Question

If $\kappa$ is an inaccessible cardinal then the tree property at $\kappa$ is equivalent to weak compactness of $\kappa$, which implies that $\square(\kappa)$ fails---that is, that every coherent sequence of clubs of length $\kappa$ can be threaded.

I am wondering about other implications involving square and the tree property, namely:

• If $\kappa$ is an inaccessible cardinal and $\square(\kappa)$ fails, must $\kappa$ have the tree property (and therefore be weakly compact?)

• If $\kappa$ is a regular cardinal, does $\neg \square(\kappa)$ imply that $\kappa$ has the tree property?

• If $\kappa$ is a regular cardinal with the tree property, must $\square(\kappa)$ fail?

(By the way, I am aware of the relative consistency result that if $\square(\kappa)$ fails for some regular cardinal $\kappa$, then $\kappa$ is weakly compact in $L$ and so in particular it has the tree property in $L$.)

Answer 1

The answers to the second and third questions are no and yes, respectively. I don't know the answer to the first question.

For the second question, let $\lambda$ be regular and let $\kappa > \lambda$ be weakly compact. Then forcing with $\mathrm{Coll}(\lambda, <\kappa$) yields a model in which $\kappa = \lambda^+$, $\square(\kappa)$ fails, and, since $\lambda^{<\lambda}=\lambda$, there is a special $\kappa$-Aronszajn tree, so the tree property fails.

For the third question, the usual construction of a special Aronszajn tree from a weak square sequence using minimal walks (see, for example, section 5.1 of Cummings' "Notes on Singular Cardinal Combinatorics") still yields a $\kappa$-Aronszajn tree when applied to a $\square(\kappa)$-sequence when $\kappa$ is regular, so $\square(\kappa)$ implies the failure of the tree property.

Answer 2

To complete the accepted answer to my question (which addresses parts 2 and 3) perhaps I should mention here that the answer to part 1 is no: If $\delta$ is a supercompact cardinal and $\kappa$ is the least inaccessible cardinal above $\delta$ then $\square(\kappa)$ fails but $\kappa$ is not weakly compact.