Does $Π^1_{2n+1} = (Σ^1_{2n+1})^{M_{2n-1}}$?

Question

Every $Π^1_1$ formula $φ$ without free second order variables can be converted into a $Σ^1_1$ $ψ$ such that $φ ⇔ ψ^\mathrm{HYP}$, and vice versa. ($\mathrm{HYP}$ is the hyperarithmetical universe, which can be identified with $L_{ω_1^\mathrm{CK}}$.) Essentially, arbitrary $Π^1_1$ statement $⇔$ well-foundedness of some recursive relation '$≺$' $⇔$ existence of a hyperarithmetical set iterating the Turing jump along '$≺$' $⇔$ arbitrary $(Σ^1_1)^\mathrm{HYP}$ statement.

My question is whether, assuming projective determinacy, an analogous correspondence holds for $Π^1_{2n+1}$; and I conjecture $Π^1_{2n+1} = (Σ^1_{2n+1})^{M_{2n-1}}$.

$M_n$ is the minimal iterable inner model with $n$ Woodin cardinals. $M_0$ is the constructible universe $L$, and $M_{-1}$ (not an inner model) would be $L_{ω_1^{\mathrm{CK}}}$. The reals in $M_n$ are precisely those that are $Δ^1_{n+2}$ in a countable ordinal. For even $n$, $M_n$ is $Σ^1_{n+2}$ correct, but this is not the case for odd $n$.

A positive answer to the question should enhance our understanding of $Π^1_{2n+1}$ prewellordering and uniformization. True $Σ^1_{2n}$ statements can be 'graded' (assuming projective determinacy) by the complexity of the least witness. For example, consider a theory $T$ such as ZFC + "there is a supercompact cardinal", and assume that $T$ has sufficiently sound models. At the $Σ^1_2$ level, we can assign $T$ an ordinal based on the least height of a transitive model of $T$; furthermore, there is a $Δ^1_2$ example of such a model. This generalizes to $Σ^1_{2n}$ and models of $T$ that are closed under $M_{2n-3}^\#$. But what kind of witnesses do we have for $Π^1_{2n+1}$ statements?

The first paragraph gives an answer for $Π^1_1$, and we can generalize it as follows. Assuming $M_{2n+1}^\#$ exists, a $Π^1_{2n+3}$ statement $T$ is true iff there is an iterable model $M$ of ZFC + "$2n+1$ Woodin cardinals" such that $M^{\mathrm{Coll}(ω,δ)}⊨T$ where $δ$ is the least Woodin cardinal in $M$. (The use of ZFC in $M$ is essentially arbitrary; also, genericity iterations allow Woodin cardinals to 'absorb' real quantifiers.) Presumably, such an $M$ can be chosen to be $Δ^1_{2n+3}$ and its existence is $(Σ^1_{2n+3})^{M_{2n+1}}$ (but how?) (Also, the least complexity of such $M$ should be connected to $Π^1_{2n+3}$ prewellordering, but the exact connection is unclear to me as the prewellordering is about sets of reals.)

Answer 1

Your conjecture is true. We assume PD throughout. The proof I see requires citing a number of facts from inner model theory and descriptive set theory. First, it uses Woodin's theorem characterizing the reals of $M_{2n-1}$ as the set $Q_{2n+1}$ of points in Baire space that are $\Delta^1_{2n+1}$ definable from a countable ordinal. In symbols:

Theorem (Woodin). $\omega^\omega\cap M_{2n-1} = Q_{2n+1}$.

It uses a correctness theorem for the odd levels:

Theorem $M_{2n-1}$ is $\Pi^1_{2n}$-correct.

A set $A\subseteq \omega^\omega$ is $\Pi^1_{2n+1}$-bounded if $\Pi^1_{2n+1} = \exists^{A} \Pi^1_{2n+1}$. We need that $Q_{2n+1}$ is $\Pi^1_{2n+1}$-bounded. In fact, something stronger is true (see Kechris-Martin-Solovay's "Introduction to $Q$-theory"):

Theorem (Kechris-Martin-Solovay). $Q_{2n+1}$ is the largest $\Pi^1_{2n+1}$-bounded subset of Baire space.

We need Moschovakis's "Spector-Gandy theorem for the odd levels" (Moschovakis, Descriptive Set Theory, 6E.7):

Theorem (Moschovakis) $\Pi^1_{2n+1}\cap\omega^\omega = \exists^{\Delta^1_{2n+1}\cap \omega^\omega}\Pi^1_{2n}\cap \omega^\omega$.

Moschovakis's theorem will actually be applied to $Q_{2n+1}$ using the $Q$-Theory Reflection Theorem:

Theorem (Kechris-Martin-Solovay) If $A\subseteq \omega^\omega$ is $\Pi^1_{2n+1}$, then $\exists x\in \Delta^1_{2n+1}\ A(x)$ if and only if $\exists x\in Q_{2n+1}\ A(x)$.

Given these facts, the calculation becomes a straightforward pointclass calculation.

By definition, $\Sigma^1_{2n+1} = \exists^{\omega^\omega}\Pi^1_{2n}$, so $(\Sigma^1_{2n+1})^{M_{2n-1}} = \exists^{\omega^\omega\cap M_{2n-1}}(\Pi^1_{2n})^{M_{2n-1}}$.

Woodin's theorem characterizing $\Pi^1_{2n+1}$ along with the $\Pi^1_{2n}$-correctness of $M_{2n-1}$ imply $\exists^{\omega^\omega\cap M_{2n-1}}(\Pi^1_{2n})^{M_{2n-1}}\cap\omega^\omega = \exists^{Q_{2n+1}}\Pi^1_{2n} \cap\omega^\omega$.

Moschovakis's Spector-Gandy Theorem along with the $Q$-Theory Reflection Theorem yields that $\exists^{Q_{2n+1}}\Pi^1_{2n}\cap \omega^\omega = \exists^{\Delta^1_{2n+1}\cap\omega^\omega}\Pi^1_{2n}\cap \omega^\omega = \Pi^1_{2n+1}\cap \omega^\omega$.

Stringing together a bunch of pointclass identities, one can conclude that $(\Sigma^1_{2n+1})^{M_{2n-1}}\mathrel{\cap} \omega^\omega = \Pi^1_{2n+1} \cap \omega^\omega$.

You might also want to look at Theorem 4.12 of John Steel's paper Projectively Well-Ordered Inner Models. I think you can use the proof to get a more inner model theoretic proof of your conjecture, but Steel's result is closely related and of independent interest.