# Absolute formulas with high complexity

This is a repost of a MSE question.

It is a standard result that $$\Sigma_1^{\mathsf{ZF}}$$-formulas are upward absolute between $$\mathsf{ZF}$$ $$\in$$-models, while $$\Pi_1^{\mathsf{ZF}}$$-formulas are downward absolute between $$\mathsf{ZF}$$ $$\in$$-models, so in particular $$\Delta_1^{\mathsf{ZF}}$$-formulas are absolute between $$\mathsf{ZF}$$ $$\in$$-models.

This is also sharp, there are some stronger results ($$\Pi_2^{\mathsf{ZF}}$$-formulas are $$H(\kappa)$$-$$V$$-downward absolute for an uncountable $$\kappa$$), but in general it's not hard to construct $$\Sigma_2^{\mathsf{ZF}}$$, $$\Pi_2^{\mathsf{ZF}}$$ and $$\Delta_2^{\mathsf{ZF}}$$-formulas for which the appropriate absoluteness result fails.

Is there a simple way to construct, given $$n\in\Bbb N$$, a formula $$\varphi$$ such that $$\varphi$$ is $$\Delta_n^{\mathsf{ZF}}$$ (and not equivalent to a simpler formula) and $$\varphi$$ is absolute between $$\mathsf{ZF}$$ $$\in$$-models?

• I think your question itself may not be absolute. If $V$ is the shortest model of $\mathsf{ZF} + V=L$ then there are no transitive set-sized models of $\mathsf{ZF}$ and no proper inner models. If $V$ is the second to shortest model of $\mathsf{ZF} + V=L$ then there is precisely one set-sized model. So depending on how you define it in both of these cases all formulas are absolute. – James Hanson Aug 8 '19 at 19:33
• On the other hand, if you give yourself enough transitive models, there is a strong negative answer. For example, if ZF proves $\varphi(x)$ is upwards absolute between transitive models of ZF then $\varphi(x)$ is equivalent over ZF + every set belongs to a transitive model of ZF to the $\Sigma_1$-formula "there is a transitive model $M\vDash \text{ZF}$ such that $x\in M$ and $M\vDash \varphi(x)$." In other words, absoluteness between transitive models of ZF is the same as being simply expressible relative to a theory that proves the existence of enough transitive models. – Gabe Goldberg Aug 8 '19 at 19:36
• @Gabe thanks, this is a very interesting result I wasn't aware of. Is there something similar for downward absoluteness? – Alessandro Codenotti Aug 9 '19 at 7:06
• Yeah: if $\varphi(x)$ is downwards absolute, then $\varphi(x)$ is equivalent to the $\Pi_1$-formula "for all transitive $M$ containing $x$, $M\vDash \varphi(x)$" (again, assuming there are enough transitive models). – Gabe Goldberg Aug 9 '19 at 19:33