If AD$\_\mathbb{R}$ holds and $\kappa < \Theta$ then every tree $T$ on $\kappa$ is weakly homogeneous (Martin–Woodin, "Weakly homogeneous trees.") I recall hearing that the hypothesis can be weakened to "AD holds and there is a Suslin cardinal above $\kappa$." Is this correct? If so, does anyone know where a proof can be found, or have the idea of the proof?


It may help to keep in mind that the main consequence of $AD\_\mathbb{R} + \kappa < \Theta$ used in Martin–Woodin is the existence of a normal fine measure on $\mathcal{P}\_{\omega\_1}(\mathcal{P}(\kappa^{<\omega}) \cup \mathrm{meas}(\kappa^{<\omega}))$, where $\mathrm{meas}(\kappa^{<\omega})$ denotes the set of countably complete measures on $\kappa^{<\omega}$ (equivalently, on $\kappa^n$ for some $n<\omega$.)


In other words, the proof uses that $\omega\_1$ is $\mathcal{P}(\kappa^{<\omega}) \cup \mathrm{meas}(\kappa^{<\omega})$-supercompact. If we only assume "AD holds and there is a Suslin cardinal above $\kappa$" then because the measures are essentially ordinals one can show that $\omega\_1$ is $\mathrm{meas}(\kappa^{<\omega})$-supercompact, but I see no way to show that $\omega\_1$ is $\mathcal{P}(\kappa^{<\omega})$-supercompact. So perhaps a more substantial modification to the Martin–Woodin proof is required.



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.


