In Descriptive Complexity, Immerman hasCorollary 7.23. The following conditions are equivalent:1. P = NP.2. Over finite, ordered structures, FO(LFP) = SO.This can be thought of as amplifying P=NP to an equivalent statement over (presumably) larger complexity classes.  Note that SO captures the polynomial-time hierarchy PH, and that FO(LFP) captures P, so this can be thought of as P=NP iff P=PH.(The interesting part of this is the statement that P=NP implies P=PH; it is trivial that P=CC implies P=NP for any class CC that contains NP.  Immerman simply remarks if P=NP then PH=NP, presumably because P=NP can be used with the oracle definition of PH to show inductively that the whole hierarchy collapses.)My question is:How much further can P=NP be amplified in this way?In particular, what is the largest known class CC' such that P=NP implies P=CC', and the smallest class CC such that P=NP implies CC=NP?  This would allow P=NP to be replaced by the equivalent question CC=CC'.  P appears to be a rather powerful class, which seems to provide little wiggle room for arguments trying to separate it from NP: how far can the wiggle room be amplified?I would of course also be interested in an argument that shows that P=PH is the limit of this approach.Edit: note the closely related question Why doesn't P=NP imply P=AP (i.e. P=PSPACE)? which focuses on the other direction, why we don't have proofs that P=PSPACE.  Answers there by Kaveh and Peter Shor argue that the number of alternations being fixed is key.  Another related question is A decision problem which is not known to be in PH but will be in P if P=NP which asks for a candidate problem; the answers there also can be used to construct answers for this question, although these classes are somewhat artificial (thanks to Tsuyoshi Ito for pointing this out).  In a more general setting, Collapsing of exptime and alternation bounded turing machine asks whether a local collapse at any level in an alternation hierarchy induces an upward collapse, as happens with the polynomial-time hierarchy.
Can one amplify P=NP beyond P=PH?
cc.complexity theory;reference request;complexity classes;conditional results;polynomial hierarchy
From  Russell Impagliazzo's comment:As a way of formalizing   what languages are in $\mathsf{P}$ if $\mathsf{P}=\mathsf{NP}$,   Regan introduced the complexity class $\mathsf{H}$.   A language $L$ is in $\mathsf{H}$ if and only if $L$ is in $\mathsf{P}^O$  relative to every oracle $O$ so that $\mathsf{P}^O=\mathsf{NP}^O$.   Thus, $L$ is in $\mathsf{H}$ if the statement   $\mathsf{P}=\mathsf{NP} \implies L\in\mathsf{P}$ relativizes.   $\mathsf{PH} \subseteq \mathsf{H} \subseteq \mathsf{AltTime}(O(\lg\lg n),\mathsf{poly})$.   From Toda's theorem, and some of the lemmas in Toda's theorem,   it is also true that $\mathsf{H} \subseteq \mathsf{P}^{\mathsf{mod}_q \mathsf{P}}$ for every $q$.   Basically, any oracle satisfying $\mathsf{P}^O=\mathsf{NP}^O$   gives a new upper bound on $\mathsf{H}$.   It is open whether $\mathsf{H}=\mathsf{PH}$.And from Lance Fortnow's comment:Let $f(n)$ be any unbounded function.   $\mathsf{H}$ is not contained in $\mathsf{AltTime}(f(n),\mathsf{poly})$ and   if you could prove   $\mathsf{P}=\mathsf{NP}$ implies $\mathsf{P}=\mathsf{AltTime}(f(n),\mathsf{poly})$ then $\mathsf{NP}$ is different than $\mathsf{L}$.For definition of $\mathsf{H}$ see definition 6.3 inKenneth W. Regan, Index sets and presentations of complexity classes, 1999