# Souslin trees and weakly compact cardinals

In Souslin trees on the first inaccessible cardinal it is asked if it is consistent that there are no $\kappa-$Souslin trees at the least inaccessible cardinal $\kappa$.

In this question I would like to ask an apparently simpler question:

Question. Is it consistent that there are no $\kappa-$Souslin trees for some inaccessible not weakly compact cardinal $\kappa$?

Remarks. (A) If $\kappa$ is weakly compact, then the tree property holds at $\kappa$, in particular there are no $\kappa-$Souslin trees,

(B) If $V=L,$ then $\kappa$ is weakly compact iff there are no $\kappa-$Souslin trees.

(C) $\Diamond_\kappa+GCH$ does not imply the existence of a $\kappa-$Souslin tree, for $\kappa$ inaccessible. For example if $\kappa$ is ineffable and $GCH$ holds, then $\Diamond_\kappa+GCH$ holds and there are no $\kappa-$Souslin trees.

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What happens if we start with a weakly compact cardinal $\kappa$ in $V$, and then force to kill the Mahloness of $\kappa$ by adding club $C\subset\kappa$ avoiding the regular cardinals below $\kappa$? Does this create a $\kappa$-Suslin tree? –  Joel David Hamkins Mar 30 '14 at 13:51
I think it does not give the answer. Note that a sufficient condition for having a $\kappa-$Souslin tree for $\kappa$ inaccessible, is square-with-built-in-diamond at $\kappa$, as is implicit in [Sh:624]. A weaker sufficient conditions is, e.g.,: there exists a $\square(\kappa)-$sequence $< C_\alpha: \alpha<\kappa >$, and a stationary set $S$ in $\kappa$, which is disjoint from $acc(C_\alpha)$ for all $\alpha<\kappa$, and $\Diamond(S)$ holds. To force the failure of these principles, we need more than weakly compact cardinals. –  Mohammad Golshani Mar 30 '14 at 14:10
Does the existence of a $\kappa-$Souslin tree for $\kappa$ inaccessible have any effects on saturated ideals on $\kappa$ or something else? Maybe this way we can reduce the problem to something that we know more about it. –  Mohammad Golshani Mar 30 '14 at 14:15
If there is a $\kappa$-closed $\kappa$ saturated ideal on $\kappa$ and $\kappa$ is not measurable then $\kappa$ is inaccessible and there is a $\kappa$-Suslin tree (by splitting the positive sets in a partial binary tree). Actually, in this case every normal $\kappa$-tree contains a Suslin subtree. –  Yair Hayut Mar 31 '14 at 17:54
@Mohammad: you don't actually need a $\square (\kappa )$ sequence, all you need is a non reflecting stationary set $S$ with diamond on it. –  Yair Hayut May 22 '14 at 17:49