# Are the following two "tree properties" equivalent?

Let $$\kappa$$ and $$\lambda$$ be cardinals. A thin $$(\kappa,\lambda)$$-list is a function $$L:[\lambda]^{<\kappa}\longrightarrow [\lambda]^{<\kappa}$$ such that for all $$x\in[\lambda]^{<\kappa}$$, $$L(x)\subseteq x$$ and $$\{L(y)\;|\;y\subseteq x\}$$ has cardinality $$<\kappa$$.

Say that $$(\kappa,\lambda)$$-STP holds iff whenever $$L$$ is a thin $$(\kappa,\lambda)$$-list, there is some $$b\subseteq\lambda$$ such that $$\{x\;|\;x\cap b=L(x)\}$$ is stationary and that $$\kappa$$ satisfies the super tree property iff $$(\kappa,\lambda)$$-STP holds for all $$\lambda\geq\kappa$$.

Say that $$(\kappa,\lambda)$$-SSTP holds iff for every sequence $$(L_{\alpha})_{\alpha\in\mu}$$ (where $$\mu<\kappa$$), if every $$L_{\alpha}$$ is a thin $$(\kappa,\lambda)$$-list, there is a sequence $$(b_{\alpha})_{\alpha\in\mu}$$ such that $$\{x\;|\;\forall\alpha\in\mu(b_{\alpha}\cap x=L_{\alpha}(x))\}$$ is stationary. Say that $$\kappa$$ satisfies the simultaneous super tree property iff $$(\kappa,\lambda)$$-SSTP holds for all $$\lambda\geq\kappa$$.

Let $$\kappa$$ be supercompact, $$\lambda\geq\kappa$$ and $$U$$ a normal $$\kappa$$-complete ultrafilter on $$[\lambda]^{<\kappa}$$. For a thin $$(\kappa,\lambda)$$-List $$L$$ one can show that there is some $$b$$ such that $$\{x\;|\;x\cap b=L(x)\}\in U$$, therefore, using $$\kappa$$-completeness, we have that supercompact cardinals satisfy the simultaneous super tree property, so the simultaneous super tree property is consistent, at least modulo the existence of a supercompact cardinal (and I suspect $$\omega_2$$ has the simultaneous super tree property if PFA holds). This begs the following questions:

1. Are the principles $$(\kappa,\lambda)$$-STP and $$(\kappa,\lambda)$$-SSTP equivalent for all $$\lambda$$ (or some fixed $$\lambda$$ depending on $$\kappa$$)?
2. Is the super tree property equivalent to the simultaneous super tree property?

The $$(\kappa,\lambda)$$-STP and $$(\kappa,\lambda)$$-SSTP are equivalent for any uncountable cardinals $$\kappa\leq\lambda$$: Let $$\mu<\kappa$$ and $$(L_\gamma)_{\gamma<\mu}$$ a sequence of thin $$(\kappa,\lambda)$$-lists. We can then "amalgamate" these lists into one list $$L$$ as follows: Let $$h:\lambda\times\mu\rightarrow\lambda$$ be a bijection. Let $$C=\{x\in[\lambda]^{<\kappa}\mid \mu\subseteq x\wedge\ x \text{ is closed under both } h \text{ and }h^{-1}\}$$ Then $$C$$ is club in $$[\lambda]^{<\kappa}$$ so for all our intents and purposes it is enough to define $$L$$ on $$C$$. $$L$$ puts the information of $$L_\gamma(x)$$ onto the "$$\gamma$$th slice" of $$x$$, i.e. for all $$\beta, \gamma$$: $$h(\beta,\gamma)\in L(x)\Leftrightarrow \beta\in L_\gamma(x)$$ If $$x\in C$$ then indeed $$L(x)\subseteq x$$ and every $$L_\gamma(x)$$, $$\gamma<\kappa$$ is coded into $$L(x)$$ in a uniform way. Using $$\mu<\kappa$$ it should be easy to see that $$L$$ is a thin $$(\kappa,\lambda)$$-list (we need here that $$\kappa$$ is inaccessible but this follows from $$(\kappa, \lambda)$$-STP, see the remark at the end).
If $$(\kappa,\lambda)$$-STP holds true, there is $$b\subseteq \lambda$$ and $$S\subseteq C$$ stationary with $$L(x)=x\cap b$$ for all $$x\in S$$. Now for $$\gamma<\mu$$ define $$b_\gamma$$ as the "$$\gamma$$th slice" of $$b$$, i.e.: $$b_\gamma=\{\beta<\lambda\mid h(\beta,\gamma)\in b\}$$ We get that for $$x\in S$$ and $$\beta\in x$$, $$\gamma<\mu$$: $$\beta\in x\cap b_\gamma\Leftrightarrow h(\beta,\gamma)\in x\cap b\Leftrightarrow h(\beta, \gamma)\in L(x)\Leftrightarrow \beta\in L_\gamma(x)$$ So $$L_\gamma(x)=x\cap b_\gamma$$.
It is also worth noting that $$(\kappa, \kappa)$$-STP (and $$(\kappa, \kappa)$$-SSTP for that matter) are both equivalent to $$\kappa$$ being ineffable.