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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.

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I think that to show that $\omega_1$ is $\mathcal{P}(\kappa^{<\omega})$-supercompact we need the "strong coding lemma" of Woodin. –  Carlo Von Schnitzel Jul 24 '12 at 1:26
    
@alephomega: Could you elaborate? The only strong coding lemma that I can find reference to is listed as an open problem in the Koellner–Woodin handbook chapter. –  Trevor Wilson Jul 24 '12 at 2:16
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2 Answers

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.

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Thanks, but I'm going to wait a bit before accepting to see if anyone can answer the second question. –  Trevor Wilson Jul 22 '12 at 13:37
    
No problem. I am actually curious myself now, I would like to see a proof of this fact. –  Carlo Von Schnitzel Jul 22 '12 at 14:25
    
By the way, I think you mean "by Kechris $S(\kappa)$ would be non-selfdual." –  Trevor Wilson Jul 22 '12 at 16:35
    
Good catch, thanks. –  Carlo Von Schnitzel Jul 22 '12 at 23:31
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up vote 3 down vote accepted

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 by Turing determinacy we get a fine, countably complete measure $U$ on the set of 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.

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