If $P=PP$ then the counting hierarchy collapses to $CH=P$. Because so many complexity classes are contained in $CH$, this causes most classes to now be contained in $P$. My question is whether this is true for all languages in $PSPACE$, or not.

What is a candidate language $L\in PSPACE$ such that we do not know that $P=PP$ implies that $L\in P$?

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    $\begingroup$ Please explain when you downvote. Note that Turbo is not asking about proven separations, but rather about candidate classes that resist collapse even when $P=CH$, if I understand correctly, which is a good question imo. $\endgroup$ Jun 22, 2019 at 14:47
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    $\begingroup$ I guess you mean besides $PSPACE$-complete languages, which is trivial. $\endgroup$
    – rus9384
    Aug 17, 2019 at 12:20

1 Answer 1


Trivially there will be space-bounded classes $P\ne Space(n)\subsetneq Space(n^2)\subsetneq PSPACE$, and $NSpace(n)\subsetneq NSpace(n^2)\subsetneq PSPACE$, which do not collapse due to the space hierarchy theorem.

More interestingly, the Complexity Zoo lists two other classes that is in PSPACE but not in CH, namely

  • $QSZK$ (Quantum Statistical Zero Knowledge)
  • $RG[\#1]$ (single-round referreed games).

Perhaps $QSZK$ satisfies your condition. The best published upper bound (that I can find) is that $QSZK\subseteq PSPACE$ by Watrous[1]. Perhaps it does not satisfy your condition; because it has a complete problem that asks to determine the trace distance between the quantum states produced by two quantum circuits given as input, and this is typically doable in $P^{\#P}$, which you have assumed is equal to $P^{\#P}=P$. I don't know, and invite people to think about whether $QSZK\subseteq CH$ or not. Since you asked specifically for a hierarchy, you can think about the hierarchy $QSZK^{{QSZK}^{\cdots}}$. It is unknown whether this class is self-low, i.e. whether $QSZK=QSZK^{QSZK}$ and the hierarchy collapses.


Known results:

$$QMA=QIP[1]\subseteq A_0PP\subseteq PP$$ $$QMA[2]=QMA[k]$$ $$QSZK\subseteq QIP[2]\subseteq QIP[3]=QIP=PSPACE.$$

Perhaps $RG[\#1]$ satisfies your condition. No upper bound is known except that $RG[\#1]\subseteq RG[\#2]=PSPACE\subseteq RG=EXP$.

[1] J. Watrous. Limits on the power of quantum statistical zero-knowledge, to appear in Proceedings of IEEE FOCS'2002

  • $\begingroup$ Would any of these languages collapse to $PSAPCE$ or to $P$ if $NC=PP$ actually held? How far can we go down below $NC$ and still have proper separation? $\endgroup$
    – Turbo
    Jul 24, 2019 at 19:06

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