# Climbing quickly up $L$

This question is motivated by Joel David Hamkins' answer to Gödel's Constructible Universe in Infinitary Logics (A Possible Approach to HOD Problem), in which he shows that, if we replace first-order definability by $\mathcal{L}_{\infty\infty}$-definability in the definition of $L$, we wind up building all of $V$.

This relies on using $\mathcal{L}_{\infty\infty}^V$; clearly if we take $$\overline{L}_{\alpha+1}=\{A\subseteq\overline{L}_\alpha: \exists \varphi\in \overline{L}_\alpha\cap\mathcal{L}_{\infty\infty}(A=\varphi^{\overline{L}_\alpha})\},$$ we just wind up with $L$ itself at the end.

Moreover, we also climb up $L$ at about the same speed as in the usual construction, since if I have an infinitary formula in $L_\alpha$, the subset of $L_\alpha$ it defines will show up in $L_{\alpha+17}$ (say).

A similar observation, however, seems to fail for second-order logic: if we interpret the second-order quantifiers as ranging over $L$, so that $$\hat{L}_{\alpha+1}=\{X\subseteq\hat{L}_\alpha:\exists \varphi\in \mathcal{L}_{II}(L\models \text{"}\varphi^{\hat{L}_\alpha}=A\text{"})\},$$ then the only obvious upper bound on the $\hat{L}$-hierarchy is $\hat{L}_{\alpha}\subseteq L_\beta\implies \hat{L}_{\alpha+1}\subseteq L_{\vert\beta\vert^+}$ (by condensation); and this is really a trivial observation. So in principle the $\hat{L}$-hierarchy could grow quite fast.

My question is:

How fast does the $\hat{L}$-hierarchy grow?

In one sense I think you almost have the answer anyway. Assume $V=L$. For any cardinal $\beta$ let $C_\beta = \{ \gamma\in (\beta,\beta^+)| \,L_\gamma \prec L_{\beta^+}\}$ ($\prec$ meaning First Order elementarily).
Let $\hat L_0 =L_\omega=HF$. Then $\hat L_1 = L_{\gamma_1}$ where $\gamma_1$ is the minimal element of $C_\omega$. Let $\gamma_2$ be the next element of $C_\omega$. Then $\hat L_2 = L_{\gamma_2}$.
This is because Second Order definability over $L_\beta$ is essentially equivalent to First Order definability over $L_{\beta^+}$; one deduces that S.O. definability over $\hat L_\tau \cup \{\hat L_\tau\}$ yields everything in $L_\delta$ where $\delta$ is least in $C_\beta$ greater than $On\cap \hat L_\tau$.
In short: if, setting $\gamma_0=\omega \, , \langle\gamma_\tau\,|\, \tau\in On\rangle$ enumerates $Card \cup \bigcup_{\beta\in Card} C_\beta$ in ascending order, $\hat L_\tau = L_{\gamma_\tau}$.