# Side condition forcing and tree property

Starting from a weakly compact cardinal $\kappa$ one can use Neeman's side condition forcing to obtain the tree property at $\aleph_2$. Also It can be obtained using other side condition constructions such as Velickovic's method. By the way in the final model $\kappa$ is $\aleph_2$ and $2^{\aleph_0}=\aleph_2$, I may note that all new reals are Cohen.

Although it is possible to obtain tree property at $\aleph_2$ and $\aleph_3$ simultaneously, but there is no hope(at the moment and as far as I know) using forcing with side condition to generalize it to more than two cardinals, and this is because of the closure properties of the forcing with side conditions which we don't have at the moment. One possible way (at least in my opinion) is to compare the forcing with Mitchell's poset originally used in his thesis. So generally speaking my question is what kind of relations are known about the generic objects of Mitchell's poset and forcing with side condition $\mathbb{P}_\kappa$ used by Neeman. More precisely I am interested in:

Suppose that $\kappa$ is inaccessible and $\alpha<\kappa$ uncountable ordinal. Is there any projection $\pi_\alpha:\mathbb{P}_{\kappa}\longrightarrow Add(\omega,\alpha)\ast A\dot{dd}(\omega_1,1)$?

• The question needs some clarification, e.g. when you say "but there is no hope(at the moment and as far as I know) to generalize it to more than two cardinals", you mean using side-condition-type posets, right? Abraham got the Tree Property for all $\aleph_n$ ($n \ge 2$) using Mitchell-type posets. – Sean Cox Jan 9 '17 at 16:45

I think the answer is no, but there are some technical issues that I don't see how to solve at the moment. Let me instead 1) use Krueger's side condition forcing instead of Neeman's; and 2) work with regular embeddings (a.k.a. complete embeddings) in the other direction, instead of projections. Some remarks about these simplifications appear at the end.

CLAIM: Let $P$ denote Krueger's side condition forcing ("adequate set forcing"). Then there is no regular embedding from $\text{Add}(\omega,\alpha) * \text{Add}(\omega_1,1)$ into $P$.

PROOF: Suppose toward a contradiction that there exists a regular embedding $e: \text{Add}(\omega,\alpha) * \text{Add}(\omega_1,1) \to P$. Let $G*H$ be generic for $\text{Add}(\omega,\alpha) * \text{Add}(\omega_1,1)$, and let $K$ be generic over $V[G*H]$ for the quotient $P/e[ G*H]$. By my first joint paper with Krueger (Quotients of strongly proper forcings) the quotient $P/e[G]$ is strongly proper on a stationary set of countable models; this uses a nice property of Krueger's poset which we call the * property (see bottom of page 6 of that paper; it basically reqires that the poset is well-met, and that if 3 conditions are pairwise compatible, then there is a lower bound for all 3). Since $P/G$ is strongly proper for a stationary set of countable models, then the pair $(V[G], V[K])$ has the $\omega_1$-approximation property. But $H$ adds a new subset of $\omega_1$ all of whose proper initial segments are in $V[G]$. Since $V[G*H] \subset V[K]$, this is a contradiction to the fact that the pair $(V[G], V[K])$ has the $\omega_1$-approximation property.

REMARK 1: Regarding the use of Krueger's poset instead of Neeman's: The * property is easy to check for Krueger's poset, but I don't know if it holds for Neeman's poset. If so, then the same proof would work for Neeman's poset too. Note: our theorem about quotients of strongly proper forcings isn't true if the * assumption is omitted; see Section 5 of that paper for a counterexample (in fact the quotient of the counterexample even fails to have the $\omega_1$-approx property). Though there are probably other alternatives to * that would suffice to get strongly proper quotients.

REMARK 2: There can be no meet-preserving" forcing projection $\pi:P \to \text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$ (by meet-preserving I mean $\pi(p_0 \wedge p_1) = \pi(p_0) \wedge \pi(p_1)$). For this, $P$ can be any well-met strongly proper poset, you don't need the * property. The meet-preservation of $\pi$ would play a similar role that the * property plays in our paper cited above. Basically if $p_0$ is compatible with $p_1$ in $P$ and both happen to be in $P/\pi^{-1}[G]$ (where $G$ is generic for $\text{Add}(\omega,\alpha)$), then meet-preservation of $\pi$ ensures that $p_0$ and $p_1$ are compatible in $P/\pi^{-1}[G]$ as well. This implies (via an argument similar to the one in our paper) that, in $V[G]$, $P/\pi^{-1}[G]$ will be strongly proper on a stationary set of countable models, and hence the proof from the claim above goes through again.

REMARK 3: Regarding regular embeddings versus projections. If $\pi:P \to \text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$ is a (not necessarily meet-preserving) projection as in your question, then there is a regular embedding from $\text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$ into the boolean completion $\text{RO}(P)$, but not necessarily into $P$ itself. Usually such distinctions are irrelevant in forcing theory, but I'm not sure here. The problem is that Boolean completions screw up the * property---in fact no nontrivial boolean algebra has the * property---and as mentioned in Remark 1, something like the * property is needed in order for the quotient to have the $\omega_1$ approximation property. If the Claim is still true when you replace the target of $e$ with $\text{RO}(P)$ ( instead of requiring the target to be $P$) then there can be no projection $\pi:P \to \text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$. But I don't know if that revised claim is true.

• could you explain with more details? – Rahman. M Jan 9 '17 at 16:51
• I just edited to add more details – Sean Cox Jan 9 '17 at 17:09
• @Rahman.M there was a technical issue that I overlooked. The answer has been edited accordingly. – Sean Cox Jan 10 '17 at 0:32
• I did not check it for Neeman's forcing, but your answer is insightful. thanks. – Rahman. M Jan 11 '17 at 20:16