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Quoting from Green-Tao, "Linear equations in primes" (especially Cor. 1.9 in https://arxiv.org/pdf/math/0606088.pdf), any system of linear forms of finite complexity and without any local obstructions will assume simultaneous prime values infinitely often.

(E.g., $(X,Y,X+Y)$ is of finite complexity, but with a local obstruction at 2, since one of these is always even. On the other hand $(X,Y,X+2Y)$ has no local obstruction, so there are infinitely many primes $x,y$ such that $x+2y$ is also prime.)

My question is: Does this result remain true when, instead of asking for prime values, one asks for values in some positive density subset of the primes? This kind of strengthening is true e.g. for the famous special case $X, X+Y, X+2Y,...$ of arithmetic progressions in primes, but I couldn't find it in the general case.

Again, to illustrate what I mean: Let's take the set $S$ of all primes $\equiv 1$ mod $4$. Then the system $(X,Y,3X+4Y)$ will be obstructed, since for $x,y\equiv 1$ mod $4$, one has $3x+4y\equiv 3$ mod $4$. On the other hand, $(X,Y,5X+4Y)$ would not be obstructed, so should be expected to have infinitely many values in $S^3$. Is this known in generality?

[I should add that for the last example, one may of course just replace X and Y by 4X+1 and 4Y+1; but of course there are positive density subsets of primes which are not just residue classes. I'm interested in those as well.]

[EDIT: To be more explicit: Let S be a Chebotarev set of primes, i.e. the set of all primes with some given Frobenius in some given Galois extension $K/\mathbb{Q}$. Then there will usually be a congruence condition (but not only!) on $S$ coming from the largest cyclotomic subfield of $K$. So there is some $N$ and some smallest possible finite union $U$ of mod $N$ residue classes containing all of $S$. A local ostruction for a system of linear forms now definitely occurs if there are no integer values simultaneously lying in $U$. I'm asking if that is the only obstruction for values to simultaneously lie in $S$.

Example: if S is the set of primes which decompose completely in the splitting field of $x^3-x^2+1$, then the largest abelian subfield is $\mathbb{Q}(\sqrt{-23})$. To be decomposed in this field means to be a square mod 23 (leaving 11 of 23 possible residue classes). I'm tempted to conjecture that a system of linear forms of finite complexity which has integer values simultaneously lying in one of these 11 classes (i.e. not obstructed at 23), and which for any other prime $p$ has integer solutions simultaneously coprime to $p$ (i.e. not obstructed at $p$) will have prime values simultaneously lying in $S$.]

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    $\begingroup$ It is not clear what would you call a local obstruction, when the subset you look at is npt a residue class. Moreover, usually the densityr of solutions of a certain system in primes is small, so you may be able just to get rid of them, keeping positive density of primes. $\endgroup$ – Ilya Bogdanov Nov 21 '18 at 7:24
  • $\begingroup$ @Ilya Bogdanov: I didn't specify what the local obstruction should mean in that case since I'm not sure about the precise result, but intuitively the only way to obstruct solutions locally at p should be when the set S intersects certain mod p^N residue classes (almost) trivially whereas the values of one of the linear forms always comes to lie in such a class. Since I'm thinking in particular of the case where S is a Chebotarev set (i.e., primes with a given Frobenius in a given Galois extension K/Q), obstructions should only come from cyclotomic subfields of K. $\endgroup$ – Joachim König Nov 21 '18 at 8:16
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    $\begingroup$ I think $(X, Y, Z, X+Y+Z)$ has finite complexity, and no local obstructions in the G-T sense, but clearly you can't solve it in primes 1 mod 3. So we definitely need a more robust sense of local obstruction. $\endgroup$ – Sean Eberhard Nov 21 '18 at 19:24
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    $\begingroup$ Your example is a bit strange. The system $(X, Y, 5X+4Y)$ in general is clearly obstructible: try solving it in primes 1 mod 100. But as you point out (via your substitution), it is solvable in primes 1 mod 4. So your desired definition of obstruction has to tie in your specific set $S$. But then it's unclear what you're asking in general unfortunately. $\endgroup$ – Sean Eberhard Nov 21 '18 at 19:35
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    $\begingroup$ @Sean: "Obstructed" of course should always mean "obstructed with respect to S". Since there is no tuple of integers $(X,Y,Z)$ such that each of $X,Y,Z,X+Y+Z$ is in $S=3\mathbb{N}+1$, that's of course obstructed. $\endgroup$ – Joachim König Nov 22 '18 at 22:52
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Roughly speaking, the transference principle used in my work with Ben shows that the obstructions to solving (finite complexity) linear equations in dense sets of primes are the same as the obstructions to solving linear equations in dense sets of integers (modulo a technical issue involving dilating the equations by a medium-size integer $W$, coming from the "$W$-trick"). Thanks to the inverse conjecture for the Gowers norms, these obstructions are classified: local obstructions such as congruence classes are certainly obstructions, but in fact every nilsystem could potentially produce an obstruction.

For instance, the circle shift $x \mapsto x + \sqrt{2} \hbox{ mod } 1$ on ${\bf R}/{\bf Z}$ provides a family of obstructions based on the distribution of values of $\sqrt{2} n \hbox{ mod } 1$ in the set $S$ of interest. If for example $S$ happens to be contained in the Bohr set $\{ n: \{ \sqrt{2} n \} \in [0, 0.01] \}$ (where $\{x\} := x - \lfloor x \rfloor$ is the fractional part function), then there will be no patterns of the form $x, x+y, x+2y+2$ in the set $S$, despite the lack of congruence obstructions, simply because $\sqrt{2} x - 2 (\sqrt{2} (x+y)) + \sqrt{2}(x+2y+2) = 2\sqrt{2} \hbox{ mod } 1$ is too far away from zero. Intersecting this Bohr set with the primes would then give a positive density set of primes that had no patterns of the form $x,x+y,x+2y+2$.

For more complicated patterns one can cook up similar examples involving for instance the quadratic Bohr set $\{ n: \{ \sqrt{2} n^2 \}\in [0,0.01]\}$ or the more exotic ``bracket quadratic'' Bohr set $\{ n: \{ \lfloor \sqrt{2} n \rfloor \sqrt{3} n \} \in [0,0.01] \}$ that is associated to a Heisenberg nilflow. For a pattern of complexity $s$ (as defined in our paper, or by using the somewhat smaller notion of "true complexity" studied in later papers by Gowers-Wolf and others), every nilsystem of nilpotency class $s$ or less is a potential obstruction. (Congruence class obstructions can be thought of as corresponding to finite systems, which by convention can be viewed as "complexity 0" obstructions.)

On the other hand, if your set $S$ is equidistributed with respect to all such Bohr sets (after taking into account the local irregularities of distributions of the primes, for instance by using the "W-trick" of our paper), then the theory in my papers with Ben (together with one paper as well with Tamar Ziegler) will allow one to compute asymptotics for these patterns.

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    $\begingroup$ That helps a lot, thanks! So, would this equidistribution property be guaranteed for Chebotarev sets? $\endgroup$ – Joachim König Nov 23 '18 at 7:15
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    $\begingroup$ I would imagine that Chebotarev sets have enough multiplicative structure that Vinogradov's method of bilinear sums would be able to apply here. A model problem would be to see if one can get any nontrivial cancellation in sums such as $\sum_{n \leq x: n \in S} e( \sqrt{2} n)$. (In our paper we ended up replacing the primes with the Mobius function as it was a bit cleaner to get cancellation for the latter, but as long as one can save a large power of $\log x$ in the estimates they should be basically equivalent.) $\endgroup$ – Terry Tao Nov 23 '18 at 16:09

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