delta = \bigcup\_{\beta<\delta} L\_\beta$ for limit $\delta$), and $J\_\alpha$ for the Jensen hierarchy ($J\_0=\varnothing$ and $J\_{\alpha+1} = \mathrm{rud}(J\_\alpha)$ is the rud-closure of $J\_\alpha \cup \{J\_\alpha\}$ and $J\_\delta = \bigcup\_{\beta<\delta} J\_\beta$ for limit $\delta$). Let $M \mathrel{\preceq\_1} N$ (if $M$ and $N$ are transitive sets mean “$(M,{\in})$ is a $\Sigma\_1$-elementary submodel of $(N,{\in})$”.


Consider the following two relations between two ordinals $\sigma<\gamma$:


* say that “$\sigma$ is $\gamma$-stable” when $L\_\sigma \mathrel{\preceq\_1} L\_\gamma$,
* say that “$\sigma$ is $\gamma$-J-stable” when $J\_\sigma \mathrel{\preceq\_1} J\_\gamma$.



> 
> **Question:** Are the above relations equivalent? If not, is there still a way to define “$\sigma$ is $\gamma$-stable” in terms of the Jensen hierarchy and/or “$\sigma$ is $\gamma$-J-stable” in terms of the Gödel hierarchy?
> 
> 
> 


Clearly, any one of the above relations implies that $L\_\sigma = J\_\sigma$ (because $\sigma$ is, at the very least, admissible $>\omega$), so the problem concerns the right-hand side, as $\gamma$ is not required to be admissible, or even a limit ordinal. Certainly $J\_\sigma \mathrel{\preceq\_1} J\_\gamma$ implies $L\_\sigma \mathrel{\preceq\_1} L\_\gamma$, so the question is whether the converse holds, or, if not, whether we can still find a way to express stability for one hierarchy in terms of the other.


I know that $\sigma$ is $(\sigma+1)$-stable iff it is $\Pi\_m$-reflecting for each $m$, for example, but I don't see whether this also applies to being $(\sigma+1)$-J-stable (or, if not, what the corresponding criterion would be).


**Bonus question:** What if we replace $1$ by $n$ (i.e., $\Sigma\_1$ by $\Sigma\_n$) throughout?



Stability for the Gödel and Jensen hierarchies
As remarked the non-trivial direction is to show $L\_\sigma\prec\_{\Sigma\_1}L\_\gamma$ implies $J\_\sigma\prec\_{\sigma\_1}J\_\gamma$. Let's take the extreme case that $\gamma=\sigma+1$. Suppose $J\_{\sigma+1}\models \exists u \varphi(u,x)$ where $\varphi\in\Sigma\_0$ and $x\in J\_\sigma = L\_\sigma$. The short answer is to copy everything onto subsets of $\sigma$ and use $\mathcal{P}(L\_\sigma)\cap J\_{\sigma+1}= L\_{\sigma+1}$. But in more detail: as $J\_{\sigma+1}=\bigcup\_{n\in\omega} S\_{\omega\sigma+n}$ (see Jensen's original Fine Structure paper, or Devlin's "Constructibility" for the definition of the $S\_\nu$-hierarchy, and these and other assertions) we have for some $n$ that $S\_{\omega\sigma+n}\models \exists u \varphi(u,x)$. (Here we have $\omega\sigma=\sigma$ as well as the fact that the Goedel pairing function $Q$ is $\Delta\_1^{L\_\sigma}$). By adding some extra `Basic Functions' to the usual list, we can assume each $S\_{\omega\sigma+n}$ is transitive. Now use the fact that there is an onto function $F:\omega\sigma \rightarrow S\_{\omega\sigma+n}$ in $J\_{\sigma+1}$. Use this to get a subset $f\subseteq \omega\sigma$ with


$$\langle \sigma, Q \mbox{''}f \rangle \cong \langle S\_{\omega\sigma+n},\in \rangle $$
and $f\in J\_{\sigma+1} $ and so also in $L\_{\sigma+1} $. We now have an existential statement of the form $L\_{\sigma+1} \models \mbox{''}\exists \sigma\exists f \subseteq \sigma \ldots $''. So reflect this to some $\tau<\sigma$ and obtain:
$$\langle \tau, Q \mbox{''} \bar f \rangle \models
\exists u \varphi(u,Q(\xi)\_1)
 $$ 


where we ensure for some $\xi \in \bar f$, $ Q(\xi)\_1 $ represents $x$ in the structure say. Taking the transitive collapse of $\langle \tau, Q \mbox{''}\bar f\rangle
 $ then yields a transitive model $U\in J\_\sigma$ of $\exists u \varphi(u,x)$ as needed.


[Edit: added to address comment. Now use this argument repeatedly to prove by induction on $\delta$ for $\sigma+1\leq\delta<\gamma$ that $L\_\sigma\prec\_{\Sigma\_1}L\_\gamma$ implies $J\_\sigma\prec\_{\sigma\_1}J\_\gamma$. The successor case of a $\delta+1$ is just a variant of the one given and the case of a limit $\delta$ is of course trivial.]


For the **Bonus Q** just note that for $n>1$, if $L\_\sigma\prec\_{\Sigma\_n}L\_\gamma$, then both $L\_\sigma$ and $L\_\gamma$ are limits of admissible ordinals, so then $L\_\sigma=J\_\sigma$ and $L\_\gamma=J\_\gamma$ and there is nothing to do!


