TITLE: Is the fusion argument on trees of uncountable height consistent?
QUESTION [8 upvotes]: In the countable context where we are given a perfect subtree $T$ of $2^{<\omega}$ and a sequence of colorings $f_i: T\to 2, i\in \omega$, it is possible to obtain a perfect subtree $T'\subset T$ and an infinite set of levels $A\in [\omega]^\omega$ such that for each $i\in \omega$ there exists $n_i\in A$ such that for all $t\in T$ with $ht(t)\in A\backslash (n_i+1)$, $f_i(t)=f_i(t\restriction n_i)$.
This is easily obtainable via 1-dimensional Halpern-Läuchli theorem (https://en.wikipedia.org/wiki/Halpern–Läuchli_theorem) along with a typical fusion argument.
My question is: is the generalization to trees on larger cardinals consistent?
More precisely, is the following consistent:
Let $\kappa$ be at least an inaccessible cardinal.
Given a perfect $<\kappa$-closed subtree $T$ of $2^{<\kappa}$ and a sequence of colorings $f_i: T\to 2, i\in \kappa$, it is possible to obtain a perfect $<\kappa$-closed subtree $T'\subset T$ and a set of levels $A\in [\kappa]^\kappa$ such that for each $i\in \kappa$ there exists $n_i\in A$ such that for all $t\in T$ with $ht(t)\in A\backslash (n_i+1)$, $f_i(t)=f_i(t\restriction n_i)$.
Some remark:

When the number of colorings is $<\kappa$, this is true (in fact stronger consequence is true, namely the exists a perfect subtree that is simultaneously homogeneous for all the colorings)
It seems the naïve generalization of classical fusion argument will just fail at the limit stage.

Any thoughts will be appreciated.

REPLY [2 votes]: The answer is yes. In fact, more general statements are true. See https://arxiv.org/abs/1704.06827 for more detail (in particular Theorem 3.1).