Before we define modular forms we recall some basic properties about quadratic forms.
We will need these afterwards in order to define Zagierβs cusp forms associated to dis-criminants. Moreover, we introduce Heegner geodesics and Heegner points, which will be crucial in the course of this thesis.
20
An integral binary quadratic form π is a homogeneous quadratic polynomial of two variables with integer coefficients, i.e.,
π(π₯, π¦) = ππ₯2+ππ₯π¦+ππ¦2
with π, π, πβZ. The discriminant of π is given by Ξ(π) := π2β4ππ. Given πas above not identically zero and of discriminant ΞβZ we can distinguish three cases:
β IfΞ>0then one can find(π₯, π¦),(π₯β², π¦β²)βZ2 such that π(π₯, π¦)>0andπ(π₯β², π¦β²)<
0. In this case πis called indefinite.
β If Ξ = 0 then either π(π₯, π¦)β₯0 or π(π₯, π¦) β€0 for all(π₯, π¦)β Z2, and π is called positive or negative semi-definite, respectively. One can further check that π is positive or negative semi-definite if and only if π, πβ₯0 or π, πβ€0, respectively.
β If Ξ<0 then either π(π₯, π¦) >0 orπ(π₯, π¦) <0 for all (π₯, π¦) βZ2 with (π₯, π¦)ΜΈ= 0, and π is called positive or negative definite, respectively. Here π is positive or negative definite if and only ifπ, π > 0or π, π <0, respectively.
From now on we will only consider quadratic forms π(π₯, π¦) = ππ₯2+ππ₯π¦+ππ¦2 where the first coefficient π is divisible by π. We denote the set of all such forms by π¬, i.e.,
π¬:={οΈ
π(π₯, π¦) =ππ₯2+ππ₯π¦+ππ¦2: π, π, πβZ, π |π}οΈ
.
For π β π¬ as above we call π½ β Z/2πZ with π½ β‘π mod 2π the class of π, and for π of class π½ we clearly haveΞ(π)β‘π½2 mod 4π.
Let π½ β Z/2πZ and Ξ β Z with Ξ β‘ π½2 mod 4π. We define π¬π½,Ξ as the set of integral binary quadratic forms of class π½ and discriminant Ξ, i.e.,
π¬π½,Ξ :=
{οΈ
π(π₯, π¦) =ππ₯2+ππ₯π¦+ππ¦2 β π¬: Ξ(π) = Ξ, π β‘π½ mod 2π }οΈ
.
Note that for π = 1 the class π½ β Z/2Z is uniquely determined by the discriminant Ξ and may thus be omitted. In this case one simply writes π¬Ξ.
The groupΞ0(π) acts onπ¬π½,Ξ from the right via π.π =ππ‘ππ
for π β π¬π½,Ξ and π β Ξ0(π), where we identify π(π₯, π¦) = ππ₯2 +ππ₯π¦ +ππ¦2 with its associated matrix
π=
(οΈ π π/2 π/2 π
)οΈ
.
We write [π] for the Ξ0(π)-orbit of π, and π βΌ πβ² if π and πβ² are equivalent modulo Ξ0(π), i.e., ifπ β[πβ²]. For ΞΜΈ= 0 the quotientπ¬π½,Ξ/Ξ0(π) is always finite as its order is essentially given as a sum of generalized class numbers. However, forΞ = 0 the group Ξ0(π) acts onπ¬π½,0 with infinitely many orbits.
Also, one easily checks that π β π¬ is indefinite, positive (semi-)definite or negative (semi-)definite if and only if π.π is for all π β Ξ0(π). Thus, if Ξ β€ 0 then the only quadratic form π of discriminant Ξ with π βΌ βπ is the zero-form π β‘ 0 which is positive and negative semi-definite at the same time.
Let now πβ π¬π½,Ξ with π(π₯, π¦) =ππ₯2+ππ₯π¦+ππ¦2 and πΜΈβ‘0. Depending on the sign of Ξ we associate to π a geodesic, a parabolic point or a point in H. These associated objects are essentially given by the roots ofπ(π₯,1), where we understand βas a root of π(π₯,1)if π(1,0) = 0.
β If Ξ > 0 we associate to π the unique geodesic ππ in H joining the two distinct roots ofπ(π₯,1)in P1(R), namely
π₯= βπ+β Ξ
2π and π₯β² = βπββ Ξ 2π
if π ΜΈ= 0, and π₯ =βππ and π₯β² = β if π = 0. The orientation of ππ is defined to go fromπ₯toπ₯β², except when π= 0 andπ <0, in which case it runs from π₯β² toπ₯. This guarantees that the geodesics associated to π and βπ have the same image in H but their orientations are inverted. As a subset of H, the geodesic ππ can also be written as
ππ={οΈ
π βH: π|π|2+πRe(π) +π= 0}οΈ
. We callππ the Heegner geodesic associated to π.
β IfΞ = 0we associate toπthe parabolic point ππ which is given as the unique root of π(π₯,1)inP1(Q), namelyππ =β2ππ ifπΜΈ= 0, andππ =β otherwise. In abuse of notation we also callππ the cusp associated to π.
β IfΞ<0we associate to π the pointππ βHwhich is the unique root of π(π₯,1)in the upper half-plane, i.e.,
ππ:=β π 2π +π
βοΈ|Ξ|
2|π| .
The point ππ is called the Heegner point or CM point associated to π. We further define the set of all Heegner points associated to the class [π] as
π»[π]:={ππβ²: πβ² β[π]}.
(2.3.1)
In abuse of notation we simply write π»π for the set π»[π], and we set π»π = β if Ξ(π) β₯ 0. Moreover, we define the set of all Heegner points of class π½ and discriminant Ξby
π»π½,Ξ:={ππβ²: πβ² β π¬π½,Ξ}.
(2.3.2)
ForΞβZ with Ξβ‘π½2 mod 4π and Ξβ₯0we again simply set π»π½,Ξ=β .
Letπβ π¬π½,Ξ with πΜΈβ‘0 and π βΞ0(π). It is easy to verify that π₯βHβͺRβͺ {β}
is a root of (π.π)(π₯,1) if and only if π π₯ is a root of π(π₯,1). Therefore, the action of Ξ0(π) onπ¬π½,Ξβ {0} is compatible with the above identifications in the sense that
ππ.π =πβ1ππ, ππ.π =πβ1ππ and ππ.π =πβ1ππ (2.3.3)
if Ξ> 0, Ξ = 0 or Ξ< 0, respectively. This also shows that the set of Heegner points associated to some class [π] defined above can indeed be written as
π»π ={π ππ: π βΞ0(π)}.
(2.3.4) 22
Here we recall that π»π=β if Ξ(π)β₯0.
Moreover, the stabilizer of the quadratic formπin Ξ0(π) agrees with the stabilizer of its associated object, i.e.,depending on the sign of Ξ the stabilizer (Ξ0(π))π equals the stabilizer (Ξ0(π))ππ, (Ξ0(π))ππ or (Ξ0(π))ππ, respectively. Here it is important that for πΜΈβ‘0 we can only have π.π =βπ for some π βΞ0(π)if Ξ(π)>0 in which case π does not fix the oriented geodesic ππ but swaps its endpoints.
Lemma 2.3.1. Let πβ π¬ with Ξ(π) > 0. Then the stabilizer of π in Ξ0(π)/{Β±1} is trivial if Ξ(π) is a square, and infinite cyclic otherwise.
Proof. LetΞ = Ξ(π), and recall that(Ξ0(π))π = (Ξ0(π))ππ as noted above. Therefore, the stabilizer (Ξ0(π))π/{Β±1} is either trivial or infinite cyclic, and every non-trivial element in this stabilizer is hyperbolic.
If Ξ is a square then the endpoints of the geodesic ππ are rational, and thus there is no hyperbolic element in Ξ0(π) fixing the oriented geodesic ππ. So (Ξ0(π))π = {Β±1}.
Conversely, suppose that Ξ is not a square. Then Pellβs equation π‘2 βΞπ’2 = 1 has a non-trivial solution (π‘, π’)βZ2. Writing π(π₯, π¦) = ππ₯2+ππ₯π¦+ππ¦2 we set
π :=
(οΈπ‘+ππ’ 2ππ’
β2ππ’ π‘βππ’ )οΈ
,
which defines an element ofΞ0(π)asπ dividesπ. Now an easy computation shows that π.π = π. Further, π ΜΈ= Β±1 since if π = π = 0 the discriminant Ξ = π2 β4ππ would be a square. Hence we have shown that the stabilizer ofπ inΞ0(π)/{Β±1}is non-trivial.
Thus it has to be infinite cyclic.
Corollary 2.3.2. Let ππ be a Heegner geodesic. Then ππ is infinite ifΞ(π) is a square, and closed otherwise. Conversely, every closed or infinite geodesic π can be realized as a Heegner geodesic ππ =π with πβ π¬ and Ξ(π)>0.
Proof. Letπβ π¬withΞ(π)>0. IfΞ(π)is a square, then the endpoints of the geodesic ππare rational, and thus the geodesicππis infinite. Further, ifΞ(π)is not a square, then the stabilizer ofπ in Ξ0(π)/{Β±1} is non-trivial by Lemma 2.3.1, and thusππ is closed.
Conversely, letπbe a closed or infinite geodesic. Ifπis closed then we findπ βΞ0(π) with π ΜΈ= Β±1 which stabilizes π. Then π = ππ for π(π₯, π¦) = Β±(ππ₯2 + (πβπ)π₯π¦βππ¦2) where we write π =(οΈπ π
π π
)οΈ, and where we choose the sign Β±1such that the orientations of π and ππ match. If on the other hand π is infinite, then we find π, π β P1(Q) with π=ππ,π. Hence π=ππ for π(π₯, π¦) =π(π₯βππ¦)(π₯βππ¦) where we choose πβπZ, π ΜΈ= 0, such that πis integral, and such that the orientations of πand ππ agree.
Sometimes it will be useful to extend the action ofΞ0(π)onπ¬to an action ofSL2(R) on the set of all not necessarily integral binary quadratic formsπ(π₯, π¦) =ππ₯2+ππ₯π¦+ππ¦2 with π, π, π β R. In this context we also extend the definitions of geodesics ππ, cusps ππ and points in the upper half-plane ππ to non-integral quadratic formsπ. One easily checks that the identities from (2.3.3) still hold, i.e., that
ππ.πΌ=πΌβ1ππ, ππ.πΌ =πΌβ1ππ and ππ.πΌ=πΌβ1ππ
for π(π₯, π¦) =ππ₯2+ππ₯π¦+ππ¦2 with π, π, πβR and πΌβSL2(R).
The following lemma shows that using scaling matrices associated to the geodesic ππ, the parabolic pointππor the pointππβHwe can transform an integral binary quadratic formπΜΈβ‘0into an (in general non-integral) binary quadratic form of one of the following standard types:
Lemma 2.3.3. Let πβ π¬ with π(π₯, π¦) =ππ₯2+ππ₯π¦+ππ¦2, πΜΈβ‘0 and π=|Ξ(π)|1/2. (a) If Ξ(π)>0 then
π.π=
(οΈ 0 π/2 π/2 0
)οΈ
(2.3.5)
for some πβSL2(R) if and only if π maps π0 to ππ preserving orientations. Here π0 is the standard geodesic from 0 to β.
(b) If Ξ(π) = 0 then
π.π=
(οΈ0 0 0 ππ
)οΈ
(2.3.6)
for π β SL2(R) if and only if π is a parabolic scaling matrix for the cusp ππ. Here ππ βQ* is given by
ππ:= sign(π+π)(π, π)(π/π, π/2) (π, π/2) , which simplifies to ππ = sign(π+π)(π/π, π) if π is squarefree.
(c) If Ξ(π)<0 then
π.π= sign(π)
(οΈπ/2 0 0 βπ/2
)οΈ
(2.3.7)
for π βSL2(R) if and only if ππ=ππ.
Proof. Firstly, let Ξ(π) > 0 and π β SL2(R) mapping π0 to ππ. Then π.π needs to be of the form (οΈ0 π
π 0
)οΈ with π β R*. Further, π is independent of the choice of π as Ξ(π.π) = Ξ(π)yields π=Β±π/2. We find that π=π/2 by choosing
π= (4|π|π)β1/2
(οΈβsign(π)(π+π) πβπ
2|π| β2π
)οΈ
if π ΜΈ= 0, π = (οΈ1βπ/π
0 1
)οΈ if π = 0 and π > 0, and π = (οΈβπ/π β1
1 0
)οΈ if π = 0 and π < 0. This proves part (a).
Next letΞ(π) = 0 and let π βSL2(R) be a parabolic scaling matrix for the cusp ππ. Then πβ =ππ, which implies π.π =(οΈ0 0
0 π
)οΈ for some π βR*. Moreover, if πβ² βSL2(R) is another scaling matrix for ππ then πβ1πβ² = (οΈΒ±1 β
0 Β±1
)οΈ for some β β R, showing that π does not depend on the choice of π. If π = 0 the statement becomes trivial by choosing π = 1. If on the other hand πΜΈ= 0, we set π = (π, π/2),π= (π/π, π)and
π =
βοΈπ π
(οΈπ/2π π½π/π
βπ/π πΏπ/π )οΈ
. 24
Here π½, πΏ are integers withππ½ +ππΏ/2 =π. Now a direct computation of π.π shows that π = πππ(π½, πΏ). Since π/π = (π/π, π)/π = (π/π, π/2)/π and π(π½, πΏ) = sign(π)(π, π) we obtain the formula given in part (b) of the lemma.
For (c) let Ξ(π) < 0 and π β SL2(R) with ππ = ππ. Then (π.π)(π,1) = 0 and thus π.π=(οΈπ 0
0 βπ
)οΈfor someπ βR*. Again,πneeds to be independent of the choice ofπ since Ξ(π.π) = Ξ(π)implies π=Β±π/2. By choosing
π= (2π|π|)β1/2
(οΈsign(π)π βπ
0 2π
)οΈ
we find that the sign ofπ equals the sign of π.
Finally, we note that the converse of the proven implications in (a), (b) and (c) are indeed trivial, which is why we did not mention them.
In particular, the previous Lemma helps us to understand the quadratic forms of dis-criminant zero, which turn out to correspond to cusps of the underlying group Ξ0(π).
More precisely, if π is squarefree there is only one class π¬π½,0 of forms of discriminant zero, namely π¬0,0, which can be identified with the set of cusps πΆ(Ξ0(π)) as follows:
Lemma 2.3.4. Let π be squarefree. Then the map
π¬0,0β {0} βP1(Q)Γ(Zβ {0}), πβ¦β(ππ, ππ)
is a bijection, where ππ = sign(π+π)(π, π) for π(π₯, π¦) = ππ π₯2 +ππ₯π¦+ππ¦2. Further, the above map is compatible with the corresponding actions of Ξ0(π) where Ξ0(π) acts trivially on Zβ {0}, i.e., for π β Ξ0(π) the element π.π is mapped to (πβ1ππ, ππ).
Thus the induced map
(π¬0,0β {0})/Ξ0(π)βπΆ(Ξ0(π))Γ(Zβ {0}), [π]β¦β([ππ], ππ) is again a bijection.
Proof. We start by noting that given π, πβ² β π¬0,0β {0} we have π.πππ =
(οΈ0 0 0 ππ
)οΈ
and πβ².πππβ² =
(οΈ0 0 0 ππβ²
)οΈ
whereπππ, ππ
πβ² βSL2(R)are parabolic scaling matrices for the cuspsππ,ππβ², respectively.
Thus, ifππ =ππβ² and ππ =ππβ² we find π=
(οΈ0 0 0 ππ
)οΈ
.ππβ1
π =
(οΈ0 0 0 ππβ²
)οΈ
.πβ1π
πβ² =πβ²
as πππ and πππβ² are both scaling matrices for the same cusp ππ =ππβ². Hence, the given map πβ¦β(ππ, ππ) is indeed injective.
Further, given π β P1(Q) and π β Z with π ΜΈ= 0 we can define π = (οΈ0 0
0π
)οΈπβ1π where ππ is a scaling matrix for the cusp π. Then π β π¬ since ππ
(οΈ1 1
0 1
)οΈππβ1 β Ξ0(π), and by construction πis mapped to (π, π). So the given map is also surjective.
In order to show that the map πβ¦β(ππ, ππ)is compatible with the actions of Ξ0(π), we only need to check thatππ.π =ππforπβ π¬0,0 withπΜΈβ‘0andπ βΞ0(π), since we
already know that ππ.π = πβ1ππ. Let πππ βSL2(R) be a scaling matrix for ππ. Then πβ1πππ is a scaling matrix for πβ1ππ and thus
(οΈ0 0 0 ππ.π
)οΈ
= (π.π).ππ.π = (π.π).(πβ1πππ) =
(οΈ0 0 0 ππ
)οΈ
proving the claim.
Given π β π¬ with π(π₯, π¦) = ππ₯2 +ππ₯π¦ +ππ¦2 we also define the real-valued rational function ππ(π)by
ππ(π) := π|π|2+ππ’+π (2.3.8) π£
for π =π’+ππ£ βH. It is easy to check that ππ(π)2 := |π(π,1)|2
π£2 βΞ(π).
(2.3.9)
In particular, we note that if Ξ(π) > 0 the geodesic ππ is given as the zero set of the rational function ππ(π) inH (without orientation).