# Just a little absoluteness might be cheaper?

Absoluteness is a wonderful thing, but expensive consistency-strength wise. My question is, when can we get large amounts of absoluteness in specific situations for much cheaper?

Specifically, fix a definable forcing notion $\mathbb{P}$ - it is reasonable to ask how hard it would be to find a model where, say, $\Pi^1_3$ statements aren't changed by forcing with $\mathbb{P}$. For example, take $\mathbb{P}$ to be Cohen forcing. Then $L$ is certainly not such a model, but any forcing extension of $L$ by $\mathbb{P}$ is. On the other hand, there are certainly definable forcing notions such that the statement "Forcing with $\mathbb{P}$ doesn't change the truth value of $\Pi^1_3$ sentences" requires consistency strength beyond ZFC.

My question is when we can make "local absoluteness assumptions" without additional consistency strength. For a specific instance of this question:

Are there reasonable hypotheses on definable forcings $\mathbb{P}$ which imply that the statement $$\text{PA(\mathbb{P}) = "Projective sentences are absolute under forcing with \mathbb{P}"}$$ has no additional consistency strength?

I would ask for sufficient and necessary conditions, but I think that's demanding far too much.

I've added the "descriptive set theory" tag because I think it may be relevant; feel free to delete if you think that's bonkers.

• Sy and I have a cute result along these lines in our joint paper. Nice question! The Tomek-Haim book should have some additional examples (at low levels), with a more recent paper by Joan and Hugh showing that there are no higher level versions of some of these results. – Andrés E. Caicedo Jun 11 '15 at 15:09
• Noah, it should be downloadable from my blog. (Recent issues of the JSL are first available through Project Euclid, with subscription, and added to JSTOR a few years later.) – Andrés E. Caicedo Jun 11 '15 at 15:19
• I don't understand your claim that after forcing by $\mathbb{P}$ over $L$, we get a model where $\Pi^1_3$ statements aren't changed by (further) forcing with $\mathbb{P}$. This doesn't seem to be true for all $\mathbb{P}$. One can build wild counterexamples by defining $\mathbb{P}$ by "if $V=L$, then $\mathbb{P}$ is like this simple thing, and otherwise it is like this other wild thing". You are assuming that the forcing relation for forcing $\Pi^1_3$ statements is the same in $V$ as it is in $V^{\mathbb{P}}$, or what? – Joel David Hamkins Jun 12 '15 at 1:17
• Also, even if $\mathbb{P}$ was literally the same poset in $L$ and the extension by $L$ of $\mathbb{P}$, if it wasn't homogeneous, then one can make the poset do totally different things under different conditions. – Joel David Hamkins Jun 12 '15 at 1:24
• By a result of Brendle, $\Sigma_4^1$-Amoeba absoluteness implies that $\aleph_1$ is inaccessible in $L$: arxiv.org/pdf/math/9209206v1.pdf So a necessary condition for a property that implies $PA(\mathbb{P})$ is that it doesn't hold for Amoeba forcing. – Haim Jun 14 '15 at 8:23

• A weird aside: Bagaria and Friedman also show that $\Sigma^1_3$-absoluteness for proper forcings implies that either $\omega_1$ is Mahlo in $L$ or $\omega_2$ is inaccessible in $L$. – Noah Schweber Jun 18 '15 at 1:36