We know that if $P=NP$ then the whole PH collapses. What if the polynomial hierarchy collapses partially ? (Or how to understand that PH could collapse above a certain point and not below ?)
In shorter words, what would be the consequences of $NP=coNP$ and $P\ne NP$ ?
To me, one of the most basic and surprising consequences of $\mathsf{NP}=\mathsf{coNP}$ is the existence of short proofs for a whole host of problems where it is very difficult to see why they should have short proofs. (This is sort of taking a step back from "What other complexity implications does this collapse have?" to "What are the very basic, down-to-earth reasons this collapse would be surprising?")
For example, if $\mathsf{NP}=\mathsf{coNP}$, then for every graph that is not Hamiltonian, there is a short proof of that fact. Similarly for graphs that are not 3-colorable. Similarly for pairs of graphs that are not isomorphic. Similarly for any propositional tautology.
In a world where $\mathsf{P} \neq \mathsf{NP} = \mathsf{coNP}$, the difficulty in proving propositional tautologies isn't that some short tautologies have long proofs - because in such a world every tautology has a polynomially short proof - but rather that there is some other reason that we are unable to find those proofs efficiently.
If we also assume $\mathsf{NP}=\mathsf{RP}$, then the hypothesis would also cause the collapse of randomized classes: $\,\,\mathsf{ZPP}=\mathsf{RP}=\mathsf{CoRP}=\mathsf{BPP}$. Although these are all conjectured to unconditionally collapse into $\mathsf{P}$, anyway, it is still open whether that indeed happens. In any case, $\mathsf{NP}=co\mathsf{NP}$ does not seem to imply in itself that these randomized classes collapse.
If they do not, that is, we at least have $\mathsf{BPP}\neq \mathsf{P}$, then, along only with the $\mathsf{NP}=co\mathsf{NP}$ hypothesis, this would have another important consequence: $\,\,\mathsf{E}\neq \mathsf{NE}$. This follows from a result of Babai, Fortnow, Nisan and Wigderson, which says that if all unary (tally) languages in $\mathsf{PH}$ fall in $\mathsf{P}$, then $\mathsf{BPP}=\mathsf{P}$. Thus, if $\mathsf{BPP}\neq \mathsf{P}$, then they cannot all fall in $\mathsf{P}$, as the $\mathsf{NP}=co\mathsf{NP}$ assumption implies $\mathsf{PH}=\mathsf{NP}$. Therefore, there must be a tally language in $\mathsf{NP}-\mathsf{P}$. Finally, the presence of a tally language in $\mathsf{NP}-\mathsf{P}$ is well known to imply $\mathsf{E}\neq \mathsf{NE}$.
The above reasoning shows the interesting effect that the $\mathsf{NP}=co\mathsf{NP}$ hypothesis, despite being a collapse, actually amplifies the separating power of $\mathsf{BPP}\neq \mathsf{P}$, as the latter alone is not known to imply $\mathsf{E}\neq \mathsf{NE}$. This "anomaly" seems to support the conjecture $\mathsf{BPP}= \mathsf{P}$.
There are two definitions for counting classes beyond ${\bf \#P}$. One was defined by Valiant and the other one was defined by Toda.
${ \rm \underline {Valiant's-Definition:}}$ For any class $C$, define $\#C =\cup_{A\in C}(\#P)^{A}$, where $({\#P}^A)$ means the functions counting the accepting paths of nondeterministic polynomial-time Turing machines having $A$'s their oracle.
By Valiant's definition we already have ${\bf \#NP} = {\bf \#CoNP}$
${ \rm \underline {Toda's-Definition:}}$ For any class $C$, define $\# .C$ to be the class of functions $f$ such that for some $C-$computable two-argument predicate $R$ and some polynomial $p$, for every string $x$ it holds that: $f(x)=||\{y|p(|x|)=|y|$ and $R(x,y)\}||$.
By Toda's definition we have ${\bf \#.NP} = {\bf \#.CoNP}$ if and only if ${\bf NP} = {\bf CoNP}$.
Then if we also assume that ${\bf P}\not = {\bf NP}$ then we would have ${\bf FP} \not = {\bf \# P}$.
Ker-i Ko Showed that there is an oracle that makes PH collapse at the k-th level. See "Ker-I Ko: Relativized Polynomial Time Hierarchies Having Exactly K Levels. SIAM J. Comput. 18(2): 392-408 (1989)".
