Level 10

0

F10+ ^{-12 -103 0 103 12}

1 10 1 10

Figure 3.56: Γ0(10)+

For the cusp ∞ We have Γ∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series associated with Γ0(10)+:

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

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

The space of modular forms Letk be an even integer k>4. We haveMk(Γ0(10)+) =CEk,10+⊕ Sk(Γ0(10)+) andSk(Γ0(10)+) = (C(E2/3,100)⁸∆10⊕C(E2/3,100)⁴(∆10)²⊕C(∆10)³)Mk−8(Γ0(10)+). Then, we have

Mk(Γ0(10)+) =E_k,10+⁰(C((E2/3,100)⁴)ⁿ⊕C((E2/3,100)⁴)ⁿ⁻¹∆10⊕ · · · ⊕C(∆10)ⁿ),

wheren= dim(Mk(Γ0(10)+))−1 =b3k/8−2(k/4− bk/4c)c, and whereE_k,10+⁰:= 1, (E_2/3,10⁰)³E_8/3,10⁰, (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+:= (E_2/3,10⁰)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 ₁₀^{3 1}₂

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.

3.10.2 Γ

0

(10) + 10 = Γ

^∗₀

(10)

Fundamental domain We have a fundamental domain for Γ^∗₀(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^iθ/√

10 → e^i(π−θ)/√

10, ¡₋₃₋₁

10 3

¢W10 : e^iθ/(3√

10) + 1/3 → e^i(π−θ)/(3√

10)−1/3, and

¡_{−9 4}

20 −9

¢: (e^iθ+ 5)/12→(e^i(π−θ)−5)/12. Then, we have Γ^∗₀(10) =h(^{1 1}_{0 1}), W10, ¡₋₃₋₁

10 3

¢, (_{20 9}^{9 4})i. (3.213)

-1 2

-3

7

-3

10 0 ³

10 3 7 1 2 1

10

Figure 3.58: Γ^∗₀(10)

Valence formula The cusps of Γ^∗₀(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 Γ^∗₀(10), which is not identically zero. We have

v∞(f) +v−1/2(f) +1

2v_i/^√₁₀(f) +1

2vρ10,1(f) +1

2vρ10,3(f) + X

p∈Γ^∗₀(10)\H p6=i/√

10,ρ10,1,ρ10,3

vp(f) = 3k

4 . (3.214)

Furthermore, the stabilizer of the elliptic point i/√

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

©±I,±¡₋₃₋₁

10 3

¢ª,©

±I,±¡_{−13 3}

30 −7

¢W10

ª).

For the cusp∞ We have Γ∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ^∗₀(10):

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

(5^k−1)(2^k−1) fork>4.

(3.215)

For the cusp −1/2 We have Γ_−1/2 =©

±¡_{10n+1 5n}

−20n −10n+1

¢; n∈Zª

and γ_−1/2 =W_10,5, and we have the Eisenstein series for the cusp−1/2 associated with Γ^∗₀(10):

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

(5^k−1)(2^k−1) fork>4.

(3.216) We also haveγ_−1/2⁻¹ Γ^∗₀(10)γ−1/2= Γ^∗₀(10).

0

The space of modular forms We define

∆^∞₁₀₊₁₀:= ∆^∞₁₀∆⁰₁₀, ∆^−1/2₁₀₊₁₀:= ∆^−1/2₁₀ ∆^−1/5₁₀ , which are 2nd semimodular forms for Γ^∗₀(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

Mk(Γ^∗₀(10)) =CE^∞_k,10+10⊕CE^−1/2_k,10+10⊕Sk(Γ^∗₀(10)),

Sk(Γ^∗₀(10)) = (C∆8,1,10+10⊕C∆8,2,10+10⊕C∆8,3,10+10⊕C∆8,4,10+10⊕C(∆10)³)Mk−8(Γ^∗₀(10)) for every even integerk>4. Then, we haveM4n+2(Γ^∗₀(10)) =E2,10+100M4n(Γ^∗₀(10)) and

M8n(Γ^∗₀(10)) =C(E^∞_8,10+10)ⁿ⊕C(E_8,10+10^∞ )ⁿ⁻¹∆8,1,10+10⊕C(E_8,10+10^∞ )ⁿ⁻¹∆8,3,10+10

⊕C(E_8,10+10^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,3,10+10(∆10)³⁽ⁿ⁻¹⁾

⊕C(E_8,10+10^−1/2 )ⁿ⊕C(E_8,10+10^−1/2 )ⁿ⁻¹∆8,2,10+10⊕C(E^−1/2_8,10+10)ⁿ⁻¹∆8,4,10+10

⊕C(E_8,10+10^−1/2 )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,4,10+10(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ, M8n+4(Γ^∗₀(10)) =E_4,10+10^∞ (C(E_8,10+10^∞ )ⁿ⊕C(E_8,10+10^∞ )ⁿ⁻¹∆8,1,10+10⊕C(E_8,10+10^∞ )ⁿ⁻¹∆8,3,10+10

⊕C(E_8,10+10^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,3,10+10(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_4,10+10^−1/2 (C(E_8,10+10^−1/2 )ⁿ⊕C(E_8,10+10^−1/2 )ⁿ⁻¹∆8,2,10+10⊕C(E_8,10+10^−1/2 )ⁿ⁻¹∆8,4,10+10

⊕C(E_8,10+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)²⊕C(∆^∞₁₀₊₁₀)²∆^−1/2₁₀₊₁₀)(∆10)³ⁿ. Furthremore, we can write

M4n(Γ^∗₀(10)) =C(∆^∞₁₀₊₁₀)³ⁿ⊕C(∆^∞₁₀₊₁₀)³ⁿ⁻¹∆^−1/2₁₀₊₁₀⊕ · · · ⊕C(∆^−1/2₁₀₊₁₀)³ⁿ.

Hauptmodul We define thehauptmodul of Γ^∗₀(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 SinceW_10,5⁻¹Γ^∗₀(10)W10,5= Γ^∗₀(10), we have E_k,10+10^−1/2 (W10,5z) = (2√

5z+√

5)^kE_k,10+10^∞ (z). (3.219)

Furthermore, we have

E_k,10+10^−1/2 (e^iθ0/(3√

10)−1/3) = (−√

5e^iθ−√

2)^kE_k,10+10^∞ (e^iθ/√ 10), E_k,10+10^−1/2 (e^iθ0/√

10) = (√

2(e^iθ+ 1)/3)^kE_k,10+10^∞ (e^iθ/(3√

10)−1/3), E_k,10+10^−1/2 (itan(θ/2)/2) = ((e^iθ+ 1)/(2√

5))^kE_k,10+10^∞ ((e^iθ−9)/20), wheree^iθ0 = (−2√

10−7 cosθ+ 3isinθ)/(7 + 2√

10 cosθ).

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

E_k,10+10^{∞ -12-37 -103 0 103 37 12}

1 10

E_k,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.

3.10.3 Γ

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^iθ+ 1)/10→(e^i(π−θ)−1)/10,¡₋₇₋₄

30 17

¢W10,5:e^iθ/(4√

5) + 1/4→e^i(π−θ)/(4√

5)−1/4, and¡₋₉₋₅

20 11

¢W10,5:e^iθ/(2√

5) + 1/2→e^i(π−θ)/(2√

5)−1/2. Then, we have Γ0(10) + 5 =h(^{1 1}_{0 1}), W10,5, (_{10 1}^{1 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 are∞and 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=ρ10,1,ρ10,2,ρ10,4

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,±W10,5},©

±I,±¡₋₃₋₂

20 13

¢W10,5

ª).

0

For the cusp∞ We have Γ∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ0(10) + 5:

E^∞_k,10+5(z) := 2^k5^k/2E_k(10z)−5^k/2E_k(5z) + 2^kE_k(2z)−E_k(z)

(5^k/2+ 1)(2^k−1) fork>4. (3.223)

For the cusp 0 We have Γ0 ={±(_10n^{1 0}₁) ;n∈Z} and γ0 =W10, and we have the Eisenstein series for the cusp 0 associated with Γ0(10) + 5:

E⁰_k,10+5(z) := −2^k/2(5^k/2Ek(10z)−5^k/2Ek(5z) +Ek(2z)−Ek(z))

(5^k/2+ 1)(2^k−1) fork>4. (3.224) We also haveγ₀⁻¹ (Γ0(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)⁵∆⁰₁₀₊₅,

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

Mk(Γ0(10) + 5) =CE^∞_k,10+5⊕CE⁰_k,10+5⊕Sk(Γ0(10) + 5),

Sk(Γ0(10) + 5) = (C∆8,1,10+5⊕C∆8,2,10+5⊕C∆8,3,10+5⊕C∆8,4,10+5⊕C(∆10)³)Mk−8(Γ0(10) + 5) for every even integerk>4. Then, we haveM_4n+2(Γ₀(10) + 5) =E_2,10+5⁰M_4n(Γ₀(10) + 5) and

M8n(Γ0(10) + 5) =C(E_8,10+5^∞ )ⁿ⊕C(E_8,10+5^∞ )ⁿ⁻¹∆8,1,10+5⊕C(E_8,10+5^∞ )ⁿ⁻¹∆8,3,10+5

⊕C(E_8,10+5^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,3,10+5(∆10)³⁽ⁿ⁻¹⁾

⊕C(E⁰_8,10+5)ⁿ⊕C(E_8,10+5⁰ )ⁿ⁻¹∆_8,2,10+5⊕C(E_8,10+5⁰ )ⁿ⁻¹∆_8,4,10+5

⊕C(E_8,10+5⁰ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,4,10+5(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ, M8n+4(Γ0(10) + 5) =E_4,10+5^∞ (C(E_8,10+5^∞ )ⁿ⊕C(E^∞_8,10+5)ⁿ⁻¹∆8,1,10+5⊕C(E_8,10+5^∞ )ⁿ⁻¹∆8,3,10+5

⊕C(E_8,10+5^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,3,10+5(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_4,10+5⁰ (C(E_8,10+5⁰ )ⁿ⊕C(E_8,10+5⁰ )ⁿ⁻¹∆8,2,10+5⊕C(E⁰_8,10+5)ⁿ⁻¹∆8,4,10+5

⊕C(E_8,10+5⁰ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,4,10+5(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

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

M4n(Γ0(10) + 5) =C(∆^∞₁₀₊₅)³ⁿ⊕C(∆^∞₁₀₊₅)³ⁿ⁻¹∆⁰₁₀₊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 + 6q−8q²+ 17q³− · · ·, (3.225) wherev∞(J10+5) =−1 andv0(J10+5) = 1. Then, we have

J10+5:∂F10+5\ {z∈H; Re(z) =±1/2} →[−6−2√

5,0]⊂R. (3.226)

Location of the zeros of the Eisenstein series SinceW₁₀⁻¹(Γ0(10) + 5)W10= Γ0(10) + 5, we have E⁰_k,10+5(W10z) = (√

10z)^kE_k,10+5^∞ (z). (3.227)

Furthermore, we have

E_k,10+5⁰ (−1/2 +i/(2 tan(θ/2))) = ((e^iθ−1)/√

10)^kE^∞_k,10+5((e^iθ−1)/10), E_k,10+5⁰ (e^iθ0/(2√

5)−1/2) = (−(√

5e^iθ−1)/(2√

2))^kE_k,10+5^∞ (e^iθ/(4√

5)−1/4), E_k,10+5⁰ (e^iθ0/(4√

5)−1/4) = (−(√

5e^iθ−1)/√

2)^kE_k,10+5^∞ (e^iθ/(2√

5)−1/2), wheree^iθ0 = (2√

5−6 cosθ+ 4isinθ)/(6−2√

5 cosθ).

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

E_k,10+5^{∞ -12 -103 -16 0 16 103 12} E_k,10+5^{0 -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.

3.10.4 Γ

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^iθ+ 1)/10→(e^i(π−θ)−1)/10,¡₋₃₋₁

10 3

¢W10,5:e^iθ/(5√

2)−1/5→e^i(π−θ)/(5√

2)−2/5, and (_{40 9}^{9 2})W10,5:e^iθ/(5√

2) + 2/5→e^i(π−θ)/(5√

2) + 1/5. Then, we have

Γ0(10) + 2 =h(^{1 1}_{0 1}), W10,2, (_{10 1}^{1 0}), (_{40 9}^{9 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 are∞and 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=ρ10,1,ρ10,5

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

ª (resp. ©

±I,±¡_{3 −1}

10−3

¢,±(^{13 3}_{30 7})W10,2,±(_{40 9}^{9 2})W10,2

ª).

For the cusp∞ We have Γ∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ0(10) + 2:

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

(5^k−1)(2^k/2+ 1) fork>4. (3.231)

For the cusp 0 We have Γ0 ={±(_10n^{1 0}₁) ;n∈Z} and γ0 =W10, and we have the Eisenstein series for the cusp 0 associated with Γ0(10) + 2:

E⁰_k,10+2(z) := −5^k/2(2^k/2Ek(10z) +Ek(5z)−2^k/2Ek(2z)−Ek(z))

(5^k−1)(2^k/2+ 1) fork>4. (3.232) We also haveγ₀⁻¹ (Γ0(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)⁵∆⁰₁₀₊₂,

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

Mk(Γ0(10) + 2) =CE^∞_k,10+2⊕CE⁰_k,10+2⊕Sk(Γ0(10) + 2),

Sk(Γ0(10) + 2) = (C∆8,1,10+2⊕C∆8,2,10+2⊕C∆8,3,10+2⊕C∆8,4,10+2⊕C(∆10)³)Mk−8(Γ0(10) + 2) for every even integerk>4. Then, we haveM8n+2(Γ0(10) + 2) =E2,10+20M8n(Γ0(10) + 2) and

M8n(Γ0(10) + 2) =C(E^∞_8,10+2)ⁿ⊕C(E_8,10+2^∞ )ⁿ⁻¹∆8,1,10+2⊕C(E_8,10+2^∞ )ⁿ⁻¹∆8,3,10+2

⊕C(E_8,10+2^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,3,10+2(∆10)³⁽ⁿ⁻¹⁾

⊕C(E_8,10+2⁰ )ⁿ⊕C(E_8,10+2⁰ )ⁿ⁻¹∆8,2,10+2⊕C(E_8,10+2⁰ )ⁿ⁻¹∆8,4,10+2

⊕C(E_8,10+2⁰ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,4,10+2(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ,

M8n+4(Γ0(10) + 2) =E_4,10+2^∞ (C(E_8,10+2^∞ )ⁿ⊕C(E^∞_8,10+2)ⁿ⁻¹∆8,1,10+2⊕C(E_8,10+2^∞ )ⁿ⁻¹∆8,3,10+2

⊕C(E_8,10+2^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,3,10+2(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_4,10+2⁰ (C(E_8,10+2⁰ )ⁿ⊕C(E_8,10+2⁰ )ⁿ⁻¹∆8,2,10+2⊕C(E⁰_8,10+2)ⁿ⁻¹∆8,4,10+2

⊕C(E_8,10+2⁰ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,4,10+2(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕(E_2/3,10⁰)²(C(∆^∞₁₀₊₂)²⊕C(∆⁰₁₀₊₂)²⊕C∆10)(∆10)³ⁿ,

M8n+6(Γ0(10) + 2) =E_6,10+2^∞ (C(E_8,10+2^∞ )ⁿ⊕C(E^∞_8,10+2)ⁿ⁻¹∆8,1,10+2⊕C(E_8,10+2^∞ )ⁿ⁻¹∆8,3,10+2

⊕C(E_8,10+2^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,3,10+2(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_6,10+2⁰ (C(E_8,10+2⁰ )ⁿ⊕C(E_8,10+2⁰ )ⁿ⁻¹∆8,2,10+2⊕C(E⁰_8,10+2)ⁿ⁻¹∆8,4,10+2

⊕C(E_8,10+2⁰ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,4,10+2(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_2/3,10⁰(C(∆^∞₁₀₊₂)⁴⊕C(∆⁰₁₀₊₂)⁴⊕C(∆^∞₁₀₊₂)³∆⁰₁₀₊₂

⊕C∆^∞₁₀₊₂(∆⁰₁₀₊₂)³⊕C(∆10)²)(∆10)³ⁿ. Here, we defineE_2/3,10+2⁰:= ³

q

E_2,10+2⁰ wherev_ρ10,1(E_2/3,10+2⁰) =v_ρ10,5(E_2/3,10+2⁰) = 1, and we can write

Mk(Γ0(10) + 2) =E_k,10+2⁰(C(∆^∞₁₀₊₂)ⁿ⊕C(∆^∞₁₀₊₂)ⁿ⁻¹∆⁰₁₀₊₂⊕ · · · ⊕C(∆⁰₁₀₊₂)ⁿ),

wheren= dim(Mk(Γ0(10) + 2))−1 =b3k/4−2(3k/8− b3k/8c)c, and whereE_k,10+2⁰:= 1, (E2/3,10+20)³, (E_2/3,10+2⁰)², andE_2/3,10+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−2−3q+ 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 SinceW₁₀⁻¹(Γ0(10) + 2)W10= Γ0(10) + 2, we have E⁰_k,10+2(W10z) = (√

10z)^kE_k,10+2^∞ (z). (3.235)

0

Furthermore, we have

E_k,10+2⁰ (−1/2 +i/(2 tanθ/2)) = ((e^iθ−1)/√

10)^kE^∞_k,10+2((e^iθ−1)/10), E_k,10+2⁰ (e^iθ0/√

2 + 1) = (−(√

2e^iθ+ 1)/√

5)^kE_k,10+2^∞ (e^iθ/(5√

2)−2/5), E_k,10+2⁰ (e^iθ00/√

2−1) = ((e^iθ+√ 2)/√

5)^kE^∞_k,10+2(e^iθ/(5√

2) + 1/5), wheree^iθ0= (−2√

2−3 cosθ+isinθ)/(3 + 2√

2 cosθ) ande^iθ00= (−2√

2 + 3 cosθ+isinθ)/(3−2√ 2 cosθ).

E_k,10+2^{∞ -12 -103 -101 0 101 103 12} E_k,10+2^{0 -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 E_k,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)

3.10.5 Γ

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^iθ+ 1)/10 →(e^i(π−θ)−1)/10, ¡₋₃₋₁

10 3

¢: (e^iθ−3)/10→ (e^i(π−θ)−3)/10, ¡_{3 −1}

10−3

¢: (e^iθ+ 3)/10→(e^i(π−θ)+ 3)/10, and¡_{−9 4}

20 −9

¢: (e^iθ+ 9)/20→(e^i(π−θ)−9)/20. Then, we have Γ0(10) =h(^{1 1}_{0 1}), (_{10 1}^{1 0}), ¡₋₃₋₁

10 3

¢, ¡_{3 −1}

10−3

¢, (_{20 9}^{9 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=ρ10,1,ρ10,5

vp(f) =3k

2 . (3.238)

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

±I,±¡₋₃₋₁

10 3

¢ª(resp. ©

±I,±¡_{3 −1}

10−3

¢ª).

-3 -2 -1

-4 -2 2 4

The zeros ofE_k,10+2^∞ for 46k640

-3 -2 -1

-4 -2 2 4

The zeros of E_1000,10+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∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ0(10):

E_k,10^∞ (z) := 10^kEk(10z)−5^kEk(5z)−2^kEk(2z) +Ek(z)

(5^k−1)(2^k−1) fork>4. (3.239)

For the cusp 0 We have Γ0 ={±(_10n^{1 0}₁) ;n∈Z} and γ0 =W10, and we have the Eisenstein series for the cusp 0 associated with Γ0(10):

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

(5^k−1)(2^k−1) 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∈Zª

and γ_−1/2 =W_10,5, and we have the Eisenstein series for the cusp−1/2 associated with Γ0(10):

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

(5^k−1)(2^k−1) 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∈Zª

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

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

(5^k−1)(2^k−1) 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:= (E_2/3,10⁰)²(∆⁰₁₀)²(∆^−1/2₁₀ )²(∆^−1/5₁₀ )²∆10, ∆8,2,10:= (E_2/3,10⁰)²(∆^∞₁₀)²(∆^−1/2₁₀ )²(∆^−1/5₁₀ )²∆10,

∆8,3,10:= (E_2/3,10⁰)²(∆^∞₁₀)²(∆⁰₁₀)²(∆^−1/5₁₀ )²∆10, ∆8,4,10:= (E_2/3,10⁰)²(∆^∞₁₀)²(∆⁰₁₀)²(∆^−1/2₁₀ )²∆10,

∆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:= ∆^∞₁₀∆⁰₁₀(∆^−1/5₁₀ )²(∆10)², ∆8,8,10:= ∆^∞₁₀∆⁰₁₀(∆^−1/2₁₀ )²(∆10)². Now, we have

Mk(Γ0(10)) =CE_k,10^∞ ⊕CE_k,10⁰ ⊕CE_k,10^−1/2⊕CE_k,10^−1/5⊕Sk(Γ0(10)), Sk(Γ0(10)) = (C∆8,1,10⊕C∆8,2,10⊕C∆8,3,10⊕C∆8,4,10⊕C∆8,5,10

⊕C∆8,6,10⊕C∆8,7,10⊕C∆8,8,10⊕C(∆10)³)Mk−8(Γ0(10))

0

for every even integerk>4. Then, we haveM8n+2(Γ0(10)) =E2,10+20M8n(Γ0(10))⊕CE2,10+100(∆10)³ⁿ⊕ CE2,10+50(∆10)³ⁿ,M8n+6(Γ0(10)) =E2,10+20M8n+4(Γ0(10))⊕C(E2,10+100)³(∆10)³ⁿ⊕C(E2,10+50)³(∆10)³ⁿ, and

M8n(Γ0(10)) =C(E_8,10^∞ )ⁿ⊕C(E_8,10^∞ )ⁿ⁻¹∆8,1,10⊕C(E^∞_8,10)ⁿ⁻¹∆8,5,10

⊕C(E_8,10^∞ )ⁿ⁻¹(∆₁₀)³⊕ · · · ⊕C∆_8,5,10(∆₁₀)³⁽ⁿ⁻¹⁾

⊕C(E_8,10⁰ )ⁿ⊕C(E⁰_8,10)ⁿ⁻¹∆8,2,10⊕C(E_8,10⁰ )ⁿ⁻¹∆8,6,10

⊕C(E_8,10⁰ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,6,10(∆10)³⁽ⁿ⁻¹⁾

⊕C(E_8,10^−1/2)ⁿ⊕C(E_8,10^−1/2)ⁿ⁻¹∆8,3,10⊕C(E^−1/2_8,10 )ⁿ⁻¹∆8,7,10

⊕C(E_8,10^−1/2)ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,7,10(∆10)³⁽ⁿ⁻¹⁾

⊕C(E_8,10^−1/5)ⁿ⊕C(E_8,10^−1/5)ⁿ⁻¹∆8,4,10⊕C(E^−1/5_8,10 )ⁿ⁻¹∆8,8,10

⊕C(E_8,10^−1/5)ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,8,10(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ, M8n+4(Γ0(10)) =E^∞_4,10(C(E_8,10^∞ )ⁿ⊕C(E_8,10^∞ )ⁿ⁻¹∆8,1,10⊕C(E^∞_8,10)ⁿ⁻¹∆8,5,10

⊕C(E_8,10^∞ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,5,10(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_4,10⁰ (C(E⁰_8,10)ⁿ⊕C(E⁰_8,10)ⁿ⁻¹∆8,2,10⊕C(E_8,10⁰ )ⁿ⁻¹∆8,6,10

⊕C(E_8,10⁰ )ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,6,10(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_4,10^−1/2(C(E_8,10^−1/2)ⁿ⊕C(E_8,10^−1/2)ⁿ⁻¹∆8,3,10⊕C(E_8,10^−1/2)ⁿ⁻¹∆8,7,10

⊕C(E_8,10^−1/2)ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,7,10(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕E_4,10^−1/5(C(E_8,10^−1/5)ⁿ⊕C(E_8,10^−1/5)ⁿ⁻¹∆8,4,10⊕C(E_8,10^−1/5)ⁿ⁻¹∆8,8,10

⊕C(E_8,10^−1/5)ⁿ⁻¹(∆10)³⊕ · · · ⊕C∆8,8,10(∆10)³⁽ⁿ⁻¹⁾⊕C(∆10)³ⁿ)

⊕(C∆^∞₁₀∆⁰₁₀⊕C∆⁰₁₀∆^−1/2₁₀ ⊕C∆^−1/2₁₀ ∆^−1/5₁₀ )(∆10)³ⁿ⁺¹.

Furthermore, we can write

Mk(Γ0(10)) =E_k,10⁰(C(∆^∞₁₀)ⁿ⊕C(∆^∞₁₀)ⁿ⁻¹∆⁰₁₀⊕ · · · ⊕C(∆⁰₁₀)ⁿ),

wheren= dim(Mk(Γ0(10)))−1 =b3k/2−2(k/4− bk/4c)c, and whereE_k,10⁰:= 1 and E_2/3,10+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 = W_10,5⁻¹Γ0(10)W10,5 = W_10,2⁻¹Γ0(10)W10,2= Γ0(10), we have

(√

10z)^−kE_k,10⁰ (W10z) = (2√ 5z+√

5)^−kE_k,10^−1/2(W10,5z) = (5√

2z+ 2√

2)^−kE_k,10^−1/5(W10,2z) =E_k,10^∞ (z).

Lower arcs of∂F10

-5 -4 -3 -2 -1

-2 -1 1 2

Figure 3.68: Image byJ10

Furthermore, we have

E_k,10⁰ (−1/2 +i/(2 tan(θ/2))) = ((e^iθ−1)/√

10)^kE_k,10^∞ ((e^iθ−1)/10), E_k,10⁰ ((e^iθ1+ 3)/8) = ((e^iθ−3)/√

10)^kE_k,10^∞ ((e^iθ−3)/10), E_k,10⁰ ((e^i(π−θ1)−3)/8) = ((3e^iθ−1)/√

10)^kE_k,10^∞ ((e^iθ+ 3)/10), E⁰_k,10((e^iθ2−9)/20) = ((5e^iθ−1)/(2√

10))^kE_k,10^∞ ((e^iθ−9)/20), E_k,10^−1/2((e^iθ3−1)/10) = ((3e^iθ+ 2)/√

5)^kE_k,10^∞ ((e^iθ−1)/10), E_k,10^−1/2((e^iθ4−1)/6) = ((e^iθ+ 2)/√

5)^kE_k,10^∞ ((e^iθ−3)/10), E_k,10^−1/2((e^i(π−θ4)−1)/6) = ((e^iθ+ 1)/√

5)^kE_k,10^∞ ((e^iθ+ 3)/10), E_k,10^−1/2(itan(θ/2)/2) = ((e^iθ+ 1)/(2√

5))^kE_k,10^∞ ((e^iθ−5)/20), E_k,10^−1/5(−1/2 +itan(θ/2)/10) = ((e^iθ+ 1)/√

2)^kE_k,10^∞ ((e^iθ−1)/10), E_k,10^−1/5(−3/10 +itan(θ/2)/10) = ((e^iθ+ 1)/√

2)^kE_k,10^∞ ((e^iθ−3)/10), E_k,10^−1/5(3/10 +itan(θ/2)/10) = ((e^iθ+ 1)/√

2)^kE_k,10^∞ ((e^iθ+ 3)/10), E_k,10^−1/5(itan(θ/2)/5) = ((e^iθ−1)/(2√

2))^kE_k,10^∞ ((e^iθ−9)/20),

where e^iθ1 = (−3 + 5 cosθ+ 4isinθ)/(5−3 cosθ), e^iθ2 = (21−29 cosθ+ 20isinθ)/(29−21 cosθ), e^iθ3 = (−12−13 cosθ+ 5isinθ)/(13 + 12 cosθ), ande^iθ4= (4 + 5 cosθ+ 3isinθ)/(5 + 4 cosθ).

E_k,10^{∞ -12 -25 -15 0 15 25 12} E_k,10^{0 -12 -25 -15 0 15 25 12}

E_k,10^{−1/2 -12 -25 -15 0 15 25 12} E_k,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 E_k,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 E_k,10^∞ for 46k640

-5 -4 -3 -2 -1

-2 -1 1 2

The zeros ofE_1000,10^∞

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

ドキュメント内 A dissertation submitted in partial fulfillment of the requirements for the degree of (ページ 108-126)