*This was previously [asked at MSE](https://math.stackexchange.com/questions/3658513/does-having-many-models-yield-complex-second-order-theories) without success.*

***

Suppose $T$ is a complete **first-order** theory with continuum-many countable models up to isomorphism. We define two sets of Turing degrees associated to $T$ via second-order logic:

 - $SecTh(T)$ is the set of Turing degrees of **second-order** theories of models of $T$.

 - $SecTh_0(T)$ is the set of Turing degrees of **second-order** theories of *countable* models of $T$.

I'm interested in how simple these sets can be. Specifically:

> Is there a $T$ such that $SecTh(T)$ is not cofinal in the Turing degrees? If not, what about $SecTh_0(T)$?

Recall that $X$ is cofinal in the Turing degrees if every Turing degree is below some element of $X$.

****

Here's what I've been able to figure out already:

 - Assuming $V=L$, the answer is negative for $SecTh_0$ (and hence $SecTh$ a fortiori). The key point is that  under $V=L$ the set of [pointwise-definable](https://arxiv.org/abs/1105.4597) levels of $L$ is unbounded in $\omega_1$. Given a countable $\mathcal{A}\models T$, let $\alpha_\mathcal{A}$ be the least index of a pointwise-definable level of $L$ containing an isomorphic copy of $\mathcal{A}$. The second-order theory of $\mathcal{A}$ computes the first-order theory of $L_{\alpha_\mathcal{A}}$, which in turn computes every real in $L_{\alpha_\mathcal{A}}$ by pointwise definability. Now simply use the fact that $\{\alpha_\mathcal{A}:\mathcal{A}\models T\}$ is unbounded in $\omega_1$. (Note that this is a "naive" version of the idea behind [mastercodes](https://projecteuclid.org/euclid.jsl/1183740554).)

 - I don't actually see an immediate argument that $SecTh(T)$ need be uncountable! A priori, $Th_2(\mathcal{A})$ may not be enough to build a concrete copy of $\mathcal{A}$ in any sense that I can see. It is *consistent* that $SecTh_0(T)$ (and hence $SecTh(T)$) is always uncountable, since it's consistent that no two countable structures are second-order elementarily equivalent (see Marek's theorem mentioned [here](https://mathoverflow.net/a/161694/8133)), but that's the best I know. 

 - Under sufficiently strong large cardinal hypotheses, $SecTh_0(T)$ is subject to [Martin's Cone Theorem](https://cstheory.stackexchange.com/a/46145/52244), and hence for an affirmative answer for the second question it would be enough to find a $T$ such that $SecTh_0(T)$ does not contain an upper cone. Under such hypotheses *and* replacing $\mathsf{ZFC}$ with $\mathsf{ZF}$ + Turing Determinacy the same would hold for $SecTh(T)$, but I don't immediately see how to get this stronger result without determinacy.