$\mathit{NP}$-hard statements which are $\mathit{NP}$-complete under the Riemann Hypothesis $\newcommand\NP{\mathit{NP}}\newcommand\SAT{\mathit{SAT}}\newcommand\CH{\mathit{CH}}\newcommand\PSPACE{\mathit{PSPACE}}$Are there $\NP$-hard problems which are $\NP$-complete under the Riemann Hypothesis corresponding to the following cases?

*

*Superpolynomial certificate length becomes polynomial sized.


*Randomized polynomial time reduction to $\SAT$ becomes deterministic polynomial time reduction.


*Reduction to $\SAT$ becomes polynomial deterministic time from superpolynomial deterministic time.


*Certificate becomes polynomial time verifiable.


*An infinitely often reduction always holds.


*The problem was originally in $\CH$ (counting hierarchy) or below and strictly in $\PSPACE$.
I think 2. and 6. are impossible unless RH is false and so might be difficult to identify while 1., 3. and 4. might be similar.
Bounty is for 2. or 6.
I would be happy to award for 2. and 6. if there is a reduction for these under the condition of failure of RH and in my opinion such a twisted reduction is allowed but just 2. and 6. are disallowed under correctness of RH.
It would be amazing if there are reductions to 2. and 6. under both truth and falseness of RH and this I think is unlikely since I think reduction under truth of RH is unlikely.
 A: Here's an example, but you may consider it to be a bit artificial Consider the following language: s is in $L$ if $S=(x,y)$ where $x$ is a 3-SAT instance, $y$ is a satisfying instance for $y$, and also satisfying that if $n$ is the number of true variables in $y$, then the $n$th non-trivial zero of the zeta function has real part equal to $1/2$. But this is a bit artificial. My guess is that you specified your five examples in order to avoid that.
Edit: The following does not work. Possibly it can be fixed, but see Emil's comment below.
But we can modify this is a little bit to cheat and get an example of 1, provided one is willing to use GRH rather than RH; my guess is that a similar construction can be done with RH, but it doesn't immediately jump out at me.  The first detail is that assuming GRH, the smallest positive primitive root $u$,  of a prime $p$ is bounded above by a power of $\log p$. In particular, Shoup showed that $u = O( (\log p)^6)$.
Given positive constants $c_1$, $c_2$, and a function $f(n)$ Let $A_{c_1,c_2, f(n)}$ be the language given by the following: $a$ is in $A$ if $a=(x,y,z,w)$ where $x$, $y$, and $z$ are defined as follows. We have $x$ is a 3-SAT instance and $y$ is a satisfying set of variables for $x$. $z$ is a prime $p$ such that $y< p < (c_1)p^{c_2}$ (where we are thinking of our list $y$ of variable assignments as just a binary number). We have $w$ is a string of 1s where the number of 1s is a primitive root of $p$, and $w$'s length is bounded by $f(p)$.
Now, we can choose a $c_1$ and $c_2$ such that there is always such a prime $p$ when there is a $y$. (Baker, Harman and Pintz showed that one can take $c_2 = 0.525$). So choose such a $c_1$ and $c_2$. Then, note that if we believe GRH, we can always find such a $w$ which is polynomial in the length of $x$, and we can choose $f(n)$ so that $w$'s length is guaranteed to be less than $f(p)$ if GRH is true.
So now our NP-hard question then is just is a given $x$ a valid $x$ for an $(x,y,z,w)$ in $A$? If GRH is true, we can always find a $y$, $z$ and $w$ which is guaranteed to be just polynomial in length when $x$ is true. But without GRH we can only get that this is NP-hard since our $w$ may be too large.
This is essentially just a padding trick. My guess is that a similar padding trick can be used with just RH as an assumption but I don't immediately see how to do it.  And this is still a pretty artificial problem. I'd be interested to see if someone has a more natural example.
