I am looking for an intuitive reason for why the Maynard-Tao weights work well to capture many primes of the form $n+h_1, \ldots , n+h_k$, where $(h_1, \ldots , h_k)$ is any admissible $k$-tuple.

For instance, where is the mass of the weights concentrated? Why is it good to concentrate the mass on these integers? Is there a explanation to why these work better than the GPY-weights (see this related question)?

For instance Proposition 4.2 of this polymath 8b paper asserts that the only integers the contribute are the $n$ such that $n+h_1, \ldots, n+h_k$ are $x^{\epsilon}$-rough, where $\epsilon = 1/(10k)$.

  • 2
    $\begingroup$ My understanding is that the key advantage is that the weight is defined on tuples of integers $n+h_1,\dots, n+h_k$ rather than on the single integer $(n+h_1)\dots(n+h_k)$. This allows greater flexibility which turns out to be a big advantage, though I don't know how exactly the weight depends on the factorization. $\endgroup$
    – Will Sawin
    Jan 27 '16 at 19:47
  • $\begingroup$ @WillSawin: Have a look at my response, I hope you find it useful. $\endgroup$
    – GH from MO
    Jan 27 '16 at 23:31

In retrospect the Maynard-Tao weights are quite natural, and they can be understood as the result of the following evolution.

1. Let us start from the naive weights $$\nu(n):=1_\text{$n+h_1,\dots,n+h_k$ are primes}$$ If we believe in Dickson's conjecture, then these weights are perfectly fine, because they concentrate on the best $n$'s. The problem is that proving the desired inequality $$ \sum_{x\leq n\leq 2x}\nu(n)\sum_{i=1}^k 1_\text{$n+h_i$ is prime}>\sum_{x\leq n\leq 2x}\nu(n)\tag{$\ast$}$$ is as hard as proving Dickson's conjecture. Indeed, this inequality implies that not all the $\nu(n)$'s are zero. A slightly refined analytic variant of these naive weights is $$\nu(n):=\sum_{d\mid P(n)}\mu(d)\log^{k+\ell}\left(\frac{P(n)}{d}\right),$$ where $P(n):=(n+h_1)\dots(n+h_k)$ is as usual. The right hand side is the convolution of $\mu$ and $\log^{k+\ell}$ evaluated at $P(n)$, hence (nontrivially) these weights are nonnegative and they are nonzero if and only if $P(n)$ has at most $k+\ell$ distinct prime factors. So, if $\ell$ is not too large (e.g. $\ell<k/2$), these weights capture $n$'s such that $n+h_1,\dots,n+h_k$ contains several primes. An advantage of these weights is that they incorporate the usual inclusion-exclusion sieve technology via the Möbius values $\mu(d)$. However, they suffer from the same problem as the original naive weights: we cannot evaluate the two sides of ($\ast$) for them. At a technical level, the source of the problem is that the divisors $d$ of $P(n)$ get too large (compared to $x$), and we lose control of the error terms.

2. A key idea of sieve theory is to remedy the above mentioned difficulty by restricting to fewer divisors, e.g. by cutting them off at some artificial bound $R$. Of course, a smooth cut-off is usually better for analytic purposes than a sharp cut-off. For example, one can try to use the weights $$\nu(n):=\left(\sum_{\substack{{d\mid P(n)}\\{d\leq R}}}\mu(d)\left(1-\frac{\log d}{\log R}\right)^{k+\ell}\right)^2,$$ which differs from the previous version (up to scaling) as follows: the restriction $d\leq R$ is in place, the fraction $P(n)/d$ became $R/d$, and the whole sum got squared. Only the last change needs an explanation: the cut-off $d\leq R$ destroys positivity of the weight, and squaring is a good analytic way to restore positivity.

3. The above squared weights are precisely the Goldston-Pintz-Yildirim weights. A key insight of Soundararajan was that it is more efficient to perform the smooth cut-off $d\leq R$ by using a more general function of $\frac{\log d}{\log R}$. That is, the Goldston-Pintz-Yildirim weights evolved to $$\nu(n):=\left(\sum_{d\mid P(n)}\mu(d)g\left(\frac{\log d}{\log R}\right)\right)^2,$$ where $g:\mathbb{R}\to\mathbb{R}$ is a smooth function supported on $[0,1]$. Morally, these weights try to imitate the event that $P(n)$ has few prime factors by sieving through the divisors $d\mid P(n)$. However, what we really want is that many of the individual factors $n+h_1,\dots,n+h_k$ are primes, not just that their product has few prime factors. So it is natural to further refine the above Soundararajan weights to $$\nu(n):=\left(\sum_{\forall i: d_i\mid n+h_i}\mu(d_1)\dots\mu(d_k)f\left(\frac{\log d_1}{\log R},\dots,\frac{\log d_k}{\log R}\right)\right)^2,$$ where $f:\mathbb{R}^k\to\mathbb{R}$ is a smooth function supported on the simplex $$\{(t_1,\dots,t_k)\in\mathbb{R}_{\geq 0}^k: t_1+\dots+t_k\leq 1\}.$$ These are precisely Maynard-Tao weights, and they really try to imitate that many of the individual factors $n+h_1,\dots,n+h_k$ are primes.

4. It is worthwhile to bear in mind that the Soundararajan weights (hence also the Goldston-Pintz-Yildirim weights) are a special case of the Maynard-Tao weights. Indeed, let us assume (without much loss of generality) that $n+h_1,\dots,n+h_k$ are pairwise coprime. Then, for $$f(t_1,\dots,t_k):=g(t_1+\dots+t_k)$$ the Maynard-Tao weights reduce to the Soundararajan weights.

  • 1
    $\begingroup$ Can you say a word on the motivation of the Soundararajan weights? It is always confusing to me - a naive calculation of the optimal parameters for the Selberg sieve is certainly along the lines of your bullet point 2, but not something as general as 3. Is there any insight on why we can allow the flexibility of a general polynomial? $\endgroup$
    – Pig
    Jan 27 '16 at 23:58
  • 1
    $\begingroup$ @user31814: This is an excellent question, and I don't have a good intuition here. One possible explanation could be based on the parameter $\ell$, which is pretty arbitrary at this point (apart from being small). One could take several different values of $\ell$ inside the sum and combine the resulting powers: this would already yield a rather general $g$ as in Soundararajan's weights, hence optimizing the outcome over all possible $g$'s seems like a reasonable idea. $\endgroup$
    – GH from MO
    Jan 28 '16 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.