2017
DOI: 10.1007/s00013-017-1072-x
|View full text |Cite
|
Sign up to set email alerts
|

Comparing Hecke eigenvalues of newforms

Abstract: Given two distinct newforms with real Fourier coefficients, we show that the set of primes where the Hecke eigenvalues of one of them dominate the Hecke eigenvalues of the other has density ≥ 1/16. Furthermore, if the two newforms do not have complex multiplication, and neither is a quadratic twist of the other, we also prove a similar result for the squares of their Hecke eigenvalues.

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
4
1

Citation Types

1
6
0

Year Published

2018
2018
2024
2024

Publication Types

Select...
6

Relationship

1
5

Authors

Journals

citations
Cited by 9 publications
(7 citation statements)
references
References 7 publications
1
6
0
Order By: Relevance
“…Remark 2. One can easily find that for i � 0, j � 1, or i � j � 1, we have ((1 + i − δ 1 )/4(ij + j + δ 2 ) 2 ) � (1/16), which are consistent with the numerical values of Chiriac's results (see eorems 1 and 1.3 in [15]). By comparison, ((1 + i − δ 1 )/4(ij + j + δ 2 ) 2 ) � (1/16[(j + 1)/2] 2 ) with i � 0, 1 ≤ j ≤ 8, and ((1 + i − δ 1 )/4(ij + j + δ 2 ) 2 ) � (1/4j (j + 1) 2 ) with 1 ≤ i � j ≤ 4.…”
Section: Introductionsupporting
confidence: 86%
See 2 more Smart Citations
“…Remark 2. One can easily find that for i � 0, j � 1, or i � j � 1, we have ((1 + i − δ 1 )/4(ij + j + δ 2 ) 2 ) � (1/16), which are consistent with the numerical values of Chiriac's results (see eorems 1 and 1.3 in [15]). By comparison, ((1 + i − δ 1 )/4(ij + j + δ 2 ) 2 ) � (1/16[(j + 1)/2] 2 ) with i � 0, 1 ≤ j ≤ 8, and ((1 + i − δ 1 )/4(ij + j + δ 2 ) 2 ) � (1/4j (j + 1) 2 ) with 1 ≤ i � j ≤ 4.…”
Section: Introductionsupporting
confidence: 86%
“…Inspired by [13], Chiriac [15] started to compare Hecke eigenvalues over prime numbers and simultaneously showed that the sets of primes for λ f (p) < λ g (p) and λ 2 f (p) < λ 2 g (p) both have analytic density at least 1/16. Notice that the pair-Sato-Tate conjecture yields a stronger result for the former set in [15] with natural density 1/2 in replace of at least 1/16 (see Proposition 2.1 (iii) in [16]). Of course, this result is also valid for the analytic density since the existence of the natural density implies that of the analytic density, and they are equal.…”
Section: Introductionmentioning
confidence: 99%
See 1 more Smart Citation
“…As it was already mentioned in the introduction, without appealing to the joint Sato-Tate distribution, one can still give a lower bound for the density of the set of primes from part (iii) of Proposition 2.1. This was done by the author in [2], where it is proved that if f = g then {p | λ f (p) > λ g (p)} has analytic density at least 1/16. Moreover, it is also shown that if f and g do not have complex multiplication, and neither is a quadratic twists of the other, then the same lower bound holds for the set {p | λ 2 f (p) > λ 2 g (p)}.…”
Section: Resultsmentioning
confidence: 99%
“…If, instead of considering the set of dominating coefficients over all positive integers, one restricts the analysis just to those indexed by prime numbers, then the joint Sato-Tate distribution readily implies that the corresponding set of primes has density 1/2. Without such a powerful tool, the author [2] obtained a lower bound of 1/16 for the analytic density of that set of primes using the holomorphy and the non-vanishing only of the first few symmetric power L-functions (see Section 3 for further discussion).…”
Section: Introductionmentioning
confidence: 99%