TITLE: Resembling the Levy Collapse
QUESTION [10 upvotes]: Suppose $\kappa$ is a weakly compact cardinal.  Is there a $\kappa$-c.c. forcing $\mathbb{P}$ such that $\mathbb{P} \subseteq V_\kappa$ and $\Vdash_{\mathbb{P}} \kappa = \aleph_1$, where $\mathbb{P}$ is provably NOT equivalent to the Levy collapse $Col(\omega,<\kappa)$?
I ask this because if $\kappa$ is weakly compact and $\mathbb{P}$ has the above properties, then one can prove:
a) There are stationarily many $\alpha < \kappa$ such that $\mathbb{P} \cap V_\alpha$ is a regular suborder of $\mathbb{P}$.
b) There are unboundedly many $\alpha < \kappa$ such that $\mathbb{P} \cap V_\alpha$ is equivalent to $Col(\omega,\alpha)$.
c) If $G$ is $\mathbb{P}$-generic over $V$, then there is a further forcing $\mathbb{Q}$ such that if $H \subseteq \mathbb{Q}$ is generic, then in $V[G][H]$, there is $G' \subseteq Col(\omega,<\kappa)$ which is generic over $V$, and $\mathbb{R}^{V[G]} = \mathbb{R}^{V[G']}$.
So in some sense $\mathbb{P}$ resembles $Col(\omega,<\kappa)$.

REPLY [2 votes]: Here is a different example: Let $\lambda<\kappa$ be regular uncountable. Then (perhaps surprisingly?) $\mathrm{Col}(\omega,{<}\kappa)$ and $\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\kappa)$ are not forcing equivalent. This is a consequence of the following fact:
(Let us write $\mathbb B\lessdot\mathbb C$ for $\mathbb B$ is a regular subalgebra of $\mathbb C$.)

Suppose $\mathbb B$ is a $\kappa$-cc atomless cBa that can be written as
$\mathbb B=\bigcup_{\alpha<\kappa}\mathbb B_\alpha$ so that

$\vert \mathbb B_\alpha\vert<\kappa$ for $\alpha<\kappa$.
$\mathbb B_\alpha\lessdot\mathbb B_\beta$ for $\alpha<\beta<\kappa$.
Any $\mu<\kappa$ is countable in $V^{\mathbb B_\alpha}$ for some $\alpha<\kappa$.

Then $\mathbb B\cong\mathrm{RO}(\mathrm{Col}(\omega,{<}\kappa))$ iff
$\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb B_\beta\lessdot\mathbb B\}$
contains a club.

We can write $\mathbb B=\mathrm{RO}(\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\kappa))$ as $\bigcup_{\alpha<\kappa}\mathbb B_\alpha$ with
$$\mathbb B_\alpha=\mathrm{RO}(\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\alpha))$$
and this satisfies 1.-3. above. However,
$$\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb B_\beta\lessdot\mathbb B\}\cap\mathrm{Lim}=\{\alpha<\kappa\mid\mathrm{cof}(\alpha)\geq\lambda\}$$
is costationary in $\kappa$ and hence $\mathbb B$ is not isomorphic to $\mathrm{RO}(\mathrm{Col}(\omega,{<}\kappa))$. The homogeneity of these two Boolean algebras implies that they are not forcing equivalent either.
This inequivalence here is an inconvenience that indeed comes up in practice. For example there is a difficult argument due to Woodin that shows, among other things, that one can force a dense ideal on $\omega_1$ from an almost huge cardinal (see Theorem 7.60 in Foreman's handbook article). The idea is to turn $\kappa$ into $\omega_1$ and $j(\kappa)$ (where $j$ is an embedding witnessing that $\kappa$ is almost huge) into $\omega_2$ and finally to show that a lift of $j$ exists after collapsing $\kappa$. To make that work, Woodin drops to a symmetric submodel of a forcing extension. In the inconsistent world where $\mathrm{Col}(\omega, {{<}j(\kappa)})\cong\mathrm{Col}(\omega, \kappa)\times\mathrm{Col}(\kappa, {<}j(\kappa))$ this would not be necessary and the easier naive approach would go through.