It is known that (light-face) $\Sigma_3^1$ generic absoluteness is consistent with $\mathsf{ZFC}$: Friedman and Bagaria showed that it holds in the $\text{Coll}(\omega, < \kappa)$ extension of $V$ where $\kappa$ is a $\Sigma_2$-correct cardinal.

My questions are about what large cardinal principles can prove $\Sigma_3^1$-generic absoluteness. In particular:

1) If $0^\sharp$ exists (or even $x^\sharp$ exists for all reals $x$), does (light-face) $\Sigma_3^1$-generic absoluteness holds.

2) What if there is a measurable cardinal, then does $\Sigma_3^1$-generic absoluteness hold?

Is there any large cardinal whose existence implies $\Sigma_3^1$ generic-absoluteness?

  • $\begingroup$ If my memory serves me right, a measurable implies $\Sigma^1_3$-absoluteness for forcing smaller than the measurable. So the existence of a proper class of measurable cardinals implies $\Sigma^1_3$-absoluteness. $\endgroup$
    – Asaf Karagila
    Aug 22, 2015 at 15:33
  • $\begingroup$ @AsafKaragila Do you know where I can find a proof of the result you mentioned? $\endgroup$
    – William
    Aug 22, 2015 at 22:32
  • $\begingroup$ It was mentioned in a course about generic absoluteness and stationary tower forcing, and was attributed to Martin-Solovay. No concrete reference was given, or else I would have probably post it as an answer. $\endgroup$
    – Asaf Karagila
    Aug 22, 2015 at 22:41
  • $\begingroup$ There is a proof (or at least a sketch) in these notes by Steel (Theorem 2.13). $\endgroup$ Aug 23, 2015 at 0:20

2 Answers 2


A proper class of measurables more than suffices.

It suffices for the generic absoluteness to have X-sharp exists for every set of ordinals X. Then the Martin-Solovay tree can be constructed throughout On.

One place to look is below. The definition of the Martin-Solovay tree (I seem to recall) is given for an arbitrary cardinal $\kappa$, but even if it is only given for $\omega_1$ just replace $\omega_1$ by $\kappa$, and take the union of the trees on all regular cardinals $\kappa$ to get a tree on $On$.

A. Kechris Homogeneous Trees and Projective Scales. Cabal seminar 77-79: Proceedings of the Caltech-UCLA Logic Seminar 1977-1979 1981,Eds.A. Kechris, D.A. Martin, and Y. Moschovakis 839, Springer Lecture Notes in Mathematics Series, Pages 33-73

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    $\begingroup$ The existence of sharps (as in Philip's second paragraph) is also necessary, so this assumption is optimal. $\endgroup$ Aug 26, 2015 at 21:42
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    $\begingroup$ @Andres: One might even say that the consistency bound is... sharp :-P $\endgroup$
    – Asaf Karagila
    Aug 28, 2015 at 7:52
  • $\begingroup$ @AndresCaicedo Is sharps really necessary? Bagaria and Friedman showed that it consistent with $\mathsf{ZFC}$ that (lightface) $\Sigma_3^1$-generic absoluteness holds, for example in the $\text{Coll}(\omega, <\lambda)$ extension where $\lambda$ is a $\Sigma_2$-correct cardinal (which is not a large cardinal). $\endgroup$
    – William
    Sep 1, 2015 at 2:31
  • $\begingroup$ @AndresCaicedo In your paper "Projective Well-Ordering of the Reals" Theorem 3 seems to be what you are referring to. But the definition of generic absoluteness is stronger something like a two-step generic absoluteness. Also what is the sharp of an arbitary set. Should the existence of $A^\sharp$ be equivalent to there is an elementary embedding of $L[A]$ to $L[A]$? Can it be some kind of mouse? $\endgroup$
    – William
    Sep 1, 2015 at 2:42
  • $\begingroup$ @William Yes.${}$ $\endgroup$ Sep 1, 2015 at 2:42

Re a proof of this result: I once wrote up an elementary proof of this which avoids the Martin-Solovay tree: See Theorem 1 of https://ivv5hpp.uni-muenster.de/u/rds/talks-barcelona.dvi

  • $\begingroup$ Quite nice, +1! $\endgroup$ Sep 14, 2017 at 12:09

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