On the zeros of Eisenstein series for $Γ_0^* (5)$ and $Γ_0^* (7)$

0.23 MB·English

by  Junichi Shigezumi

ON THE ZEROS OF THE EISENSTEIN SERIES
∗ ∗
FOR Γ (5) AND Γ (7)
0 0
Junichi Shigezumi
Abstract. We locate almost all the zeros of the Eisenstein series associated with the Fricke groups of
level 5 and 7 in their fundamental domains by applying and extending the method of F. K. C. Rankin
7
and H. P. F. Swinnerton-Dyer (1970). We also use the arguments of some terms of the Eisenstein series
0
in order to improve existing error bounds.
0
2

n 1 Introduction
a
J
F. K. C. Rankin and H. P. F. Swinnerton-Dyer considered the problem of locating the zeros of the
1
Eisenstein series E (z) in the standard fundamental domain F [RSD]. They proved that all of the zeros
1 k
ofEk(z)inFlieontheunitcircle. Theyalsostatedtowardstheendoftheirstudythat“Thismethodcan

] equally well be applied to Eisenstein series associated with subgroups of the modular group.” However,
T
it seems unclear how widely this claim holds.
N
Subsequently, T. Miezaki, H. Nozaki, and the present author considered the same problem for the
h. Fricke group Γ∗0(p) (See [K], [Q]), and proved that all of the zeros of the Eisenstein series Ek∗,p(z) in a
t certain fundamental domain lie on a circle whose radius is equal to 1/√p, p=2,3 [MNS].
a
The Fricke group Γ (p) is not a subgroup of SL (Z), but it is commensurable with SL (Z). For a
m ∗0 2 2
fixed prime p, we define Γ (p):=Γ (p) Γ (p)W , where Γ (p) is a congruence subgroup of SL (Z).
[ ∗0 0 ∪ 0 p 0 2
Let k >4 be an even integer. For z H:= z C; Im(z)>0 , let
∈ { ∈ }
2
v E (z):= 1 pk/2E (pz)+E (z) (1)
9 k∗,p pk/2+1 k k
0 (cid:16) (cid:17)
4 be the Eisenstein series associated with Γ (p). (cf. [SG])
∗0
7
Henceforth, we assume that p=5 or 7. The region
0
6
0 F∗(p):={|z|>1/√p, |z|>1/(2√p), −1/26Re(z)60}
h/ z >1/√p, z >1/(2√p), 06Re(z)<1/2 (2)
{| | | | }
t
a [
m is a fundamental domain for Γ∗0(p). (cf. [SH], [SE]) Define A∗p := F∗(p)∩{z ∈ C; |z| = 1/√p or |z| =
1/(2√p) .
: }
v Inthepresentpaper,wewillapplythemethodofF.K.C.RankinandH.P.F.Swinnerton-Dyer(RSD
i Method) to the Eisenstein series associated with Γ (5) and Γ (7). We have the following conjectures:
X ∗0 ∗0
r Conjecture 1.1. Let k >4 be an even integer. Then all of the zeros of E (z) in F (5) lie on the arc
a k∗,5 ∗
A .
∗5
Conjecture 1.2. Let k >4 be an even integer. Then all of the zeros of E (z) in F (7) lie on the arc
k∗,7 ∗
A .
∗7
First, we prove that all but at most 2 zeros of E (z) in F (p) lie on the arc A (See Subsection 4.1
k∗,p ∗ ∗p
and 5.1). Second, if (24/(p+1)) k, we prove that all of the zeros of E (z) in F (p) lie on A (See
| k∗,p ∗ ∗p
Subsection 4.2 and 5.2).
We can then prove that if (24/(p+1)) ∤ k, all but one of the zeros of E (z) in F (p) lie on A .
k∗,p ∗ ∗p
Furthermore, let α [0,π] (resp. α [0,π]) be the angle which satisfies tanα = 2 (resp. tanα =
5 7 5 7
∈ ∈
5/√3), and let α [0,π] be the angle which satisfies α k(π/2+α )/2 (mod π). Then, since
p,k p,k p
∈ ≡
α is an irrational multiple of π, α appear in the interval [0,π] uniformly for all even integers k > 4.
p p,k
In Subsection 4.3, we prove that all of the zeros of E (z) in F (5) are on A if α < (116/180)π
k∗,5 ∗ ∗5 5,k
or (117/180)π < α . That is, we prove about 179/180 of Conjecture 1.1. Similarly, in Subsection
5,k
5.3 and 5.4, we prove that all of the zeros of E (z) in F (7) are on A if “α < (127.68/180)π or
k∗,7 ∗ ∗7 7,k
2000 Mathematics Subject Classification: Primary 11F11.
Key Words and Phrases: Eisenstein series; Fricke group; locating zeros; modular forms.
2 J. Shigezumi
(128.68/180)π < α for k 2 (mod 6)” or “α < (108.5/180)π or (109.5/180)π < α for k 4
7,k 7,k 7,k
≡ ≡
(mod 6)”. Thus we can also prove about 179/180of Conjecture 1.2.
In [RSD], we considered a bound for the error terms R (See (9)) in terms only of their absolute
1
values. However, in the present paper, we also use the arguments of some terms in the series. We can
then approach the exact value of the Eisenstein series.
A more detailed account of the material in the present study may be found in [SJ].
2 General Theory
2.1 Preliminaries
Let v (f) be the order of a modular function f at a point p.
p
2.1.1 Γ (5)
∗0
We define
A := z; z =1/√5, π/2<Arg(z)<π/2+α ,
∗5,1 { | | 5}
A := z; z+1/2 =1/(2√5), α <Arg(z)<π/2 .
∗5,2 { | | 5 }
Then, A =A A i/√5, ρ , ρ , where ρ := 1/2+i/ 2√5 and ρ := 2/5+i/5.
∗5 ∗5,1∪ ∗5,2∪{ 5,1 5,2} 5,1 − 5,2 −
Letf be a modularformfor Γ (5)ofweightk,andlet k be anevenintegersuchthatk 2 (mod 4),
∗0 (cid:0) (cid:1) ≡
then
f(i/√5)=f(W i/√5)=ikf(i/√5)= f(i/√5).
5
−
Thus, we have f(i/√5) = 0. Similarly, we have f(ρ ) = f(ρ ) = 0. Thus, we have v (f) > 1,
5,1 5,2 i/√5
v (f)>1, and v (f)>1.
ρ5,1 ρ5,2
On the other hand, if k 0 (mod 4), then we have v (E )=v (E )=v (E )=0.
≡ i/√5 k∗,5 ρ5,1 k∗,5 ρ5,2 k∗,5
2.1.2 Γ (7)
∗0
We define
A := z; z =1/√7, π/2<Arg(z)<π/2+α ,
∗7,1 { | | 7}
A := z; z+1/2 =1/(2√7), α π/6<Arg(z)<π/2 .
∗7,2 { | | 7− }
Then, A =A A i/√7, ρ , ρ , where ρ := 1/2+i/ 2√7 and ρ := 5/14+√3i/14.
∗7 ∗7,1∪ ∗7,2∪{ 7,1 7,2} 7,1 − 7,2 −
Let f be a modular form for Γ (7) of weight k. If k 2 (mod 4), then we have v (f) > 1 and
∗0 ≡ (cid:0) (cid:1) i/√7
v (f)>1. On the other hand, if k 0 (mod 4), then we have v (E )=v (E )=0.
ρ7,1 ≡ i/√7 k∗,7 ρ7,1 k∗,7
Similarly, if k 0 (mod 6), then we have v (f) > 1, while if k 0 (mod 6), then we have
6≡ ρ7,2 ≡
v (E )=0.
ρ7,2 k∗,7
2.2 Valence Formula
In order to determine the location of zeros of E (z) in F (p), we require the valence formula for Γ (p).
k∗,p ∗ ∗0
Proposition 2.1. Let f be a modular function of weight k for Γ (5), which is not identically zero. We
∗0
have
1 1 1 k
v (f)+ v (f)+ v (f)+ v (f)+ v (f)= . (3)
∞ 2 i/√5 2 ρ5,1 2 ρ5,2 p 4
p∈ΓX∗0(5)\H
p6=i/√5,ρ5,1,ρ5,2
Proposition 2.2. Let f be a modular function of weight k for Γ (7), which is not identically zero. We
∗0
have
1 1 1 k
v (f)+ v (f)+ v (f)+ v (f)+ v (f)= . (4)
∞ 2 i/√7 2 ρ7,1 3 ρ7,2 p 3
p∈ΓX∗0(7)\H
p6=i/√7,ρ7,1,ρ7,2
The proofs of the above propositions are very similar to that for the valence formula for SL (Z) (cf.
2
[SE]).
Eisenstein series for Γ (5) and Γ (7) 3
∗0 ∗0
2.3 Some Eisenstein series of low weights
By means of a straightforwardcalculation, we have the following propositions:
Proposition 2.3. The location of the zeros of the Eisenstein series E in F (5) for 4 > k > 10 are
k∗,5 ∗
given by the following table:
k v v v v V V
∞ i/√5 ρ5,1 ρ5,2 5∗,1 5∗,2
4 0 0 0 0 1 0
6 0 1 1 1 0 0
8 0 0 0 0 1 1
10 0 1 1 1 1 0
where V denotes the number of simple zeros of the Eisenstein series E on the arc A for n=1,2.
5∗,n k∗,5 ∗5,n
Proposition 2.4. The location of the zeros of the Eisenstein series E in F (7) for k = 4,6, and 12
k∗,7 ∗
are given by the following table:
k v v v v V
∞ i/√7 ρ7,1 ρ7,2 7∗
4 0 0 0 1 1
6 0 1 1 0 1
12 0 0 0 0 4
where V denotes the number of simple zeros of the Eisenstein series E on A A .
7∗ k∗,7 ∗7,1∪ ∗7,2
2.4 The space of modular forms
Let M be the space of modular forms for Γ (p) of weight k, and let M 0 be the space of cusp forms
k∗,p ∗0 k∗,p
forΓ (p)ofweightk. Uponconsideringthe mapM f f( ) C, itisclearthatthe kernelofthis
∗0 k∗,p ∋ 7→ ∞ ∈
map is given by M 0. So dim(M /M 0) 6 1, and M = CE M 0. Let η(z) be the Dedekind’s
k∗,p k∗,p k∗,p k∗,p k∗,p⊕ k∗,p
η-function.
2.4.1 Γ (5)
∗0
Note that ∆ =η4(z)η4(5z) is a cusp form for Γ (5) of weight 4. We have the following theorem:
5 ∗0
Theorem 2.1. Let k be an even integer.
(1) For k <0 and k =2, M =0.
k∗,5
(2) For k =0 and 6, we have M 0 =0, and dim(M )=1 with a base E .
k∗,5 k∗,5 k∗,5
(3) M 0 =∆ M .
k∗,5 5 k∗ 4,5
−
The proof of this theorem is very similar to that for SL (Z). Furthermore, we have
2
M =C(E )n C(E )n 1∆ C∆n,
4∗n,5 4∗,5 ⊕ 4∗,5 − 5⊕···⊕ 5
M4∗n+6,5 =E6∗,5((E4∗,5)n⊕C(E4∗,5)n−1∆5⊕···⊕C∆n5).
Thus, we have the following proposition:
Proposition 2.5. Let k >4 be an even integer. For every f M , we have
∈ k∗,5
v (f)>s , v (f)>s , v (f)>s
i/√5 k ρ5,1 k ρ5,2 k (5)
(s =0,1such that2s k (mod 4)).
k k
≡
2.4.2 Γ (7)
∗0
We define ∆7 :=η6(z)η6(7z) and E2,7′(z):=(7E2(7z) E2(z))/6. We then have the following theorem:
−
Theorem 2.2. Let k be an even integer, and we define ∆7,4 :=(5/16)((E2,7′)2−E4∗,7) and
∆0 :=(559/690)((41065/137592)(E E E ) E ∆ )
7,10 4∗,7 6∗,7− 1∗0,7 − 6∗,7 7,4
(1) M 0 =M 0 M .
k∗,7 1∗2,7 k∗ 12,7
(2) For k <0 and k−=2, M =0. We have M =C.
k∗,7 0∗,7
4 J. Shigezumi
(3) We have M 0 =C∆ , M 0 =C∆0 /∆ ,
4∗,7 7,4 6∗,7 7,10 7,4
M 0 =C(∆ )2 CE ∆ , M 0 =C∆0 CE ∆ ,
8∗,7 7,4 ⊕ 4∗,7 7,4 1∗0,7 7,10⊕ 6∗,7 7,4
M 0 =C(∆ )2 C(∆ )3 CE (∆ )2 C(E )2∆ , and
1∗2,7 7 ⊕ 7,4 ⊕ 4∗,7 7,4 ⊕ 4∗,7 7,4
M 0 =C∆ ∆0 CE (∆ )2 CE E ∆ .
1∗4,7 7,4 7,10⊕ 6∗,7 7,4 ⊕ 4∗,7 6∗,7 7,4
We thus have the following table:
k f v v v v V
∞ i/√7 ρ7,1 ρ7,2 7∗
4 (E2,7′)2 0 0 0 4 0
∆ 1 0 0 1 0
7,4
10 ∆0 2 1 1 1 0
7,10
12 (∆ )2 4 0 0 0 0
7
Furthermore, we have
M =E C(E )3n (E )3(n 1)M 0 (E )3(n 2)(M 0 )2 (M 0 )n
k∗,7 k∗−12n,7 4∗,7 ⊕ 4∗,7 − 1∗2,7⊕ 4∗,7 − 1∗2,7 ⊕···⊕ 1∗2,7
M 0n (M 0 )n. o
⊕ k∗−12n,7 1∗2,7
Thus, we have the following proposition:
Proposition 2.6. Let k >4 be an even integer. For every f M , we have
∈ k∗,7
v (f)>s , v (f)>s (s =0,1such that2s k (mod 4)),
i/√7 k ρ7,1 k k k ≡ (6)
v (f)>t (s =0,1,2such that 2t k (mod 6)).
ρ7,2 k k − k ≡
3 The method of Rankin and Swinnerton-Dyer (RSD Method)
3.1 RSD Method
Let k >4 be an even integer. For z H, we have
∈
1
Ek(z)= (cz+d)−k. (7)
2
(cX,d)=1
Moreover,we have F= z >1, 1/26Re(z)60 z >1, 06Re(z)<1/2 .
{| | − }∪{| | }
At the beginning of their proof in [RSD], Rankin and Swinnerton-Dyer considered the following
function:
F (θ):=eikθ/2E eiθ , (8)
k k
which is real for all θ [0,π]. Considering the four terms w(cid:0)ith(cid:1)c2+d2 =1, they proved that
∈
F (θ)=2cos(kθ/2)+R , (9)
k 1
where R denotes the remaining terms of the series. Moreover they showed that R <2 for all k >12.
1 1
| |
If cos(kθ/2) is +1 or 1, then F (2mπ/k) is positive or negative, respectively, and we can show the
k
−
existence of the zeros. In addition, we can prove that all of the zeros of E (z) in F lie on the unit circle
k
using the Valence Formula and the theory on the space of modular forms for SL (Z).
2
∗
3.2 The function: F
k,p,n
WeexpectallofthezerosoftheEisensteinseriesEk∗,p(z)inF∗(p)tolieonthearcseiθ/√pandeiθ/(2√p)−
1/2, which form the boundary of the fundamental domain defined by the equation (2).
We define
Fk∗,p,1(θ):=eikθ/2Ek∗,p eiθ/√p , (10)
Fk∗,p,2(θ):=eikθ/2Ek∗,p eiθ(cid:0)/2√p−(cid:1)1/2 . (11)
(cid:0) (cid:1)
Eisenstein series for Γ (5) and Γ (7) 5
∗0 ∗0
We can write
1 1
Fk∗,p,1(θ)= 2 (ceiθ/2+√pde−iθ/2)−k+ 2 (ce−iθ/2+√pdeiθ/2)−k, (12)
(cX,d)=1 (cX,d)=1
p∤c p∤c
1 k 1 k
Fk∗,p,2(θ)= 2 ceiθ/2+d√pe−iθ/2 − + 2 ce−iθ/2+d√peiθ/2 −
(cX,d)=1(cid:16) (cid:17) (cX,d)=1(cid:16) (cid:17)
p∤c2cd p∤c2cd
| | (13)
2k k 2k k
+ ceiθ/2+d√pe−iθ/2 − + ce−iθ/2+d√peiθ/2 − .
2 2
(cX,d)=1(cid:16) (cid:17) (cX,d)=1(cid:16) (cid:17)
p∤c2∤cd p∤c2∤cd
Hence we can use these expressions as definitions. Note that (ceiθ/2+√pde−iθ/2)−k and (ce−iθ/2+
√pdeiθ/2)−k are conjugates of each other for any pair (c, d). Thus, we have the following proposition:
Proposition 3.1. F (θ) is real, for all θ [0,π].
k∗,p,1 ∈
Proposition 3.2. F (θ) is real, for all θ [0,π].
k∗,p,2 ∈
Now, we define
F (θ) π/26θ 6π/2+α
F (θ)= k∗,5,1 5 .
k∗,5 (Fk∗,5,2(θ−π/2) π/2+α5 6θ 6π
Then, F is continuous in the interval [π/2,π]. Note that F (π/2+α )=ei(π/2)k/2F (α ).
k∗,5 k∗,5,1 5 k∗,5,2 5
Similarly,
F (θ) π/26θ 6π/2+α
F (θ)= k∗,7,1 7 .
k∗,7 (Fk∗,7,2(θ−2π/3) π/2+α7 6θ 67π/6
whereuponF iscontinuousintheinterval[π/2,7π/6]. NotealsothatF (π/2+α )=ei(2π/3)k/2F (α
k∗,7 k∗,7,1 7 k∗,7,2 7−
π/6).
3.3 Application of RSD Method
We introduce N :=c2+d2. First, we consider the case N =1. For this case, we can write
F (θ)=2cos(kθ/2)+R , (14)
k∗,p,1 p∗,1
Fk∗,p,2(θ)=2cos(kθ/2)+Rp∗,2, (15)
where R and R denote the terms satisfying N >1 of F and F , respectively.
p∗,1 p∗,2 k∗,p,1 k∗,p,2
3.3.1 For Γ (5)
∗0
For R , we will consider the following cases: N =2, 5, 10, 13, 17, and N >25. Considering 2/√56
5∗,1 −
cosθ 60, we have
1 k/2 1 k/2 1 k 384√6 1 k
R 62+4 +2 + +2 + . (16)
| 5∗,1| 2 3 ··· 9 k 3 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) − (cid:18) (cid:19)
Similarly,for R , we will consider the cases: N =2, 5, 10,13,17, 25,26,29, andN >34. For these
5∗,2
cases we have
2 k/2 1 k/2 1 k/2 2112√33 8 k/2
R 62+2 +2 + +2 + . (17)
| 5∗,2| 3 2 ··· 129 k 3 33
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) − (cid:18) (cid:19)
We want to show that R < 2 and R < 2. Note that the case for which (c,d) = (2,1) (resp.
| 5∗,1| | 5∗,2| ±
(c,d)= (1, 1)) yields a bound equal to 2 for R (resp. R ).
± − | 5∗,1| | 5∗,2|
6 J. Shigezumi
3.3.2 Γ (7)
∗0
For R , we will consider the cases: N =2, 5, 10, ..., 61, and N >65. Then, we have
7∗,1
1 k/2 1 k/2 1 k/2 28160 11 k/2
|R7∗,1|64+6 3 +4 7 +···+2 352 + 7(k 3) 64 . (18)
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) − (cid:18) (cid:19)
Similarly, for R , we will consider the cases: N =2, 5, 10, ..., 89, and N >97. For these cases we
7∗,2
have
1 k/2 1 k/2 1 k/2 62464√6 1 k/2
R 64+2 +2 + +2 + . (19)
| 7∗,2| 2 3 ··· 571 21(k 3) 8
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) − (cid:18) (cid:19)
Note that the cases (c,d)= (2,1) and (3,1) (resp. (c,d)= (1, 1) and (3, 1)) yield a bound
± ± ± − ± −
equal to 4 for R (resp. R ).
| 7∗,1| | 7∗,2|
3.4 Arguments of some terms
3.4.1 Γ (5)
∗0
In the previous subsection, the important point was the fact that the cases (c,d) = (2,1) and (c,d) =
±
(1, 1) do not yield good bounds for R and R , respectively.
± − | 5∗,1| | 5∗,2|
Let θ1′ :=2Arg 2eiθ1/2+√5e−iθ1/2 and θ2′ :=2Arg eiθ2/2+√5e−iθ2/2 , then we have
−
(cid:8) (cid:9) (cid:8) (cid:9)
√5 2 √5+1
tanθ1′/2= − tanθ1/2, tanθ2′/2= tanθ2/2.
−√5+2 −√5 1
−
Furthermore, it is easy to show that
√5 2
tan( π+π/2+α +d (tπ/k))/2< − tan(π/2+α (tπ/k))/2
5 1 5
− −√5+2 −
<tan( π+π/2+α +(tπ/k))/2,
5
−
where d <1/(1+4tan(t/2)(π/k)). Thus,
1
θ =π/2+α (tπ/k)
1 5
−
π+π/2+α5+d1(tπ/k)<θ1′ < π+π/2+α5+(tπ/k),
⇒− −
kθ /2=k(π/2+α )/2 (t/2)π
1 5
−
(k/2)π+k(π/2+α5)/2+d1(t/2)π<kθ1′/2< (k/2)π+k(π/2+α5)/2+(t/2)π.
⇒− −
Similarly,
kθ /2=kα /2+(t/2)π
2 5
(k/2)π+kα5/2 (t/2)π <kθ2′/2< (k/2)π+kα5/2 d2(t/2)π,
⇒− − − −
where d <1/(1+tan(t/2)(π/k)).
2
Note that
0 (k 0 (mod 4))
(k/2)π ≡ (mod 2π),
− ≡(π (k 2 (mod 4))
≡
and both d and d tend to 1 in the limit as k tends to , or in the limit as t tends to 0.
1 2
∞
Recall that α k(π/2+α )/2 (mod π), then we can write k(π/2+α )/2 = α +mπ for some
5,k 5 5 5,k
≡
integer m. We define α5,k′ kθ1′/2 mπ (mod 2π) for θ1 =π/2+α5 (tπ/k).
≡ − −
Similarly, we define β5,k kα5/2 (mod π) and β5,k′ kθ2′/2 (kα5/2 β5,k) (mod 2π) for θ2 =
≡ ≡ − −
α +(tπ/k).
5
Eisenstein series for Γ (5) and Γ (7) 7
∗0 ∗0
3.4.2 Γ (7)
∗0
Similarly to the previous subsection, we consider the arguments of some terms such that (c,d)= (2,1)
±
and (3,1) for R , and (c,d)= (1, 1) and (3, 1) for R .
± | 7∗,1| ± − ± − | 7∗,2|
Let θ1,1′ :=2Arg 2eiθ1/2+√7e−iθ1/2 , θ1,2′ :=2Arg 3eiθ1/2+√7e−iθ1/2 ,
θ2,1′ :=2Arg −eiθ2/2(cid:8)+√7e−iθ2/2 , and(cid:9)θ2,2′ :=2Arg −(cid:8)3eiθ2/2+√7e−iθ2/2 (cid:9).
We have
(cid:8) (cid:9) (cid:8) (cid:9)
kθ /2=k(π/2+α )/2 (t/2)π
1 7
−
(k/3)π+k(π/2+α )/2+d (t/2)π
7 1,1
 <kθ1,1′/2<(k/3)π+k(π/2+α7)/2+(3t/2)π,
⇒−(k/3)π+k(π/2+α7)/2−tπ
<kθ1,2′/2< (k/3)π+k(π/2+α7)/2 d1,2(t/2)π,
− −
kθ2/2=k(α7 π/6)/2+(t/2)π
−
(k/3)π+k(α π/6)/2 (3t/4)π
7
− − −
 <kθ2,1′/2<−(k/3)π+k(α7−π/6)/2−d2,1(t/2)π,
⇒(k/3)π+k(α7−π/6)/2+d2,2(t/2)π
<kθ2,2′/2<(k/3)π+k(α7 π/6)/2+(t/4)π,
−
whered <3/(1+2√3tan(t/2)(π/k)),d <2/(1+√3tan(t/2)(π/k)),d <3/(2+√3tan(t/2)(π/k)),
1,1 1,2 2,1
and d <1/(2+3√3tan(t/2)(π/k)).
2,2
Note that
0 (k 0 (mod 6)) 0 (k 0 (mod 6))
≡ ≡
(k/3)π 2π/3 (k 2 (mod 6)), (k/3)π 4π/3 (k 2 (mod 6)),
 
≡ ≡ − ≡ ≡
4π/3 (k 4 (mod 6)) 2π/3 (k 4 (mod 6))
≡ ≡
modulo 2π. Furthermore, in the limit as k tends to or in the limit as t tends to 0, d1,1 (resp. d1,2,
∞
d , and d ) tends to 3 (resp. 2, 3/2, and 1/2).
2,1 2,2
Recallthatα7,k k(π/2+α7)/2 (mod π). Then,wedefineα7,k,n′ kθ1,n′/2 (k(π/2+α7)/2 α7,k)
≡ ≡ − −
(mod 2π) for n=1,2 and for θ =π/2+α (tπ/k).
1 7
−
Similarly, we define β7,k k(α7 π/6)/2 (mod π) and β7,k,n′ kθ2,n′/2 (k(α7 π/6)/2 β7,k)
≡ − ≡ − − −
(mod 2π) for n=1,2 and for θ =α π/6+(tπ/k).
2 7
−
3.5 Algorithm
In this subsection, we consider the bound
R <2c for everyk >k , (20)
| p∗,n| 0 0
for some c > 0 and an even integer k . Furthermore, we will detail an algorithm that can be used to
0 0
derive the above bound.
Let Λ be an index set, and, applying the RSD method, let us write
R 62 ek v (c ,d ,θ),
| p∗,n| λ Λ λ k λ λ
∈
where the factor “2” comes from the relationPv (c,d,θ) = v ( c, d,θ). Furthermore, let I be a finite
k k
− −
subset of Λ such that ek v (c ,d ,θ) does not tend to 0 in the limit as k tends to for all i I, and
assume I ⊂N. Then, wei dekfinie Xii :=e−i 2 vk(ci,di,θ)−2/k for every i∈I. ∞ ∈
Assume that for every i I and k >k and for some c and u ,
0 i′ i
∈
k
Re eki cieiθ/2+√pdie−iθ/2 − 6ci′Xi−k/2,
Xi−k/2 >(cid:12)(cid:12)(cid:12)(cid:12) 1+(cid:26)ui((cid:16)π/k), 2 λ Λ I ekλ(cid:17)vk((cid:27)c(cid:12)(cid:12)(cid:12)(cid:12)λ,dλ,θ)6b(1/s)k/2,
∈ \
and let the number t>0 be given. P
8 J. Shigezumi
Step 1. “Determine the number a .”
1
First, in order to show the bound (20), we wish to use the bound
i I ci′Xi−k/2 <c0−a1(tπ/k)2 (21)
∈
for every i I and k>k and for soPme a >0.
0 1
∈
To show the bound (20) by the above bound (21), we need b(1/s)k/2 < a (tπ/k)2 for every k > k .
1 0
Define f(k):=sk/2/b k2/(2a t2π2). If we have k logs>4 and
1 0
−
a >(bk2)/(2sk0/2t2π2), (22)
1 0
then we have f(k ) > 0, f (k ) > 0, and f (k ) > 0. In the present paper, we always have k logs > 4.
0 ′ 0 ′′ 0 0
Thus, it is enough to consider the bound (22).
Step 2. “Determine the number c and a .”
0,i 1,i
Second, to show bound (21), we wish to use the bounds
ci′Xi−k/2 <c0,i−a1,i(tπ/k)2 (23)
for every i I and k>k and for some c >0, a >0.
0 0,i 1,i
∈
We determine c and a such that
0,i 1,i
c >0, a >0, c = c , and a = a . (24)
0,i 1,i 0 i I 0,i 1 i I 1,i
∈ ∈
P P
Step 3. “Determine a discriminant Y for every i I.”
i
∈
Finally, for the bound (23), we consider following sufficient conditions:
Xk/2 >c +a (tπ/k)2, X >a +a (tπ/k)2.
i i 2,i i 3,i 4,i
For the former bound, it is enough to show that
c =c /c , a >c2(a /c )/ 1 c (a /c )(tπ/k )2 , (25)
i i′ 0,i 2,i i 1,i i′ { − i 1,i i′ 0 }
while for the latter bound, it is enough to show that
a =c2/k0, a =((2a )/(c k ))c2/k0.
3,i i 4,i 2,i i 0 i
Because we have
c2/k 61+2(logc )/k+2(logc )2c2/k/k2,
i i i i
π 2 1 1 2a t2π2 1
X a +a t > u π 2logc 2(logc )2c2/k 2,i c2/k0
i− 3,i 4,i k k i − i− i i k − c i k2
(cid:18) (cid:16) (cid:17) (cid:19) (cid:26) 0 i 0(cid:27)
1
=: Y . (26)
i
k ×
In conclusion, if we have Y >0, then the bounds (23), (21), and (20) hold.
i
Note that the above bounds are sufficient conditions; they are not always necessary.
4 Γ (5) (For Conjecture 1.1)
∗0
The proof of Conjecture 1.1 is significantly more difficult than the proof of the theorems for Γ (2) and
∗0
Γ (3). The most difficult point concerns the argument Arg(ρ ), which is not a rational multiple of π.
∗0 5,2
Eisenstein series for Γ (5) and Γ (7) 9
∗0 ∗0
4.1 All but at most 2 zeros
We have the following lemma:
Lemma 4.1. We have the following bounds:
“We have R <2cos(c π) for θ [π/2,π/2+α tπ/k]”
| 5∗,1| 0′ 1 ∈ 5−
(1) For k >12, (c ,t)=(1/3,1/6).
0′
(2) For k >58, (c ,t)=(33/80,9/40).
0′
“We have R <2cos(c π) for θ [α +tπ/k,π/2]”
| 5∗,2| 0′ 2 ∈ 5
(3) For k >12, (c ,t)=(0,1/2).
0′
(4) For k >22, (c ,t)=(1/3,1/2).
0′
(5) For k >46, (c ,t)=(7/30,1/5).
0′
Proof. (3) Let k >12 and x=π/(2k), then 06x6π/24, and so 1 cosx>(32/33)x2. Thus, we have
−
1 1 16
eiθ/2 √5e iθ/2 2 > (6 2√5cos(α +x))>1+ x2,
− 5
4| − | 4 − 11
1 96
eiθ/2+√5e iθ/2 k >1+ x2 (k >12),
−
2k| | 11
288π2 1
2k 2v (1,1,θ)62 .
· k − π2+66k2
In inequality(17), replace 2 with the bound 2 288π2 1 . Then
− π2+66k2
288π2 1 2 k/2 1 k/2 2112√33 8 k/2
|R5∗,2|62− π2+66k2 +2 3 +···+2 129 + k 3 33 .
(cid:18) (cid:19) (cid:18) (cid:19) − (cid:18) (cid:19)
Furthermore, (2/3)k/2 decreases more rapidly in k than 1/k2, and for k >12, we have
|R5∗,2|61.9821...
(1),(2),(4),(5)WewillusethealgorithmintheSubsection3.5. Furthermore,wehaveX =v (2,1,θ ) 2/k >
1 k 1 −
1+4t(π/k) in the proof of (1) and (2), and we have X = (1/4) v (1, 1,θ ) 2/k > 1+t(π/k) in the
1 k 2 −
−
proof of (4) and (5). We have c =c 6cos(c π). We can show Y >0 for every item.
0 0,1 0′ 1
When 4 k, by the valence formula for Γ (5) and Proposition 2.5, we have at most k/4 zeros on the
| ∗0
arc A . We have k/4+1 integer points (i.e. cos(kθ/2) = 1) in the interval [π/2,π]. By the above
∗5 ±
lemma’sconditions(1)and(3),wecanprove R <2or R <2atallbutatmostoneintegerpoint.
| 5∗,1| | 5∗,2|
Then, we have all but at most 2 zeros on A .
∗5
On the other hand, when 4 ∤ k, we have at most (k 6)/4 zeros on the arc A . Similarly to the
− ∗5
previous case, we have all but at most 2 zeros on A .
∗5
Thus, we have the following proposition:
Proposition 4.1. Let k >4 be an even integer. Then all but at most 2 of the zeros of E (z) in F (5)
k∗,5 ∗
lie on the arc A .
∗5
4.2 The case 4 k
|
For π/12 < α < 3π/4, by Lemma 4.1 (1) and (3), we can prove R < 2 or R < 2 at all of the
5,k | 5∗,1| | 5∗,2|
integer points.
Now, we can write
Fk∗,5,1(θ1)=2cos(kθ1/2)+2Re(2e−iθ1/2+√5eiθ1/2)−k+R5∗,1′,
Fk∗,5,2(θ2)=2cos(kθ2/2)+2k·2Re(e−iθ2/2−√5eiθ2/2)−k+R5∗,2′.
For0<α <π/12,thelastintegerpointofF (θ )isintheinterval[π/2+α π/(6k),π/2+α ].
5,k k∗,5,1 1 5− 5
We have |R5∗,1′|< 2 for θ1 ∈ [π/2,π/2+α5]. Furthermore, because 0 <α5,k′ < π/6 for 0< t < 1/6, we
have Sign cos(kθ /2) =Sign Re(2e iθ1/2+√5eiθ1/2) k for θ [π/2+α π/(6k),π/2+α ].
1 − − 1 5 5
{ } { } ∈ −
For 3π/4<α <π, the first integer point of F (θ ) is in the interval [α ,α +π/(2k)]. We have
5,k k∗,5,2 2 5 5
|R5∗,2′|<2 and Sign{cos(kθ2/2)}=Sign{Re(e−iθ2/2−√5eiθ2/2)−k} for θ2 ∈[α5,α5+π/(2k)].
Thus, we have the following proposition:
10 J. Shigezumi
Proposition 4.2. Let k >4 be an integer which satisfies 4 k. Then all of the zeros of E (z) in F (5)
| k∗,5 ∗
lie on the arc A .
∗5
4.3 The case 4 ∤ k
4.3.1 The case 0<α <π/2.
5,k
Now, at most two zeros remain. At the point such that kθ /2 = k(π/2+α )/2 α π/3, we have
1 5 5,k
− −
R < 1 by Lemma 4.1 (1), and we have 2cos(kθ /2)= 1. Then, we have at least one zero between
| 5∗,1| 1 ±
the second to last integer point for A and the point kθ /2. Similarly, by Lemma 4.1 (4), we have at
∗5,1 1
least one zero between the second integer point and the point kθ /2=kα /2+β +π/3.
2 5 5,k
4.3.2 The case π/2<α <π.
5,k
For this case, we expect one more zero between the last integer point for A and the first one for A .
∗5,1 ∗5,2
We consider the following cases:
(i) “The case 7π/10<α <π”
5,k
For 3π/4<α <π, we can use Lemma 4.1 (1).
5,k
•
For 7π/10<α <3π/4, we can use Lemma 4.1 (2).
5,k
•
For each case, we consider the point such that kθ /2 = k(π/2 + α )/2 α +π c π. We have
1 5 5,k 0′
− −
α π+c π > (t/2)π and R < 2cos(c π), and we have 2cos(kθ /2) = 2cos(c π). Then, we
5,k − 0′ | 5∗,1| 0′ 1 ± 0′
have at least one zero between the second to last integer point for A and the point kθ /2.
∗5,1 1
(ii) “The case π/2<α <19π/30”
5,k
For π/2<α <7π/12, we can use Lemma 4.1 (4).
5,k
•
For 7π/12<α <19π/30,we can use Lemma 4.1 (5).
5,k
•
Similar to the case (i) above, we consider the point such that kθ /2=kα /2 β +c π for each case.
2 5 5,k 0′
−
(iii) “The case 13π/20<α <7π/10”
5,k
We have X = v (2,1,θ ) 2/k > 1+4t(π/k), and let cos(c π) = cos((x/180)π (t/2)π). Then,
1 k 1 − 0′
− −
using the algorithm of Subsection 3.5, we prove “For (x/180)π < α < (y/180)π, we have R <
5,k | 5∗,1|
2cos(c π) for θ =π/2+α tπ/k.” for ten cases, namely, (x,y,t)=(121,126,3/20),(120,121,1/10),
0′ 1 5
−
(118.8,120,1/10), (118.1,118.8,2/25), (117.7,118.1,1/15), (117.45,117.7,3/50), (117.27,117.45,1/20),
(117.15,117.27,9/200),(117.06,117.15,1/25),(117,117.06,1/25).
Foreachcase,weconsiderthepointsuchthatkθ /2=k(π/2+α )/2 (t/2)π. Wehaveα π+c π >
1 5 5,k 0′
− −
(t/2)π and R <2cos(c π), andwe have 2cos(kθ /2) >2cos(c π). Then,we haveatleastonezero
| 5∗,1| 0′ | 1 | 0′
between the second to last integer point for A and the point kθ /2.
∗5,1 1
(iv) “The case 19π/30<α <29π/45”
5,k
We have X = (1/4) v (1, 1,θ ) 2/k > 1+t(π/k) and cos(c π) = cos((y/180)π π/2+(t/2)π).
1 k 2 − 0′
− −
Then, we prove “For (x/180)π <α <(y/180)π, we have R <2cos(c π) for θ =α +tπ/k.” for
5,k | 5∗,2| 0′ 2 5
three cases, namely, (x,y,t)=(114,115.4,4/25),(115.4,115.8,3/25),(115.8,116,1/10).
Similar to the case (iii) above,we consider the point such that kθ /2=kα /2+(t/2)π for each case.
2 5
In conclusion, we have the following proposition:
Proposition 4.3. Let k > 4 be an integer which satisfies 4 ∤ k, and let α [0,π] be the angle which
5,k
∈
satisfies α k(π/2+α )/2 (mod π). If we have α <29π/45or 13π/20<α , then all of the zeros
5,k 5 5,k 5,k
≡
of E (z) in F (5) lie on the arc A . Otherwise, all but at most one zero of E (z) in F (5) lie on A
k∗,5 ∗ ∗5 k∗,5 ∗ ∗5
4.4 The remaining case “4 ∤ k and 29π/45 < α5,k < 13π/20”
In the previous subsection, we left one zero between the last integer point for A and the first one for
∗5,1
A for the case of “4 ∤ k and 29π/45 < α < 13π/20”. For the cases of 13π/20 < α < 7π/10 and
∗5,2 5,k 5,k
19π/30<α <29π/45, the width x y becomes smaller as the intervals of the bounds approach the
5,k
| − |
interval [29π/45,13π/20]. It seems that the width x y needs to be smaller still if we are to prove
| − |