Universal $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set

Question

Does anyone know of a reference for the fact that if $\lambda$ is a limit of Woodin cardinals, then the pointclass $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ is $\omega$-parameterized? By this I mean that there is a $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $U \subset \omega \times \omega^\omega$ that is universal in the sense that every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $A \subset \omega^\omega$ is equal to the section $U_n$ for some $n < \omega$.

As far as I am concerned it would be enough to have a complete $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $U \subset \omega^\omega$ such that every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $A \subset \omega^\omega$ is equal to the pre-image $f^{-1}(C)$ by some recursive function $f$.

In either case, this seems not entirely trivial to me. Suppose that $A \in (\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$, say $$x \in A \iff \exists B \in \text{Hom}_{\mathord{<}\lambda}\,(HC; \in, A) \models \varphi[x].$$ It seems like we need to bound the complexity of the formulas $\varphi$ we are considering. One way to do this is to show that if there is a set $B \in \text{Hom}_{\mathord{<}\lambda}$ with $(HC; \in, B) \models \varphi[x]$ then there is also a set $B' \in \text{Hom}_{\mathord{<}\lambda}$ with $(HC; \in, B') \models \psi[x]$, where $\psi$ is the Skolem normal form of $\varphi$. We can get $B'$ using the fact that $\text{Hom}_{\mathord{<}\lambda}$ is projectively closed and every $\text{Hom}_{\mathord{<}\lambda}$ set admits a $\text{Hom}_{\mathord{<}\lambda}$ uniformization.

This might be more familiar in the equivalent context of $\mathsf{AD}^+$ where we would talk about $\Sigma^2_1$ rather than $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$, but I don't recall having seen any argument for it anywhere. Is there a published reference for this? Or perhaps an obvious argument that is simpler than the one I started to sketch above?

By the way, my motivation for asking this is the following. If $\lambda$ is a limit of Woodin cardinals and the derived model at $\lambda$ satisfies $\theta_0 < \Theta$ (or equivalently, every $\Pi^2_1$ set is Suslin) then every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set becomes $\lambda$-universally Baire in some $\mathord{<}\lambda$-generic extension $V[g]$. (More precisely, its re-interpretation in $V[g]$ becomes $\lambda$-universally Baire.) To get a single $\mathord{<}\lambda$-generic extension $V[g]$ in which every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set is universally Baire, it seems like the simplest way would be to consider a universal set.

Answer 1

I don't know if this answer will address your specific problem directly, but under AD, Wadge's Lemma implies that every non-selfdual pointclass has a universal set. There is an argument in Jackson's article in the Handbook (p.1761): basically if you have a $\Gamma$-complete set $A$, then define for every product space $X=X_0 \times X_1 \times ...\times X_n$ some sets $U_X \subseteq \omega^{\omega} \times X$ by $U_x(y, x_0,...,x_n) \leftrightarrow f_y(<x_0,...,x_n>) \in A$ where $y \in \mathbb{R}$ code real-valued Lipschitz continuous functions on $\mathbb{R}$, i.e $y$ computes the output value of $f_y(a_0,...,a_n)$ given the input $(a_0,...,a_n)$ by $(y(<a_0>),...,y(<a_0,...,a_n>))$. Then the sets $U_x$ are in $\Gamma$ because we assume $\Gamma$ is at least adequate (so we get recursive substitution) and whenever you take some set $B$ in $\Gamma$ then by Wadge's Lemma, it is Lipschitz reducible to $A$ so for some real $y$, $(U_X)_y=B$.

Now since $\Sigma^2_1$ is obviously adequate, and it is nonselfdual, then it has a universal set. I guess that makes it a Spector pointclass (it is normed, even has the scale property).