Some elements of $L$ become constructible only in levels higher than its rank level. So I ask:


Let $V$ be such that $V = L$.
For which ordinals $\alpha$ do we have $V\_\alpha = L\_\alpha$?


Indeed, we have this for $\omega$ and if $\alpha$ is the class of all ordinals. But do we have this for other types of ordinals? Can we, for instance, have $V\_\alpha = L\_\alpha$ for a singular cardinal?



For which ordinals do we have $V_\alpha = L_\alpha$?
Yes; in fact the first $\alpha>\omega$ with $V\_\alpha=L\_\alpha$ has cofinality $\omega$. To obtain this $\alpha$, define $f:Ord\to Ord$ by $f(\xi)=$ the smallest $\eta$ such that $V\_\xi\subseteq L\_\eta$. Such an $\eta$ exists because of the assumption that $V=L$. Write $f^n$ for the $n$-fold iterate of $f$. Then $\sup\_{n\in\omega}f^n(\omega+1)$ is the desired $\alpha$.


More generally, the ordinals you asked about are exactly the fixed points of $f$. Since $f$ is a normal function (increasing and continuous), its fixed points constitute a closed unbounded class of ordinals.


