Identifying cardinals in $\alpha$-recursion theory Throughout, we work in $V=L$.
Fix an uncountable cardinal $\kappa$. $\kappa$-recursion theory is the natural generalization of recursion theory from $\omega$ to $\kappa$, using the following analogy:


*

*Finite = element of $L_\omega$ $\approx$ element of $L_\kappa$

*C.e. = $\Sigma_1$ over $L_\omega$ $\approx$ $\Sigma_1$ over $L_\kappa$

*Computable = $\Delta_1$ over $L_\omega$ $\approx$ $\Delta_1$ over $L_\kappa$.
The vast majority of computability-theoretic concepts - e.g. productivity, immunity, etc. - generalize naturally to this setting. However, the converse is not true: there are natural notions at the $\kappa$ level which have no analogues, or no nontrivial analogues, on $\omega$.
My question is about one of these - namely, the cardinality predicate:

Is the relation "is a cardinal" computable in the sense of $L_\kappa$?

It is easy to show that it is $\Pi_1$, but I don't see a $\Sigma_1$ definition, and indeed I don't think there is one. But I don't immediately see how to show that there is none . . .

If the cardinals were $\Sigma_1$-definable, then taking any countable elementary submodel $M$ when $\kappa>\omega_1$, we get that there are some uncountable cardinals in $M$.
Collapsing $M$ gives us some $L_\gamma$, for a countable $\gamma$. But now the collapse of those cardinals also satisfy being a cardinal in $L_\gamma$, and being a $\Sigma_1$ property, this is upwards absolute.
But now we run into a bit of a problem, since countable ordinals are not usually cardinals.

Of course, for $\kappa=\omega_1$, the set of cardinals is $\omega+1$ which is indeed $\Sigma_1$ definable.
