This is my third question on this site regarding Montgomery's conjecture -- and I apologize if this is too much -- but I am still not understanding well why this conjecture is believed to be true.

The conjecture I am talking about is as follows (I am giving the slightly corrected version of Freidlander-Granville). et $q>1$ be an integer, $a$ an integer coprime to $q$, $\psi(q,a,x) = \sum_{p^\alpha < x, p^\alpha \equiv a \pmod{q}} \Lambda(n)$. Then:

Conjecture for $x>q$, one has $\psi(x,q) = \frac{x}{\phi(q)} + O(x^{1/2+\epsilon} q^{-1/2})$, with an implied constant depending only on $\epsilon$.

In his answer to my preceding question, Matt Young gives the following heuristic: For $\chi$ a non-principal Dirichlet character of $(\mathbb Z/q\mathbb Z)^\ast$, one has under GRH $\psi(\chi,x) = O(x^{1/2+\epsilon})$. Now $\psi(q,a,x)$ is the arithmetic average of the (approximately $q$) terms $\chi^{-1}(a) \psi(\chi,x)$, and if those (weighted) terms are in random position one should expect their sum to have a norm roughly $q^{1/2}$ the norm of the individual term (by Einstein's Brownian motion theorem if you like), giving the conjecture.

This heuristic helped me a lot then but now I'd like to go further.

Is there a good reason to believe that the $\chi^{-1}(a) \psi(\chi,x)$ are kind of randomly distributed, for every $a$ relatively prime to $q$? Could this expectation have some link, through the explicit formulas, with the conjectures on the position of zeros on the critical line of Dirichlet L-functions (assuming GRH)?

  • $\begingroup$ I think Katz has something in his book on Kloostermann sums, that they are equidistributed in some suitable sense (look at all Kloos to given a modulus, similar to looking at all characters). This is then a function field analogue perhaps. Maybe this is a bit of stretch though. $\endgroup$
    – v08ltu
    Commented Jun 5, 2013 at 18:53
  • $\begingroup$ Is this "equidistribution in some suitable sense" of Kloosterman sums related to Chebotarev's theorem? $\endgroup$ Commented Jun 8, 2013 at 21:20

1 Answer 1


This is not exactly what you asked, but I believe that there is a simpler heuristic for Montgomery's conjecture which does not involve characters.

Let $A$ be the arithmetic progression $a\pmod q$, where $a$ and $q$ are coprime to each other. We consider a sequence of Bernoulli random variables $(X_n)_{n\in A}$ such that

$$\textbf{Prob}(X_n=1)=\frac{q}{\phi(q)\log n} \quad\text{and}\quad \textbf{Prob}(X_n=0)=1-\frac{q}{\phi(q)\log n}.$$

The event $X_n=1$ models the event that $n$ is prime. (There are about $\le x/q$ elements in $A$ up to $x$ and about $x/(\phi(q)\log x)$ of them are primes.) Let $\Psi(x)=\sum_{n\le x} X_n\log n$. This is a random model for $\psi(x)$ (or rather $\theta(x)=\sum_{p\le x}\log p$, but this is merely a technicality). Note that

$$ \mathbb{E}[\Psi(x)] = \frac{q}{\phi(q)}\sum_{n\le x,\,n\equiv a\pmod q}1 = \frac{q}{\phi(q)}(x/q+O(1))=\frac{x}{\phi(q)} + O(q/\phi(q)) $$ and $$ \text{Var}[\Psi(x)] = \frac{q}{\phi(q)} \sum_{n\le x,\, n\equiv a\pmod q}\log n \sim \frac{x\log x}{\phi(q)}. $$ The Central Limit Theorem implies that the random variable $(\Psi(x) - x/\phi(q))/\sqrt{x\log x/\phi(q)}$ tends to the standard normal distribution. In particular, almost surely as $x\to\infty$, we have that $$ \left|\Psi(x) - \frac{x}{\phi(q)} \right| \ll x^\epsilon \sqrt{\frac{x\log x}{\phi(q)}}. $$ (In fact, one should expect this to become accurate as soon as $x/q$ becomes a bit large, that is to say there are enough summands in $\Psi(x)$.) So we deduced Montgomery's conjecture for $\Psi(x)$, the random model of $\psi(x)$.

  • $\begingroup$ Edited to correct a mistake in the probability that $X_n=1$ and $X_n=0$. $\endgroup$ Commented Jun 10, 2013 at 13:30
  • $\begingroup$ Dear Dimitris, thanks for your answer. Even if it is not exactly what I asked, this is more or less what I needed. Actually, I recently noticed that some non-abelian generalizations of Montgomery's conjecture (suggered by Murty and Murty) are false (I didn't want then to be false, I swear-- I wanted to apply them: but they gave me too strong results to be true). Now I am trying to understand what goes wrong, in order to get a conjecture that might be correct. I will try to work out what this heuristics give in the non-abelian case, and see what comes out. $\endgroup$
    – Joël
    Commented Jun 11, 2013 at 1:01

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