Whether $\textbf{PH}$ collapses to $\Sigma^p_1$ when $\overline{3SAT}\in \textbf{BP$\cdot$NP}$

Question

I'm doing ex. 7.8 in Arora and Barak

Show that if $\overline{3SAT}\in \textbf{BP$\cdot$NP}$, then $\textbf{PH}$ collapses to $\Sigma_3^p$.

This is definition of $\textbf{NP}/poly$:

A nondeterministic circuit has two inputs $x, y$. We say that $C$ accepts $x$ iff there exists $y$ such that $C(x, y) = 1$. The size of the circuit is measured as a function of |x|. Let $\textbf{NP}/poly$ be the languages that are decided by polynomial size nondeterministic circuits.

We know that $\textbf{BP$\cdot$NP}\subseteq \textbf{NP}/poly$, so we need to adapt from Karp-Lipton theorem to show that $\textbf{PH}$ will collapses to $\Sigma_3^p$. On the other hand, I show that $\textbf{PH}$ collapes to $\Sigma_1^p$.

We have $$\varphi\in \overline{3SAT}\iff \forall u\in\{0,1\}^{|\varphi|}\varphi(u)=0\tag{1}.$$ Because $\overline{3SAT}\in \textbf{BP$\cdot$NP}$, we infer that there exists a $poly(n)$-sized circuit family $\{C_n\}_{n\ge 1}$ s.t.$$\varphi\in\overline{3SAT}\iff\exists w,y\in \{0,1\}^{poly(|\varphi|)} \text{$w$ describes the circuit $C_{|\varphi|}$ and $C_{|\varphi|}(\varphi,y)=1$}.\tag{2}$$

When $\varphi\notin \overline{3SAT}$, there are no $y$ such that $C(\varphi,y)=1$ since $C$ decides $\varphi$ by definition. Finally, we construct a TM $M$, whose parameters are $\varphi, w,y$, evaluating $C(\varphi,y)$. Thus,$$\varphi\in\overline{3SAT}\iff\exists w,y\in \{0,1\}^{poly(|\varphi|)} \text{$w$ describes a circuit and $M(\varphi,w,y)=1$}.$$ In other words, $\textbf{PH}$ collapses to $\Sigma_1^p.$

The current result, however, is $\textbf{PH}$ collapses to $\Sigma_3^p$ as in the exercise. I do worry about whether the (2) is correct or not. Because it is the core of the proof, the false probability of (2) is quite high!

Answer 1

$\def\cosat{\overline{\mathrm{3SAT}}}\DeclareMathOperator\poly{poly}$The condition $$\exists w,y\in \{0,1\}^{\poly(|\varphi|)} \text{ ($w$ describes a circuit $C$ and $C(\varphi,y)=1$)}$$ does not define $\cosat$. In fact, it holds for all formulas $\varphi$, as you can take for $w$ a decription of the constant-$1$ circuit.

What you need to do is to insert into the condition a conjunct ensuring that the nondeterministic circuit does, actually, compute $\cosat$ as claimed, rather than something completely different: $$\forall\psi\in\{0,1\}^{|\varphi|}\:(\psi\in\cosat\leftrightarrow\exists y\in\{0,1\}^{\poly(|\varphi|)}\:C(\psi,y)).$$ This is as a $\mathbf{\Pi^p_2}$ condition, blowing the overall complexity of the condition to $\mathbf{\Sigma^p_3}$.

Of course, then it becomes pointless to express $\cosat$ like this, as it basically says “$\varphi$ is in $\cosat$ iff it is accepted by a complicated device that accepts $\cosat$”. What you really need to do is to express an arbitrary $\mathbf{\Pi^p_3}$ predicate $P$ in this way, using an outer $\exists w$ quantifier for a description of a nondeterministic circuit that is then used to turn the inside $\mathbf{\Pi^p_1}$ part of the description of $P$ to $\mathbf{\Sigma^p_1}$ (with parameter $w$), showing that $\mathbf{\Pi^p_3=\Sigma^p_3}$.

This is really the exact same issue as in the standard proof of the Karp–Lipton theorem, only one level of the hierarchy higher. It would be really helpful if you try to make sure you really understand the original Karp–Lipton theorem first, as it does not seem you do at the moment.

A side note: the class $\mathbf{BP\cdot NP}$ is more commonly called $\mathbf{AM}$, and the assumption $\mathbf{\cosat\in BP\cdot NP}$, that is $\mathbf{coNP\subseteq AM}$, actually implies $$\mathbf{PH=AM=coAM=\Sigma^p_2=\Pi^p_2.}$$ As a brief sketch, this can be proved as follows: using amplification of the success probability of the $\mathbf{BP}$ part to make the error exponentially small, show that $\mathbf{\exists\cdot BP\cdot NP=BP\cdot NP}$ (i.e., $\mathbf{MAM=AM}$). Assuming $\mathbf{coNP\subseteq BP\cdot NP}$, we also have $\mathbf{co\cdot BP\cdot NP=BP\cdot coNP\subseteq BP\cdot BP\cdot NP=BP\cdot NP}$. Thus, $\mathbf{BP\cdot NP=AM}$ contains $\mathbf{NP}$ and it is closed under both $\exists$ and complement, whence it contains the whole polynomial hierarchy. Finally, use the (nontrivial) fact that $\mathbf{AM=coR\cdot NP\subseteq\Pi^p_2}$. ($\mathbf{coR\cdot NP}$ is usually called “$\mathbf{AM}$ with perfect completeness”.)

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UPDATE: The question was now changed so that the condition reads $$\varphi\in\cosat\iff\exists w,y\in \{0,1\}^{\poly(|\varphi|)} \text{ ($w$ describes the circuit $C_{|\varphi|}$ and $C_{|\varphi|}(\varphi,y)=1$).}$$ This expression for $\cosat$ is not a $\Sigma^p_1$ predicate, or even a $\Sigma^p_3$ predicate. The sequence of circuits $\{C_n\}_{n\in\mathbb N}$ is nonuniform, hence the check “$w$ describes the circuit $C_{|\varphi|}$” is, a priori, not even computable at all, let alone within the polynomial hierarchy.