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# Level 10

0

F10+ -12 -103 0 103 12

1 10 1 10

Figure 3.56: Γ0(10)+

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

Ek,10+(z) :=10k/2Ek(10z) + 5k/2Ek(5z) + 2k/2Ek(2z) +Ek(z)

(5k/2+ 1)(2k/2+ 1) fork>4. (3.209)

The space of modular forms Letk be an even integer k>4. We haveMk0(10)+) =CEk,10+ Sk0(10)+) andSk0(10)+) = (C(E2/3,100)810⊕C(E2/3,100)4(∆10)2⊕C(∆10)3)Mk−80(10)+). Then, we have

Mk0(10)+) =Ek,10+0(C((E2/3,100)4)nC((E2/3,100)4)n−110⊕ · · · ⊕C(∆10)n),

wheren= dim(Mk0(10)+))−1 =b3k/8−2(k/4− bk/4c)c, and whereEk,10+0:= 1, (E2/3,100)3E8/3,100, (E2/3,100)2, andE2/3,100E8/3,100, whenk≡0, 2, 4, and 6 (mod 8), respectively.

Hauptmodul We define thehauptmodul of Γ0(10)+:

J10+:= (E2/3,100)4/∆10=1

q+ 4 + 22q+ 56q2+ 177q3+· · ·, (3.210) wherev(J10+) =−1 andvρ10,1(J10+) = 4. Then, we have

J10+:∂F10+\ {z∈H; Re(z) =±1/2} →[−4,16]R. (3.211)

Location of the zeros of the Eisenstein series Fork6800, we can prove that all of the zeros of Ek,10+ lie on the lower arcs of∂F10+ by numerical calculation.

-1

2

-3

10 0 103 12

1 10 1 10

Figure 3.57: Location of the zeros of the Eisenstein series

Location of the zeros of Hecke type Faber Polynomial Form 6200, we can prove that all of the zeros ofFm,10+ lie on the lower arcs of∂F10+ by numerical calculation.

0

0

### (10)

Fundamental domain We have a fundamental domain for Γ0(10) as follows:

F10+10={|z+ 9/20|>1/20, −1/26Re(z)<−3/7}

n

|z+ 1/3|>1/(3

10),−3/76Re(z)<−3/10 o [ n|z|>1/

10, −3/106Re(z)60o [ n

|z|>1/

10,0< Re(z)63/10o [ n|z−1/3|>1/(3

10),3/10< Re(z)63/7o [

{|z−9/20|>1/20,3/7< Re(z)<1/2},

(3.212)

where W10 : e/√

10 ei(π−θ)/√

10, ¡−3−1

10 3

¢W10 : e/(3√

10) + 1/3 ei(π−θ)/(3√

10)1/3, and

¡−9 4

20 −9

¢: (e+ 5)/12(ei(π−θ)5)/12. Then, we have Γ0(10) =h(1 10 1), W10, ¡−3−1

10 3

¢, (20 99 4)i. (3.213)

-1 2

-3

7

-3

10 0 3

10 3 7 1 2 1

10

Figure 3.58: Γ0(10)

Valence formula The cusps of Γ0(10) are and−1/2, and the elliptic points arei/√

10 andρ10,5=

−3/7 +i/(5√

10). Letf be a modular function of weightkfor Γ0(10), which is not identically zero. We have

v(f) +v−1/2(f) +1

2vi/10(f) +1

2vρ10,1(f) +1

2vρ10,3(f) + X

p∈Γ0(10)\H p6=i/

10,ρ10,110,3

vp(f) = 3k

4 . (3.214)

Furthermore, the stabilizer of the elliptic point i/√

10 (resp. ρ10,1, ρ10,3) is {±I,±W10} (resp.

10 3

±I,±¡−13 3

30 −7

¢W10

ª).

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

Ek,10+10 (z) := (10k/2+ 1)(10k/2Ek(10z) +Ek(z))(5k/2+ 2k/2)(5k/2Ek(5z) + 2k/2Ek(2z))

(5k1)(2k1) fork>4.

(3.215)

For the cusp −1/2 We have Γ−1/2

±¡10n+1 5n

−20n −10n+1

¢; n∈

and γ−1/2 =W10,5, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(10):

Ek,10+10−1/2 (z) := −(5k/2+ 2k/2)(10k/2Ek(10z) +Ek(z)) + (10k/2+ 1)(5k/2Ek(5z) + 2k/2Ek(2z))

(5k1)(2k1) fork>4.

(3.216) We also haveγ−1/2−1 Γ0(10)γ−1/2= Γ0(10).

0

The space of modular forms We define

10+10:= ∆10010,−1/210+10:= ∆−1/210−1/510 , which are 2nd semimodular forms for Γ0(10) of weight 2. Furthermore, we define

8,1,10+10:= ∆10+10(∆−1/210+10)5,8,2,10+10:= (∆10+10)5−1/210+10,

8,3,10+10:= (∆10+10)2(∆−1/210+10)4,8,4,10+10:= (∆10+10)4(∆−1/210+10)2. Now, we have

Mk0(10)) =CEk,10+10CE−1/2k,10+10⊕Sk0(10)),

Sk0(10)) = (C∆8,1,10+10C∆8,2,10+10C∆8,3,10+10C∆8,4,10+10C(∆10)3)Mk−80(10)) for every even integerk>4. Then, we haveM4n+20(10)) =E2,10+100M4n0(10)) and

M8n0(10)) =C(E8,10+10)nC(E8,10+10 )n−18,1,10+10C(E8,10+10 )n−18,3,10+10

C(E8,10+10 )n−1(∆10)3⊕ · · · ⊕C∆8,3,10+10(∆10)3(n−1)

C(E8,10+10−1/2 )nC(E8,10+10−1/2 )n−18,2,10+10C(E−1/28,10+10)n−18,4,10+10

C(E8,10+10−1/2 )n−1(∆10)3⊕ · · · ⊕C∆8,4,10+10(∆10)3(n−1)C(∆10)3n, M8n+40(10)) =E4,10+10 (C(E8,10+10 )nC(E8,10+10 )n−18,1,10+10C(E8,10+10 )n−18,3,10+10

C(E8,10+10 )n−1(∆10)3⊕ · · · ⊕C∆8,3,10+10(∆10)3(n−1)C(∆10)3n)

⊕E4,10+10−1/2 (C(E8,10+10−1/2 )nC(E8,10+10−1/2 )n−18,2,10+10C(E8,10+10−1/2 )n−18,4,10+10

C(E8,10+10−1/2 )n−1(∆10)3⊕ · · · ⊕C∆8,4,10+10(∆10)3(n−1)C(∆10)3n)

(C∆10+10(∆−1/210+10)2C(∆10+10)2−1/210+10)(∆10)3n. Furthremore, we can write

M4n0(10)) =C(∆10+10)3nC(∆10+10)3n−1−1/210+10⊕ · · · ⊕C(∆−1/210+10)3n.

Hauptmodul We define thehauptmodul of Γ0(10):

J10+10:= ∆−1/210+10/∆10+10(=η−6(z)η6(2z)η−6(5z)η6(10z)) = 1

q+ 6 + 21q+ 62q2+ 162q3+· · ·, (3.217) wherev(J10+10) =−1 andv−1/2(J10+10) = 1. Then, we have

J10+10:∂F10+10\ {z∈H; Re(z) =±1/2} →[0,9 + 4

5]R. (3.218)

Location of the zeros of the Eisenstein series SinceW10,5−1Γ0(10)W10,5= Γ0(10), we have Ek,10+10−1/2 (W10,5z) = (2√

5z+

5)kEk,10+10 (z). (3.219)

Furthermore, we have

Ek,10+10−1/2 (e0/(3√

10)1/3) = (−

5e−√

2)kEk,10+10 (e/√ 10), Ek,10+10−1/2 (e0/√

10) = (

2(e+ 1)/3)kEk,10+10 (e/(3√

10)1/3), Ek,10+10−1/2 (itan(θ/2)/2) = ((e+ 1)/(2

5))kEk,10+10 ((e9)/20), wheree0 = (−2

107 cosθ+ 3isinθ)/(7 + 2√

10 cosθ).

Fork6500, we can prove that all of the zeros ofEk,10+10lie on the lower arcs of∂F10+10by numerical calculation.

Ek,10+10 -12-37 -103 0 103 37 12

1 10

Ek,10+10−1/2 -12-37 -103 0 103 37 12

1 10

Figure 3.59: Location of the zeros of the Eisenstein series

Location of the zeros of Hecke type Faber Polynomial For every odd integer m 6 200 but m= 7,9,11, we can prove that all of the zeros of Fm,10+10 lie on the lower arcs of∂F10+10by numerical calculation. On the other hand, by numerical calculation, form= 7,9,11, we can prove that all but two of the zeros ofFm,10+10 lie on the lower arcs of∂F10+10, and two of the zeros ofFm,10+10 do not lie on

∂F10+10. For the other cases wheremis even andm6200, by numerical calculation, we can prove that all but one of the zeros ofFm,10+10lie on the lower arcs of∂F10+10, and one of the zeros ofFm,10+10 lies on∂F10+10but does not on the lower arcs.

0

### (10) + 5

Fundamental domain We have a fundamental domain for Γ0(10) + 5 as follows:

F10+5= n

|z+ 1/2|>1/(2

5),−1/26Re(z)<−3/10o [ n

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

5), −3/106Re(z)<−1/6 o [{|z+ 1/10|>1/10,−1/66Re(z)60}[

{|z−1/10|>1/10,0< Re(z)61/6}

[ n|z−1/4|>1/(4

5), 1/6< Re(z)63/10o [ n

|z−1/2|>1/(2

5),3/10< Re(z)<1/2 o

, (3.220) where¡−1 0

10−1

¢: (e+ 1)/10(ei(π−θ)1)/10,¡−7−4

30 17

¢W10,5:e/(4√

5) + 1/4→ei(π−θ)/(4√

5)1/4, and¡−9−5

20 11

¢W10,5:e/(2√

5) + 1/2→ei(π−θ)/(2√

5)1/2. Then, we have Γ0(10) + 5 =h(1 10 1), W10,5, (10 11 0), ¡−3−1

10 3

¢i. (3.221)

-1

2

-3

10

-1

6 0 1

6 3 10

1 2

Figure 3.60: Γ0(10) + 5

Valence formula The cusps of Γ0(10)+5 areand 0, and the elliptic points areρ10,1=−3/10+i/10, ρ10,2 = −1/2 +i/(2√

5), and ρ10,4 = −1/6 +i/(6√

5). Let f be a modular function of weight k for Γ0(10) + 5, which is not identically zero. We have

v(f) +v0(f) +1

2vρ10,1(f) +1

2vρ10,2(f) +1

2vρ10,4(f) + X

p∈Γ0(10)+5\H p6=ρ10,110,210,4

vp(f) = 3k

4 . (3.222) Furthermore, the stabilizer of the elliptic point ρ10,1 (resp. ρ10,2, ρ10,4) is ©

±I,±¡−3−1

10 3

¢ª (resp.

±I,±¡−3−2

20 13

¢W10,5

ª).

0

For the cusp We have Γ=(10 1n) ; n∈Z}, and we have the Eisenstein series for the cusp associated with Γ0(10) + 5:

Ek,10+5(z) := 2k5k/2Ek(10z)5k/2Ek(5z) + 2kEk(2z)−Ek(z)

(5k/2+ 1)(2k1) fork>4. (3.223)

For the cusp 0 We have Γ0 =(10n1 01) ;n∈Z} and γ0 =W10, and we have the Eisenstein series for the cusp 0 associated with Γ0(10) + 5:

E0k,10+5(z) := −2k/2(5k/2Ek(10z)5k/2Ek(5z) +Ek(2z)−Ek(z))

(5k/2+ 1)(2k1) fork>4. (3.224) We also haveγ0−10(10) + 5)γ0= Γ0(10) + 5.

The space of modular forms We define

10+5:= ∆10−1/210 ,010+5:= ∆010−1/510 ,

which are 2nd semimodular forms for Γ0(10) + 5 of weight 2. Furthermore, we define

8,1,10+5:= ∆10+5(∆010+5)5,8,2,10+5:= (∆10+5)5010+5,

8,3,10+5:= (∆10+5)2(∆010+5)4,8,4,10+5:= (∆10+5)4(∆010+5)2. Now, we have

Mk0(10) + 5) =CEk,10+5CE0k,10+5⊕Sk0(10) + 5),

Sk0(10) + 5) = (C∆8,1,10+5C∆8,2,10+5C∆8,3,10+5C∆8,4,10+5C(∆10)3)Mk−80(10) + 5) for every even integerk>4. Then, we haveM4n+20(10) + 5) =E2,10+50M4n0(10) + 5) and

M8n0(10) + 5) =C(E8,10+5 )nC(E8,10+5 )n−18,1,10+5C(E8,10+5 )n−18,3,10+5

C(E8,10+5 )n−1(∆10)3⊕ · · · ⊕C∆8,3,10+5(∆10)3(n−1)

C(E08,10+5)nC(E8,10+50 )n−18,2,10+5C(E8,10+50 )n−18,4,10+5

C(E8,10+50 )n−1(∆10)3⊕ · · · ⊕C∆8,4,10+5(∆10)3(n−1)C(∆10)3n, M8n+40(10) + 5) =E4,10+5 (C(E8,10+5 )nC(E8,10+5)n−18,1,10+5C(E8,10+5 )n−18,3,10+5

C(E8,10+5 )n−1(∆10)3⊕ · · · ⊕C∆8,3,10+5(∆10)3(n−1)C(∆10)3n)

⊕E4,10+50 (C(E8,10+50 )nC(E8,10+50 )n−18,2,10+5C(E08,10+5)n−18,4,10+5

C(E8,10+50 )n−1(∆10)3⊕ · · · ⊕C∆8,4,10+5(∆10)3(n−1)C(∆10)3n)

(C∆10+5(∆010+5)2C(∆10+5)2010+5)(∆10)3n. Furthremore, we can write

M4n0(10) + 5) =C(∆10+5)3nC(∆10+5)3n−1010+5⊕ · · · ⊕C(∆010+5)3n.

Hauptmodul We define thehauptmodul of Γ0(10) + 5:

J10+5:= ∆010+5/∆10+5(=η4(z)η−4(2z)η4(5z)η−4(10z)) = 1

q 4 + 6q8q2+ 17q3− · · ·, (3.225) wherev(J10+5) =−1 andv0(J10+5) = 1. Then, we have

J10+5:∂F10+5\ {z∈H; Re(z) =±1/2} →[−62

5,0]R. (3.226)

Location of the zeros of the Eisenstein series SinceW10−10(10) + 5)W10= Γ0(10) + 5, we have E0k,10+5(W10z) = (√

10z)kEk,10+5 (z). (3.227)

Furthermore, we have

Ek,10+50 (−1/2 +i/(2 tan(θ/2))) = ((e1)/

10)kEk,10+5((e1)/10), Ek,10+50 (e0/(2√

5)1/2) = (−(

5e1)/(2

2))kEk,10+5 (e/(4√

5)1/4), Ek,10+50 (e0/(4√

5)1/4) = (−(

5e1)/

2)kEk,10+5 (e/(2√

5)1/2), wheree0 = (2

56 cosθ+ 4isinθ)/(6−2

5 cosθ).

Fork6450, we can prove that all of the zeros ofEk,10+5 lie on the lower arcs of∂F10+5 by numerical calculation.

Ek,10+5 -12 -103 -16 0 16 103 12 Ek,10+50 -12 -103 -16 0 16 103 12 Figure 3.61: Location of the zeros of the Eisenstein series

Location of the zeros of Hecke type Faber Polynomial Form 6200, we can prove that all of the zeros ofFm,10+5 lie on the lower arcs of∂F10+5 by numerical calculation.

0

### (10) + 2

Fundamental domain We have a fundamental domain for Γ0(10) + 2 as follows:

F10+2= n

|z+ 2/5|>1/(5

2),−1/26Re(z)6−3/10o [ n

|z+ 1/5|>1/(5

2),−3/10< Re(z)<−1/10 o [{|z+ 1/10|>1/10,06Re(z)60}[

{|z−1/10|>1/10,0< Re(z)<1/10}

[ n|z−1/5|>1/(5

2),1/106Re(z)63/10o [ n

|z−2/5|>1/(5

2),3/10< Re(z)<1/2 o

, (3.228) where¡−1 0

10−1

¢: (e+ 1)/10(ei(π−θ)1)/10,¡−3−1

10 3

¢W10,5:e/(5√

2)1/5→ei(π−θ)/(5√

2)2/5, and (40 99 2)W10,5:e/(5√

2) + 2/5→ei(π−θ)/(5√

2) + 1/5. Then, we have

Γ0(10) + 2 =h(1 10 1), W10,2, (10 11 0), (40 99 2)i. (3.229)

-1

2

-3

10

-1

10 0 1

10 3 10

1 2

Figure 3.62: Γ0(10) + 2

0

Valence formula The cusps of Γ0(10) + 2 areand 0, and the elliptic points areρ10,1=−3/10 +i/10 andρ10,5= 3/10 +i/10. Let f be a modular function of weightkfor Γ0(10) + 2, which is not identically zero. We have

v(f) +v0(f) +1

4vρ10,1(f) +1

4vρ10,5(f) + X

p∈Γ0(10)+2\H p6=ρ10,110,5

vp(f) =3k

4 . (3.230)

Furthermore, the stabilizer of the elliptic pointρ10,1(resp.ρ10,5) is©

±I,±¡−3−1

10 3

¢,±W10,2¡−3−1

10 3

¢W10,2

±I,±¡3 −1

10−3

¢(13 330 7)W10,2(40 99 2)W10,2

ª).

For the cusp We have Γ=(10 1n) ; n∈Z}, and we have the Eisenstein series for the cusp associated with Γ0(10) + 2:

Ek,10+2(z) := 2k/25kEk(10z) + 5kEk(5z)2k/2Ek(2z)−Ek(z)

(5k1)(2k/2+ 1) fork>4. (3.231)

For the cusp 0 We have Γ0 =(10n1 01) ;n∈Z} and γ0 =W10, and we have the Eisenstein series for the cusp 0 associated with Γ0(10) + 2:

E0k,10+2(z) := −5k/2(2k/2Ek(10z) +Ek(5z)2k/2Ek(2z)−Ek(z))

(5k1)(2k/2+ 1) fork>4. (3.232) We also haveγ0−10(10) + 2)γ0= Γ0(10) + 2.

The space of modular forms We define

10+2:= ∆10−1/510 ,010+2:= ∆010−1/210 ,

which are 2nd semimodular forms for Γ0(10) + 2 of weight 2. Furthermore, we define

8,1,10+2:= ∆10+2(∆010+2)5,8,2,10+2:= (∆10+2)5010+2,

8,3,10+2:= (∆10+2)2(∆010+2)4,8,4,10+2:= (∆10+2)4(∆010+2)2. Now, we have

Mk0(10) + 2) =CEk,10+2CE0k,10+2⊕Sk0(10) + 2),

Sk0(10) + 2) = (C∆8,1,10+2C∆8,2,10+2C∆8,3,10+2C∆8,4,10+2C(∆10)3)Mk−80(10) + 2) for every even integerk>4. Then, we haveM8n+20(10) + 2) =E2,10+20M8n0(10) + 2) and

M8n0(10) + 2) =C(E8,10+2)nC(E8,10+2 )n−18,1,10+2C(E8,10+2 )n−18,3,10+2

C(E8,10+2 )n−1(∆10)3⊕ · · · ⊕C∆8,3,10+2(∆10)3(n−1)

C(E8,10+20 )nC(E8,10+20 )n−18,2,10+2C(E8,10+20 )n−18,4,10+2

C(E8,10+20 )n−1(∆10)3⊕ · · · ⊕C∆8,4,10+2(∆10)3(n−1)C(∆10)3n,

M8n+40(10) + 2) =E4,10+2 (C(E8,10+2 )nC(E8,10+2)n−18,1,10+2C(E8,10+2 )n−18,3,10+2

C(E8,10+2 )n−1(∆10)3⊕ · · · ⊕C∆8,3,10+2(∆10)3(n−1)C(∆10)3n)

⊕E4,10+20 (C(E8,10+20 )nC(E8,10+20 )n−18,2,10+2C(E08,10+2)n−18,4,10+2

C(E8,10+20 )n−1(∆10)3⊕ · · · ⊕C∆8,4,10+2(∆10)3(n−1)C(∆10)3n)

(E2/3,100)2(C(∆10+2)2C(∆010+2)2C∆10)(∆10)3n,

M8n+60(10) + 2) =E6,10+2 (C(E8,10+2 )nC(E8,10+2)n−18,1,10+2C(E8,10+2 )n−18,3,10+2

C(E8,10+2 )n−1(∆10)3⊕ · · · ⊕C∆8,3,10+2(∆10)3(n−1)C(∆10)3n)

⊕E6,10+20 (C(E8,10+20 )nC(E8,10+20 )n−18,2,10+2C(E08,10+2)n−18,4,10+2

C(E8,10+20 )n−1(∆10)3⊕ · · · ⊕C∆8,4,10+2(∆10)3(n−1)C(∆10)3n)

⊕E2/3,100(C(∆10+2)4C(∆010+2)4C(∆10+2)3010+2

C∆10+2(∆010+2)3C(∆10)2)(∆10)3n. Here, we defineE2/3,10+20:= 3

q

E2,10+20 wherevρ10,1(E2/3,10+20) =vρ10,5(E2/3,10+20) = 1, and we can write

Mk0(10) + 2) =Ek,10+20(C(∆10+2)nC(∆10+2)n−1010+2⊕ · · · ⊕C(∆010+2)n),

wheren= dim(Mk0(10) + 2))−1 =b3k/4−2(3k/8− b3k/8c)c, and whereEk,10+20:= 1, (E2/3,10+20)3, (E2/3,10+20)2, andE2/3,10+20, whenk≡0, 2, 4, and 6 (mod 8), respectively.

Hauptmodul We define thehauptmodul of Γ0(10) + 2:

J10+2:= ∆010+2/∆10+2(=η2(z)η2(2z)η−2(5z)η−2(10z)) = 1

q−23q+ 6q2+ 2q3− · · · , (3.233) wherev(J10+2) =−1 andv0(J10+2) = 1. Then, we have

J10+2: {|z+ 1/10|= 1/10,−1/106Re(z)60} →[−1,0]R, n

|z+ 2/5|= 1/(5

2),−1/26Re(z)6−3/10 o

→ {−36Re(z)6−1,06Im(z)64}, n

|z−1/5|= 1/(5

2), 1/106Re(z)63/10 o

→ {−36Re(z)6−1,−46Im(z)60}.

(3.234)

Thus,J10+2 does not take real value on some arcs of∂F10+2.

Lower arcs of∂F10+2

-3 -2 -1

-4 -2 2 4

Figure 3.63: Image byJ10+2

Location of the zeros of Eisenstein series SinceW10−10(10) + 2)W10= Γ0(10) + 2, we have E0k,10+2(W10z) = (√

10z)kEk,10+2 (z). (3.235)

0

Furthermore, we have

Ek,10+20 (−1/2 +i/(2 tanθ/2)) = ((e1)/

10)kEk,10+2((e1)/10), Ek,10+20 (e0/√

2 + 1) = (−(

2e+ 1)/

5)kEk,10+2 (e/(5√

2)2/5), Ek,10+20 (e00/√

21) = ((e+ 2)/

5)kEk,10+2(e/(5√

2) + 1/5), wheree0= (−2

2−3 cosθ+isinθ)/(3 + 2√

2 cosθ) ande00= (−2

2 + 3 cosθ+isinθ)/(3−2 2 cosθ).

Ek,10+2 -12 -103 -101 0 101 103 12 Ek,10+20 -12 -103 -101 0 101 103 12

Figure 3.64: Neighborhood of location of the zeros of the Eisenstein series

Now, we can observe that the zeros of Ek,10+2 do not lie on the arcs of ∂F10+2 for small weight k by numerical calculation. However, when the weightk increases, then the location of the zeros seems to approach to lower arcs of∂F10+2. (See Figure 3.65)

Location of the zeros of Hecke type Faber Polynomial We can observe that some zeros ofFm,10+2

do not lie on the lower arcs of∂F10+2 for small weight mby numerical calculation. However, when the weightmincreases, then the location of the zeros seems to approach to lower arcs of∂F10+2. (see Figure 3.66)

0

### (10)

Fundamental domain We have a fundamental domain for Γ0(10) as follows:

F10={|z+ 9/20|>1/20, −1/26Re(z)<−2/5}[

{|z+ 3/10|>1/10,−2/56Re(z)6−3/10}

[{|z+ 3/10|>1/10,−3/10< Re(z)<−1/5}[

{|z+ 1/10|>1/10,−1/56Re(z)60}

[{|z−1/10|>1/10,0< Re(z)<1/5}[

{|z−3/10|61/10,1/56Re(z)63/10}

[{|z−3/10|>1/10,3/10< Re(z)62/5}[

{|z−9/20|>1/20, 2/5< Re(z)<1/2},

(3.236)

where ¡−1 0

10 −1

¢: (e+ 1)/10 (ei(π−θ)1)/10, ¡−3−1

10 3

¢: (e3)/10 (ei(π−θ)3)/10, ¡3 −1

10−3

¢: (e+ 3)/10(ei(π−θ)+ 3)/10, and¡−9 4

20 −9

¢: (e+ 9)/20(ei(π−θ)9)/20. Then, we have Γ0(10) =h(1 10 1), (10 11 0), ¡−3−1

10 3

¢, ¡3 −1

10−3

¢, (20 99 4)i. (3.237)

Valence formula The cusps of Γ0(10) are ∞, 0, −1/2, and −1/5. Let f be a modular function of weightk for Γ0(10), which is not identically zero. We have

v(f) +v0(f) +v−1/2(f) +v−1/5(f) +1

2vρ10,1(f) +1

2vρ10,5(f) + X

p∈Γ0(10)\H p6=ρ10,110,5

vp(f) =3k

2 . (3.238)

Furthermore, the stabilizer of the elliptic pointρ10,1(resp.ρ10,5) is©

±I,±¡−3−1

10 3

±I,±¡3 −1

10−3

¢ª).

-3 -2 -1

-4 -2 2 4

The zeros ofEk,10+2 for 46k640

-3 -2 -1

-4 -2 2 4

The zeros of E1000,10+2

Figure 3.65: Image byJ10+2

0

-3 -2 -1

-4 -2 2 4

The zeros ofFm,10+2 form640

-3 -2 -1

-4 -2 2 4

The zeros ofF200,10+2

Figure 3.66: Image byJ10+2

-12 -25 -15 0 1 5

2 5

1 2

Figure 3.67: Γ0(10)

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

Ek,10 (z) := 10kEk(10z)5kEk(5z)2kEk(2z) +Ek(z)

(5k1)(2k1) fork>4. (3.239)

For the cusp 0 We have Γ0 =(10n1 01) ;n∈Z} and γ0 =W10, and we have the Eisenstein series for the cusp 0 associated with Γ0(10):

Ek,100 (z) := 10k/2(Ek(10z)−Ek(5z)−Ek(2z) +Ek(z))

(5k1)(2k1) fork>4. (3.240) We also haveγ0−1 Γ0(10)γ0= Γ0(10).

For the cusp −1/2 We have Γ−1/2

±¡10n+1 5n

−20n −10n+1

¢; n∈

and γ−1/2 =W10,5, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(10):

Ek,10−1/2(z) := −5k/2(2kEk(10z)−Ek(5z)2kEk(2z) +Ek(z))

(5k1)(2k1) fork>4. (3.241) We also haveγ−1/2−1 Γ0(10)γ−1/2= Γ0(10).

For the cusp −1/5 We have Γ−1/5

±¡10n+1 2n

−50n −10n+1

¢; n∈

and γ−1/5 =W10,2, and we have the Eisenstein series for the cusp−1/5 associated with Γ0(10):

Ek,10−1/5(z) := −2k/2(5kEk(10z)5kEk(5z)−Ek(2z) +Ek(z))

(5k1)(2k1) fork>4. (3.242) We also haveγ−1/5−1 Γ0(10)γ−1/5= Γ0(10).

The space of modular forms We define

8,1,10:= (E2/3,100)2(∆010)2(∆−1/210 )2(∆−1/510 )210,8,2,10:= (E2/3,100)2(∆10)2(∆−1/210 )2(∆−1/510 )210,

8,3,10:= (E2/3,100)2(∆10)2(∆010)2(∆−1/510 )210,8,4,10:= (E2/3,100)2(∆10)2(∆010)2(∆−1/210 )210,

8,5,10:= (∆010)2−1/210−1/510 (∆10)2,8,6,10:= (∆10)2−1/210−1/510 (∆10)2,

8,7,10:= ∆10010(∆−1/510 )2(∆10)2,8,8,10:= ∆10010(∆−1/210 )2(∆10)2. Now, we have

Mk0(10)) =CEk,10 CEk,100 CEk,10−1/2CEk,10−1/5⊕Sk0(10)), Sk0(10)) = (C∆8,1,10C∆8,2,10C∆8,3,10C∆8,4,10C∆8,5,10

C∆8,6,10C∆8,7,10C∆8,8,10C(∆10)3)Mk−80(10))

0

for every even integerk>4. Then, we haveM8n+20(10)) =E2,10+20M8n0(10))⊕CE2,10+100(∆10)3n CE2,10+50(∆10)3n,M8n+60(10)) =E2,10+20M8n+40(10))⊕C(E2,10+100)3(∆10)3n⊕C(E2,10+50)3(∆10)3n, and

M8n0(10)) =C(E8,10 )nC(E8,10 )n−18,1,10C(E8,10)n−18,5,10

C(E8,10 )n−1(∆10)3⊕ · · · ⊕C∆8,5,10(∆10)3(n−1)

C(E8,100 )nC(E08,10)n−18,2,10C(E8,100 )n−18,6,10

C(E8,100 )n−1(∆10)3⊕ · · · ⊕C∆8,6,10(∆10)3(n−1)

C(E8,10−1/2)nC(E8,10−1/2)n−18,3,10C(E−1/28,10 )n−18,7,10

C(E8,10−1/2)n−1(∆10)3⊕ · · · ⊕C∆8,7,10(∆10)3(n−1)

C(E8,10−1/5)nC(E8,10−1/5)n−18,4,10C(E−1/58,10 )n−18,8,10

C(E8,10−1/5)n−1(∆10)3⊕ · · · ⊕C∆8,8,10(∆10)3(n−1)C(∆10)3n, M8n+40(10)) =E4,10(C(E8,10 )nC(E8,10 )n−18,1,10C(E8,10)n−18,5,10

C(E8,10 )n−1(∆10)3⊕ · · · ⊕C∆8,5,10(∆10)3(n−1)C(∆10)3n)

⊕E4,100 (C(E08,10)nC(E08,10)n−18,2,10C(E8,100 )n−18,6,10

C(E8,100 )n−1(∆10)3⊕ · · · ⊕C∆8,6,10(∆10)3(n−1)C(∆10)3n)

⊕E4,10−1/2(C(E8,10−1/2)nC(E8,10−1/2)n−18,3,10C(E8,10−1/2)n−18,7,10

C(E8,10−1/2)n−1(∆10)3⊕ · · · ⊕C∆8,7,10(∆10)3(n−1)C(∆10)3n)

⊕E4,10−1/5(C(E8,10−1/5)nC(E8,10−1/5)n−18,4,10C(E8,10−1/5)n−18,8,10

C(E8,10−1/5)n−1(∆10)3⊕ · · · ⊕C∆8,8,10(∆10)3(n−1)C(∆10)3n)

(C∆10010C∆010−1/210 C∆−1/210−1/510 )(∆10)3n+1.

Furthermore, we can write

Mk0(10)) =Ek,100(C(∆10)nC(∆10)n−1010⊕ · · · ⊕C(∆010)n),

wheren= dim(Mk0(10)))1 =b3k/2−2(k/4− bk/4c)c, and whereEk,100:= 1 and E2/3,10+20, when k≡0 and 2 (mod 4), respectively.

Hauptmodul We define thehauptmodul of Γ0(10):

J10:= ∆010/∆10(=η3(z)η−1(2z)η(5z)η−3(10z)) = 1

q 3 +q+ 2q2+ 2q3− · · · , (3.243) wherev(J10) =−1 andv0(J10) = 1. Then, we have

J10: {|z+ 1/10|= 1/10,−1/106Re(z)60} →[−4,0]R,

{|z+ 3/10|= 1/10,−2/56Re(z)6−3/10} → {−46Re(z)<−3.8,06Im(z)62}, {|z−3/10|= 1/10,1/56Re(z)63/10} → {−46Re(z)<−3.8,−26Im(z)60}, {|z+ 9/20|= 1/20,−1/26Re(z)6−2/5} →[−5,−4]⊂R.

(3.244)

Thus,J10 does not take real value on some arcs of∂F10.

Location of the zeros of the Eisenstein series Since W10−1Γ0(10)W10 = W10,5−1Γ0(10)W10,5 = W10,2−1Γ0(10)W10,2= Γ0(10), we have

(

10z)−kEk,100 (W10z) = (2√ 5z+

5)−kEk,10−1/2(W10,5z) = (5√

2z+ 2

2)−kEk,10−1/5(W10,2z) =Ek,10 (z).

Lower arcs of∂F10

-5 -4 -3 -2 -1

-2 -1 1 2

Figure 3.68: Image byJ10

Furthermore, we have

Ek,100 (−1/2 +i/(2 tan(θ/2))) = ((e1)/

10)kEk,10 ((e1)/10), Ek,100 ((e1+ 3)/8) = ((e3)/

10)kEk,10 ((e3)/10), Ek,100 ((ei(π−θ1)3)/8) = ((3e1)/

10)kEk,10 ((e+ 3)/10), E0k,10((e29)/20) = ((5e1)/(2

10))kEk,10 ((e9)/20), Ek,10−1/2((e31)/10) = ((3e+ 2)/

5)kEk,10 ((e1)/10), Ek,10−1/2((e41)/6) = ((e+ 2)/

5)kEk,10 ((e3)/10), Ek,10−1/2((ei(π−θ4)1)/6) = ((e+ 1)/

5)kEk,10 ((e+ 3)/10), Ek,10−1/2(itan(θ/2)/2) = ((e+ 1)/(2

5))kEk,10 ((e5)/20), Ek,10−1/5(−1/2 +itan(θ/2)/10) = ((e+ 1)/

2)kEk,10 ((e1)/10), Ek,10−1/5(−3/10 +itan(θ/2)/10) = ((e+ 1)/

2)kEk,10 ((e3)/10), Ek,10−1/5(3/10 +itan(θ/2)/10) = ((e+ 1)/

2)kEk,10 ((e+ 3)/10), Ek,10−1/5(itan(θ/2)/5) = ((e1)/(2

2))kEk,10 ((e9)/20),

where e1 = (−3 + 5 cosθ+ 4isinθ)/(5−3 cosθ), e2 = (2129 cosθ+ 20isinθ)/(29−21 cosθ), e3 = (−1213 cosθ+ 5isinθ)/(13 + 12 cosθ), ande4= (4 + 5 cosθ+ 3isinθ)/(5 + 4 cosθ).

Ek,10 -12 -25 -15 0 15 25 12 Ek,100 -12 -25 -15 0 15 25 12

Ek,10−1/2 -12 -25 -15 0 15 25 12 Ek,10−1/5 -12 -25 -15 0 15 25 12

Figure 3.69: Location of the zeros of the Eisenstein series

Now, we can observe that the zeros of Ek,10 do not lie on the arcs of ∂F10 for small weight k by numerical calculation. However, when the weight k increases, then the location of the zeros seems to

0

approach to lower arcs of∂F10. (See Figure 3.70)

Location of the zeros of Hecke type Faber Polynomial We can observe that some zeros ofFm,10

do not lie on the lower arcs of ∂F10 for small weight m by numerical calculation. However, when the weight mincreases, then the location of the zeros seems to approach to lower arcs of∂F10. (see Figure 3.71)

-5 -4 -3 -2 -1

-2 -1 1 2

The zeros of Ek,10 for 46k640

-5 -4 -3 -2 -1

-2 -1 1 2

The zeros ofE1000,10

Figure 3.70: Image byJ10

0

-5 -4 -3 -2 -1

-2 -1 1 2

The zeros of Fm,10 form640

-5 -4 -3 -2 -1

-2 -1 1 2

The zeros ofF200,10

Figure 3.71: Image byJ10

Outline