Level 12 - A dissertation submitted in partial fulfillment of the requirements for the degree o

-¹₂ -¹₄ 0 ¹

4 1 2 1

43 1 23

Figure 3.78: Γ0(12)+

For the cusp∞ We have Γ∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ0(12)+ of weightk>4:

E_k,12+^∞ (z) :=2^k3^k/2Ek(12z)−2 3^k/2Ek(6z) + 2^kEk(4z) + 3^k/2Ek(3z)−2Ek(2z) +Ek(z)

(3^k/2+ 1)(2^k−1) . (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:

E_k,12+^−1/2(z) :=

−2 2^k/2(2^k3^k/2Ek(12z)−3^k/2(2^k+ 1)Ek(6z) + 2^kEk(4z) + 3^k/2Ek(3z)−(2^k+ 1)Ek(2z) +Ek(z))

(3^k/2+ 1)(2^k−1) .

(3.264) We also haveγ_−1/2⁻¹ Γ0(12) + γ_−1/2= Γ0(12)+.

The space of modular forms We define

∆^∞₁₂₊:= ∆^∞₁₂∆⁰₁₂∆^−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 haveMk(Γ0(12)+) =CE_k,12+^∞ ⊕CE_k,12+^−1/2⊕Sk(Γ0(12)+) andSk(Γ0(12)+) = ∆12+Mk−4(Γ0(12)+) for every even integerk>4. Then, we haveM4n+2(Γ0(12)+) =E2,12+0M4n(Γ0(12)+) and

M4n(Γ0(12)+) =C(E_4,12+^∞ )ⁿ⊕C(E_4,12+^∞ )ⁿ⁻¹∆12+⊕ · · · ⊕CE_4,12+^∞ (∆12+)ⁿ⁻¹

⊕C(E_4,12+^−1/2)ⁿ⊕C(E_4,12+^−1/2)ⁿ⁻¹∆12+⊕ · · · ⊕CE_4,12+^−1/2(∆12+)ⁿ⁻¹⊕C(∆12+)ⁿ. Furthremore, we can write

M4n(Γ0(12)+) =C(∆^∞₁₂₊)²ⁿ⊕C(∆^∞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 SinceW_12+,6⁻¹ (Γ0(12)+)W12+,6= Γ0(12)+, we have E_k,12+^−1/2(W12+,6z) = (√

6z+√

6/2)^kE_k,12+^∞ (z). (3.267)

Furthermore, we have

E_k,12+^−1/2(e^iθ⁰/(2√

3)) = ((√

3e^iθ−1)/√

2)^kE^∞_k,12+(e^iθ/(2√ 3)), E^−1/2_k,12+(i/(2 tan(θ/2))) = ((e^iθ+ 1)/√

6)^kE_k,12+^∞ (e^iθ/6−1/3), wheree^iθ⁰ = (−√

3−2 cosθ+isinθ)/(2 +√ 3 cosθ).

Now, recall that Γ0(12)+ =T_1/2⁻¹(Γ0(6) + 3)T_1/2. Then, fork 6600, since we can prove that all of the zeros of E^∞_k,6+3 lie on the lower arcs of ∂F6+3 by numerical calculation, we have all of the zeros of E_k,12+^∞ in the lower arcs of∂F12+.

E_k,12+^∞ ^-¹² ^-¹⁴ ⁰ ¹⁴ ¹²

1 43 1 23

E_k,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+.

3.12.2 Γ

(12) + 12 = Γ

^∗₀

(12)

Fundamental domain We have a fundamental domain for Γ^∗₀(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^iθ/(2√

3) → e^i(π−θ)/(2√

3), ¡_{−7 2}

24 −7

¢ : (e^iθ+ 7)/24 → (e^i(π−θ)−7)/24, and ¡_{−5 2}

12−5

¢ : (e^iθ+ 5)/12→(e^i(π−θ)−5)/12. Then, we have

Γ^∗₀(12) =h(^{1 1}_{0 1}), W12, (_{12 5}^{5 2}), (_{24 7}^{7 2})i. (3.269)

-1

3 0 ¹₃ ¹₂

1 23

Figure 3.80: Γ^∗₀(12)

Valence formula The cusps of Γ^∗₀(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 Γ^∗₀(12), which is not identically zero. We have

v∞(f) +v_−1/3(f) +v_−1/2(f) +1

2v_i/(2^√₃₎(f) +1

2vρ12,2(f) + X

p∈Γ^∗₀(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∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ^∗₀(12) of weightk>4:

E_k,12+12^∞ (z) :=

2^k3^kE_k(12z) + 3^k/2(3^k/2−1)E_k(6z)−2^k3^k/2E_k(4z)−3^kE_k(3z)−(3^k/2−1)E_k(2z) +E_k(z)

(3^k−1)(2^k−1) . (3.271)

For the cusp −1/3 We have Γ−1/3 =©

±¡_12n+1 _4n

−36n −12n+1

¢; n∈Zª

and γ−1/3 =W12,4, and we have the Eisenstein series for the cusp−1/3 associated with Γ^∗₀(12) of weightk>4:

E_k,12+12^−1/3 (z) :=

−(2^k3^k/2Ek(12z) + 3^k/2(3^k/2−1)Ek(6z)−2^k3^k/2Ek(4z)−3^kEk(3z) + (3^k/2−1)Ek(2z) +Ek(z))

(3^k−1)(2^k−1) .

(3.272) We also haveγ_−1/3⁻¹ Γ^∗₀(12)γ−1/3= Γ^∗₀(12).

For the cusp −1/2 We have Γ_−1/2 =©

±¡_6n+1 _3n

−12n−6n+1

¢ ; n∈Zª

and γ_−1/2 =W12−,6, and we have the Eisenstein series for the cusp−1/2 associated with Γ^∗₀(12) of weightk>4:

E_k,12+12^−1/2 (z) :=

−(2^k3^k/2Ek(12z)−3^k/2(2^k+ 1)Ek(6z) + 2^kEk(4z) + 3^k/2Ek(3z)−(2^k+ 1)Ek(2z) +Ek(z))

(3^k/2+ 1)(2^k−1) . (3.273)

Note that we haveE_k,12+12^−1/2 = 2⁻¹2^−k/2E^−1/2_k,12+. Note thatγ⁻¹_−1/2 Γ^∗₀(12)γ_−1/26= Γ^∗₀(12).

The space of modular forms We define

∆^∞₁₂₊₁₂:= ∆^∞₁₂∆⁰₁₂, ∆^−1/3₁₂₊₁₂:= ∆^−1/3₁₂ ∆^−1/4₁₂ , ∆^−1/2₁₂₊₁₂:= ∆^−1/2₁₂ ∆^−1/6₁₂ , which are 4th semimodular forms for Γ^∗₀(12) of weight 1. Furthermore, we define

∆12,1,12+12:= (E1,12+120)⁵E2,12+0∆^−1/3₁₂₊₁₂∆^−1/2₁₂₊₁₂∆12, ∆12,2,12+12:= (E1,12+120)⁵E2,12+0∆^∞₁₂₊₁₂∆^−1/2₁₂₊₁₂∆12,

∆12,3,12+12:= (E1,12+120)⁵E2,12+0∆^∞₁₂₊₁₂∆^−1/3₁₂₊₁₂∆12, ∆12,4,12+12:= (E1,12+120)⁴∆^−1/3₁₂₊₁₂∆^−1/2₁₂₊₁₂(∆12)²,

∆12,5,12+12:= (E1,12+120)⁴∆^∞₁₂₊₁₂∆^−1/2₁₂₊₁₂(∆12)², ∆12,6,12+12:= (E1,12+120)⁴∆^∞₁₂₊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+120∆12(C∆^∞₁₂₊₁₂∆^−1/2₁₂₊₁₂⊕C∆^−1/3₁₂₊₁₂∆^−1/2₁₂₊₁₂⊕C∆^∞₁₂₊₁₂∆^−1/3₁₂₊₁₂), A2= (∆12)²(C∆^∞₁₂₊₁₂∆^−1/2₁₂₊₁₂⊕C∆^−1/3₁₂₊₁₂∆^−1/2₁₂₊₁₂⊕C∆^∞₁₂₊₁₂∆^−1/3₁₂₊₁₂),

B1=C(E_12,12+12^∞ )ⁿ⊕C(E_12,12+12^∞ )ⁿ⁻¹∆12,1,12+12⊕C(E_12,12+12^∞ )ⁿ⁻¹∆12,4,12+12

⊕C(E_12,12+12^∞ )ⁿ⁻¹∆12,7,12+12⊕C(E_12,12+12^∞ )ⁿ⁻¹(∆12)⁴⊕C(E_12,12+12^∞ )ⁿ⁻²∆12,1,12+12(∆12)⁴

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

B2=C(E_12,12+12^−1/3 )ⁿ⊕C(E_12,12+12^−1/3 )ⁿ⁻¹∆12,2,12+12⊕C(E_12,12+12^−1/3 )ⁿ⁻¹∆12,5,12+12

⊕C(E_12,12+12^−1/3 )ⁿ⁻¹∆12,8,12+12⊕C(E_12,12+12^−1/3 )ⁿ⁻¹(∆12)⁴⊕C(E_12,12+12^−1/3 )ⁿ⁻²∆12,2,12+12(∆12)⁴

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

B3=C(E_12,12+12^−1/2 )ⁿ⊕C(E_12,12+12^−1/2 )ⁿ⁻¹∆12,3,12+12⊕C(E_12,12+12^−1/2 )ⁿ⁻¹∆12,6,12+12

⊕C(E_12,12+12^−1/2 )ⁿ⁻¹∆12,9,12+12⊕C(E_12,12+12^−1/2 )ⁿ⁻¹(∆12)⁴⊕C(E_12,12+12^−1/2 )ⁿ⁻²∆12,3,12+12(∆12)⁴

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

Mk(Γ^∗₀(12)) =CE^∞_k,12+12⊕CE^−1/3_k,12+12⊕CE^−1/2_k,12+12⊕Sk(Γ^∗₀(12)),

Sk(Γ^∗₀(12)) = (C∆12,1,12+12⊕C∆12,2,12+12⊕ · · · ⊕C∆12,9,12+12⊕C(∆12)⁴)Mk−12(Γ^∗₀(12)) for every even integerk>4. Then, we have

M12n(Γ^∗₀(12)) =B1⊕B2⊕B3⊕C(∆12)⁴ⁿ,

M12n+2(Γ^∗₀(12)) =E2,12+0M12n(Γ^∗₀(12))⊕CE1,12+120∆^∞₁₂₊₁₂(∆12)⁴ⁿ,

M12n+4(Γ^∗₀(12)) =E_4,12+12^∞ (B1⊕C(∆12)⁴ⁿ)⊕E_4,12+12^−1/3 (B2⊕C(∆12)⁴ⁿ)⊕E_4,12+12^−1/2 (B3⊕C(∆12)⁴ⁿ)

⊕(CE_1,12+12⁰⊕C∆^∞₁₂₊₁₂)(∆₁₂)⁴ⁿ⁺¹,

M_12n+6(Γ^∗₀(12)) =E_6,12+12^∞ (B₁⊕C(∆₁₂)⁴ⁿ)⊕E_6,12+12^−1/3 (B₂⊕C(∆₁₂)⁴ⁿ)⊕E_6,12+12^−1/2 (B₃⊕C(∆₁₂)⁴ⁿ)

⊕A1(∆12)⁴ⁿ,

M12n+8(Γ^∗₀(12)) =E_8,12+12^∞ (B1⊕C(∆12)⁴ⁿ)⊕E_8,12+12^−1/3 (B2⊕C(∆12)⁴ⁿ)⊕E_8,12+12^−1/2 (B3⊕C(∆12)⁴ⁿ)

⊕E2,12+0A1(∆12)⁴ⁿ⊕A2(∆12)⁴ⁿ,

M12n+10(Γ^∗₀(12)) =E_10,12+12^∞ (B1⊕C(∆12)⁴ⁿ)⊕E_10,12+12^−1/3 (B2⊕C(∆12)⁴ⁿ)⊕E_10,12+12^−1/2 (B3⊕C(∆12)⁴ⁿ)

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

Mk(Γ^∗₀(12)) =E_k,12+12⁰(C(∆^∞₁₂₊₁₂)ⁿ⊕C(∆^∞₁₂₊₁₂)ⁿ⁻¹∆^−1/3₁₂₊₁₂⊕ · · · ⊕C(∆^−1/3₁₂₊₁₂)ⁿ),

wheren= dim(Mk(Γ^∗₀(12)))−1 =bk−2(k/4− bk/4c)c, and whereE_k,12+12⁰ := 1 andE1,12+120, when k≡0 and 2 (mod 4), respectively.

Hauptmodul We define thehauptmodul of Γ^∗₀(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)

Location of the zeros of the Eisenstein series SinceW_12,4⁻¹Γ^∗₀(12)W12,4= Γ^∗₀(12), we have

E_k,12+12^−1/3 (W12,4z) = (6z−2)^kE_k,12+12^∞ (z). (3.276) Furthermore, we have

E_k,12+12^−1/3 (−1/2 +i/(6 tan(θ/2))) = (−(e^iθ−1)/2)^kE_k,12+12^∞ ((e^iθ−5)/12), E_k,12+12^−1/3 (itan(θ/2)/3) = (−(e^iθ+ 1)/4)^kE_k,12+12^∞ ((e^iθ−7)/24),

E_k,12+12^−1/3 (e^iθ⁰/(2√

3)) = (−(√

3e^iθ+ 2))^kE_k,12+12^∞ (e^iθ/(2√ 3)), where e^iθ⁰ = (4√

3 + 7 cosθ+isinθ)/(7 + 4√

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

E_k,12+12^−1/2 (e^i(π−θ⁰⁾/(2√

3)) = (2e^iθ+√

3)^kE^−1/2_k,12+12(e^iθ/(2√ 3)), E_k,12+12^−1/2 ((e^iθ−7)/24) = (√

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

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

E_k,12+12^∞ ^-¹² ^-¹³ ⁰ ¹³ ¹²

1 23

E_k,12+12^−1/3 ^-¹² ^-¹³ ⁰ ¹³ ¹²

1 23

E_k,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.

3.12.3 Γ

₀

(12) + 4

We have Γ0(12) + 4 =T_1/2⁻¹Γ0(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^iθ+ 1)/12→(e^i(π−θ)−1)/12 andW12,4: 1/3 +e^iθ/6→ −1/3 +e^i(π−θ)/6. Then, we have

Γ0(12) + 4 =h−I, (^{1 1}_{0 1}), W12,4, (_{12 1}^{1 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∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ0(12) + 4:

E_k,12+4^∞ (z) := 2^k3^kEk(12z)−2 3^kEk(6z)−2^kEk(4z) + 3^kEk(3z) +Ek(2z)−Ek(z)

(3^k−1)(2^k−1) fork>4.

(3.280)

For the cusp 0 We have Γ0 ={±(_12n^{1 0}₁) ;n∈Z} and γ0 =W12, and we have the Eisenstein series for the cusp 0 associated with Γ0(12) + 4:

E⁰_k,12+4(z) := 3^k/2(2^kEk(12z)−2Ek(6z)−2^kEk(4z) +Ek(3z) + 2Ek(2z)−Ek(z))

(3^k−1)(2^k−1) fork>4. (3.281) We also haveγ₀⁻¹(Γ0(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:

E_k,12+4^−1/2 (z) :=

2 2^k/23^k/2(2^k3^kEk(12z)−3^k(2^k+ 1)Ek(6z)−2^kEk(4z) + 3^kEk(3z) + (2^k+ 1)Ek(2z)−Ek(z))

(3^k−1)(2^k−1) .

(3.282) We also haveγ_−1/2⁻¹ (Γ0(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:

E_k,12+4^−1/6 (z) :=

−2 2^k/2(2^k3^k/2Ek(12z)−3^k/2(2^k+ 1)Ek(6z) + 2^kEk(4z) + 3^k/2Ek(3z)−(2^k+ 1)Ek(2z) +Ek(z))

(3^k−1)(2^k−1) .

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

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 Γ^∗₀(12) of weight 1.

We have Mk(Γ0(12) + 4) = CE_k,12+4^∞ ⊕CE⁰_k,12+4 ⊕CE_k,12+4^−1/2 ⊕CE_k,12+4^−1/6 ⊕Sk(Γ0(12) + 4) and Sk(Γ0(12)+4) = ∆12+Mk−4(Γ0(12)+4) for every even integerk>4. Then, we haveM4n+2(Γ0(12)+4) = E2,12+0M4n(Γ0(12) + 4)⊕C(E1,12+120)²(∆12+)ⁿ⊕C(E1,12+30)²(∆12+)ⁿ and

M4n(Γ0(12) + 4) =C(E_4,12+4^∞ )ⁿ⊕C(E_4,12+4^∞ )ⁿ⁻¹∆12+⊕ · · · ⊕CE_4,12+4^∞ (∆12+)ⁿ⁻¹

⊕C(E_4,12+4⁰ )ⁿ⊕C(E⁰_4,12+4)ⁿ⁻¹∆12+⊕ · · · ⊕CE_4,12+4⁰ (∆12+)ⁿ⁻¹

⊕C(E_4,12+4^−1/2 )ⁿ⊕C(E^−1/2_4,12+4)ⁿ⁻¹∆12+⊕ · · · ⊕CE_4,12+4^−1/2 (∆12+)ⁿ⁻¹

⊕C(E_4,12+4^−1/6 )ⁿ⊕C(E^−1/6_4,12+4)ⁿ⁻¹∆₁₂₊⊕ · · · ⊕CE_4,12+4^−1/6 (∆₁₂₊)ⁿ⁻¹⊕C(∆₁₂₊)ⁿ. Furthermore, we can write

M2n(Γ0(12) + 4) =C(∆^∞₁₂₊₄)²ⁿ⊕C(∆^∞₁₂₊₄)²ⁿ⁻¹∆⁰₁₂₊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))

q−4 + 6q−4q²−3q³+· · ·, (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 W₁₂⁻¹(Γ0(12) + 4)W12 = W_12+,6⁻¹ (Γ0(12) + 4)W12+,6=W_12+,2⁻¹ (Γ0(12) + 4)W12+,2= Γ0(12) + 4, we have

(2√

3z)^−kE_k,12+4⁰ (W12z) = (√ 6z+√

6/2)^−kE_k,12+4^−1/2 (W12+,6z) = (3√

2z+ 1/√

2)^−kE_k,12+4^−1/6 (W12+,2z) =E_k,12+4^∞ (z).

Furthermore, we have

E⁰_k,12+4(−1/2 +i/(2 tan(θ/2))) = ((e^iθ−1)/(2√

3))^kE_k,12+4^∞ ((e^iθ−1)/12), E_k,12+4⁰ (−1/3 +e^iθ⁰/6) = (−(2e^iθ−1)/√

3)^kE_k,12+4^∞ (−1/3 +e^iθ/6), E^−1/2_k,12+4((e^iθ⁰⁰−1)/12) = ((5e^iθ+ 1)/(2√

6))^kE_k,12+4^∞ ((e^iθ−1)/12), E_k,12+4^−1/2 (itan(θ/2)/2) = ((e^iθ+ 1)/√

6)^kE_k,12+4^∞ (−1/3 +e^iθ/6), E_k,12+4^−1/6 (−1/2 +itan(θ/2)/3) = ((e^iθ+ 1)/(2√

2))^kE_k,12+4^∞ ((e^iθ−1)/12), E_k,12+4^−1/6 (i/(6 tan(θ/2))) = ((e^iθ−1)/√

2)^kE_k,12+4^∞ (−1/3 +e^iθ/6),

wheree^iθ⁰ = (4−5 cosθ+ 3isinθ)/(5−4 cosθ) ande^iθ⁰⁰= (−5−13 cosθ+ 12isinθ)/(13 + 5 cosθ).

Now, recall that Γ₀(12) + 4 =T_1/2⁻¹Γ₀(6)T_1/2. Then, for k6500, since we can prove that all of the zeros ofE^∞_k,6lie on the lower arcs of ∂F6by numerical calculation, we have all of the zeros ofE_k,12+4^∞ in the lower arcs of∂F12+4.

E_k,12+4^∞ ^-¹² ^-¹⁶ ⁰ ¹⁶ ¹² E_k,12+4⁰ ^-¹² ^-¹⁶ ⁰ ¹⁶ ¹²

E_k,12+4^−1/2 ^-¹² ^-¹⁶ ⁰ ¹⁶ ¹² E_k,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.

3.12.4 Γ

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^iθ/(4√

3) → −1/4 +e^i(π−θ)/(4√

3), (_{24 7}^{7 2})W12,3 : 1/4 +e^iθ/(4√

3) → 1/4 + e^i(π−θ)/(4√

3),¡_{−1 0}

12−1

¢: (e^iθ+ 1)/12→(e^i(π−θ)−1)/12, and¡_{−5 2}

12 −5

¢: (e^iθ+ 5)/12→(e^i(π−θ)−5)/12.

Then, we have

Γ0(12) + 3 =h(^{1 1}_{0 1}), W12,3, (_{12 1}^{1 0}), (_{12 5}^{5 2})i. (3.287)

-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=ρ12,1,ρ12,3

vp(f) =k. (3.288)

Furthermore, the stabilizer of the elliptic pointρ12,1(resp.ρ12,3) is{±I,±W12,3}(resp. {±I,±(_{24 7}^{7 2})W12,3}).

For the cusp∞ We have Γ∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ0(12) + 3:

E_k,12+3^∞ (z) := 2^k3^k/2Ek(12z)−3^k/2Ek(6z) + 2^kEk(4z)−Ek(2z)

(3^k/2+ 1)(2^k−1) fork>4. (3.289) Note that we haveE_k,12+3^∞ (z) =E_k,6+3^∞ (2z).

For the cusp 0 We have Γ0 ={±(_12n^{1 0}₁) ;n∈Z} and γ0 =W12, and we have the Eisenstein series for the cusp 0 associated with Γ0(12) + 3:

E⁰_k,12+3(z) := −(3^k/2Ek(6z)−3^k/2Ek(3z) +Ek(2z)−Ek(z))

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

For the cusp −1/2 We have Γ−1/2 =©

±¡_6n+1 _3n

−12n−6n+1

¢ ; n∈Zª

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:

E_k,12+3^−1/2 (z) :=

−(2^k3^k/2Ek(12z)−3^k/2(2^k+ 1)Ek(6z) + 2^kEk(4z) + 3^k/2Ek(3z)−(2^k+ 1)Ek(2z) +Ek(z))

(3^k/2+ 1)(2^k−1) . (3.291)

Note that we have E_k,12+3^−1/2 (z) = E_k,12+12^−1/2 (z) = 2⁻¹2^−k/2E_k,12+^−1/2(z). We also have γ_−1/2⁻¹ (Γ0(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+0∆⁰₁₂₊₃∆^−1/2₁₂₊₃∆12, ∆12,2,12+3:= (E1,12+30)⁵E2,12+0∆^∞₁₂₊₃∆^−1/2₁₂₊₃∆12,

∆12,3,12+3:= (E1,12+30)⁵E2,12+0∆^∞₁₂₊₃∆⁰₁₂₊₃∆12, ∆12,4,12+3:= (E1,12+30)⁴∆⁰₁₂₊₃∆^−1/2₁₂₊₃(∆12)²,

∆12,5,12+3:= (E1,12+30)⁴∆^∞₁₂₊₃∆^−1/2₁₂₊₃(∆12)², ∆12,6,12+3:= (E1,12+30)⁴∆^∞₁₂₊₃∆⁰₁₂₊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)²∆⁰₁₂₊₃(∆12)³. In addition, we denote

A1=E1,12+30∆12(C∆^∞₁₂₊₃∆^−1/2₁₂₊₃⊕C∆⁰₁₂₊₃∆^−1/2₁₂₊₃⊕C∆^∞₁₂₊₃∆⁰₁₂₊₃), A2= (∆12)²(C∆^∞₁₂₊₃∆^−1/2₁₂₊₃⊕C∆⁰₁₂₊₃∆^−1/2₁₂₊₃⊕C∆^∞₁₂₊₃∆⁰₁₂₊₃),

B1=C(E^∞_12,12+3)ⁿ⊕C(E_12,12+3^∞ )ⁿ⁻¹∆12,1,12+3⊕C(E_12,12+3^∞ )ⁿ⁻¹∆12,4,12+3

⊕C(E_12,12+3^∞ )ⁿ⁻¹∆12,7,12+3⊕C(E_12,12+3^∞ )ⁿ⁻¹(∆12)⁴⊕C(E_12,12+3^∞ )ⁿ⁻²∆12,1,12+3(∆12)⁴

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

B2=C(E⁰_12,12+3)ⁿ⊕C(E_12,12+3⁰ )ⁿ⁻¹∆12,2,12+3⊕C(E_12,12+3⁰ )ⁿ⁻¹∆12,5,12+3

⊕C(E_12,12+3⁰ )ⁿ⁻¹∆12,8,12+3⊕C(E_12,12+3⁰ )ⁿ⁻¹(∆12)⁴⊕C(E_12,12+3⁰ )ⁿ⁻²∆12,2,12+3(∆12)⁴

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

B3=C(E^−1/2_12,12+3)ⁿ⊕C(E_12,12+3^−1/2 )ⁿ⁻¹∆12,3,12+3⊕C(E_12,12+3^−1/2 )ⁿ⁻¹∆12,6,12+3

⊕C(E_12,12+3^−1/2 )ⁿ⁻¹∆12,9,12+3⊕C(E_12,12+3^−1/2 )ⁿ⁻¹(∆12)⁴⊕C(E_12,12+3^−1/2 )ⁿ⁻²∆12,3,12+3(∆12)⁴

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

Now, we have

Mk(Γ0(12) + 3) =CE^∞_k,12+3⊕CE⁰_k,12+3⊕CE^−1/2_k,12+3⊕Sk(Γ0(12) + 3),

Sk(Γ0(12) + 3) = (C∆12,1,12+3⊕C∆12,2,12+3⊕ · · · ⊕C∆12,9,12+3⊕C(∆12)⁴)Mk−12(Γ0(12) + 3) for every even integerk>4. Then, we have

M12n(Γ0(12) + 3) =B1⊕B2⊕B3⊕C(∆12)⁴ⁿ,

M12n+2(Γ0(12) + 3) =E2,12+0M12n(Γ0(12) + 3)⊕CE1,12+30∆^∞₁₂₊₃(∆12)⁴ⁿ,

M12n+4(Γ0(12) + 3) =E^∞_4,12+3(B1⊕C(∆12)⁴ⁿ)⊕E_4,12+3⁰ (B2⊕C(∆12)⁴ⁿ)⊕E_4,12+3^−1/2 (B3⊕C(∆12)⁴ⁿ)

⊕(CE1,12+30⊕C∆^∞₁₂₊₃)(∆12)⁴ⁿ⁺¹,

M12n+6(Γ0(12) + 3) =E^∞_6,12+3(B1⊕C(∆12)⁴ⁿ)⊕E_6,12+3⁰ (B2⊕C(∆12)⁴ⁿ)⊕E_6,12+3^−1/2 (B3⊕C(∆12)⁴ⁿ)

⊕A1(∆12)⁴ⁿ,

M12n+8(Γ0(12) + 3) =E^∞_8,12+3(B1⊕C(∆12)⁴ⁿ)⊕E_8,12+3⁰ (B2⊕C(∆12)⁴ⁿ)⊕E_8,12+3^−1/2 (B3⊕C(∆12)⁴ⁿ)

⊕E2,12+0A1(∆12)⁴ⁿ⊕A2(∆12)⁴ⁿ,

M12n+10(Γ0(12) + 3) =E^∞_10,12+3(B1⊕C(∆12)⁴ⁿ)⊕E_10,12+3⁰ (B2⊕C(∆12)⁴ⁿ)⊕E_10,12+3^−1/2 (B3⊕C(∆12)⁴ⁿ)

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

Furthermore, we can write

Mk(Γ0(12) + 3) =E_k,12+3⁰(C(∆^∞₁₂₊₃)ⁿ⊕C(∆^∞₁₂₊₃)ⁿ⁻¹∆⁰₁₂₊₃⊕ · · · ⊕C(∆⁰₁₂₊₃)ⁿ),

wheren= dim(Mk(Γ0(12) + 3))−1 =bk−2(k/4− bk/4c)c, and whereE_k,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+ 7q³−9q⁵+· · · , (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 W₁₂⁻¹(Γ0(12) + 3)W12 = W_12−,6⁻¹ (Γ0(12) + 3)W12−,6= Γ0(12) + 3, we have

(2√

3z)^−kE⁰_k,12+3(W12z) = (2√ 3z+√

3)^−kE_k,12+3^−1/2 (W12−,6z) =E_k,12+3^∞ (z). (3.294)

Lower arcs of∂F12+3

-4 -3 -2 -1

-4 -2 2 4

Figure 3.85: Image byJ12+3

Furthermore, we have

E_k,12+3⁰ (−1/2 +i/(2 tan(θ/2))) = ((e^iθ−1)/(2√

3))^kE^∞_k,12+3((e^iθ−1)/12), E_k,12+3⁰ (−1/2 +i/(6 tan(θ/2))) = (−(e^iθ−1)/2)^kE^∞_k,12+3((e^iθ−5)/12),

E_k,12+3⁰ (1/2 +e^iθ⁰/(2√

3)) = (−(√

3e^iθ+ 1)/2)^kE^∞_k,12+3(−1/4 +e^iθ/(4√ 3)), E_k,12+3⁰ (−1/2 +e^i(π−θ⁰⁾/(2√

3)) = ((e^iθ+√

3)/2)^kE_k,12+3^∞ (1/4 +e^iθ/(4√ 3)), E_k,12+3^−1/2 (itan(θ/2)/6) = ((e^iθ+ 1)/2)^kE_k,12+3^∞ ((e^iθ−1)/12), E_k,12+3^−1/2 (itan(θ/2)/2) = ((e^iθ+ 1)/(2√

3))^kE^∞_k,12+3((e^iθ−5)/12), E_k,12+3^−1/2 (e^i(π−θ⁰⁾/(2√

3)) = ((e^iθ+√

3)/2)^kE_k,12+3^∞ (−1/4 +e^iθ/(4√ 3)), E_k,12+3^−1/2 (e^iθ⁰/(2√

3)) = (−(√

3e^iθ+ 1)/2)^kE^∞_k,12+3(1/4 +e^iθ/(4√ 3)),

wheree^iθ⁰ = (−√

3−2 cosθ+isinθ)/(2 +√ 3 cosθ).

E_k,12+3^∞ ^-¹² ^-¹⁴ ⁰ ¹⁴ ¹²

1 43

E_k,12+3⁰ ^-¹² ^-¹⁴ ⁰ ¹⁴ ¹²

1 43

E_k,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 E_k,12+3^∞ (z) =E^∞_k,6+3(2z). Fork6600, we can prove that all of the zeros ofE_k,6+3^∞ lie on the lower arcs of ∂F6+3 by numerical calculation, then we have all of the zeros of E_k,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.

3.12.5 Γ

(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^iθ+ 1)/12→(e^i(π−θ)−1)/12, ¡_{−5 1}

24 −5

¢: (e^iθ+ 5)/24→(e^i(π−θ)−5)/24, ¡_{−7 2}

24−7

¢: (e^iθ+ 7)/24→(e^i(π−θ)−7)/24, and¡_{−5 2}

12 −5

¢: (e^iθ+ 5)/12→(e^i(π−θ)−5)/12. Then, we have

Γ0(12) =h−I, (^{1 1}_{0 1}), (_{12 1}^{1 0}), (_{12 5}^{5 2}), (_{24 5}^{5 1}), (_{24 7}^{7 2})i. (3.296)

-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∞={±(¹_{0 1}ⁿ) ; n∈Z}, and we have the Eisenstein series for the cusp∞ associated with Γ0(12):

E^∞_k,12(z) :=2^k3^kEk(12z)−3^kEk(6z)−2^kEk(4z) +Ek(2z)

(3^k−1)(2^k−1) fork>4. (3.298) Note that we haveE_k,12^∞ (z) =E_k,6^∞(2z).

For the cusp 0 We have Γ0 ={±(_12n^{1 0}₁) ;n∈Z} and γ0 =W12, and we have the Eisenstein series for the cusp 0 associated with Γ0(12):

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

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

±¡_12n+1 _4n

−36n −12n+1

¢; n∈Zª

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

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

(3^k−1)(2^k−1) fork>4. (3.300) Note that we haveE_k,12^−1/3(z) = 2^−k/2E_k,6^−1/3(z). We also haveγ_−1/3⁻¹ Γ0(12)γ−1/3= Γ0(12).

±¡_12n+1 _3n

−48n −12n+1

¢; n∈Zª

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

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

(3^k−1)(2^k−1) fork>4. (3.301) Note that we haveE_k,12^−1/4(z) =E_k,6^−1/4(2z). We also haveγ⁻¹_−1/4 Γ0(12)γ_−1/4= Γ0(12).

±¡_6n+1 _3n

−12n−6n+1

¢ ; n∈Zª

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

E_k,12^−1/2(z) :=

3^k/2(2^k3^kEk(12z)−3^k(2^k+ 1)Ek(6z)−2^kEk(4z) + 3^kEk(3z) + (2^k+ 1)Ek(2z)−Ek(z))

(3^k−1)(2^k−1) . (3.302)

Note that we haveE_k,12^−1/2(z) = 2⁻¹2^−k/2E_k,12+4^−1/2 (z). We also haveγ_−1/2⁻¹ Γ0(12)γ−1/2= Γ0(12).

±¡_6n+1 _n

−36n−6n+1

¢ ; n∈Zª

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

E_k,12^−1/6(z) :=

−(2^k3^k/2Ek(12z)−3^k/2(2^k+ 1)Ek(6z) + 2^kEk(4z) + 3^k/2Ek(3z)−(2^k+ 1)Ek(2z) +Ek(z))

(3^k−1)(2^k−1) . (3.303)

Note that we haveE_k,12^−1/6(z) = 2⁻¹2^−k/2E_k,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−:= ∆^∞₁₂∆⁰₁₂(∆^−1/4₁₂ )²∆^−1/2₁₂ ∆^−1/6₁₂ , ∆6,4,12−:= ∆^∞₁₂∆⁰₁₂(∆^−1/3₁₂ )²∆^−1/2₁₂ ∆^−1/6₁₂ ,

∆6,5,12−:= ∆^∞₁₂∆⁰₁₂∆^−1/3₁₂ ∆^−1/4₁₂ (∆^−1/6₁₂ )², ∆6,6,12−:= ∆^∞₁₂∆⁰₁₂∆^−1/3₁₂ ∆^−1/4₁₂ (∆^−1/2₁₂ )². Now, we have

Mk(Γ0(12)) =CE_k,12^∞ ⊕CE_k,12⁰ ⊕CE_k,12^−1/3⊕CE_k,12^−1/4⊕E_k,12^−1/2⊕CE_k,12^−1/6⊕Sk(Γ0(12)), Sk(Γ0(12)) = (C∆6,1,12−⊕C∆6,2,12−⊕ · · · ⊕C∆6,6,12−⊕C(∆12)²)Mk−6(Γ0(12)), for every even integerk>4. Then, we have

M_6n(Γ₀(12)) =C(E_6,12^∞ )ⁿ⊕C(E_6,12^∞ )ⁿ⁻¹∆_6,1,12−⊕C(E_6,12^∞ )ⁿ⁻¹(∆₁₂)²⊕ · · · ⊕C∆_6,1,12−(∆₁₂)²⁽ⁿ⁻¹⁾

⊕C(E_6,12⁰ )ⁿ⊕C(E_6,12⁰ )ⁿ⁻¹∆6,2,12−⊕C(E_6,12⁰ )ⁿ⁻¹(∆12)²⊕ · · · ⊕C∆6,2,12−(∆12)²⁽ⁿ⁻¹⁾

⊕C(E_6,12^−1/3)ⁿ⊕C(E_6,12^−1/3)ⁿ⁻¹∆_6,3,12−⊕C(E_6,12^−1/3)ⁿ⁻¹(∆₁₂)²⊕ · · · ⊕C∆_6,3,12−(∆₁₂)²⁽ⁿ⁻¹⁾

⊕C(E_6,12^−1/4)ⁿ⊕C(E_6,12^−1/4)ⁿ⁻¹∆6,4,12−⊕C(E_6,12^−1/4)ⁿ⁻¹(∆12)²⊕ · · · ⊕C∆6,4,12−(∆12)²⁽ⁿ⁻¹⁾

⊕C(E_6,12^−1/2)ⁿ⊕C(E_6,12^−1/2)ⁿ⁻¹∆6,5,12−⊕C(E_6,12^−1/2)ⁿ⁻¹(∆12)²⊕ · · · ⊕C∆6,5,12−(∆12)²⁽ⁿ⁻¹⁾

⊕C(E_6,12^−1/6)ⁿ⊕C(E_6,12^−1/6)ⁿ⁻¹∆6,6,12−⊕C(E_6,12^−1/6)ⁿ⁻¹(∆12)²⊕ · · · ⊕C∆6,6,12−(∆12)²⁽ⁿ⁻¹⁾

⊕C(∆12)²ⁿ, M6n+2(Γ0(12)) =E2,12+0M6n(Γ0(12))

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

M6n+4(Γ0(12)) =E_4,12^∞ (C(E_6,12^∞ )ⁿ⊕C(E_6,12^∞ )ⁿ⁻¹∆6,1,12−⊕C(E_6,12^∞ )ⁿ⁻¹(∆12)²⊕ · · · ⊕C(∆12)²ⁿ)

⊕E_4,12⁰ (C(E_6,12⁰ )ⁿ⊕C(E_6,12⁰ )ⁿ⁻¹∆6,2,12−⊕C(E_6,12⁰ )ⁿ⁻¹(∆12)²⊕ · · · ⊕C(∆12)²ⁿ)

⊕E_4,12^−1/3(C(E_6,12^−1/3)ⁿ⊕C(E_6,12^−1/3)ⁿ⁻¹∆6,3,12−⊕C(E^−1/3_6,12 )ⁿ⁻¹(∆12)²⊕ · · · ⊕C(∆12)²ⁿ)

⊕E_4,12^−1/4(C(E_6,12^−1/4)ⁿ⊕C(E_6,12^−1/4)ⁿ⁻¹∆6,4,12−⊕C(E^−1/4_6,12 )ⁿ⁻¹(∆12)²⊕ · · · ⊕C(∆12)²ⁿ)

⊕E_4,12^−1/2(C(E_6,12^−1/2)ⁿ⊕C(E_6,12^−1/2)ⁿ⁻¹∆6,5,12−⊕C(E^−1/2_6,12 )ⁿ⁻¹(∆12)²⊕ · · · ⊕C(∆12)²ⁿ)

⊕E_4,12^−1/6(C(E_6,12^−1/6)ⁿ⊕C(E_6,12^−1/6)ⁿ⁻¹∆6,6,12−⊕C(E^−1/6_6,12 )ⁿ⁻¹(∆12)²⊕ · · · ⊕C(∆12)²ⁿ)

⊕(C(∆^∞₁₂)⁴(∆⁰₁₂)⁴⊕C(∆^−1/3₁₂ )⁴(∆^−1/4₁₂ )⁴)(∆12)²ⁿ⊕CE1,12+120(∆12)²ⁿ⁺¹.

Furthermore, we can write

M2n(Γ0(12)) =C(∆^∞₁₂)⁴ⁿ⊕C(∆^∞₁₂)⁴ⁿ⁻¹∆⁰₁₂⊕ · · · ⊕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))

q−3 + 2q+q³−2q⁷− · · · , (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 W₁₂⁻¹(Γ0(12))W12 =W_12,4⁻¹(Γ0(12))W12,4 = W_12,3⁻¹(Γ0(12))W12,3= Γ0(12), we have

(2√

3z)^−kE⁰_k,12(W12z) = (6z−2)^−kE^−1/2_k,12 (W6,3z) = (4√ 3z+√

3)^−kE^−1/6_k,12 (W12,4z) =E_k,12^∞ (z).

Furthermore, we have

E_k,12⁰ (−1/2 +i/(2 tan(θ/2))) = ((e^iθ−1)/(2√

3))^kE_k,12^∞ ((e^iθ−1)/12), E_k,12⁰ ((e^iθ¹−5)/12) = (−(5e^iθ−1)/(4√

3))^kE_k,12^∞ ((e^iθ−5)/24), E_k,12⁰ ((e^iθ²−7)/24) = (−(7e^iθ−1)/(4√

3))^kE_k,12^∞ ((e^iθ−7)/24), E_k,12⁰ ((e^iθ¹−5)/24) = (−(5e^iθ−1)/(2√

3))^kE_k,12^∞ ((e^iθ−5)/12), E_k,12^−1/3((e^iθ³−5)/24) = (−(3e^iθ+ 1)/2)^kE_k,12^∞ ((e^iθ−1)/12), E_k,12^−1/3((e^iθ³−1)/12) = ((3e^iθ+ 1)/4)^kE_k,12^∞ ((e^iθ−5)/24),

E_k,12^−1/3(itan(θ/2)/3) = (−(e^iθ+ 1)/4)^kE_k,12^∞ ((e^iθ−7)/24), E_k,12^−1/3(−1/2 +i/(6 tan(θ/2))) = (−(e^iθ−1)/2)^kE_k,12^∞ ((e^iθ−5)/24),

E_k,12^−1/4((e^iθ⁴−5)/24) = ((e^iθ+ 2)/√

3)^kE_k,12^∞ ((e^iθ−1)/12), E_k,12^−1/4(−1/2 +itan(θ/2)/4) = ((e^iθ+ 1)/(2√

3))^kE_k,12^∞ ((e^iθ−5)/24), E_k,12^−1/4(i/(4 tan(θ/2))) = ((e^iθ−1)/(2√

3))^kE_k,12^∞ ((e^iθ−7)/24), E^−1/4_k,12 ((e^iθ⁴⁰−1)/12) = ((e^iθ−2)/√

3)^kE_k,12^∞ ((e^iθ−5)/12), E^−1/2_k,12 ((e^iθ¹⁰−7)/24) = ((5e^iθ+ 1)/(2√

3))^kE_k,12^∞ ((e^iθ−1)/12), E^−1/2_k,12 ((e^iθ²⁰−5)/24) = ((7e^iθ+ 1)/(4√

3))^kE_k,12^∞ ((e^iθ−5)/24), E^−1/2_k,12 ((e^iθ¹⁰−1)/12) = ((5e^iθ+ 1)/(4√

3))^kE_k,12^∞ ((e^iθ−7)/24), E_k,12^−1/2(itan(θ/2)/2) = ((e^iθ+ 1)/(2√

3))^kE_k,12^∞ ((e^iθ−5)/12), E_k,12^−1/6(itan(θ/2)/6) = (−(e^iθ+ 1)/2)^kE_k,12^∞ ((e^iθ−1)/12), E_k,12^−1/6(−1/2 +i/(3 tan(θ/2))) = (−(e^iθ−1)/4)^kE_k,12^∞ ((e^iθ−5)/24), E^−1/6_k,12 ((e^iθ³⁰−5)/12) = (−(3e^iθ−1)/4)^kE_k,12^∞ ((e^iθ−7)/24), E^−1/6_k,12 ((e^iθ³⁰−7)/24) = (−(3e^iθ−1)/2)^kE_k,12^∞ ((e^iθ−5)/12),

wheree^iθ¹= (5−13 cosθ+ 12isinθ)/(13−5 cosθ),e^iθ¹⁰ = (−5−13 cosθ+ 12isinθ)/(13 + 5 cosθ),e^iθ² = (7−25 cosθ+24isinθ)/(25−7 cosθ),e^iθ²⁰ = (−7−25 cosθ+24isinθ)/(25+7 cosθ),e^iθ³= (−3−5 cosθ+ 4isinθ)/(5 + 3 cosθ),e^iθ³⁰ = (3−5 cosθ+ 4isinθ)/(5−3 cosθ),e^iθ⁴ = (4 + 5 cosθ+ 3isinθ)/(5 + 4 cosθ), ande^iθ⁴⁰ = (−4 + 5 cosθ+ 3isinθ)/(5−4 cosθ).

Now, recall thatE_k,12^∞ (z) =E_k,6^∞(2z). Then, fork6500, since we can prove that all of the zeros of E_k,6^∞ lie on the lower arcs of∂F6 by numerical calculation, we have all of the zeros ofE^∞_k,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.

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E_k,12^∞ ^-¹² ^-¹³^-¹⁴^-¹⁶ ⁰ ¹⁶ ¹⁴ ¹³ ¹² E_k,12⁰ ^-¹² ^-¹³ ^-¹⁴^-¹⁶ ⁰ ¹⁶ ¹⁴ ¹³ ¹²

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

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

Figure 3.88: Location of the zeros of the Eisenstein series

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Chapter 4

A note on

zeros of Eisenstein series

for genus zero Fuchsian group

Let Γ⊆SL2(R) be a genus zero Fuchsian group of the first kind having∞as a cusp, and letE_2k^Γ be the holomorphic Eisenstein series associated with Γ for the∞cusp 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 ofE_2k^Γ 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 E_2k^Γ. We can prove most of the zeros ofE_2k^Γ inF lie on its lower arcs under the same assumption.

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