TITLE: On the independence of the Kurepa Hypothesis
QUESTION [5 upvotes]: Kurepa Hypothesis says there is a Kurepa tree, which is a $\omega_1$-tree has at least $\omega_2$ many branches.
It is known that beginning from a model with an inaccessible cardinal $\kappa$, after collapes $\kappa$ to $\omega_2$ using the Levy collape, then in the generic extension, Kurepa Hypothesis fails.
In above generic extension, $\omega_2$ is equal to $\kappa$ and by a counting argument for nice names, $2^{\omega_1}=\omega_2$. My question is that "is it consistent that Kurepa Hypothesis fails and $2^{\omega_1}>\omega_2$?"
(The reason I think this question: The biggest possible value of the number of branches is $2^{\omega_1}$, so in the environment of $2^{\omega_1}=\omega_2$, it is most difficult for the living of a Kurepa tree. So I want to know whether this requiement is necessary.)

REPLY [2 votes]: Another solution can be obtained by adding many reals and using a lemma of Spencer Unger that generalizes the lemma used by Silver.

Lemma (Unger) Suppose $\mu \leq \kappa$ are regular, there is $\tau \leq \mu$ such that $2^\tau \geq \kappa$, and $\mathbb{P}$ is $\mu$-c.c. and $\mathbb{Q}$ is $\mu$-closed.  Then $\Vdash_\mathbb{P} \mathbb{Q}$ does not add branches to $\kappa$-trees.

MR2945572 Unger, Spencer. Fragility and indestructibility of the tree property. Arch. Math. Logic 51 (2012), no. 5-6, 635–645. 
Start with Silver's model of $\neg KH$, in which an inaccessible $\theta$ is Levy collapsed to $\omega_2$ by $\mathbb{Q} = Col(\omega_1,<\kappa)$.  Then add at least $\omega_3$ many Cohen or Random reals, call that $\mathbb{P}$.  Any $\omega_1$-tree $T$ in $V^{\mathbb{Q} \times \mathbb{P}}$ is captured in some intermediate extension by a subforcing $\mathbb{Q}_0 \times \mathbb{P}_0$ of size $<\kappa$, where $\mathbb{Q}_0$ is rank initial segment of $\mathbb{Q}$.  By the well known fact about Knaster posets $\mathbb{P} / \mathbb{P}_0$ adds no branches to $T$, and by Unger's lemma, $\mathbb{Q} / \mathbb{Q}_0$ does not add branches over $V^{\mathbb{P} \times \mathbb{Q}_0}$.  Finally, since $\kappa$ is inaccessible in $V$, the number of branches of $T$ in $V^{\mathbb{Q}_0 \times \mathbb{P}_0}$ is $<\kappa$.