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.