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# Level 12

0

-12 -14 0 1

4 1 2 1

43 1 23

Figure 3.78: Γ0(12)+

For the cusp We have Γ=(10 1n) ; n∈Z}, and we have the Eisenstein series for the cusp associated with Γ0(12)+ of weightk>4:

Ek,12+ (z) :=2k3k/2Ek(12z)2 3k/2Ek(6z) + 2kEk(4z) + 3k/2Ek(3z)2Ek(2z) +Ek(z)

(3k/2+ 1)(2k1) . (3.263)

For the cusp −1/2 We have Γ−1/2=n

±³

3n+1 3n/2

−6n −3n+1

´

; n∈Zo

andγ−1/2 =W12+,6, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(12)+ of weightk>4:

Ek,12+−1/2(z) :=

−2 2k/2(2k3k/2Ek(12z)3k/2(2k+ 1)Ek(6z) + 2kEk(4z) + 3k/2Ek(3z)(2k+ 1)Ek(2z) +Ek(z))

(3k/2+ 1)(2k1) .

(3.264) We also haveγ−1/2−1 Γ0(12) + γ−1/2= Γ0(12)+.

The space of modular forms We define

12+:= ∆12012−1/312−1/412 ,−1/212+ := (∆−1/212 )2(∆−1/612 )2, which are 2nd semimodular forms for Γ0(12)+ of weight 2.

Now, we haveMk0(12)+) =CEk,12+ ⊕CEk,12+−1/2⊕Sk0(12)+) andSk0(12)+) = ∆12+Mk−40(12)+) for every even integerk>4. Then, we haveM4n+20(12)+) =E2,12+0M4n0(12)+) and

M4n0(12)+) =C(E4,12+ )nC(E4,12+ )n−112+⊕ · · · ⊕CE4,12+ (∆12+)n−1

C(E4,12+−1/2)nC(E4,12+−1/2)n−112+⊕ · · · ⊕CE4,12+−1/2(∆12+)n−1C(∆12+)n. Furthremore, we can write

M4n0(12)+) =C(∆12+)2nC(∆12+)2n−1−1/212+ ⊕ · · · ⊕C(∆−1/212+ )2n.

Hauptmodul We define thehauptmodul of Γ0(12)+:

J12+:= ∆12+−1/2/∆12+(=η−6(z)η12(2z)η−6(3z)η−6(4z)η12(6z)η−6(12z))

= 1

q+ 6 + 15q+ 32q2+ 87q3+· · ·, (3.265)

wherev(J12+) =−1 andv−1/2(J12+) = 1. Then, we have

J12+:∂F12+\ {z∈H; Re(z) =±1/2} →[0,16]R. (3.266)

Location of the zeros of the Eisenstein series SinceW12+,6−10(12)+)W12+,6= Γ0(12)+, we have Ek,12+−1/2(W12+,6z) = (√

6z+

6/2)kEk,12+ (z). (3.267)

Furthermore, we have

Ek,12+−1/2(e0/(2√

3)) = ((

3e1)/

2)kEk,12+(e/(2√ 3)), E−1/2k,12+(i/(2 tan(θ/2))) = ((e+ 1)/

6)kEk,12+ (e/6−1/3), wheree0 = (−

32 cosθ+isinθ)/(2 +√ 3 cosθ).

Now, recall that Γ0(12)+ =T1/2−10(6) + 3)T1/2. Then, fork 6600, since we can prove that all of the zeros of Ek,6+3 lie on the lower arcs of ∂F6+3 by numerical calculation, we have all of the zeros of Ek,12+ in the lower arcs of∂F12+.

Ek,12+ -12 -14 0 14 12

1 43 1 23

Ek,12+−1/2 -12 -14 0 14 12

1 43

1 23

Figure 3.79: Location of the zeros of the Eisenstein series

Location of the zeros of Hecke type Faber Polynomial Similarly to the Eisenstein series, for m6200, since we can prove that all of the zeros ofFm,6+3 lie on the lower arcs of∂F6+3 by numerical calculation, we have all of the zeros ofFm,12+ in the lower arcs of∂F12+.

0

0

### (12)

Fundamental domain We have a fundamental domain for Γ0(12) as follows:

F12+12={|z+ 5/12|>1/12, −1/26Re(z)<−1/3}[

{|z+ 7/24|>1/24, −1/36Re(z)<−2/7}

[ n|z|>1/(2

3),−2/76Re(z)60o [ n

|z|>1/(2

3),0< Re(z)62/7 o [{|z−7/24|>1/24,2/7< Re(z)61/3}[

{|z−5/12|>1/12,1/3< Re(z)<1/2},

(3.268)

where W12 : e/(2√

3) ei(π−θ)/(2√

3), ¡−7 2

24 −7

¢ : (e+ 7)/24 (ei(π−θ)7)/24, and ¡−5 2

12−5

¢ : (e+ 5)/12(ei(π−θ)5)/12. Then, we have

Γ0(12) =h(1 10 1), W12, (12 55 2), (24 77 2)i. (3.269)

-1

2

-1

3 0 13 12

1 23

Figure 3.80: Γ0(12)

0

Valence formula The cusps of Γ0(12) are∞,−1/3, and−1/2, and the elliptic points arei/(2√ 3) and ρ12,2 =−2/7 +i/(14√

3). Letf be a modular function of weight k for Γ0(12), which is not identically zero. We have

v(f) +v−1/3(f) +v−1/2(f) +1

2vi/(23)(f) +1

2vρ12,2(f) + X

p∈Γ0(12)\H p6=i/(2

3),ρ12,2

vp(f) =k. (3.270)

Furthermore, the stabilizer of the elliptic pointi/(2√

3) (resp. ρ12,2) is{±I,±W12}(resp. ©

±I,±¡ 7 −2

−24 7

¢W12

ª).

For the cusp We have Γ=(10 1n) ; n∈Z}, and we have the Eisenstein series for the cusp associated with Γ0(12) of weightk>4:

Ek,12+12 (z) :=

2k3kEk(12z) + 3k/2(3k/21)Ek(6z)2k3k/2Ek(4z)3kEk(3z)(3k/21)Ek(2z) +Ek(z)

(3k1)(2k1) . (3.271)

For the cusp −1/3 We have Γ−1/3

±¡12n+1 4n

−36n −12n+1

¢; n∈

and γ−1/3 =W12,4, and we have the Eisenstein series for the cusp−1/3 associated with Γ0(12) of weightk>4:

Ek,12+12−1/3 (z) :=

−(2k3k/2Ek(12z) + 3k/2(3k/21)Ek(6z)2k3k/2Ek(4z)3kEk(3z) + (3k/21)Ek(2z) +Ek(z))

(3k1)(2k1) .

(3.272) We also haveγ−1/3−1 Γ0(12)γ−1/3= Γ0(12).

For the cusp −1/2 We have Γ−1/2

±¡6n+1 3n

−12n−6n+1

¢ ; n∈

and γ−1/2 =W12−,6, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(12) of weightk>4:

Ek,12+12−1/2 (z) :=

−(2k3k/2Ek(12z)3k/2(2k+ 1)Ek(6z) + 2kEk(4z) + 3k/2Ek(3z)(2k+ 1)Ek(2z) +Ek(z))

(3k/2+ 1)(2k1) . (3.273)

Note that we haveEk,12+12−1/2 = 2−12−k/2E−1/2k,12+. Note thatγ−1−1/2 Γ0(12)γ−1/26= Γ0(12).

The space of modular forms We define

12+12:= ∆12012,−1/312+12:= ∆−1/312−1/412 ,−1/212+12:= ∆−1/212−1/612 , which are 4th semimodular forms for Γ0(12) of weight 1. Furthermore, we define

12,1,12+12:= (E1,12+120)5E2,12+0−1/312+12−1/212+1212,12,2,12+12:= (E1,12+120)5E2,12+012+12−1/212+1212,

12,3,12+12:= (E1,12+120)5E2,12+012+12−1/312+1212,12,4,12+12:= (E1,12+120)4−1/312+12−1/212+12(∆12)2,

12,5,12+12:= (E1,12+120)412+12−1/212+12(∆12)2,12,6,12+12:= (E1,12+120)412+12−1/312+12(∆12)2,

12,7,12+12:= (∆−1/312+12)2−1/212+12(∆12)3,12,8,12+12:= ∆12+12(∆−1/212+12)2(∆12)3,

12,9,12+12:= (∆12+12)2−1/312+12(∆12)3.

In addition, we denote

A1=E1,12+12012(C∆12+12−1/212+12C∆−1/312+12−1/212+12C∆12+12−1/312+12), A2= (∆12)2(C∆12+12−1/212+12C∆−1/312+12−1/212+12C∆12+12−1/312+12),

B1=C(E12,12+12 )nC(E12,12+12 )n−112,1,12+12C(E12,12+12 )n−112,4,12+12

C(E12,12+12 )n−112,7,12+12C(E12,12+12 )n−1(∆12)4C(E12,12+12 )n−212,1,12+12(∆12)4

⊕ · · · ⊕C∆12,7,12+12(∆12)4(n−1)C(∆12)4n,

B2=C(E12,12+12−1/3 )nC(E12,12+12−1/3 )n−112,2,12+12C(E12,12+12−1/3 )n−112,5,12+12

C(E12,12+12−1/3 )n−112,8,12+12C(E12,12+12−1/3 )n−1(∆12)4C(E12,12+12−1/3 )n−212,2,12+12(∆12)4

⊕ · · · ⊕C∆12,8,12+12(∆12)4(n−1)C(∆12)4n,

B3=C(E12,12+12−1/2 )nC(E12,12+12−1/2 )n−112,3,12+12C(E12,12+12−1/2 )n−112,6,12+12

C(E12,12+12−1/2 )n−112,9,12+12C(E12,12+12−1/2 )n−1(∆12)4C(E12,12+12−1/2 )n−212,3,12+12(∆12)4

⊕ · · · ⊕C∆12,8,12+12(∆12)4(n−1)C(∆12)4n. Now, we have

Mk0(12)) =CEk,12+12CE−1/3k,12+12CE−1/2k,12+12⊕Sk0(12)),

Sk0(12)) = (C∆12,1,12+12C∆12,2,12+12⊕ · · · ⊕C∆12,9,12+12C(∆12)4)Mk−120(12)) for every even integerk>4. Then, we have

M12n0(12)) =B1⊕B2⊕B3C(∆12)4n,

M12n+20(12)) =E2,12+0M12n0(12))CE1,12+12012+12(∆12)4n,

M12n+40(12)) =E4,12+12 (B1C(∆12)4n)⊕E4,12+12−1/3 (B2C(∆12)4n)⊕E4,12+12−1/2 (B3C(∆12)4n)

(CE1,12+120C∆12+12)(∆12)4n+1,

M12n+60(12)) =E6,12+12 (B1C(∆12)4n)⊕E6,12+12−1/3 (B2C(∆12)4n)⊕E6,12+12−1/2 (B3C(∆12)4n)

⊕A1(∆12)4n,

M12n+80(12)) =E8,12+12 (B1C(∆12)4n)⊕E8,12+12−1/3 (B2C(∆12)4n)⊕E8,12+12−1/2 (B3C(∆12)4n)

⊕E2,12+0A1(∆12)4n⊕A2(∆12)4n,

M12n+100(12)) =E10,12+12 (B1C(∆12)4n)⊕E10,12+12−1/3 (B2C(∆12)4n)⊕E10,12+12−1/2 (B3C(∆12)4n)

C(E1,12+120)4A1(∆12)4n⊕E2,12+0A2(∆12)4n+1CE1,12+120(∆12)4n+3. Furthermore, we can write

Mk0(12)) =Ek,12+120(C(∆12+12)nC(∆12+12)n−1−1/312+12⊕ · · · ⊕C(∆−1/312+12)n),

wheren= dim(Mk0(12)))1 =bk−2(k/4− bk/4c)c, and whereEk,12+120 := 1 andE1,12+120, when k≡0 and 2 (mod 4), respectively.

Hauptmodul We define thehauptmodul of Γ0(12):

J12+12:= ∆−1/312+12/∆12+12(=η−4(z)η4(3z)η4(4z)η−4(12z)) = 1

q+ 4 + 14q+ 36q2+ 85q3+· · · , (3.274) wherev(J12+12) =−1 andv−1/3(J12+12) = 1. Then, we have

J12+12:∂F12+12\ {z∈H; Re(z) =±1/2} →[−1,7 + 4

3]R. (3.275)

0

Location of the zeros of the Eisenstein series SinceW12,4−1Γ0(12)W12,4= Γ0(12), we have

Ek,12+12−1/3 (W12,4z) = (6z−2)kEk,12+12 (z). (3.276) Furthermore, we have

Ek,12+12−1/3 (−1/2 +i/(6 tan(θ/2))) = (−(e1)/2)kEk,12+12 ((e5)/12), Ek,12+12−1/3 (itan(θ/2)/3) = (−(e+ 1)/4)kEk,12+12 ((e7)/24),

Ek,12+12−1/3 (e0/(2√

3)) = (−(

3e+ 2))kEk,12+12 (e/(2√ 3)), where e0 = (4

3 + 7 cosθ+isinθ)/(7 + 4√

3 cosθ). On the other hand, recall that Ek,12+12−1/2 (z) = 2−12−k/2Ek,12+ (z). Moreover, by the transformation withW12,3forEk,12+−1/2, we have

Ek,12+12−1/2 (ei(π−θ0)/(2√

3)) = (2e+

3)kE−1/2k,12+12(e/(2√ 3)), Ek,12+12−1/2 ((e7)/24) = (

3(1 +i/tan(θ/2)))kEk,12+12−1/2 (i/(4 tan(θ/2))).

Fork6500, we can prove that all of the zeros ofEk,12+12lie on the lower arcs of∂F12+12by numerical calculation.

Ek,12+12 -12 -13 0 13 12

1 23

Ek,12+12−1/3 -12 -13 0 13 12

1 23

Ek,12+12−1/2 -12 -13 0 13 12

1 23

Figure 3.81: Location of the zeros of the Eisenstein series

Location of the zeros of Hecke type Faber Polynomial For every integerm6200 such thatm6≡

2,4 (mod 6) butm= 3,6,12,13,15, we can prove that all of the zeros ofFm,12+12 lie on the lower arcs of

∂F12+12 by numerical calculation. On the other hand, by numerical calculation, for m= 3,6,12,13,15, we can prove that all but two of the zeros ofFm,12+12 lie on the lower arcs of ∂F12+12, and two of the zeros of Fm,12+12 do not lie on ∂F12+12. For the other cases where m is m 6200 such that m 2,4 (mod 6), by numerical calculation, we can prove that all but one of the zeros ofFm,12+12lie on the lower arcs of∂F12+12, and one of the zeros ofFm,12+12 lies on∂F12+12 but does not on the lower arcs.

0

### (12) + 4

We have Γ0(12) + 4 =T1/2−1Γ0(6)T1/2.

Fundamental domain We have a fundamental domain for Γ0(12) + 4 as follows:

F12+4={|z+ 1/3|>1/6,−1/26Re(z)<−1/6}[

{|z+ 1/12|>1/12,−1/66Re(z)60}

[{|z−1/12|>1/12,0< Re(z)61/6}[

{|z−1/3|>1/6, 1/6< Re(z)<1/2}, (3.277) where¡−1 0

12 −1

¢: (e+ 1)/12(ei(π−θ)1)/12 andW12,4: 1/3 +e/6→ −1/3 +ei(π−θ)/6. Then, we have

Γ0(12) + 4 =h−I, (1 10 1), W12,4, (12 11 0)i. (3.278)

-12 -16 0 1 6

1 2

Figure 3.82: Γ0(12) + 4

Valence formula The cusps of Γ0(12) + 4 are∞, 0,−1/2, and−1/6. Letf be a modular function of weightk for Γ0(12) + 4, which is not identically zero. We have

v(f) +v0(f) +v−1/2(f) +v−1/6(f) + X

p∈Γ0(12)+4\H

vp(f) =k. (3.279)

For the cusp We have Γ=(10 1n) ; n∈Z}, and we have the Eisenstein series for the cusp associated with Γ0(12) + 4:

Ek,12+4 (z) := 2k3kEk(12z)2 3kEk(6z)2kEk(4z) + 3kEk(3z) +Ek(2z)−Ek(z)

(3k1)(2k1) fork>4.

(3.280)

For the cusp 0 We have Γ0 =(12n1 01) ;n∈Z} and γ0 =W12, and we have the Eisenstein series for the cusp 0 associated with Γ0(12) + 4:

E0k,12+4(z) := 3k/2(2kEk(12z)2Ek(6z)2kEk(4z) +Ek(3z) + 2Ek(2z)−Ek(z))

(3k1)(2k1) fork>4. (3.281) We also haveγ0−10(12) + 4)γ0= Γ0(12) + 4.

For the cusp −1/2 We have Γ−1/2= n

±

³3n+1 3n/2

−6n −3n+1

´

; n∈Z o

andγ−1/2 =W12+,6, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(12) + 4 of weightk>4:

Ek,12+4−1/2 (z) :=

2 2k/23k/2(2k3kEk(12z)3k(2k+ 1)Ek(6z)2kEk(4z) + 3kEk(3z) + (2k+ 1)Ek(2z)−Ek(z))

(3k1)(2k1) .

(3.282) We also haveγ−1/2−10(12) + 4)γ−1/2= Γ0(12) + 4.

For the cusp −1/6 We have Γ−1/6 = n

±

³3n+1 n/2

−18n−3n+1

´

; n∈Z o

andγ−1/6 =W12+,2, and we have the Eisenstein series for the cusp−1/6 associated with Γ0(12) + 4 of weightk>4:

Ek,12+4−1/6 (z) :=

−2 2k/2(2k3k/2Ek(12z)3k/2(2k+ 1)Ek(6z) + 2kEk(4z) + 3k/2Ek(3z)(2k+ 1)Ek(2z) +Ek(z))

(3k1)(2k1) .

(3.283) We also haveγ−1/6−10(12) + 4)γ−1/6= Γ0(12) + 4.

0

The space of modular forms We define

12+4:= ∆12−1/312 ,012+4:= ∆012−1/412 ,

−1/212+4:= (∆−1/212 )2,−1/612+4:= (∆−1/612 )2, which are 4th semimodular forms for Γ0(12) of weight 1.

We have Mk0(12) + 4) = CEk,12+4 CE0k,12+4 CEk,12+4−1/2 CEk,12+4−1/6 ⊕Sk0(12) + 4) and Sk0(12)+4) = ∆12+Mk−40(12)+4) for every even integerk>4. Then, we haveM4n+20(12)+4) = E2,12+0M4n0(12) + 4)C(E1,12+120)2(∆12+)nC(E1,12+30)2(∆12+)n and

M4n0(12) + 4) =C(E4,12+4 )nC(E4,12+4 )n−112+⊕ · · · ⊕CE4,12+4 (∆12+)n−1

C(E4,12+40 )nC(E04,12+4)n−112+⊕ · · · ⊕CE4,12+40 (∆12+)n−1

C(E4,12+4−1/2 )nC(E−1/24,12+4)n−112+⊕ · · · ⊕CE4,12+4−1/2 (∆12+)n−1

C(E4,12+4−1/6 )nC(E−1/64,12+4)n−112+⊕ · · · ⊕CE4,12+4−1/6 (∆12+)n−1C(∆12+)n. Furthermore, we can write

M2n0(12) + 4) =C(∆12+4)2nC(∆12+4)2n−1012+4⊕ · · · ⊕C(∆012+4)2n.

Hauptmodul We define thehauptmodul of Γ0(12) + 4:

J12+4:= ∆012+4/∆12+4(=η4(z)η−4(2z)η−4(3z)η4(4z)η4(6z)η−4(12z))

=1

q−4 + 6q4q23q3+· · ·, (3.284)

wherev(J12+4) =−1 andv0(J12+4) = 1. Then, we have

J12+4:∂F12+4\ {z∈H; Re(z) =±1/2} →[−9,0]R. (3.285)

Location of the zeros of the Eisenstein series Since W12−10(12) + 4)W12 = W12+,6−10(12) + 4)W12+,6=W12+,2−10(12) + 4)W12+,2= Γ0(12) + 4, we have

(2

3z)−kEk,12+40 (W12z) = (√ 6z+

6/2)−kEk,12+4−1/2 (W12+,6z) = (3√

2z+ 1/

2)−kEk,12+4−1/6 (W12+,2z) =Ek,12+4 (z).

Furthermore, we have

E0k,12+4(−1/2 +i/(2 tan(θ/2))) = ((e1)/(2

3))kEk,12+4 ((e1)/12), Ek,12+40 (−1/3 +e0/6) = (−(2e1)/

3)kEk,12+4 (−1/3 +e/6), E−1/2k,12+4((e001)/12) = ((5e+ 1)/(2

6))kEk,12+4 ((e1)/12), Ek,12+4−1/2 (itan(θ/2)/2) = ((e+ 1)/

6)kEk,12+4 (−1/3 +e/6), Ek,12+4−1/6 (−1/2 +itan(θ/2)/3) = ((e+ 1)/(2

2))kEk,12+4 ((e1)/12), Ek,12+4−1/6 (i/(6 tan(θ/2))) = ((e1)/

2)kEk,12+4 (−1/3 +e/6),

wheree0 = (45 cosθ+ 3isinθ)/(5−4 cosθ) ande00= (−513 cosθ+ 12isinθ)/(13 + 5 cosθ).

Now, recall that Γ0(12) + 4 =T1/2−1Γ0(6)T1/2. Then, for k6500, since we can prove that all of the zeros ofEk,6lie on the lower arcs of ∂F6by numerical calculation, we have all of the zeros ofEk,12+4 in the lower arcs of∂F12+4.

Ek,12+4 -12 -16 0 16 12 Ek,12+40 -12 -16 0 16 12

Ek,12+4−1/2 -12 -16 0 16 12 Ek,12+4−1/6 -12 -16 0 16 12

Figure 3.83: Location of the zeros of the Eisenstein series

Location of the zeros of Hecke type Faber Polynomial Similarly to the Eisenstein series, for m 6 200, since we can prove that all of the zeros of Fm,6 lie on the lower arcs of ∂F6 by numerical calculation, we have all of the zeros ofFm,12+4 in the lower arcs of∂F12+4.

0

### (12) + 3

Fundamental domain We have a fundamental domain for Γ0(12) + 3 as follows:

F12+3={|z+ 5/12|>1/12,−1/26Re(z)6−3/8}[ n

|z+ 1/4|>1/(4

3), −3/8< Re(z)6−1/4o [ n|z+ 1/4|>1/(4

3),−1/4< Re(z)<−1/8o [

{|z+ 1/12|>1/12,−1/86Re(z)60}

[{|z−1/12|>1/12, 0< Re(z)<1/8}[ n

|z−1/4|>1/(4

3),1/86Re(z)61/4 o [ n|z−1/4|>1/(4

3),1/4< Re(z)63/8o [

{|z−5/12|>1/12,3/8< Re(z)<1/2},

(3.286)

where W12,3 : −1/4 +e/(4√

3) → −1/4 +ei(π−θ)/(4√

3), (24 77 2)W12,3 : 1/4 +e/(4√

3) 1/4 + ei(π−θ)/(4√

3),¡−1 0

12−1

¢: (e+ 1)/12(ei(π−θ)1)/12, and¡−5 2

12 −5

¢: (e+ 5)/12(ei(π−θ)5)/12.

Then, we have

Γ0(12) + 3 =h(1 10 1), W12,3, (12 11 0), (12 55 2)i. (3.287)

-1

2

-1

4 0 14 12

1 43

Figure 3.84: Γ0(12) + 3

Valence formula The cusps of Γ0(12) + 3 are∞, 0, and−1/2, and the elliptic points are ρ12,1 and ρ12,3 = 1/4 +i/(4√

3). Letf be a modular function of weightk for Γ0(12) + 3, which is not identically zero. We have

v(f) +v−1/3(f) +v−1/2(f) +1

2vρ12,1(f) +1

2vρ12,3(f) + X

p∈Γ0(12)+3\H p6=ρ12,112,3

vp(f) =k. (3.288)

0

Furthermore, the stabilizer of the elliptic pointρ12,1(resp.ρ12,3) is{±I,±W12,3}(resp. {±I,±(24 77 2)W12,3}).

For the cusp We have Γ=(10 1n) ; n∈Z}, and we have the Eisenstein series for the cusp associated with Γ0(12) + 3:

Ek,12+3 (z) := 2k3k/2Ek(12z)3k/2Ek(6z) + 2kEk(4z)−Ek(2z)

(3k/2+ 1)(2k1) fork>4. (3.289) Note that we haveEk,12+3 (z) =Ek,6+3 (2z).

For the cusp 0 We have Γ0 =(12n1 01) ;n∈Z} and γ0 =W12, and we have the Eisenstein series for the cusp 0 associated with Γ0(12) + 3:

E0k,12+3(z) := −(3k/2Ek(6z)3k/2Ek(3z) +Ek(2z)−Ek(z))

(3k/2+ 1)(2k1) fork>4. (3.290) Note that we haveEk,12+30 (z) = 2−k/2Ek,6+30 (z). We also haveγ0−10(12) + 3)γ0= Γ0(12) + 3.

For the cusp −1/2 We have Γ−1/2

±¡6n+1 3n

−12n−6n+1

¢ ; n∈

and γ−1/2 =W12−,6, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(12) + 3 of weightk>4:

Ek,12+3−1/2 (z) :=

−(2k3k/2Ek(12z)3k/2(2k+ 1)Ek(6z) + 2kEk(4z) + 3k/2Ek(3z)(2k+ 1)Ek(2z) +Ek(z))

(3k/2+ 1)(2k1) . (3.291)

Note that we have Ek,12+3−1/2 (z) = Ek,12+12−1/2 (z) = 2−12−k/2Ek,12+−1/2(z). We also have γ−1/2−10(12) + 3)γ−1/2= Γ0(12) + 3.

The space of modular forms We define

12+3:= ∆12−1/412 ,012+3:= ∆012−1/312 ,−1/212+3 := ∆−1/212−1/612 , which are 4th semimodular forms for Γ0(12) + 3 of weight 1. Furthermore, we define

12,1,12+3:= (E1,12+30)5E2,12+0012+3−1/212+312,12,2,12+3:= (E1,12+30)5E2,12+012+3−1/212+312,

12,3,12+3:= (E1,12+30)5E2,12+012+3012+312,12,4,12+3:= (E1,12+30)4012+3−1/212+3(∆12)2,

12,5,12+3:= (E1,12+30)412+3−1/212+3(∆12)2,12,6,12+3:= (E1,12+30)412+3012+3(∆12)2,

12,7,12+3:= (∆012+3)2−1/212+3(∆12)3,12,8,12+3:= ∆12+3(∆−1/212+3)2(∆12)3,

12,9,12+3:= (∆12+3)2012+3(∆12)3. In addition, we denote

A1=E1,12+3012(C∆12+3−1/212+3C∆012+3−1/212+3C∆12+3012+3), A2= (∆12)2(C∆12+3−1/212+3C∆012+3−1/212+3C∆12+3012+3),

B1=C(E12,12+3)nC(E12,12+3 )n−112,1,12+3C(E12,12+3 )n−112,4,12+3

C(E12,12+3 )n−112,7,12+3C(E12,12+3 )n−1(∆12)4C(E12,12+3 )n−212,1,12+3(∆12)4

⊕ · · · ⊕C∆12,7,12+3(∆12)4(n−1)C(∆12)4n,

B2=C(E012,12+3)nC(E12,12+30 )n−112,2,12+3C(E12,12+30 )n−112,5,12+3

C(E12,12+30 )n−112,8,12+3C(E12,12+30 )n−1(∆12)4C(E12,12+30 )n−212,2,12+3(∆12)4

⊕ · · · ⊕C∆12,8,12+3(∆12)4(n−1)C(∆12)4n,

B3=C(E−1/212,12+3)nC(E12,12+3−1/2 )n−112,3,12+3C(E12,12+3−1/2 )n−112,6,12+3

C(E12,12+3−1/2 )n−112,9,12+3C(E12,12+3−1/2 )n−1(∆12)4C(E12,12+3−1/2 )n−212,3,12+3(∆12)4

⊕ · · · ⊕C∆12,8,12+3(∆12)4(n−1)C(∆12)4n.

Now, we have

Mk0(12) + 3) =CEk,12+3CE0k,12+3CE−1/2k,12+3⊕Sk0(12) + 3),

Sk0(12) + 3) = (C∆12,1,12+3C∆12,2,12+3⊕ · · · ⊕C∆12,9,12+3C(∆12)4)Mk−120(12) + 3) for every even integerk>4. Then, we have

M12n0(12) + 3) =B1⊕B2⊕B3C(∆12)4n,

M12n+20(12) + 3) =E2,12+0M12n0(12) + 3)CE1,12+3012+3(∆12)4n,

M12n+40(12) + 3) =E4,12+3(B1C(∆12)4n)⊕E4,12+30 (B2C(∆12)4n)⊕E4,12+3−1/2 (B3C(∆12)4n)

(CE1,12+30C∆12+3)(∆12)4n+1,

M12n+60(12) + 3) =E6,12+3(B1C(∆12)4n)⊕E6,12+30 (B2C(∆12)4n)⊕E6,12+3−1/2 (B3C(∆12)4n)

⊕A1(∆12)4n,

M12n+80(12) + 3) =E8,12+3(B1C(∆12)4n)⊕E8,12+30 (B2C(∆12)4n)⊕E8,12+3−1/2 (B3C(∆12)4n)

⊕E2,12+0A1(∆12)4n⊕A2(∆12)4n,

M12n+100(12) + 3) =E10,12+3(B1C(∆12)4n)⊕E10,12+30 (B2C(∆12)4n)⊕E10,12+3−1/2 (B3C(∆12)4n)

C(E1,12+30)4A1(∆12)4n⊕E2,12+0A2(∆12)4n+1CE1,12+30(∆12)4n+3.

Furthermore, we can write

Mk0(12) + 3) =Ek,12+30(C(∆12+3)nC(∆12+3)n−1012+3⊕ · · · ⊕C(∆012+3)n),

wheren= dim(Mk0(12) + 3))1 =bk−2(k/4− bk/4c)c, and whereEk,12+30:= 1 andE1,12+30, when k≡0 and 2 (mod 4), respectively.

Hauptmodul We define thehauptmodul of Γ0(12) + 3:

J12+3:= ∆012+3/∆12+3(=η2(z)η2(3z)η−2(4z)η−2(12z)) = 1

q−2−q+ 7q39q5+· · · , (3.292) wherev(J12+3) =−1 andv0(J12+3) = 1. Then, we have

J12+3: {|z+ 1/12|= 1/12,−1/86Re(z)60} → −2 + [0,2]R, {|z+ 5/12|= 1/12,−1/26Re(z)6−3/8} → −2−[0,2]R, n

|z+ 1/4|= 1/(4

3),−3/86Re(z)6−1/4 o

→ −2 + 2√ 3[0,1], n

|z−1/4|= 1/(4

3),1/86Re(z)61/4 o

→ −2−2 3[0,1].

(3.293)

Location of the zeros of the Eisenstein series Since W12−10(12) + 3)W12 = W12−,6−10(12) + 3)W12−,6= Γ0(12) + 3, we have

(2

3z)−kE0k,12+3(W12z) = (2√ 3z+

3)−kEk,12+3−1/2 (W12−,6z) =Ek,12+3 (z). (3.294)

0

Lower arcs of∂F12+3

-4 -3 -2 -1

-4 -2 2 4

Figure 3.85: Image byJ12+3

Furthermore, we have

Ek,12+30 (−1/2 +i/(2 tan(θ/2))) = ((e1)/(2

3))kEk,12+3((e1)/12), Ek,12+30 (−1/2 +i/(6 tan(θ/2))) = (−(e1)/2)kEk,12+3((e5)/12),

Ek,12+30 (1/2 +e0/(2√

3)) = (−(

3e+ 1)/2)kEk,12+3(−1/4 +e/(4√ 3)), Ek,12+30 (−1/2 +ei(π−θ0)/(2√

3)) = ((e+

3)/2)kEk,12+3 (1/4 +e/(4√ 3)), Ek,12+3−1/2 (itan(θ/2)/6) = ((e+ 1)/2)kEk,12+3 ((e1)/12), Ek,12+3−1/2 (itan(θ/2)/2) = ((e+ 1)/(2

3))kEk,12+3((e5)/12), Ek,12+3−1/2 (ei(π−θ0)/(2√

3)) = ((e+

3)/2)kEk,12+3 (−1/4 +e/(4√ 3)), Ek,12+3−1/2 (e0/(2√

3)) = (−(

3e+ 1)/2)kEk,12+3(1/4 +e/(4√ 3)),

wheree0 = (−

32 cosθ+isinθ)/(2 +√ 3 cosθ).

Ek,12+3 -12 -14 0 14 12

1 43

Ek,12+30 -12 -14 0 14 12

1 43

Ek,12+3−1/2 -12 -14 0 14 12

1 43

Figure 3.86: Location of the zeros of the Eisenstein series

Now, recall that Ek,12+3 (z) =Ek,6+3(2z). Fork6600, we can prove that all of the zeros ofEk,6+3 lie on the lower arcs of ∂F6+3 by numerical calculation, then we have all of the zeros of Ek,12+3 in the lower arcs of∂F12+3. Similarly to Γ0(9), this case is interesting.

Location of the zeros of Hecke type Faber Polynomial Form 6200, we can prove that all of the zeros ofFm,12+3 lie on the lower arcs of∂F12+3 by numerical calculation.

0

### (12)

Fundamental domain We have a fundamental domain for Γ0(12) as follows:

F12={|z+ 5/12|>1/12,−1/26Re(z)<−1/3}[

{|z+ 7/24|>1/24,−1/36Re(z)<−1/4}

[{|z+ 5/24|>1/24, −1/46Re(z)<−1/6}[

{|z+ 1/12|>1/12, −1/66Re(z)60}

[{|z−1/12|>1/12, 0< Re(z)61/6}[

{|z−5/24|>1/24,1/6< Re(z)61/4}

[{|z−7/24|>1/24, 1/4< Re(z)61/3}[

{|z−5/12|>1/12,1/3< Re(z)<1/2},

(3.295)

where ¡−1 0

12 −1

¢ : (e+ 1)/12(ei(π−θ)1)/12, ¡−5 1

24 −5

¢: (e+ 5)/24(ei(π−θ)5)/24, ¡−7 2

24−7

¢: (e+ 7)/24(ei(π−θ)7)/24, and¡−5 2

12 −5

¢: (e+ 5)/12(ei(π−θ)5)/12. Then, we have

Γ0(12) =h−I, (1 10 1), (12 11 0), (12 55 2), (24 55 1), (24 77 2)i. (3.296)

-1

2

-1

3

-1

4

-1

6 0 1

6 1 4 1 3

1 2

Figure 3.87: Γ0(12)

Valence formula The cusps of Γ0(12) are ∞, 0,−1/3,−1/4, −1/2, and −1/6. Letf be a modular function of weightkfor Γ0(12), which is not identically zero. We have

v(f) +v0(f) +v−1/3(f) +v−1/4(f) +v−1/2(f) +v−1/6(f) + X

p∈Γ0(12)\H

vp(f) = 2k. (3.297)

For the cusp We have Γ=(10 1n) ; n∈Z}, and we have the Eisenstein series for the cusp associated with Γ0(12):

Ek,12(z) :=2k3kEk(12z)3kEk(6z)2kEk(4z) +Ek(2z)

(3k1)(2k1) fork>4. (3.298) Note that we haveEk,12 (z) =Ek,6(2z).

For the cusp 0 We have Γ0 =(12n1 01) ;n∈Z} and γ0 =W12, and we have the Eisenstein series for the cusp 0 associated with Γ0(12):

Ek,120 (z) :=3k/2(Ek(6z)−Ek(3z)−Ek(2z) +Ek(z))

(3k1)(2k1) fork>4. (3.299) Note that we haveEk,120 (z) = 2−k/2Ek,60 (z). We also haveγ0−1 Γ0(12)γ0= Γ0(12).

For the cusp −1/3 We have Γ−1/3

±¡12n+1 4n

−36n −12n+1

¢; n∈

and γ−1/3 =W12,4, and we have the Eisenstein series for the cuspassociated with Γ0(12):

Ek,12−1/3(z) :=−(3kEk(6z)3kEk(3z)−Ek(2z) +Ek(z))

(3k1)(2k1) fork>4. (3.300) Note that we haveEk,12−1/3(z) = 2−k/2Ek,6−1/3(z). We also haveγ−1/3−1 Γ0(12)γ−1/3= Γ0(12).

0

For the cusp −1/4 We have Γ−1/4

±¡12n+1 3n

−48n −12n+1

¢; n∈

and γ−1/4 =W12,3, and we have the Eisenstein series for the cusp−1/4 associated with Γ0(12):

Ek,12−1/4(z) :=−3k/2(2kEk(12z)−Ek(6z)2kEk(4z) +Ek(2z))

(3k1)(2k1) fork>4. (3.301) Note that we haveEk,12−1/4(z) =Ek,6−1/4(2z). We also haveγ−1−1/4 Γ0(12)γ−1/4= Γ0(12).

For the cusp −1/2 We have Γ−1/2

±¡6n+1 3n

−12n−6n+1

¢ ; n∈

and γ−1/2 =W12−,6, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(12) of weightk>4:

Ek,12−1/2(z) :=

3k/2(2k3kEk(12z)3k(2k+ 1)Ek(6z)2kEk(4z) + 3kEk(3z) + (2k+ 1)Ek(2z)−Ek(z))

(3k1)(2k1) . (3.302)

Note that we haveEk,12−1/2(z) = 2−12−k/2Ek,12+4−1/2 (z). We also haveγ−1/2−1 Γ0(12)γ−1/2= Γ0(12).

For the cusp −1/6 We have Γ−1/6

±¡6n+1 n

−36n−6n+1

¢ ; n∈

and γ−1/6 =W12−,2, and we have the Eisenstein series for the cusp−1/6 associated with Γ0(12) of weightk>4:

Ek,12−1/6(z) :=

−(2k3k/2Ek(12z)3k/2(2k+ 1)Ek(6z) + 2kEk(4z) + 3k/2Ek(3z)(2k+ 1)Ek(2z) +Ek(z))

(3k1)(2k1) . (3.303)

Note that we haveEk,12−1/6(z) = 2−12−k/2Ek,12+4−1/6 (z). We also haveγ−1/6−1 Γ0(12)γ−1/6= Γ0(12).

The space of modular forms We define

6,1,12−:= (∆012)2−1/312−1/412−1/212−1/612 ,6,2,12−:= (∆12)2−1/312−1/412−1/212−1/612 ,

6,3,12−:= ∆12012(∆−1/412 )2−1/212−1/612 ,6,4,12−:= ∆12012(∆−1/312 )2−1/212−1/612 ,

6,5,12−:= ∆12012−1/312−1/412 (∆−1/612 )2,6,6,12−:= ∆12012−1/312−1/412 (∆−1/212 )2. Now, we have

Mk0(12)) =CEk,12 CEk,120 CEk,12−1/3CEk,12−1/4⊕Ek,12−1/2CEk,12−1/6⊕Sk0(12)), Sk0(12)) = (C∆6,1,12−C∆6,2,12−⊕ · · · ⊕C∆6,6,12−C(∆12)2)Mk−60(12)), for every even integerk>4. Then, we have

M6n0(12)) =C(E6,12 )nC(E6,12 )n−16,1,12−C(E6,12 )n−1(∆12)2⊕ · · · ⊕C∆6,1,12−(∆12)2(n−1)

C(E6,120 )nC(E6,120 )n−16,2,12−C(E6,120 )n−1(∆12)2⊕ · · · ⊕C∆6,2,12−(∆12)2(n−1)

C(E6,12−1/3)nC(E6,12−1/3)n−16,3,12−C(E6,12−1/3)n−1(∆12)2⊕ · · · ⊕C∆6,3,12−(∆12)2(n−1)

C(E6,12−1/4)nC(E6,12−1/4)n−16,4,12−C(E6,12−1/4)n−1(∆12)2⊕ · · · ⊕C∆6,4,12−(∆12)2(n−1)

C(E6,12−1/2)nC(E6,12−1/2)n−16,5,12−C(E6,12−1/2)n−1(∆12)2⊕ · · · ⊕C∆6,5,12−(∆12)2(n−1)

C(E6,12−1/6)nC(E6,12−1/6)n−16,6,12−C(E6,12−1/6)n−1(∆12)2⊕ · · · ⊕C∆6,6,12−(∆12)2(n−1)

C(∆12)2n, M6n+20(12)) =E2,12+0M6n0(12))

(C(E1,12+120)2C(E1,12+30)2C(∆12)4C(∆012)4)(∆12)2n,

M6n+40(12)) =E4,12 (C(E6,12 )nC(E6,12 )n−16,1,12−C(E6,12 )n−1(∆12)2⊕ · · · ⊕C(∆12)2n)

⊕E4,120 (C(E6,120 )nC(E6,120 )n−16,2,12−C(E6,120 )n−1(∆12)2⊕ · · · ⊕C(∆12)2n)

⊕E4,12−1/3(C(E6,12−1/3)nC(E6,12−1/3)n−16,3,12−C(E−1/36,12 )n−1(∆12)2⊕ · · · ⊕C(∆12)2n)

⊕E4,12−1/4(C(E6,12−1/4)nC(E6,12−1/4)n−16,4,12−C(E−1/46,12 )n−1(∆12)2⊕ · · · ⊕C(∆12)2n)

⊕E4,12−1/2(C(E6,12−1/2)nC(E6,12−1/2)n−16,5,12−C(E−1/26,12 )n−1(∆12)2⊕ · · · ⊕C(∆12)2n)

⊕E4,12−1/6(C(E6,12−1/6)nC(E6,12−1/6)n−16,6,12−C(E−1/66,12 )n−1(∆12)2⊕ · · · ⊕C(∆12)2n)

(C(∆12)4(∆012)4C(∆−1/312 )4(∆−1/412 )4)(∆12)2nCE1,12+120(∆12)2n+1.

Furthermore, we can write

M2n0(12)) =C(∆12)4nC(∆12)4n−1012⊕ · · · ⊕C(∆012)4n.

Hauptmodul We define thehauptmodul of Γ0(12):

J12:= ∆012/∆12(=η3(z)η−2(2z)η−1(3z)η(4z)η2(6z)η−3(12z))

=1

q−3 + 2q+q32q7− · · · , (3.304)

wherev(J12) =−1 andv0(J12) = 1. Then, we have

J12:∂F12\ {z∈H; Re(z) =±1/2} →[−6,0]R. (3.305)

Location of the zeros of the Eisenstein series Since W12−10(12))W12 =W12,4−10(12))W12,4 = W12,3−10(12))W12,3= Γ0(12), we have

(2

3z)−kE0k,12(W12z) = (6z−2)−kE−1/2k,12 (W6,3z) = (4√ 3z+

3)−kE−1/6k,12 (W12,4z) =Ek,12 (z).

0

Furthermore, we have

Ek,120 (−1/2 +i/(2 tan(θ/2))) = ((e1)/(2

3))kEk,12 ((e1)/12), Ek,120 ((e15)/12) = (−(5e1)/(4

3))kEk,12 ((e5)/24), Ek,120 ((e27)/24) = (−(7e1)/(4

3))kEk,12 ((e7)/24), Ek,120 ((e15)/24) = (−(5e1)/(2

3))kEk,12 ((e5)/12), Ek,12−1/3((e35)/24) = (−(3e+ 1)/2)kEk,12 ((e1)/12), Ek,12−1/3((e31)/12) = ((3e+ 1)/4)kEk,12 ((e5)/24),

Ek,12−1/3(itan(θ/2)/3) = (−(e+ 1)/4)kEk,12 ((e7)/24), Ek,12−1/3(−1/2 +i/(6 tan(θ/2))) = (−(e1)/2)kEk,12 ((e5)/24),

Ek,12−1/4((e45)/24) = ((e+ 2)/

3)kEk,12 ((e1)/12), Ek,12−1/4(−1/2 +itan(θ/2)/4) = ((e+ 1)/(2

3))kEk,12 ((e5)/24), Ek,12−1/4(i/(4 tan(θ/2))) = ((e1)/(2

3))kEk,12 ((e7)/24), E−1/4k,12 ((e401)/12) = ((e2)/

3)kEk,12 ((e5)/12), E−1/2k,12 ((e107)/24) = ((5e+ 1)/(2

3))kEk,12 ((e1)/12), E−1/2k,12 ((e205)/24) = ((7e+ 1)/(4

3))kEk,12 ((e5)/24), E−1/2k,12 ((e101)/12) = ((5e+ 1)/(4

3))kEk,12 ((e7)/24), Ek,12−1/2(itan(θ/2)/2) = ((e+ 1)/(2

3))kEk,12 ((e5)/12), Ek,12−1/6(itan(θ/2)/6) = (−(e+ 1)/2)kEk,12 ((e1)/12), Ek,12−1/6(−1/2 +i/(3 tan(θ/2))) = (−(e1)/4)kEk,12 ((e5)/24), E−1/6k,12 ((e305)/12) = (−(3e1)/4)kEk,12 ((e7)/24), E−1/6k,12 ((e307)/24) = (−(3e1)/2)kEk,12 ((e5)/12),

wheree1= (5−13 cosθ+ 12isinθ)/(13−5 cosθ),e10 = (−5−13 cosθ+ 12isinθ)/(13 + 5 cosθ),e2 = (7−25 cosθ+24isinθ)/(25−7 cosθ),e20 = (−7−25 cosθ+24isinθ)/(25+7 cosθ),e3= (−3−5 cosθ+ 4isinθ)/(5 + 3 cosθ),e30 = (35 cosθ+ 4isinθ)/(5−3 cosθ),e4 = (4 + 5 cosθ+ 3isinθ)/(5 + 4 cosθ), ande40 = (−4 + 5 cosθ+ 3isinθ)/(5−4 cosθ).

Now, recall thatEk,12 (z) =Ek,6(2z). Then, fork6500, since we can prove that all of the zeros of Ek,6 lie on the lower arcs of∂F6 by numerical calculation, we have all of the zeros ofEk,12 in the lower arcs of∂F12.

Location of the zeros of Hecke type Faber Polynomial Form 6200, we can prove that all of the zeros ofFm,12 lie on the lower arcs of∂F12 by numerical calculation.

### Bibliography

[ACMS] D. Alexander, C. Cummins J. McKay, and C. Simons, Completely replicable functions. In:

Groups, combinatorics & geometry (M. Liebeck and J. Saxl eds.), 87–98, London Math. Soc.

Lecture Note Ser., No. 165, Cambridge Univ. Press, Cambridge, 1992. (Proceedings of the L.M.S.

Symposium on Groups and Combinatorics, Durham, 1990.)

[AKN] T. Asai, M. Kaneko, and H. Ninomiya, Zeros of certain modular functions and an application, Comment. Math. Univ. St. Paul. 46 (1997), 93–101.

Ek,12 -12 -13-14-16 0 16 14 13 12 Ek,120 -12 -13 -14-16 0 16 14 13 12

Ek,12−1/3 -12 -13-14-16 0 16 14 13 12 Ek,12−1/4 -12 -13 -14-16 0 16 14 13 12

Ek,12−1/2 -12 -13-14-16 0 16 14 13 12 Ek,12−1/6 -12 -13 -14-16 0 16 14 13 12

Figure 3.88: Location of the zeros of the Eisenstein series

[BKM] E. Bannai, K. Kojima, and T. Miezaki,On the zeros of Hecke type Faber polynomial, Kyushu. J.

Math. 62(2007), 15–61.

[CN] J. H. Conway, S. P. Norton,Monstrous moonshine, Bull. London Math. Soc.,11(1979), 308–339.

[Ge] J. Getz,A generalization of a theorem of Rankin and Swinnerton-Dyer on zeros of modular forms, Proc. Amer. Math. Soc.,132(2004), No. 8, 2221–2231.

[H] H. Hahn, On zeros of Eisenstein series for genus zero Fuchsian groups, Proc. Amer. Math. Soc.

135(2007), No. 8, 2391–2401.

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[Kr] A. Krieg, Modular Forms on the Fricke Group., Abh. Math. Sem. Univ. Hamburg, 65(1995), 293–299.

[MNS] T. Miezaki, H. Nozaki, and J. Shigezumi,On the zeros of Eisenstein series for Γ0(2) andΓ0(3), J. Math. Soc. Japan,59(2007), 693–706.

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[RSD] F. K. C. Rankin, H. P. F. Swinnerton-Dyer, On the zeros of Eisenstein Series, Bull. London Math. Soc., 2(1970), 169–170.

[Se] J. -P. Serre,A Course in Arithmetic, Graduate Texts in Mathematics, No. 7, Springer-Verlag, New York-Heidelberg, 1973. (Translation ofCours d’arithm´etique(French), Presses Univ. France, Paris, 1970.)

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0

[SJ1] J. Shigezumi, On the zeros of Eisenstein series for Γ0(p) and Γ0(p) of low levels, M.S. thesis, Kyushu University, 2006.

[SJ2] J. Shigezumi,On the zeros of the Eisenstein series forΓ0(5)andΓ0(7), Kyushu. J. Math. 61(2007), 527–549.

## for genus zero Fuchsian group

Let ΓSL2(R) be a genus zero Fuchsian group of the first kind havingas a cusp, and letE2kΓ be the holomorphic Eisenstein series associated with Γ for thecusp that does not vanish at but vanishes at all the other cusps. In the paper “On zeros of Eisenstein series for genus zero Fuchsian groups”, under assumptions on Γ, and on a certain fundamental domainF, H. Hahn proved that all but at mostc(Γ,F) (a constant) of the zeros ofE2kΓ lie on a certain subset of{z∈H : jΓ(z)R}.

In this note, we consider a small generalization of Hahn’s result on the domain locating the zeros of E2kΓ. We can prove most of the zeros ofE2kΓ inF lie on its lower arcs under the same assumption.

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