This question is motivated by Joel David Hamkins' answer to [Gödel's Constructible Universe in Infinitary Logics (A Possible Approach to HOD Problem)](https://mathoverflow.net/questions/156940/godels-constructible-universe-in-infinitary-logics-a-possible-approach-to-hod), 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?
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> 
> 



Climbing quickly up $L$
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, $\hat L\_\tau = L\_{\gamma\_\tau}$.


