On interlacing of zeros of certain family of modular forms
OON INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULARFORMS
EKATA SAHA AND N. SARADHA
Abstract.
Let k = 12 m ( k ) + s ≥
12 for s ∈ { , , , , , } , be an even integer and f be anormalised modular form of weight k with real Fourier coefficients, written as f = E k + m ( k ) (cid:88) j =1 a j E k − j ∆ j . Under suitable conditions on a j (rectifying an earlier result of Getz), we show that all the zeros of f ,in the standard fundamental domain for the action of SL (2 , Z ) on the upper half plane, lies on thearc A := (cid:8) e iθ : π ≤ θ ≤ π (cid:9) . Further, extending a result of Nozaki, we show that for certain family { f k } k of normalised modular forms, the zeros of f k and f k +12 interlace on A ◦ := (cid:8) e iθ : π < θ < π (cid:9) . Introduction
Let H denote the complex upper half plane. Then the full modular group SL (2 , Z ) acts on H by the transformation law z (cid:55)→ az + bcz + d for (cid:16) ac bd (cid:17) ∈ SL (2 , Z ). The standard fundamental domain for this action of SL (2 , Z ) on H is thefollowing subset of H , F := (cid:26) | z | ≥ , − ≤ (cid:60) ( z ) ≤ (cid:27) ∪ (cid:26) | z | > , < (cid:60) ( z ) < (cid:27) . Throughout this article we take k ≥ z ∈ H , the Eisenstein seriesof weight k for the full modular group SL (2 , Z ) is defined by the following absolutely convergentseries, E k ( z ) := 12 (cid:88) c,d ∈ Z ( c,d )=1 cz + d ) k . The Eisenstein series of weight 0 is defined by E := 1. The Eisenstein series E k is a modular form ofweight k for SL (2 , Z ). It is classical that the space of modular forms of weight k is generated by theEisenstein series E k and cusp forms of weight k . We write k = 12 m ( k )+ s with s ∈ { , , , , , } .We will use this notation for k throughout the article without any further mention. The uniquenormalised cusp form of weight 12, denoted by ∆, is defined as follows:∆ := E − E . Date : October 9, 2018.2010
Mathematics Subject Classification.
Key words and phrases. modular forms, location of zeros, interlacing of zeros. a r X i v : . [ m a t h . N T ] F e b EKATA SAHA AND N. SARADHA
Rankin and Swinnerton-Dyer [8] proved that for k ≥
4, all the zeros of the Eisenstein series E k lie in the arc A := (cid:26) | z | = 1 , − ≤ (cid:60) ( z ) ≤ (cid:27) = (cid:26) e iθ : π ≤ θ ≤ π (cid:27) . In 2004, extending the arguments of Rankin and Swinnerton-Dyer, Getz [4] gave a criterion for anormalised modular form of weight k for SL (2 , Z ), written as f = E k + (cid:80) m ( k ) j =1 a j E k − j ∆ j , to haveall its zeros on the arc A , in terms of a j ’s. However, there is a rectifiable error in his proof. Whileestimating H ( θ ) in [4, p. 2225, eq. (2.5)], he used an upper bound for R k − j , ≤ j ≤ m ( k ) from[4, p. 2224, eq. (2.3)] which is valid if k − j ≥
12. But k − m ( k ) is always less than 12, unlessit is 14. We present below a corrected version of his theorem.Let us define(1) δ t = − t = 0 , .
009 if t = 4 , .
304 if t = 6 , .
122 if t = 8 , .
051 if t = 10 , .
022 if t ≥ . Theorem 1.
Let k ≥ and f be a normalised modular form of weight k , written as f = E k + m ( k ) (cid:88) j =1 a j E k − j ∆ j , with a j ∈ R for ≤ j ≤ m ( k ) . Let (cid:15) := sup z ∈ A | ∆( z ) | . Suppose that (2) (3 + δ ) m ( k ) − (cid:88) j =1 | a j | (cid:15) j + (3 + δ s ) | a m ( k ) | (cid:15) m ( k ) ≤ − δ . Then f has m ( k ) zeros (other than possible zeros at i, ρ := e πi/ ) in the fundamental domain F and they all lie on the arc A . Remark 1.
Note that δ above is smaller than the δ of [4, Theorem 1]. This better value isdue to a more accurate estimation of a finite sum using computation. See §
2, Lemma 1. Getz [4]computed (cid:15) ∼ . . . . .Apart from this kind of normalised modular forms there are other examples of families of modularforms, which have been shown to have their zeros on the arc A (see [9, 1, 2]). Now for the zeros ofthese families of modular forms, one interesting question is to ask about their possible interlacing property. Definition 1.
Let α < β . Suppose that f, g are two complex valued functions with simple zeros inthe open interval ( α, β ) . Let α < t < · · · < t m < β and α < t ∗ < · · · < t ∗ m +1 < β be the zeros of f and g , respectively. We say that zeros of f and g interlace in ( α, β ) if t ∗ j < t j < t ∗ j +1 for ≤ j ≤ m. N INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULAR FORMS 3
For example, the zeros of cos( nθ ) and cos(( n + 1) θ ) interlace in (0 , π ) for an integer n ≥ E k has m ( k ) simple zeros in theopen arc A ◦ := (cid:8) e iθ : π < θ < π (cid:9) . For this, they considered the function(3) F k ( θ ) := e ikθ/ E k ( e iθ ) . This is a real valued function for θ real and it has m ( k ) zeros in the interval ( π/ , π/
3) and so does E k ( e iθ ) in A ◦ . Note that m ( k + 12) = m ( k ) + 1. Hence one may look for the interlacing propertyfor the zeros of F k ( θ ) and F k +12 ( θ ) for θ ∈ ( π/ , π/ E k ( e iθ ) and E k +12 ( e iθ ) interlace in A ◦ . This was predicted by Gekeler [3] and proved by Nozaki [7].For the zeros of certain families of weakly holomorphic modular forms considered by Asai,Kaneko and Ninomiya [1], their interlacing property was established by Jermann [6]. Similarproperties for the zeros of the weakly holomorphic modular forms, studied by Duke and Jenkins[2], were proved by Jenkins and Pratt [5]. Here we establish the interlacing property of the zerosof certain family of normalised modular forms that were considered in Theorem 1. Theorem 2.
For each k ≥ , let ( a ( k ) j ) ≤ j ≤ m ( k ) be real numbers such that (4) (3 + δ ) m ( k ) − (cid:88) j =1 | a ( k ) j | (cid:15) j + (3 + δ s ) | a ( k ) m ( k ) | (cid:15) m ( k ) ≤ (cid:18) (cid:19) k/ , where δ s is as in (1) . Then for the family of normalised modular forms ( f k ) k for SL (2 , Z ) definedby f k := E k + m ( k ) (cid:88) j =1 a ( k ) j E k − j ∆ j , the zeros of f k in the fundamental domain F lie on the arc A . Further, the zeros of f k and thoseof f k +12 interlace in A ◦ for each k ≥ . Remark 2.
By (4), we see that (2) is satisfied. Hence by Theorem 1, all the zeros of ( f k ) k lie onthe arc A , thus giving the first assertion of Theorem 2. Remark 3.
Nozaki’s result is a special case of Theorem 2 when a ( k ) j = 0 for all 1 ≤ j ≤ m ( k ).For proving the interlacing of the zeros of the Eisenstein series E k ( z ), Nozaki showed that F k ( θ )as defined in (3), is very well approximated by 2 cos( kθ/
2) for θ ∈ ( π/ , π/ G k ( θ ) := e ikθ/ f k ( e iθ )is also well approximated by 2 cos( kθ/
2) for θ ∈ ( π/ , π/ § m ( k ) zeros in ( π/ , π/ α is a zero of cos( kθ/
2) for θ ∈ ( π/ , π/ α , say ( α − (cid:15), α + (cid:15) ), containing exactly one zero α ∗ of G k ( θ ). It can be easilyseen that, the zeros of cos( kθ/
2) and cos(( k + 12) θ/
2) interlace in ( π/ , π/
3) (see § β, γ of cos(( k + 12) θ/
2) with β < α < γ.
EKATA SAHA AND N. SARADHA
Again, there exist intervals of the form ( β − δ, β + δ ) and ( γ − µ, γ + µ ), each containing exactlyone zero of cos(( k + 12) θ/ β ∗ , γ ∗ respectively. Thus if β + δ < α − (cid:15) < α + (cid:15) < γ − µ, then we obtain that β ∗ < α ∗ < γ ∗ . This argument is used to show that the zeros of G k ( θ ) and G k +12 ( θ ) interlace in ( π/ , π/
36) (see § π/
3. For proving the interlacing property inthe remaining interval we consider the interval (19 π/ , π/ π/ , π/ § A lemma and Proof of Theorem 1
We begin this section with the following lemma. This will be used in the proof of both theTheorems 1 and 2.
Lemma 1.
Let k ≥ and F k ( θ ) := e ikθ/ E k ( e iθ ) for θ ∈ [ π/ , π/ . Then F k ( θ ) = 2 cos( kθ/
2) + (cid:18)
12 cos( θ/ (cid:19) k + (cid:18) i sin( θ/ (cid:19) k + P k ( θ ) , where | P k ( θ ) | < . if k = 4 , . if k = 6 , . if k = 8 , . if k = 10 , . ) k/ if k ≥ . Remark 4.
From Lemma 1 we obtain thatsup θ ∈ [ π/ , π/ | F k ( θ ) | = sup z ∈ A | E k ( z ) | < .
009 if k = 4 , .
304 if k = 6 , .
122 if k = 8 , .
051 if k = 10 , . ) k/ if k ≥ . In particular, we shall use the following bounds for the proof of Theorem 1. For k ≥ θ ∈ [ π/ , π/ | F k ( θ ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12) i sin( θ/ (cid:12)(cid:12)(cid:12)(cid:12) k + | P k ( θ ) | < δ k , where δ k is as in (1). N INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULAR FORMS 5
Proof of Lemma 1.
For N ≥
1, let us define σ N ( θ ) := 12 (cid:88) c,d ∈ Z c d N ( c,d )=1 ce iθ/ + de − iθ/ ) k . In the above definition, whenever empty sum appears, it is assumed to be 0. Now F k ( θ ) = 12 (cid:88) c,d ∈ Z ( c,d )=1 ce iθ/ + de − iθ/ ) k , for θ ∈ [ π/ , π/ k ≥ F k ( θ ) = (cid:88) N ≥ σ N ( θ ) . One can easily see that σ ( θ ) = 2 cos( kθ/
2) and σ ( θ ) = (cid:18)
12 cos( θ/ (cid:19) k + (cid:18) i sin( θ/ (cid:19) k . Hence(6) F k ( θ ) = 2 cos( kθ/
2) + (cid:18)
12 cos( θ/ (cid:19) k + (cid:18) i sin( θ/ (cid:19) k + P k ( θ ) , where P k ( θ ) := (cid:88) N ≥ σ N ( θ ) . Now we split the above sum and write(7) P k ( θ ) = (cid:88) ≤ N ≤ A σ N ( θ ) + (cid:88) N ≥ B σ N ( θ ) , for some integer A ≥ B , the least integer larger than A which can be written as sum ofsquares of two co-prime integers.Note that for any two real numbers c, d one has | cd | ≤ c + d . Since − ≤ θ ≤ θ ∈ [ π/ , π/ | ce iθ/ + de − iθ/ | = c + d + 2 cd cos θ ≥ c + d − | cd | ≥ c + d . Thus we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) ≤ N ≤ A σ N ( θ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:88) ≤ N ≤ A (cid:88) c,d ∈ Z c d N ( c,d )=1 c + d − | cd | ) k/ . EKATA SAHA AND N. SARADHA
For any natural number N , there are at most 2(2 N / +1) pairs ( c, d ) of integers such that c + d = N . Hence for the second sum in (7), using (8) we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) N ≥ B σ N ( θ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:88) N ≥ B (cid:18) N (cid:19) k/ (2 N / + 1) ≤ (cid:18) √ B (cid:19) k/ (cid:88) N ≥ B N (1 − k ) / ≤ (cid:18) √ B (cid:19) k/ (cid:90) ∞ B − x (1 − k ) / dx = (cid:18) √ B (cid:19) ( k +2) / k − (cid:18) B − (cid:19) ( k − / . So we have(9) | P k ( θ ) | ≤ (cid:88) ≤ N ≤ A (cid:88) c,d ∈ Z c d N ( c,d )=1 c + d − | cd | ) k/ + (cid:18) √ B (cid:19) ( k +2) / k − (cid:18) B − (cid:19) ( k − / . Let us denote the right hand side of (9) by p k . Using a C++ programming we obtain optimalvalues of p k as given in Table 1. The values of p k is best possible up to second decimal place. Ourchoice of A is also indicated in the table. k A p k × . . . . . (cid:0) (cid:1) k/ Table 1The computation of the values of p k , when k ≥ p , theprogramme ran for about two hours. Our proof is now complete. (cid:3) Proof of Theorem 1.
It is easy to see that F k ( θ ) is real valued for θ ∈ [ π/ , π/ e iθ ) e iθ = F ( θ ) − F ( θ )1728 ∈ R for θ ∈ [ π/ , π/ . Now, G ( θ ) := e ikθ/ f ( e iθ ) = e ikθ/ E k ( e iθ ) + m ( k ) (cid:88) j =1 a j e i ( k − j ) θ/ E k − j ( e iθ )∆( e iθ ) j e ijθ = F k ( θ ) + m ( k ) (cid:88) j =1 a j F k − j ( θ )∆( e iθ ) j e ijθ . N INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULAR FORMS 7
Since a j ∈ R for 1 ≤ j ≤ m ( k ), we see that G ( θ ) is real valued for θ ∈ [ π/ , π/ G ( θ ) = 2 cos( kθ/
2) + (cid:18)
12 cos( θ/ (cid:19) k + (cid:18) i sin( θ/ (cid:19) k + P k ( θ ) + m ( k ) (cid:88) j =1 a j F k − j ( θ )∆( e iθ ) j e ijθ . We show that(10) | G ( θ ) − kθ/ | < . By Remark 4, | G ( θ ) − kθ/ | < δ + m ( k ) (cid:88) j =1 | a j || F k − j ( θ ) | (cid:15) j ≤ δ + (3 + δ ) m ( k ) − (cid:88) j =1 | a j | (cid:15) j + (3 + δ s ) | a m ( k ) | (cid:15) m ( k ) . Thus, | G ( θ ) − kθ/ | < kθ/ G ( θ ). Now the extremumpoints of cos( kθ/
2) for θ ∈ [ π/ , π/
3] are given by πnk , where k ≤ n ≤ k i.e. 3 m ( k ) + s ≤ n ≤ m ( k ) + s . So we have m ( k ) + 1 such n ’s. Therefore G ( θ ) has m ( k ) zeros on ( π/ , π/ (cid:3) Properties of G k ( θ )3.1. Approximation of G k ( θ ) by kθ/ . We write G k ( θ ) = e ikθ/ f k ( e iθ ) as G k ( θ ) = 2 cos( kθ/
2) + S k ( θ ) , where by Lemma 1, S k ( θ ) = (cid:18)
12 cos( θ/ (cid:19) k + (cid:18) i sin( θ/ (cid:19) k + P k ( θ ) + m ( k ) (cid:88) j =1 a ( k ) j F k − j ( θ )∆( e iθ ) j e ijθ , with | P k ( θ ) | < . (cid:0) (cid:1) k/ . Let(11) Q k ( θ ) := (cid:18) i sin( θ/ (cid:19) k + P k ( θ ) + m ( k ) (cid:88) j =1 a ( k ) j F k − j ( θ )∆( e iθ ) j e ijθ . Then by Lemma 1, Remark 4 and our hypothesis we get(12) | Q k ( θ ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) i sin( θ/ (cid:19) k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + | P k ( θ ) | + m ( k ) (cid:88) j =1 | a ( k ) j || F k − j ( θ ) | (cid:15) j < . (cid:18) (cid:19) k/ . Thus G k ( θ ) = 2 cos( kθ/
2) + S k ( θ ) = 2 cos( kθ/
2) + (cid:18)
12 cos( θ/ (cid:19) k + Q k ( θ )with | Q k ( θ ) | < . (cid:0) (cid:1) k/ . By following the proof of [7, Lemma 4.1], we observe that(13) 0 < S k ( θ ) < , whenever k ≥
24 and θ ∈ [19 π/ , π/ − π/ k ] . EKATA SAHA AND N. SARADHA
Further, we argue as in the proof of [7, Lemma 4.4] to obtain that for θ ∈ [19 π/ , α m ( k ) ], thefunction g ( θ ) := (2 cos( θ/ − k − (2 cos( θ/ − ( k +12) is minimised at θ = 19 π/
32. Hence for k ≥ S k ( θ ) − S k +12 ( θ ) = g ( θ ) + Q k ( θ ) − Q k +12 ( θ ) > g (cid:18) π (cid:19) − . k/ − . ( k +12) / > (cid:18) (cid:18) π (cid:19)(cid:19) − k (cid:32) − (cid:18) (cid:18) π (cid:19)(cid:19) − (cid:33) − . k/ > . . k − . . k > . Thus for k ≥
24 and θ ∈ [19 π/ , α m ( k ) ],(14) S k ( θ ) > S k +12 ( θ ) . Zeros of G k ( θ ) and G k +12 ( θ ) . Let α < · · · < α m ( k ) and β < · · · < β m ( k )+1 denote the zerosof cos( kθ/
2) and cos(( k + 12) θ/
2) in ( π/ , π/ α j = ( + j − k ) π if k ≡ , ( + jk ) π if k ≡ , for 1 ≤ j ≤ m ( k ) and(16) β j = ( + j − k +12 ) π if k ≡ , ( + jk +12 ) π if k ≡ , for 1 ≤ j ≤ m ( k ) + 1. We observe the following properties: for 1 ≤ j ≤ m ( k ),(17) β j < α j < β j +1 (18) min ≤ j ≤ m ( k ) { α j − β j , β j +1 − α j } ≥ πk ( k + 12)and(19) α j +1 − α j = 2 πk for 1 ≤ j ≤ m ( k ) − . Hence we have,(20) cos (cid:18) k (cid:18) α j − πk ( k + 12) (cid:19)(cid:19) cos (cid:18) k (cid:18) α j + 6 πk ( k + 12) (cid:19)(cid:19) < (cid:18) k (cid:16) α j − π k (cid:17)(cid:19) cos (cid:18) k (cid:16) α j + π k (cid:17)(cid:19) < . Remark 5.
From (15) and (16) it follows that k ≥
24, whenever α j + πk ( k +12) > π or α j ≥ π + π k or β j + π ( k +12)( k +24) > π . N INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULAR FORMS 9
We now associate the zeros of G k ( θ ) with the zeros of cos( kθ/ Lemma 2.
Let k ≥ and α ∗ < · · · < α ∗ m ( k ) be the zeros of G k ( θ ) in ( π/ , π/ . Then α ∗ j ∈ (cid:18) α j − πk ( k + 12) , α j + 6 πk ( k + 12) (cid:19) for all j such that α j + πk ( k +12) ≤ π and α ∗ j ∈ (cid:16) α j − π k , α j + π k (cid:17) for all j such that α j ≥ π/
32 + π/ k . In particular, α ∗ j ∈ (cid:0) α j − π k , α j + π k (cid:1) for ≤ j ≤ m ( k ) . Proof of Lemma 2.
Observe that (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) k (cid:18) α j − πk ( k + 12) (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) k (cid:18) α j + 6 πk ( k + 12) (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = 2 sin (cid:18) πk + 12 (cid:19) . We first claim that 2 sin (cid:16) πk +12 (cid:17) > | S k ( θ ) | for θ ∈ ( π/ , π/ (cid:18) πk + 12 (cid:19) > (cid:32) πk + 12 − (cid:18) πk + 12 (cid:19) (cid:33) > . k + 12 . Hence it is enough to show that18 . k + 12 − | S k ( θ ) | = 18 . k + 12 − (cid:18)
12 cos( θ/ (cid:19) k − | Q k ( θ ) | > . k + 12 − (cid:18)
12 cos( θ/ (cid:19) k − . k/ > . Now this is true if2 k/ (cid:18) . θ/ k − ( k + 12) − . k + 12)2 k/ (2 cos( θ/ k (cid:19) > , In particular if 10 . θ/ k − ( k + 12) > . The quantity k (cid:113) k +1210 . is maximum for k = 12, and therefore k (cid:113) k +1210 . < . θ ∈ ( π/ , π/ θ/ > . (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) k (cid:18) α j ± πk ( k + 12) (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) > | S k ( θ ) | for θ ∈ ( π/ , π/ G k (cid:18) α j ± πk ( k + 12) (cid:19) = 2 cos (cid:18) k (cid:18) α j ± πk ( k + 12) (cid:19)(cid:19) + S k (cid:18) α j ± πk ( k + 12) (cid:19) have same sign as 2 cos (cid:16) k (cid:16) α j ± πk ( k +12) (cid:17)(cid:17) . Then from (20) we get that G k (cid:18) α j − πk ( k + 12) (cid:19) G k (cid:18) α j + 6 πk ( k + 12) (cid:19) < . This implies that there exists a zero of G k ( θ ) in the interval (cid:16) α j − πk ( k +12) , α j + πk ( k +12) (cid:17) for all j such that α j + πk ( k +12) ≤ π . It is easy to see from (15) that α m ( k ) ≤ π − πk . Hence we get α m ( k ) + π k ≤ π − π k . Sincecos( kα j /
2) = 0, we have (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) k (cid:16) α j − π k (cid:17)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) k (cid:16) α j + π k (cid:17)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = 1 . As G k ( θ ) = 2 cos( kθ/
2) + S k ( θ ), by (13) we get that the sign of G k (cid:0) α j ± π k (cid:1) is same as that of2 cos (cid:0) k (cid:0) α j ± π k (cid:1)(cid:1) , whenever α j ≥ π + π k . Hence it follows from (21) that G k (cid:16) α j − π k (cid:17) G k (cid:16) α j + π k (cid:17) < . This implies that there exists a zero of G k ( θ ) in the interval (cid:0) α j − π k , α j + π k (cid:1) for all j such that α j ≥ π + π k .From Remark 5 we deduce that a zero α j of cos( kθ/
2) in ( π/ , π/
3) satisfies α j + πk ( k +12) ≤ π or α j ≥ π + π k . Moreover, (cid:18) α j − πk ( k + 12) , α j + 6 πk ( k + 12) (cid:19) ⊂ (cid:16) α j − π k , α j + π k (cid:17) . By (19) all the intervals of the form (cid:0) α j − π k , α j + π k (cid:1) are disjoint for 1 ≤ j ≤ m ( k ). We haveshown above that each of them contains at least one zero of G k ( θ ). We also know that G k ( θ ) hasexactly m ( k ) zeros in ( π/ , π/ kθ/
2) and G k ( θ ) have been labelled asper the increasing order of magnitude, the assertion of the lemma follows. (cid:3) Proof of Theorem 2
We closely follow the arguments of Nozaki [7]. However at several places our arguments aresimpler. We know that a zero α j of cos( kθ/
2) in ( π/ , π/
3) satisfies α j + πk ( k +12) ≤ π or α j ≥ π + π k . Hence we prove Theorem 2 for these two cases.4.1. Case I.
Let α j ∈ (cid:16) π , π − πk ( k +12) (cid:105) . Then we prove the following:(i) β ∗ j < α ∗ j .(ii) α ∗ j < β ∗ j +1 if β j +1 + π ( k +12)( k +24) ≤ π .4.1.1. Proof of (i).
Since α j + πk ( k +12) ≤ π , we have β j + π ( k +12)( k +24) ≤ π . Applying Lemma 2for G k ( θ ) and G k +12 ( θ ), we get α ∗ j ∈ (cid:18) α j − πk ( k + 12) , α j + 6 πk ( k + 12) (cid:19) and β ∗ j ∈ (cid:18) β j − π ( k + 12)( k + 24) , β j + 6 π ( k + 12)( k + 24) (cid:19) . Therefore by (18), we have β ∗ j < β j + 6 π ( k + 12)( k + 24) < β j + 6 πk ( k + 12) ≤ α j − πk ( k + 12) < α ∗ j . N INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULAR FORMS 11
Proof of (ii).
When β j +1 + π ( k +12)( k +24) ≤ π , by Lemma 2 β ∗ j +1 ∈ (cid:18) β j +1 − π ( k + 12)( k + 24) , β j +1 + 6 π ( k + 12)( k + 24) (cid:19) . Therefore by (18), we have α ∗ j < α j + 6 πk ( k + 12) ≤ β j +1 − πk ( k + 12) < β j +1 − π ( k + 12)( k + 24) < β ∗ j +1 . Case II.
Let α j ∈ (cid:2) π + π k , π (cid:1) . Then we prove the following:(iii) β ∗ j < α ∗ j if β j ≥ π + π k +12) .(iv) α ∗ j < β ∗ j +1 .For an integer n , cos( nπ/
2) = n ≡ , − n ≡ , . Further cos( nθ ) is decreasing at π/ n ≡ π/ n ≡ α denotes the first zero of cos( nθ ) in ( π/ , ∞ ), then cos( nθ ) is decreasing at α if n ≡ , α if n ≡ , β denotes the first zero of cos(( n + 6) θ ) in ( π/ , ∞ ),then cos(( n + 6) θ ) is increasing at β if n ≡ , β if n ≡ , ≤ j ≤ m ( k ), if cos( kθ/
2) is increasing (resp. decreasing) at α j , thencos(( k + 12) θ/
2) is decreasing (resp. increasing) at β j . We consider two subcases.(a) The function cos( kθ/
2) is increasing at α j .(b) The function cos( kθ/
2) is decreasing at α j .Note that by Remark 5, we have k ≥
24. Now for k ≥
24 and θ ∈ [19 π/ , π/ − π/ k ], S k ( θ ) > α ∗ j ≶ α j according as cos( kθ/
2) is increasing or decreasing at α j .4.2.1. Proof of (iii) and (iv) when (a) occurs.
It follows by our earlier analysis that cos(( k +12) θ/
2) is increasing at β j +1 . Also β j +1 > α j > π + π k +12) . Hence(22) α ∗ j < α j and β ∗ j +1 < β j +1 . Further β j < β ∗ j if β j ≥ π
32 + π k + 12) . The Case (a) is described pictorially below.
In the above one and other picture below, the respective arrows indicate the direction in which β ∗ j , α ∗ j and β ∗ j +1 lie. We first prove (iii). In fact, we show that the intersection point of the twocosine curves between β j and α j separates β ∗ j and α ∗ j . The function2 cos(( k + 12) θ/ − kθ/
2) = − k + 6) θ/
2) sin(3 θ ) , has a zero between β j and α j , say γ j . For 1 ≤ j ≤ m ( k ),(23) γ j = ( + j − k +6 ) π if k ≡ , ( + jk +6 ) π if k ≡ . We prove that β ∗ j ∈ ( β j , γ j ) and α ∗ j ∈ ( γ j , α j ). By (13), G k ( α j ) > G k +12 ( β j ) >
0. Thus, weare led to show that G k ( γ j ) , G k +12 ( γ j ) <
0. As in the proof of [7, Lemma 4.2], it follows that | kγ j / | > S k ( θ ) and | k + 12) γ j / | > S k +12 ( θ ) , for any j such that γ j ≥ π and θ ∈ [19 π/ , π/ kθ/
2) is increasing at α j and cos( kα j /
2) = 0, we get cos( kγ j / <
0. Hence G k ( γ j ) < G k +12 ( γ j ) <
0. This completes the proof of (iii) when (a) occurs.Next we show (iv) i.e. α ∗ j < β ∗ j +1 . Now if α j ≤ β j +1 − π k +12) , then by (22) and Lemma 2, α ∗ j < α j ≤ β j +1 − π k + 12) < β ∗ j +1 , which proves (iv). Hence we need to consider only β j +1 − π k + 12) < α j and also α ∗ j , β ∗ j +1 ∈ (cid:18) β j +1 − π k + 12) , α j (cid:19) . From (14) we have, S k ( θ ) > S k +12 ( θ ) for θ ∈ [ β j +1 − π/ k + 12) , α j ]. Further, the functioncos( kθ/ − cos(( k + 12) θ/
2) takes positive value in the interval ( γ j , γ j +1 ), where γ j ’s are as in (23).It is easy to check that [ β j +1 − π/ k + 12) , α j ] ⊂ ( γ j , γ j +1 ). Therefore, we obtain that G k ( θ ) − G k +12 ( θ ) = 2 cos( kθ/ − k + 12) θ/
2) + S k ( θ ) − S k +12 ( θ ) > θ ∈ [ β j +1 − π/ k + 12) , α j ]. Taking θ = α ∗ j , we find that G k +12 ( α ∗ j ) <
0. On the other hand, G k +12 ( β j +1 ) >
0. Hence α ∗ j < β ∗ j +1 . This completes the proof of Case (a).4.2.2. Proof of (iii) and (iv) when (b) occurs.
In this case cos(( k + 12) θ/
2) is decreasing at β j +1 . Hence(24) α ∗ j < α j and β ∗ j +1 < β j +1 . Further(25) β j < β ∗ j if β j ≥ π
32 + π k + 12) . N INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULAR FORMS 13
From (24) and (25) we easily get β ∗ j < α ∗ j if β j ≥ π
32 + π k + 12) , which proves (iii). Thus we proceed to prove (iv) i.e. α ∗ j < β ∗ j +1 . If α j + π k ≤ β j +1 , then byLemma 2 and (24), we have α ∗ j < α j + π k ≤ β j +1 < β ∗ j +1 . Hence we can assume that β j +1 < α j + π k and also α ∗ j , β ∗ j +1 ∈ (cid:16) β j +1 , α j + π k (cid:17) . Note that β j +1 < α j + π k implies(26) j > (5 k + 24) /
72 if k ≡ k − /
72 if k ≡ . Here the aim is to show that(27) G k +12 ( θ ) > G k ( θ ) for θ ∈ [ β j +1 , α j + π/ k ] . If (27) holds, then G k ( β ∗ j +1 ) <
0. Since G k ( α j ) >
0, we get α j < α ∗ j < β ∗ j +1 as required. Thus itremains to prove (27). Note that G k +12 ( θ ) − G k ( θ ) =2 cos(( k + 12) θ/ − kθ/ θ/ − ( k +12) − (2 cos( θ/ − k + Q k +12 ( θ ) − Q k ( θ ) . Let r ( θ ) := 2 cos(( k +12) θ/ − kθ/ r ( θ ) in [ β j +1 , α j + π/ k ].For this we show that r ( θ ) is decreasing in this interval. Note that r ( θ ) = − k +6) θ/
2) sin(3 θ ).Now − sin(3 θ ) is decreasing from 1 to 0 in ( π/ , π/ r ( θ ) > θ ∈ ( γ j , γ j +1 ). Thussin(( k + 6) θ/ > θ ∈ ( γ j , γ j +1 ). As γ j and γ j +1 are two consecutive zeros of sin(( k + 6) θ/ k + 6) θ/
2) is decreasing in (cid:16) γ j + γ j +1 , γ j +1 (cid:17) . From (15), (16) and (23),we deduce that γ j + γ j +1 < β j +1 and α j + π k < γ j +1 . Hence sin(( k + 6) θ/
2) is decreasing in [ β j +1 , α j + π/ k ]. As both the functions sin(( k + 6) θ/
2) and − sin(3 θ ) are positive and decreasing, we get that r ( θ ) is decreasing in [ β j +1 , α j + π/ k ]. Therefore r ( θ ) is minimised at α j + π/ k for θ ∈ [ β j +1 , α j + π/ k ]. Note that 2 cos (cid:0) k (cid:0) α j + π k (cid:1)(cid:1) = −
1. Now α j + π k = β j +1 + j − k − k ( k +12) π if k ≡ ,β j +1 + j − k +123 k ( k +12) π if k ≡ . Hence (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) k + 122 (cid:16) α j + π k (cid:17)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) (cid:16) j − k − k π (cid:17)(cid:12)(cid:12)(cid:12) if k ≡ , (cid:12)(cid:12)(cid:12) (cid:16) j − k +126 k π (cid:17)(cid:12)(cid:12)(cid:12) if k ≡ . From (26) and the condition j ≤ ( k − s ) /
12, we get that if k ≡ < j − k − k ≤ k − s − k ≤ − k and if k ≡ < j − k + 126 k ≤ k − s + 126 k ≤ − k . Therefore, (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) k + 122 (cid:16) α j + π k (cid:17)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) π − πk (cid:19) . The equation of the tangent line of the sine curve at π/ y = 12 + √ (cid:16) x − π (cid:17) . Hence 2 sin (cid:0) π − πk (cid:1) < − √ πk .Next let t ( θ ) := (2 cos( θ/ − ( k +12) − (2 cos( θ/ − k . Then t (cid:48) ( θ ) = sin( θ/ θ/ − ( k +1) (cid:0) ( k + 12)(2 cos( θ/ − − k (cid:1) . So t ( θ ) is minimised when (2 cos( θ/ − = kk +12 . We thus get, G k +12 ( θ ) − G k ( θ ) > − (cid:32) − √ πk (cid:33) + (cid:18) kk + 12 (cid:19) k/ (cid:18) kk + 12 − (cid:19) − . (cid:18) (cid:19) k/ > √ πk − k + 12 − . (cid:18) (cid:19) k/ = (cid:18) k − k + 12 (cid:19) + 4 √ π − k − . (cid:18) (cid:19) k/ > . This completes the proof of Case (b).4.3.
Remaining cases.
By Cases I and II, in order to complete the proof of Theorem 2, we needto prove the following two cases.(v) α ∗ j < β ∗ j +1 when α j ∈ (cid:16) π , π − πk ( k +12) (cid:105) and β j +1 + π ( k +12)( k +24) > π .(vi) β ∗ j < α ∗ j when α j ∈ (cid:2) π + π k , π (cid:1) and β j < π + π k +12) . N INTERLACING OF ZEROS OF CERTAIN FAMILY OF MODULAR FORMS 15
Proof of (v).
We show that α j ≥ π + π k . Hence by § α ∗ j < β ∗ j +1 . Suppose that α j < π + π k . Then using (15) we get that(28) 27 k > j −
384 if k ≡ , j −
96 if k ≡ . Since β j +1 + π ( k +12)( k +24) > π , it follows from Remark 5 and (16) that k ≥
24 and(29) 0 < ( k + 24)(72 j − k −
24) + 216 if k ≡ , ( k + 24)(72 j − k + 12) + 216 if k ≡ < ( k + 24)( − k + 192) + 1728 if k ≡ , ( k + 24)( − k ) + 1728 if k ≡ . This is a contradiction since k ≥ Proof of (vi).
As in (v), we can deduce that α j + πk ( k +12) ≤ π . Hence β ∗ j < α ∗ j by § (cid:3) Acknowledgement:
We would like to thank Biswajyoti Saha and Jhansi Bhavani V. for helpingus with the computations in Lemma 1.
References [1] T. Asai, M. Kaneko and H. Ninomiya, Zeros of certain modular functions and an application,
Comment. Math.Univ. St. Paul. (1997), no. 1, 93–101.[2] W. Duke and P. Jenkins, On the zeros and coefficients of certain weakly holomorphic modular forms, PureAppl. Math. Q. (2008), no. 4, Special Issue: In honor of Jean-Pierre Serre, Part 1, 1327–1340.[3] E.-U. Gekeler, Some observations on the arithmetic of Eisenstein series for the modular group SL(2 , Z ), Arch.Math. (Basel) (2001), 5–21.[4] J. Getz, A generalization of a theorem of Rankin and Swinnerton-Dyer on zeros of modular forms, Proc. Amer.Math. Soc. (2004), no. 8, 2221–2231.[5] P. Jenkins and K. Pratt, Interlacing of zeros of weakly holomorphic modular forms,
Proc. Amer. Math. Soc.Ser. B (2014), 63–77.[6] J. Jermann, Interlacing property of the zeros of j n ( τ ), Proc. Amer. Math. Soc. (2012), no. 10, 3385–3396.[7] H. Nozaki, A separation property of the zeros of Eisenstein series for SL(2 , Z ), Bull. Lond. Math. Soc. (2008),no. 1, 26–36.[8] F. K. C. Rankin and H. P. F. Swinnerton-Dyer, On the zeros of Eisenstein series, Bull. London Math. Soc. (1970), 169–170.[9] R. A. Rankin, The zeros of certain Poincar´e series, Compositio Math. (1982), no. 3, 255–272. Ekata Saha and N. Saradha,School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha Road, Navy Nagar,Mumbai, 400 005, India
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