How complicated can infinitary satisfiability problems be? We work in $\mathsf{ZFC+V=L}$.

For $\alpha$ a possibly uncountable admissible ordinal, let $Sat_\alpha$ be the set of $\mathcal{L}_{\infty,\omega}\cap L_\alpha$-sentences which have a model (not necessarily in $L_\alpha$). I'm curious about the complexity of $Sat_\alpha$ relative to $L_\alpha$. There are a couple easy cases where we get $\Pi_1$ and $\Sigma_1$ respectively:

*

*Suppose $\alpha$ is a countable admissible ordinal. By Barwise's completeness theorem, $\varphi\in Sat_\alpha$ iff there is no "proof" (in the appropriate sense) of $\neg\varphi$ in $L_\alpha$; this gives $Sat_\alpha\in \Pi_1(L_\alpha)$.


*On the other side of things, suppose $\kappa$ is an uncountable cardinal and $\mathcal{N}\models\varphi\in \mathcal{L}_{\kappa,\omega}$. Pick a sufficiently large $\theta$ and let $X\prec L_\theta$ with $\vert X\vert<\kappa$, $tc(\{\varphi\})\subseteq X$, and $\mathcal{N}\in X$. Let $\mu: X\rightarrow Y$ be the Mostowski collapse map for $X$; then $Y\subset L_\kappa$ and $Y\ni \mu(\mathcal{N})\models\varphi$. Since checking satisfaction of an $\mathcal{L}_{\kappa,\omega}$-sentence in a structure is sufficiently simple, this means that $Sat_\kappa$ is $\Sigma_1(L_\kappa)$.
I'm curious whether these are the only two possible complexities:

Is there an uncountable admissible ordinal $\alpha$ such that $Sat_\alpha$ is neither $\Sigma_1$ or $\Pi_1$ (with parameters) over $L_\alpha$?

I suspect that the answer is, or follows trivially from, a well-known fact about uncountable admissible ordinals, but I don't see it at the moment. It's not even clear to me whether $Sat_\alpha$ is always first-order definable over $L_\alpha$ (I suspect that it isn't).
 A: (We're working in ZFC + $V=L$.) Let $\alpha$ be the least admissible $>\omega_1$. Then $\mathrm{Sat}_\alpha$ is not definable from parameters over $L_\alpha$.
Lemma: (Assuming ZFC + $V=L$.) Let $\kappa$ be a cardinal such that $\mathrm{cof}(\kappa)>\omega$. Let $M$ be a model of "$V=L$" with $\kappa\subseteq$ the wellfounded part of $M$ and $n<\omega$ and $x\in M$ be such that $M$ is the $\Sigma_n$-hull (over $M$) of elements in $\kappa\cup\{x\}$. Suppose there is no strictly decreasing $\omega$-sequence of $M$-ordinals which is $\Sigma_n$-definable from parameters in $M$. Then $M$ is wellfounded.
Proof: Suppose $M$ is illfounded and let $\left<\alpha_i\right>_{i<\omega}$ be strictly descending in $M$'s ordinals. We can pick a sequence $\left<\varphi_i,\beta_i\right>_{i<\omega}$ such that $\varphi_i$ is a $\Sigma_n$ formula which defines $\alpha_i$ from $(\beta_i,x)$. But the sequence $\left<\varphi_i,\beta_i\right>_{i<\omega}\in L_\kappa\subseteq M$, and therefore $\left<\alpha_i\right>_{i<\omega}$ is $\Sigma_n$-definable from the parameter $\left<\varphi_i,\beta_i\right>_{i<\omega}$ over $M$, contradiction.
Now fix $\alpha$ the least admissible $>\omega_1$. Then $L_\alpha$ is the $\Sigma_1$-hull of $\omega_1+1$. So the full $L_\alpha$-theory of parameters in $\omega_1+1$ is not definable from parameters over $L_\alpha$. Let $\lambda=\omega_1+1$. Let $T$ be the  $\mathscr{L}_{\infty,\omega}\cap L_\alpha$-theory, using $\in$, $=$ and constant symbols $c$ for each $c\in\lambda$, asserting the conjunction of (i) KP + $V=L$, (ii) I am wellfounded through $\lambda$ (i.e. "$0,1,\ldots,\lambda$ are an initial segment of my ordinals"), (iii) there is no proper segment of me beyond $\lambda$ which satisfies KP,
(iv) I am the $\Sigma_1$-hull of $\lambda$, and (v) there is no infinite strictly decreasing sequence through my ordinals which is $\Sigma_1$-definable from parameters. Note that by the lemma, $L_\alpha$ is the unique model of $T$.
But now for each $\vec{\xi}\in(\omega_1+1)^{<\omega}$ and formula $\varphi$, the theory $T+\varphi(\vec{\xi})$ is satisfiable iff $L_\alpha\models\varphi(\vec{\xi})$. Therefore $L_\alpha$ cannot define satisfiability from parameters.
Taking this a bit further, we get:
Theorem: (Assume ZFC + $V=L$.) Let $\alpha$ be an infinite ordinal which is not a cardinal, such that $\mathrm{card}(\alpha)$ has cofinality $>\omega$. Then $\mathrm{Sat}_\alpha$ is definable from parameters over $L_\alpha$ iff $L_\alpha$ is closed under models witnessing its satisfiable formulas, i.e. iff \begin{equation} \mathrm{Sat}_\alpha(T)\iff L_\alpha\models
\text{"there is a model of }T\text{"}.\end{equation}
Proof Sketch: Suppose there is $T\in L_\alpha$ which is satisfiable but which has no model in $L_\alpha$. Like mentioned in the original question, $T$ has a model in $L_{\alpha^+}$. Therefore there is $\beta$ such that $\alpha\leq\beta<\alpha^+$
and $L_\beta$ projects to $\kappa=\mathrm{card}(\alpha)$ (i.e. there is $n<\omega$ and $x\in L_\beta$ such that $L_\beta$ is the $\Sigma_n$-hull of parameters in $\kappa\cup\{x\}$) and there is a model of $T$ in $L_\beta$. But since $T\in L_\alpha$, then arguing much as before, we can construct a theory $T'\in L_\alpha$ whose unique model is the least such $L_\beta$, which implies that $L_\alpha$ cannot define satisfiability, much as before. QED.
This leaves the uncountable non-cardinals $\alpha$ where $\mathrm{card}(\alpha)$ has countable cofinality, which the above obviously doesn't touch.
