What sort of cardinal number is the Löwenheim-Skolem number for second-order logic?

In their paper "On Löwenheim-Skolem-Tarski numbers for extensions of first order logic", Magidor and Väänänen make the following statement:

"For second order logic, $LS(L^{2})$ [the Löwenheim-Skolem number for second order logic--my comment] is the supremum of $\Pi_{2}$-definable ordinals..., which means that it exceeds the first measurable, the first $\kappa^{+}$-supercompact $\kappa$, and the first huge cardinal if they exist ["the Löwenheim-Skolem number $LS(L^{2})$ of second order logic $L^{2}$ is the smallest cardinal $\kappa$ such that if a theory $T$$\subset$$L^{2}$ has a model, it has a model of cardinality$\le$max($\kappa$,$|T|$)", and "$L^{2}$ extends first order logic with quantifiers of the form $\exists$$R$$\phi$($R$,$x_0$,...,$x_{{n}-1}$), where the second order variable $R$ ranges over n-ary relations on the universe for some fixed n"--my comment also but substantially quoting the authors]".

Assume that the first measurable, the first $\kappa^{+}$-supercompact $\kappa$, and the first huge cardinals exist. What type of large cardinal, then, is $LS(L^{2})$? If the answer is known, please provide the reference.

• Magidor has results on this. – Monroe Eskew Jul 19 '15 at 3:49
• Why did this get a -1? Seems a fine question to me. – Noah Schweber Jul 19 '15 at 4:23
• Thomas, in your definition of the Löwenheim-Skolem number, shouldn't it be $\kappa\leq\text{max}(\kappa,|T|)$, rather than $<$? – Joel David Hamkins Jul 19 '15 at 12:20
• Strange. Wikipedia defines it as I suggest, and that definition makes sense to me: en.wikipedia.org/wiki/L%C3%B6wenheim_number#Extensions. – Joel David Hamkins Jul 20 '15 at 1:10
• @JoelDavidHamkins is right. You can't reasonably expect to get models of $T$ that are smaller than $|T|$. – Andreas Blass Jul 20 '15 at 15:10

Theorem 1. Is $\kappa$ is a strong cardinal, then $LS(L^2) < \kappa.$
The proof if easy. Let $T \subseteq L^2$ be a theory and let $A$ be a model for $T$. e may assume the universe is some cardinal $\delta.$ Take some cardinal $\beta > \beth_{\omega}(\delta)$, and let $j: V \to M$ witness $\kappa$ is $\beta$-strong. It is easily seen $M\models$$A \models T$'', so $M \models \exists B( B \models T, |B| < j(\kappa))$. By elementarity in $V$, $T$ has a model of size $< \kappa.$
Also note that for any theory $T \subseteq L^2,$ there is a least $\delta_T$ such that if $T$ has a model, then it has a model of size $< \delta_T.$ Then $LS(L^2)=\sup \{\delta_T: T$ as above $\}$, so $LS(T^2)$ can be singular, even though it can be above some very large cardinals.