Recall that $T$ is said to be a $\kappa$-Kurepa tree if $T$ is a tree of height $\kappa$, whose levels $T_\alpha$ has size $\le|\alpha|$ for co-boundedly many $\alpha<\kappa$, and such that the set of branches of $T$ has size $>\kappa$.
Recall also that an uncountable cardinal $\kappa$ is said to be ineffable if for every sequence $\langle A_\alpha\mid\alpha<\kappa\rangle\in\prod_{\alpha<\kappa}\mathcal P(\alpha)$, there exists some $A\subseteq\kappa$ for which the set $\{ \alpha<\kappa\mid A\cap\alpha=A_\alpha\}$ is stationary. Note that every measurable cardinal is ineffable, and that every ineffable cardinal is weakly compact.
Let us point out that these concepts are mutually contradictory.
Proposition (Jensen-Kunen). If $\kappa$ is an ineffable cardinal, then there is no $\kappa$-Kurepa tree.
Proof. Suppose that $\langle \kappa,\le_T\rangle$ is a Kurepa tree, and let $\mathcal F$ denote the set of its branches. Then $|\mathcal F|>\kappa$, while $C:=\{\alpha<\kappa\mid \bigcup_{\beta<\alpha}T_\beta=\alpha\ \&\ |\{ b\cap\alpha\mid b\in\mathcal F\}|\le|\alpha|\}$ is a club in $\kappa$.
For all $\alpha\in C$, we let $\{ b^i_\alpha\mid i<\alpha\}$ be some enumeration of $\{ b\cap\alpha\mid b\in\mathcal F\}$, and then put $X_\alpha:=\bigcup_{i<\alpha}(\{i\}\times b_\alpha^i)$. Next, Fix a bijection $\pi:\kappa\times\kappa\leftrightarrow\kappa$, and define $$A_\alpha:=\begin{cases}\pi[X_\alpha]\cap\alpha,&\alpha\in C\\\emptyset,&\text{otherwise}\end{cases}.$$
Now, as $\kappa$ is ineffable, we may find some set $A\subseteq\kappa$ such that the following set is stationary: $$S=\{ \alpha<\kappa\mid A\cap\alpha=A_\alpha\}.$$
For all $i<\kappa$, denote $$b^i:=\{ \beta<\kappa\mid (i,\beta)\in \pi^{-1}[A]\}.$$ As $|\mathcal F|>\kappa$, let us fix some $b\in\mathcal F\setminus\{ b^i\mid i<\kappa\}$. If follows that we may define a function $f:\kappa\rightarrow\kappa$, by letting $$f(i):=\min\{\alpha<\kappa\mid b^i\cap\alpha\neq b\cap\alpha\}.$$
Since $S$ is stationary, we may fix an $\alpha\in S\cap C$ such that $\pi[\alpha\times\alpha]=\alpha$ and $f[\alpha]\subseteq\alpha$. Let $i<\alpha$ be such that $b^i_\alpha=b\cap\alpha$. By the choice of $\alpha$ and since $\pi$ is a bijection, we get that $$\pi^{-1}[A]\cap(\alpha\times\alpha)=\pi^{-1}[A\cap\alpha]=\pi^{-1}[A_\alpha]=\pi^{-1}[\pi[X_\alpha]\cap\pi[\alpha\times\alpha]]=X_\alpha.$$In particular, $$b^i\cap\alpha=\{\beta<\kappa\mid (i,\beta)\in X_\alpha\}=\{\beta<\kappa\mid (i,\beta)\in\bigcup_{j<\alpha}(\{j\}\times b^j_\alpha)\}=b^i_\alpha.$$Altogether, we got that $b^i\cap\alpha=b\cap\alpha$, contradicting the fact that $f(i)<\alpha$. $\square$
Assaf, what you call a $kappa$-Kurepa tree is what I know as a *slim* $kappa$-Kurepa tree. The forcing to create a slim $kappa$-Kurepa tree is ${<}kappa$-closed, but as you point out, definitely destroys the measurability of $kappa$. This shows that in Laver's indestructibilty theorem, where a supercompact cardinal $kappa$ is made indestructible by all ${<}kappa$-directed closed forcing, one cannot improve it to indestructibility by all ${<}kappa$-closed forcing.
Thanks for this info, Joel! I wonder what is the more general definition of a $kappa$-Kurepa tree?
The more general definition that I have seen is just a $kappa$ tree (meaning all levels have size less than $kappa$) with at least $kappa^+$ many branches. This is trivial when $kappa$ is a strong limit, for example, since the full binary tree has the property, and this may be the reason you dont’ consider it. The slimness property I know is essentially the extra property you mention, that the $alpha$-th level should have size $alpha$, when $alpha$ is infinite (slight difference from what you say: except for boundedly many $alpha$).
Indeed, any definition that omits a slimness condition is likely to trivialize the matter.
Furthermore, Thomas Johnstone, Vika Gitman and I proved that one may make any strongly unfoldable cardinal indestructible by the forcing to add a slim $kappa$-Kurepa tree. (This was generalized in later work: see http://jdh.hamkins.org/indestructiblestrongunfoldability/.) So one can have a strongly unfoldable cardinal with a slim $kappa$-Kurepa tree, and this provides a bound on improvements to the theorem you mention.
That’s very interesting!
Do you perhaps know of an upper bound for the non-existence of an $aleph_{omega+1}$-Kurepa tree? Dima Sinapova and myself realized this week that a supercompact cardinal and an almost huge cardinal above it, suffices. (This follows from Foreman’s theorem 4.11 in here.)
I assume you mean the $aleph_omega$ and not $aleph_{omega+1}$.
Thanks, Mohammad. I did mean $aleph_{omega+1}$, but later indeed realized that the classic Silver argument will do the job.
Yes, that was why I asked the questio, as the ordinary Levy collapse works foe successor cardinals.