Ordinals definable over $L_\kappa$ Suppose $\kappa$ is an uncountable cardinal, with $L_\kappa$ an admissible set (i.e. a model of Kripke–Platek set theory). Let $<_\gamma \subseteq \kappa \times \kappa$ denote a wellordering of $\kappa$ (equivalently, of $L_\kappa$) such that $\mathrm{ot}(<_\gamma) = \gamma$.
What is the least ordinal $\delta$ such that $<_\delta$ is not first order definable (i.e. in $\mathcal{L}_\in$) over $L_\kappa$?

Variations on this question include:


*

*What is least such ordinal when we consider $L_\alpha$ for any admissible $\alpha$?

*How is the answer affected by restricting definability, e.g. to $\Delta^0_1$, or strengthening it to $\mathcal{L}^2_\in$?

*This is all parameter-free, does allowing parameters substantially change the answer?



I'm afraid I don't know much about $L$ or admissibility so these may be naïve questions, sorry!
 A: A good reference is K. Devlin Constructibilty, Lemma 3.1 and the following discussion, where he proves a $\Sigma^1_1$ well ordering of $L$ (but it can be adapted to admissible $L_k$). So for all $\delta \lt \kappa$ $\lt_\delta$ is definable with parameters in $L_\kappa$ (and this cannot be done for $\delta \gt \kappa$ as $\delta \notin L_\kappa$.)
A: Eran has answered the main question and (in parentheses) the variation concerning admissible $\alpha$ rather than only uncountable cardinals, but you also asked about a couple of other variations.  Restricting to $\Delta^0_1$ definitions makes no difference, once you have a $\Sigma^0_1$ definition, since $\xi<_\gamma\eta$ if and only if $\xi,\eta<\gamma$ and $\eta\not<_\gamma\xi$ and $\xi\neq\eta$; this gives a $\Pi^0_1$ definition of $\xi<_\gamma\eta$ if you plug in a $\Sigma^0_1$ definition of $\eta<_\gamma\xi$. 
As far as parameters are concerned, $<_\kappa$ is definable without parameters in $L_\kappa$, and therefore $<_\delta$ will also be definable without parameters in $L_\kappa$ whenever $\delta<\kappa$ and $\delta$ is definable without parameters in $L_\kappa$.  On the other hand, if $\delta$ is not definable without parameters, then neither is $<_\delta$.  More generally, any parameters that suffice to define $\delta$ also suffice to define $<_\delta$ and vice versa.
