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Background Basic Problem The P2-path Some examples References
Background Basic Problem The P2-path Some examples References
P N umber T heory Olivier Ramar A dditive C ombinatorics - - PowerPoint PPT Presentation
P N umber T heory Olivier Ramar A dditive C ombinatorics - - PowerPoint PPT Presentation
Background Basic Problem The P 2 -path Some examples References S ome in P N umber T heory Olivier Ramar A dditive C ombinatorics in M arseille 2020
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Background Basic Problem The P2-path Some examples References
Also in Number Fields
Also different groups!
Together with Sanoli Gun and Jyothsnaa Sivaraman
Theorem Let K be a number field, ∃c(K)/
For any q integral ideal, each class of Hq(K) (the narrow ray class group) contains a p1p2p3
◮ NK/Qp1, p2, p3 ≤ c(K)NK/Q(q)16/3 +ε ◮ p1, p2, p3 of degree one.
In progress – Dependence on d(K)? To be compared with (Zaman, 2016) (ray class field, exceptional case) and with (Zaman, 2017) (Tchebotarev)
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Background Basic Problem The P2-path Some examples References
Basic Problem
Let G be a finite abelian group, Let A ⊂ G be such that |A|/G ≥ η > 1/3, Find conditions so that 3A = G. What kind of conditions?? Conditions:
◮ [C0] A generates G. ◮ [C1] A intersects every subgroup of G
- f index 2.
How do we verify these conditions? Brun-Titchmarsh inequality and its variants. [p ≤ qα and η = η(α) decreases with α].
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Background Basic Problem The P2-path Some examples References
Special case of Kneser’s Theorem:
Lemma Let A ⊂ G (G finite ABELIAN group).
Let H subgroup of G that stabilizes A + A. Suppose A meets λ cosets of H.
|A + A| ≥ (2λ − 1)|H|
Two extreme cases: Generic case: H = {0} −→ |2A| ≥ 2|A| − 1,
WIN!
A subgroup −→ |2A| = |A| ←−
No increase in size! Large Subgroup Problem
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Background Basic Problem The P2-path Some examples References
◮ 2A = A + A is a union of H-cosets. ◮ 3A is a union of H-cosets. ◮ G∗ = G/H, A∗ = A/H, it is enough to
show that 3A∗ = G∗.
◮ 2A∗ has a trivial stabilizer in G∗. ◮ Set Y = |G/H| = |G∗| and λ = |A/G|. ◮ λ|H| ≥ |A| ≥ η|G| λ ≥ ⌈ηY⌉. ◮ If λ + (2λ − 1) > |G∗| then 3A∗ = G∗.
When 3⌈ηY⌉ ≥ Y + 2, we are done! When Y ≥ 5 3η − 1, we are done! If Y = 3k or Y = 3k + 1, we are done!
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Background Basic Problem The P2-path Some examples References
A first result
Theorem Assume [C0] and [C1].
If |A/G| > 2/5 = 0.4 then 3A = G. Difficulty with G∗ = Z/5Z and A∗ = {0, 1}. We have 3A∗ G∗. Recall
Conditions:
◮ [C0] A generates G. ◮ [C1] A intersects every subgroup of G of index 2.
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Background Basic Problem The P2-path Some examples References
An unusual condition!
Conditions:
◮ [C2] A meets every subgroup of index 5.
But [possibly] NOT every coset!!
◮ [C3(Y0)]
Given any subgroup H ⊂ G of index Y ≤ Y0, Given any coset u + H,
A ∪ 2A intersects u + H.
Where does it come from? Again from sieve! From the weighted sieve in fact. It is a reformulation of Brun-Titchmarsh inequality for cosets but very different consequences!! Unclear for ray class group, work in progress –
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Background Basic Problem The P2-path Some examples References
A new (easy!) lemma
Lemma Let G∗ finite abelian.
Assume A∗ ∪ 2A∗ = G∗ and 3A∗ G∗. Then 1. 0 A∗ and A∗ = −A∗,
- 2. The stabilizer of A∗ is 0,
- 3. A∗ ∩ 2A∗ = ∅,
Sum-free set!
- 4. |A∗| ≤ (|G∗| + 1)/3,
- 5. 3A∗ = G \ {0}.
The bound |A∗| ≤ (|G∗| + 1)/3 is optimal in general. But in specific cases? Large sum-free sets −→ (Yap, 1975), (Green & Ruzsa, 2005).
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Background Basic Problem The P2-path Some examples References
Consequences
Theorem Assume [C0], [C1], [C2] and [C3(5)].
If |A/G| > 3/8 = 0.375 then 3A = G.
Theorem Assume [C0], [C1], [C2] and [C3(5)].
Assume the 2-part of G is ≃ (Z/2Z)r. If |A/G| > 4/11 = 0.3636 . . . then 3A = G.
That’s [p|q =⇒ p ≡ 3[4]] for (Z/qZ)× or (Z/4qZ)×.
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Background Basic Problem The P2-path Some examples References
Elements of proof
◮ Y = |G/H| = 8 is the difficulty. ◮ A∗ = A/G has exactly three elements. ◮ G∗ can be Z/8Z, Z/2Z × Z/4Z or (Z/2Z)3. ◮ Rule out (Z/2Z)3: A∗ is a basis of G∗. ◮ Condition of 2-part: rules out other cases.
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Background Basic Problem The P2-path Some examples References
In general
◮ In G∗ = Z/8Z, A∗ = {1, 4, 7} and A∗ = {3, 4, 5} have to be ruled out. ◮ In G∗ = Z/2Z × Z/4Z, A∗ = {(0, 1), (0, 3), (1, 0)}, A∗ = {(1, 1), (1, 3), (1, 0)}, A∗ = {(1, 1), (1, 3), (0, 2)}, A∗ = {(1, 1), (1, 3), (1, 0)}. ◮ The second does not touch Z/2Z × {0, 2}. ◮ The fourth does not touch {0} × Z/4Z.
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Background Basic Problem The P2-path Some examples References
Examples
For the upper bound: G∗ = Z/(3k + 2)Z, A∗ = [k + 1, 2k + 1], Then 2A∗ = [2k + 2, 3k + 1] ∪ [0, k],
|A∗| = k + 1 = (|G∗| + 1)/3.
For the lower bound:
Numerical experiments when G is Z/ℓZ. Here are the sizes of the possible sets A∗:
ℓ
8 11 17 18 19 20 21 22 23 24 25
|A∗|
3 4 6 6 6 6,7 6 7 8 7,8 8
ℓ
26 27 28 29
|A∗|
7,8,9 8 8,9 8,10
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Background Basic Problem The P2-path Some examples References
For the lower bound: Some examples:
A∗ = {1, 4, 7} mod 8, A∗ = {1, 3, 5, 12, 14, 16} mod 17, A∗ = {1, 3, 7, 12, 16, 21, 25, 27} mod 28, A∗ = {1, 3, 5, 12, 14, 16, 23, 25, 27} mod 28.
A large cardinality example is given by C ∪ (−C) where C = {1, 3, 5, 7, 9, 20, 22, 24} mod 48. A large cardinality example is given by C ∪ (−C) where C = {1, 3, 5, 7, 9, 11, 17} mod 49.
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Background Basic Problem The P2-path Some examples References Green, Ben, & Ruzsa, Imre Z. 2005. Sum-free sets in abelian groups. Israel J. Math., 147, 157–188. Ramaré, O., & Srivastav, Priyamvad. 2019. Products of primes in arithmetic progressions. To appear in Int. Journal of Number Theory, 17 pp. Appendix by O. Serra. Ramaré, O., & Walker, Aled. 2018. Products of primes in arithmetic progressions: a footnote in parity breaking.
- J. Number Theory of Bordeaux, 30(1), 219–225.
Yap, H. P . 1975. Maximal sum-free sets in finite abelian groups. V.
- Bull. Austral. Math. Soc., 13(3), 337–342.
Zaman, Asif. 2016. On the least prime ideal and Siegel zeros.
- Int. J. Number Theory, 12(8), 2201–2229.
Zaman, Asif. 2017. Bounding the least prime ideal in the Chebotarev density theorem.
- Funct. Approx. Comment. Math., 57(1), 115–142.