Robin's theorem (1984) states that

$$ \sigma(n) < e^\gamma n \log \log n$$ for all $n > 5040$ if and only if the Riemann hypothesis is true.

Recall that $γ$ is the Euler–Mascheroni constant and $σ(n)$ is the divisor function, given by $$\sigma(n) = \sum_{d\mid n} d.$$

To formulate a quantum Riemann hypothesis, we will use Robin's theorem and the following facts:

For a justification of the qualifier "quantum", see the following article:

Jones, Vaughan. On the origin and development of subfactors and quantum topology. Bull. Amer. Math. Soc. (N.S.) 46 (2009), no. 2, 309--326.

Galois correspondences:

  • the divisors $d\mid n$ are $1$-$1$ with the subgroups $H \subseteq C_n$,
  • the subgroups $H \subseteq G$ are $1$-$1$ with the intermediate subfactors $R \subseteq K \subseteq R \rtimes G$,
  • the intermediate subfactors $N \subseteq K \subseteq M$ are $1$-$1$ with the biprojections $b \in [e_1,id]$.

The notations match as follows:

  • $n = |G| = [M:N] = |id : e_1|$,
  • $d = |H| = [K:N] = |b : e_1|$.

The equality $|G| = |G:H| \cdot |H|$ extends to $|id : e_1| = |id:b| \cdot |b:e_1|$.

In general, $|id : e_1|$ is not necessarily an integer, but (by Jones' theorem) can be any element in
$$\{4\cos^2(\pi /n)|n=3,4,5,...\}\cup [4,+\infty).$$

Let $\mathcal{P}$ be an irreducible subfactor planar algebra. We define the analog of the set of divisors by $$D(\mathcal{P}) := \{|b : e_1| \text{ with } b \in [e_1,id] \},$$ (which is finite by Watatani's theorem) and the analog of the divisor function by
$$\sigma(\mathcal{P}) := \sum_{\beta \in D(\mathcal{P})} \beta.$$

Quantum Riemann Hypothesis (of depth $n$)

There is $\alpha_n>0$ such that for every irreducible depth $n$ subfactor planar algebra $\mathcal{P}$ with $\alpha:=|id : e_1|> \alpha_n$, we have $$\sigma(\mathcal{P}) < e^\gamma \alpha \log \log \alpha.$$

Of course, a proof of this quantum Riemann Hypothesis (QRH) is not expected as an answer of this post, because it implies the usual Riemann Hypothesis (RH).

For the group case, QRH follows from RH, because for $\sigma(G):=\sigma(\mathcal{P}(R \subseteq R \rtimes G))$, we have $\sigma(G) \le \sigma(|G|)$ by Lagrange's theorem. Idem if $|id:b|$ and $|b:e_1|$ are integers $\forall b \in [e_1,id]$, like the irreducible depth $2$ case, because then $\sigma(\mathcal{P}) \le \sigma(|id : e_1|)$.

Let's denote QRH of depth $n$ by QRH$_n$. Then, QRH$_2 \Leftrightarrow$ RH, and we can take $\alpha_2 = 5040$.

Question: Does RH imply QRH$_n \ \forall n \ge 2$? Or, do you see a counterexample for some $n$?

Bonus question: Assuming QRH$_n$ true $ \forall n \ge 2$, can the sequence $(\alpha_n)$ be bounded?

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    $\begingroup$ Is there an analog of$$\sigma(p_1^{\nu_1}\cdots p_k^{\nu_k})=(1+p_1+p_1^2+...+p_1^{\nu_1})\cdots(1+p_k+p_k^2+...+p_k^{\nu_k})?$$ $\endgroup$ – მამუკა ჯიბლაძე Aug 14 '18 at 19:35
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    $\begingroup$ Why is 5040 correct for the QRH? Maybe there are additional counterexamples with small $\alpha$. $\endgroup$ – Stopple Aug 14 '18 at 22:42
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    $\begingroup$ @მამუკაჯიბლაძე: First of all, the analog of a prime number is called a maximal subfactor (i.e. when $[e_1,id] = \{ e_1,id\}$). Unfortunately, in general, a subfactor does not decompose into maximal ones as well as a number decompose into prime numbers. There exist several type of compositions (like tensor product or free product, and many others). $\endgroup$ – Sebastien Palcoux Aug 15 '18 at 9:21
  • $\begingroup$ @მამუკაჯიბლაძე: Now, if a subfactor planar algebra $\mathcal{P}$ has index $\beta^n$ and if all its maximal intermediates have index $\beta$, then $\sigma(\mathcal{P}) = 1+\beta+ \dots + \beta^n$. Moreover, if two subfactor planar algebras $\mathcal{P}_1$ and $\mathcal{P}_2$ has no isomorphic maximal intermediates (analog of coprime), then $\sigma(\mathcal{P}_1 \otimes \mathcal{P}_2) \le \sigma(\mathcal{P}_1) \cdot \sigma(\mathcal{P}_2) $. Finally, if moreover $|D(\mathcal{P}_1)D(\mathcal{P}_2)| = |D(\mathcal{P}_1)| \cdot |D(\mathcal{P}_2)|$, then the equality holds. $\endgroup$ – Sebastien Palcoux Aug 15 '18 at 9:35
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    $\begingroup$ Stopple: You are right! If there are only finitely many counterexamples and if some of them have index $α>5040$, then we would have to replace $5040$ by the maximum of such α. Now, if I'm not mistaken, Robin's theorem stays true if we replace $5040$ by any greater number. I modified the post for (in particular) avoiding the problem you pointed out. $\endgroup$ – Sebastien Palcoux Aug 16 '18 at 17:50

Can you clarify whether there exists a notion of direct product in this setting with the desired properties?

If so, the asymptotics you are predicting only seem consistent with the hypothesis that there are no such objects besides groups. (I guess they do actually exist or you wouldn't ask this question.) The first version of this post suggested taking the direct sum of many factors of size less than $4$, and the OP said that this was invalid because most of them didn't have "bounded depth." Fair enough. But if you take direct products of many subfactors of small indices which have the property that they are "coprime" then it seems you can easily exceed your desired bound unless the only new objects also have rational orders. Here's a very weak explicit version of this idea using a single new object.

It's know that $\displaystyle{\sup \frac{\sigma(n)}{e^{\gamma} n \log \log n} = 1}$. Take an $\alpha$ for which the ratio is very close to $1$. How close will be clear below.

Let $X$ be the object coming from the cyclic group $C_{\alpha}$. Now let $Y$ be any other object whose index is some number $\eta$ which is not a rational number (assuming it exists), and let $Z = X \times Y$ (assuming it exists), and let $\beta = \alpha \eta$ be the index of $Z$. Then it would seem (by considering the subgroups of $C_{\alpha}$ and those subgroups direct sum with $Y$), and using that $\eta$ is not rational, that

$$\sigma(Z) \ge \sigma(X) + \eta \sigma(X) \sim (1 + \eta) e^{\gamma} \alpha \log \log(\alpha) \sim \left(1 + \frac{1}{\eta}\right) e^{\gamma} \beta \log \log(\beta),$$

and the QRH is false.

  • $\begingroup$ I thought about that and the depth $n$ assumption permits to avoid this problem. At index $<4$, there are three series of subfactors: (1) $A_n$ of depth $n-1$ and index $4 \cos^2(\pi/(n+1))$, $n \ge 2$, (2) $D_{2n}$ of depth $n$ and index $4 \cos^2(\pi/(4n-2))$, $n \ge 2$, (3) $E_n$ of depth $n-2$ and index $4 \cos^2(\pi/(9n-42))$, $n=6,8$. It follows that for a fixed depth $n$ and at index $<4$, there is no more an accumulation point at $4$. I asked whether it is also true for all indices in: mathoverflow.net/q/308367/34538 $\endgroup$ – Sebastien Palcoux Aug 21 '18 at 6:35
  • $\begingroup$ There exists a notion of tensor product. What are the properties you desire? There exist irrational index subfactors (see Jones' theorem). I don't understand where you use that $η$ is irrational, could you explain in details? If your last paragraph worked, then it would also contradict the usual RH by taking $Y$ coming from the cyclic group $C_η$ with $α$ and $η$ coprime, so why not? $\endgroup$ – Sebastien Palcoux Aug 21 '18 at 15:12
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    $\begingroup$ The point is that the (integral) divisors $d_i$ of $\alpha$ are distinct from the multiples $\eta d_i$, because $\eta$ is irrational. Reasons why this doesn't contradict RH: The numbers for which $\sigma(\alpha)$ is very close to $e^{\gamma} \alpha \log \log(\alpha)$ are necessary going to be divisible by all of the small primes. So there will be no (fixed) integer $\eta$ for which the numbers $d_i$ and $\eta d_i$ do not coincide. If you have to make $\eta$ bigger and bigger then $1 + \eta^{-1} \sim 1$. $\endgroup$ – user127955 Aug 21 '18 at 16:52
  • $\begingroup$ Assuming your last comment, the existence of a sequence $(n_r)_r$ omitting a single prime factor and such that $\lim_r \sigma(n_r)/ e^{\gamma}n_r \log \log n_r = 1$ would contradict RH. Very interesting !! Is the non-existence of such a sequence known without assuming RH? If yes, any reference? If no, do some people tried to find such a sequence for contradicting RH? $\endgroup$ – Sebastien Palcoux Aug 21 '18 at 18:49
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    $\begingroup$ It's pretty much obvious that, if you only consider integers which are co-prime to $S$, then the limsup will be $\prod_{p|S}(1 -1/p)$. (The difficulties are all related to inequalities for smaller numbers rather than limits.) $\endgroup$ – user127955 Aug 21 '18 at 20:03

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