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 . . .
 A: 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.
A: Let me add a coda to Asaf's answer, which treated $\Sigma_1$ definability without parameters; what happens when we allow parameters? (This is in fact the situation relevant to $\alpha$-recursion theory.)
Well, the same general idea holds, but it has a bit of a twist: if $\kappa$ is an uncountable cardinal, then the set of cardinals $<\kappa$ is $\kappa$-recursive iff $\kappa$ is a successor cardinal.
Proof. First, suppose $\kappa=\lambda^+$. Then for any $\alpha<\kappa$, $\alpha$ is a cardinal iff $\alpha=\lambda$ or $\alpha<\lambda$ and $L_\lambda\models$ "$\alpha$ is a cardinal"; this is $\Sigma_1$ in $\lambda$.
Now suppose $\kappa$ is a limit and $\varphi$ is a $\Sigma_1$-definition of the set of cardinals $<\kappa$, with parameters $a_1, . . . , a_n\in L_\kappa$. We may find a cardinal $\mu<\kappa$ with $a_1, . . . , a_n\in L_\mu$. Now take an elementary submodel $M$ of $L_\lambda$ such that $L_\mu\subseteq M$, $\vert M\vert=\mu$, and $\mu^+\in M$ (this last since $\kappa$ is a limit - so $\mu^+\in L_\kappa$); and let $N$ be the transitive collapse of $M$. Then the image $\theta$ of $\mu^+$ under the collapse map is strictly less than $\mu^+$, so is not a cardinal; but $\theta$ satisfies the $\Sigma_1$ definition in $N$ since the parameters in $\varphi$ aren't moved when passing to $N$, so it satisfies it in $L_\lambda$, a contradiction.
