We know that $$\mathsf{NP\subseteq P/Poly \iff coNP\subseteq P/Poly\implies PH=\Sigma_2^P}=\mathsf{\Pi_2^P}$$ $$\mathsf{NP\subseteq P/Log\iff coNP\subseteq P/Log\implies PH=\Sigma_0^P=\Pi_0^P}$$
Is there any circuit result that could imply $$\mathsf{PH=\Sigma_1^P=\Pi_1^P}$$ but still $$\mathsf{PH\neq\Sigma_0^P=\Pi_0^P}$$ is a possibility?
Is there a possible similar collapse result from non-circuit containment?
Conjecture:
Such result does not exist.
It is perhaps reasonable to think that circuits that people would come up with would come along with easily verifiable proofs of their correctness. (This was indirectly inspired by Dietrich & Wilson's notion of "groups of black box type", and is in the spirit of Hartmanis's "provable complexity.") Towards this end, we consider:
Definition: "Certified $\mathsf{P/poly}$" or "$\mathsf{Cert}$-$\mathsf{P/poly}$" consists of those languages $L$ such that $L$ is in $\mathsf{P/poly}$ and there is a poly-bounded certification of this fact. More precisely, $L \in \mathsf{Cert}\text{-}\mathsf{P/poly}$ if there exists a poly-time $V$, and for all $n$, there exists $C_n, \pi_n$ with $\max\{|C_n|,|\pi_n|\} \leq poly(n)$ such that
$V$ is a Cook-Reckhow proof system for the language $\{C : C \text{ correctly computes } L \text{ at length } \ell=\text{# inputs to C}\}$, that is (a) If $C(x) = L(x)$ for all $x \in \Sigma^\ell$, where $\ell$ is the number of inputs to $C$, then there exists a $\pi$ such that $V(C,\pi)=1$ and (b) If there is $x \in \Sigma^\ell$ such that $C(x) \neq L(x)$, then there is no $\pi$ such that $V(C,\pi)=1$.
For all $n$, $C_n$ has $n$ inputs and $V(C_n,\pi_n)=1$.
Then $\mathsf{Cert}\text{-}\mathsf{P/poly}$ contains $\mathsf{NPSV}_t$-uniform $\mathsf{P/poly}$, so it is probably not just $\mathsf{P}$. And we have the desired statement:
Proposition: $\mathsf{NP} \subseteq \mathsf{Cert}\text{-}\mathsf{P/poly} \Longleftrightarrow \mathsf{coNP} \subseteq \mathsf{Cert}\text{-}\mathsf{P/poly} \Longrightarrow \mathsf{PH} = \mathsf{\Sigma_1 P} = \mathsf{\Pi_1 P}$.
Proof: First we show that $\mathsf{Cert}\text{-}\mathsf{P/poly} \subseteq \mathsf{NP}$, giving the final "$\Longrightarrow$" in the statement. Suppose $L \in \mathsf{Cert}\text{-}\mathsf{P/poly}$ with verifier $V$ and $V$-verified circuit family $\{(C_n,\pi_n)\}_{n=1,2,3,...}$. The following is a nondeterministic poly-time algorithm for $L$: on input $x$ of length $n$, guess $C_n,\pi_n$, and run $V(C_n,\pi_n)$; if it accepts, then output $C_n(x)$.
Next, we show that Certified $\mathsf{P/poly}$ is closed under complement, giving the first "$\Longleftrightarrow$" in the statement. Suppose $L \in \mathsf{Cert}\text{-}\mathsf{P/poly}$, with verifier $V$ as in the above definition. Since $L \in \mathsf{NP}$ by the preceding paragraph, there is some $c$ such that $L \in \mathsf{NTIME}(n^c) \subseteq \mathsf{DTIME}(2^{n^c})$. Let $V'(C',\pi)$ compute as follows: if $C'$ has a negation gate as its output gate, then write $C' = \neg C$, and define $V'(C',\pi) = V(C,\pi)$. Otherwise, $V'(C',\pi)$ checks whether $|\pi|=2^{\ell + \ell^c}$ (where $\ell$ is the number of inputs to $C'$); if not, then $V'$ rejects, and if so, then $V'$ simply enumerates all strings $y$ of length $\ell$, and checks whether $C'(y)=L(y)$, using the $2^{\ell^c}$ deterministic algorithm for $L$, which takes time $poly(|\pi|)$. (This latter yoga was needed to make sure that $V'$ satisfies property 1 in the definition.) If $\{(C_n,\pi_n)\}$ is a certified family of circuits for $L$ with verifier $V$, then $\{(\neg C_n, \pi_n)\}$ is a certified family of circuits for $L^c$ with verifier $V'$. QED
