Weakly homogeneous trees under AD

I hope it's okay to post an answer to my own question. I am essentially repeating Woodin's proof that he just showed me. Any errors were probably introduced by me.

Recall that under AD all measures on ordinals are countably complete and ordinal-definable. By the coding of measures theorem of Kechris assuming AD and that there is a Suslin cardinal above $\kappa$, there are fewer than $\Theta$ many measures on $\kappa$ (or $\kappa^{<\omega}$ for that matter.) So we may push forward Martin's measure on $\mathcal{P}_{\omega_1}(\mathbb{R})$ to get a fine, countably complete measure $U$ on $\mathcal{P}_{\omega_1}(\operatorname{meas}(\kappa^{<\omega}))$, where $\operatorname{meas}(\kappa^{<\omega})$ is the set of all measures on $\kappa^{<\omega}$.

Fix a tree $T$ on $\omega\times \kappa$. As usual, given a real $x \in \omega^\omega$ we let $T_x = \lbrace s \in \kappa^{<\omega} : (x \restriction |s|, s) \in T\rbrace$.

Claim: $U$-almost all $\sigma$ witness the weak homogeneity of $T$.

Proof: For each $\sigma$ we define a game $G_{\sigma}$, closed for Player I, for which Player I has a winning strategy iff $\sigma$ does not witness the weak homogeneity of $T$.

• I plays: $(x_0, \alpha_0, \beta_0)$, $(x_1, \alpha_1,\beta_1),\ldots$
• II plays: $\mu_0$, $\mu_1,\ldots$

Let $x$, $\vec{\alpha}$, $\vec{\beta}$, and $\vec{\mu}$ denote the resulting sequence of moves.

• Rules for I: $\vec{\alpha} \in [T_x]$ and the sequence $\vec{\beta} \in \mathrm{Ord}^\omega$ continuously witnesses that the tower $\vec{\mu}$ is illfounded.
• Rules for II: $\vec{\mu}$ is a tower of measures in $\sigma$ concentrating on $T_x$.

If both players follow the rules until the end, we say that player I wins.

If $\sigma$ does not witness that $T$ is weakly homogeneous, say $x \in p[T]$ but there is no wellfounded tower of measures in $\sigma$ concentrating on $T_x$, then there is a continuous witness to the illfoundedness of towers of measures in $\sigma$ concentrating on $T_x$. (This is proved by an argument similar to what follows but using a fine, countably complete measure on the set of subsets of $\kappa^{<\omega}$.)

So if $\sigma$ does not witness that $T$ is weakly homogeneous, then player I has a winning strategy in $G_\sigma$. (The $x$ and $\vec{\alpha}$ are fixed in advance and the $\vec{\beta}$ comes from $\vec{\mu}$ via the continuous witness mentioned above.) Assume toward a contradiction that for $U$-almost every $\sigma$, player I has a winning strategy in $G_\sigma$. The game is closed, the moves are ordinals and measures, and all measures are ordinal-definable, so for such $\sigma$ player I has a winning strategy $F(\sigma)$ of playing the least move leading to a subgame where he or she still has a winning strategy.

• Define the integer $x_0$ to be the one played by $F(\sigma)$ on the first turn for $U$-almost every $\sigma$.
• Define a measure $\mu_0$ on $\kappa$ by $A \in \mu_0 \iff \forall^*_U \sigma\; (\alpha^{\sigma}_0 \in A)$ where $\alpha^{\sigma}_0$ is the ordinal $\alpha_0$ played by $F(\sigma)$ on the first turn.
• Define the integer $x_1$ to be the one played by $F(\sigma)$ on the second turn against $\mu_0$ for $U$-almost every $\sigma$.
• Define a measure $\mu_1$ on $\kappa^2$ by $A \in \mu_1 \iff \forall^*_U \sigma\; ((\alpha^{\sigma}_0,\alpha^{\sigma}_1) \in A)$ where $\alpha^{\sigma}_1$ is the ordinal $\alpha_1$ played by $F(\sigma)$ against $\mu_0$ on the second turn.

Continuing in this way, we get a real $x \in \omega^\omega$ and a sequence of measures $\vec{\mu}$. One can easily check that $\vec{\mu}$ is a tower of measures. Each $\mu_i$ concentrates on $T_x$ because $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in T_x$. It is a wellfounded tower because if $A_i \in \mu_i$ for all $i<\omega$ then by countable completeness of $U$ there is a $\sigma$ such that $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in A_i$ for all $i<\omega$. However, by countable completeness of $U$ there is a $\sigma$ such that $\vec{\mu}$ is a legal play by player II against player I's winning strategy $F(\sigma)$, so player I's moves $\beta^\sigma_i$ continuously witness the illfoundedness of $\vec{\mu}$. Contradiction.

You indeed don't need $AD_{\mathbb{R}}$ and $\kappa < \Theta$ and it is correct this can be weakened. The argument was improved by Woodin : Assume $AD$ and assume that $\kappa$ is less than the supremum of the Suslin cardinals if there is one (this assumption is necessary) then every tree on $\omega \times \kappa$ is weakly homogeneous. (Note that if $\kappa$ itself is the largest Suslin then there is a problem: by Kechris $S(\kappa)$ would be non-selfdual and by a Martin-Solovay tree argument, the projection of tree on $\omega \times \kappa$ which is a $S(\kappa)$-complete set of reals can't be weakly homogeneously Suslin as then the complement of that set of reals would be Suslin.) I've never seen a source with a proof. Maybe you could ask Woodin?

Woodin has also proved the $ZFC$ counterpart to this theorem, namely that if $\kappa$ is a Woodin cardinal and if $T$ is a tree on $\omega \times \alpha$, $\alpha$ some ordinal, then there is a $\beta < \kappa$ such that in the generic extension $V[G]$, $G$ generic for $coll(\omega, \beta)$, $T$ is $<\kappa$ weakly homogeneous.