Dimension Formulas
3.8 More on cusps
3.7.6.(a) Show that the number of solutions to the congruencex2−x+ 1≡ 0 (modN) is given by the formula for ε3(Γ0(N)) in Corollary 3.7.2. (A hint for this exercise is at the end of the book.)
(b) Show that the orbits Γ0(N) [n1 01] (µ3) for n= 0, . . . , N −1 such that n2−n+ 1≡0 (modN) are distinct inX0(N).
(c) Confirm formula (3.14).
(d) Similarly show that the period 2 elliptic points ofΓ0(N) are given by formula (3.15).
3.7.7.Letpe andM be positive integers withpprime,e≥1, andpM. Let m = p−1(mod M), i.e., mp ≡ 1 (modM) and 0 ≤ m < M. Consider the matrices
αj= 1 0
M j1
, 0≤j < pe and
βj=
1 (mjp−1)/M M mjp
, mj=m+jM, 0≤j < pe−1. Show that
Γ0(peM)\Γ0(M) =
⎛
⎝4
j
Γ0(peM)αj
⎞
⎠∪
⎛
⎝4
j
Γ0(peM)βj
⎞
⎠.
Iterating this construction gives a set of representatives for Γ0(N)\SL2(Z) whereN =
pe, including the representatives used at the end of the section to find the elliptic points. For example, if representatives forΓ0(peqf)\Γ0(qf) are {α0, . . . , αpe−1, β0, . . . , βpe−1−1} and representatives for Γ0(qf)\SL2(Z) are{α0, . . . , αqf−1, β0, . . . , βqf−1−1}then thepe(1 + 1/p)qf(1 + 1/q) products {αjαj,αjβj,βjαj,βjβj}are a set of representatives forΓ0(peqf)\SL2(Z).
Proof. Assume thatc= 0 andc= 0. Then forγ= [p qr t]∈SL2(Z), s =γ(s) ⇐⇒ a
c = pa+qc ra+tc.
Both fractions on the right side are in lowest terms (Exercise 3.8.1(a)), and so the result follows in this case. The cases when at least one ofc, c is zero
are left as Exercise 3.8.1(b).
The notation from the lemma will remain in effect throughout the section.
Thus s = a/c and s = a/c are elements of Q∪ {∞} expressed in lowest terms, so that [ac] anda
c
are elements ofZ2with gcd(a, c) = gcd(a, c) = 1.
The gcd condition automatically excludes the zero vector. If [ac] =γa
c
for some γ ∈ Γ(N) then taking the relation modulo N gives the congruence [ac]≡a
c
(modN). The next result, that the converse holds as well, is the small technical point needed to describe the cusps.
Lemma 3.8.2.Let [ac]anda
c
be as above. Then a
c
=γ a
c
for someγ∈Γ(N) ⇐⇒
a c
≡ a
c
(mod N).
Proof. “ =⇒” is immediate, as just noted. For “⇐= ” first assume [ac] = [10].
Thena≡1 (modN). Take integersβ andδsuch thataδ−cβ = (1−a)/N and letγ=
a βN c 1+δN
. Thenγ∈Γ(N) anda
c
=γ[ac] in this case.
In the general case there exist integersbanddsuch thatad−bc= 1. Let α=a b
c d
∈SL2(Z). Since α[10] = [ac], it follows that α−1a
c
≡α−1[ac] = [10] working moduloN. Thusα−1a
c
=γα−1[ac] for someγ ∈Γ(N), as in the preceding paragraph. SinceΓ(N) is normal in SL2(Z), settingγ=αγα−1
completes the proof.
Proposition 3.8.3.Let s=a/cands=a/c be elements ofQ∪ {∞} with gcd(a, c) = gcd(a, c) = 1. Then
Γ(N)s=Γ(N)s ⇐⇒
a c
≡ ± a
c
(modN), and
Γ1(N)s=Γ1(N)s ⇐⇒
a c
≡ ±
a+jc c
(modN)for somej, and
Γ0(N)s=Γ0(N)s ⇐⇒
ya c
≡
a+jc yc
(mod N)for somej, y.
In the second and third equivalences j can be any integer, and in the third equivalencey is any integer relatively prime to N.
Proof. For the first equivalence,
Γ(N)s=Γ(N)s ⇐⇒ s=γ(s) for someγ∈Γ(N)
⇐⇒ a
c
=±γ[ac] for someγ∈Γ(N), by Lemma 3.8.1
⇐⇒ a
c
≡ ±[ac] (modN), by Lemma 3.8.2.
The decomposition Γ1(N) =
jΓ(N)1j
0 1
reduces the second equivalence to the first, cited at the last step of the computation
Γ1(N)s =Γ1(N)s ⇐⇒ s∈Γ1(N)s
⇐⇒ s∈Γ(N)1j
0 1
sfor somej
⇐⇒ Γ(N)s=Γ(N)(s+j) for somej
⇐⇒ a
c
≡ ±[a+jcc ] (modN) for somej.
Similarly, the decompositionΓ0(N) =
yΓ1(N)x k
N y
, taken overyrelatively prime toN and xy−kN = 1, reduces the third equivalence to the second, cited at the third step of the computation
Γ0(N)s=Γ0(N)s ⇐⇒ s∈Γ1(N)x k
N y
sfor some y
⇐⇒ Γ1(N)s=Γ1(N)xa+kc
N a+yc for somey
⇐⇒ a
c
≡ ±xa+kc+jyc
yc
(modN) for somej,y
⇐⇒
ya c
≡a+jc
yc
(mod N) for some j,y,
where the “±” has been absorbed intoyand some constants intoj at the last
step.
Thus the cusps ofΓ(N) areΓ(N)s, s=a/c, for all pairs±[ac] (modN) where gcd(a, c) = 1. Condition (3) of the next lemma provides a characteri- zation purely moduloN for these representatives.
Lemma 3.8.4.Let the integersa, chave imagesa, cin Z/NZ. Then the fol- lowing are equivalent:
(1) [ac]has a lifta
c
∈Z2 with gcd(a, c) = 1, (2) gcd(a, c, N) = 1,
(3) [ac]has orderN in the additive group (Z/NZ)2.
Proof. If condition (1) holds thenk(a+sN) +l(c+tN) = 1 for some integers k, l, s, t, and soka+lc+ (ks+lt)N= 1, giving condition (2).
If condition (2) holds thenad−bc+kN = 1 for someb,d, andk, and the matrixγ=a b
c d
∈M2(Z) reduces moduloN into SL2(Z/NZ). Since SL2(Z) surjects to SL2(Z/NZ), there is a liftab
c d
∈SL2(Z), giving condition (1).
(The reader may remember this argument from the proof of Theorem 1.5.1.)
Condition (3) is the implication (k[ac] = 0 in (Z/NZ)2 =⇒ N | k), or equivalently (N | kaand N | kc =⇒ N | k), or equivalently (N | kgcd(a, c) =⇒ N | k), or equivalently gcd(gcd(a, c), N) = 1, or equiva-
lently gcd(a, c, N) = 1, i.e., condition (2).
By the equivalence of conditions (1) and (3), the cusps ofΓ(N) are now described by the pairs±[ac] of orderN in (Z/NZ)2. The bijection is
±[ac]→Γ(N)(a/c), wherea
c
is a lift of [ac] with gcd(a, c) = 1.
Using condition (2) of Lemma 3.8.4 we can count the cusps ofΓ(N). For each c ∈ {0,1, . . . , N −1} let d = gcd(c, N). Then the number of values a ∈ {0,1, . . . , N −1} such that gcd(a, c, N) = 1 is the number of values such that gcd(a, d) = 1, or (N/d)φ(d) (Exercise 3.8.2(a)). So the number of elements of order N in (Z/NZ)2 is (considering only positive divisors here and elsewhere in this section)
d|N
(N/d)φ(d)· |{c: 0≤c < N, gcd(c, N) =d}|
=
d|N
(N/d)φ(d)φ(N/d)
=N2
p|N
(1−1/p2) (Exercise 3.8.2(b)).
ForN >2 each pair ±[ac] of order N elements in (Z/NZ)2 has two distinct members since an element equal to its negative has order 2. For N = 2 the
“±” has no effect. Thus the number of cusps ofΓ(N) is ε∞(Γ(N)) =
(1/2)N2
p|N(1−1/p2) ifN >2,
3 ifN= 2.
Deriving this formula doesn’t really require first describing the cusps of Γ(N) explicitly. Since Γ(N) is normal in SL2(Z), its cusps all have the same ramification degree over SL2(Z)∞, most easily computed by applying the formula from Section 3.1 at∞,
eΓ(N)∞= [SL2(Z)∞:{±I}Γ(N)∞] = [± [1 10 1]:± [10 1N]] =N.
Since−I∈SL2(Z) and−I∈Γ(N) only forN = 2 (recall that N >1), and since [SL2(Z) :Γ(N)] =N3
p|N(1−1/p2), the projection of modular curves X(N)−→X(1) has degree
dN = [SL2(Z) :{±I}Γ(N)] =
(1/2)N3
p|N(1−1/p2) ifN >2,
6 ifN = 2,
cf. Exercise 1.2.2(b). Now the formula ε∞(Γ(N)) = dN/N reproduces the result.
But since Γ1(N) is not normal in SL2(Z), its cusps must be counted di- rectly. The representatives are pairs±[ac] of order N vectors modulo N, but now by Proposition 3.8.3 a cusp determines the upper entry a modulo the lower entryc, so in factais determined modulo gcd(c, N). This time, for each c∈ {0,1, . . . , N−1}, lettingd= gcd(c, N), the number of values a(modd) such that gcd(a, d) = 1 is justφ(d). So the number of elements of in (Z/NZ)2 that pair to describe cusps ofΓ1(N) is
d|Nφ(d)φ(N/d).
a= 0 1 2 3 4 5 6 7 8 9 10 11 c= 0 · ⊗ · · · ⊗ · ⊗ · · · ⊗
1⊗ × × × × × × × × × × × 2 · ⊗ · × · × · × · × · × 3 · ⊗ ⊗ · × × · × × · × × 4 · ⊗ · ⊗ · × · × · × · × 5⊗ × × × × × × × × × × × c= 6 · ⊗ · · · × · × · · · ⊗ 7⊗ × × × × × × × × × × × 8 · × · × · × · × · ⊗ · ⊗ 9 · × × · × × · × × · ⊗ ⊗ 10 · × · × · × · × · × · ⊗ 11⊗ × × × × × × × × × × × Figure 3.2. The cusps ofΓ(12) and ofΓ1(12)
The table in Figure 3.2 shows all of this forN = 12. The elements [ac] of order 12, indicated by “×”, pair under negation modulo 12 to represent the cusps ofΓ(12). In each row, one element is circled for each value ofamodulo gcd(c, N), describing the cusps ofΓ1(12). The circled elements from the top row, wherec = 0, pair in an obvious way, and similarly for the left column, where a= 0. When ac = 0 the circled elements are taken from the left cell of length gcd(c, N) on rowcwhen c <6 and from the right cell whenc >6, and these visibly pair. On the middle row, wherec = 6 (and more generally wherec=N/2 whenN is even), the pairing of circled elements is a bit more subtle: the leftφ(N/2)/2 elements in the left half of the row pair with the rightφ(N/2)/2 in the right half unlessφ(N/2) is odd. Butφ(N/2) is odd only forN = 2, when the whole pairing process collapses anyway, and forN = 4, when only four of the five “×”’s in the relevant table pair off, leaving three representatives (Exercise 3.8.3). Thus the number of cusps ofΓ1(N) is
ε∞(Γ1(N)) =
⎧⎪
⎨
⎪⎩
2 ifN = 2,
3 ifN = 4,
1 2
d|Nφ(d)φ(N/d) ifN = 3 orN >4.
The sum doesn’t have as tidy a product form as the corresponding sum for Γ(N) despite appearing simpler (this reflects that Γ1(N) is not normal in SL2(Z)), so we leave it as is.
To count the cusps of Γ0(N) recall from Proposition 3.8.3 that for this group, vectors [ac] and a
c
with gcd(a, c) = gcd(a, c) = 1 represent the same cusp when
ya c
≡a+jc
yc
(modN) for somej andywith gcd(y, N) = 1. The bottom condition, c ≡ yc(modN) for some such y, is equivalent to gcd(c, N) = gcd(c, N), in which case letting d = gcd(c, N) and letting y0 ∈ Zsatisfy y0 ≡ cc−1 (mod N) makes the condition equivalent to y ≡ y0+iN/d(modN) for somei (confirming the calculations in the paragraph is Exercise 3.8.4). For any divisord of N, pick one value c moduloN such that gcd(c, N) =d. Then any cusp ofΓ0(N) represented by some vectora
c
with gcd(c, N) = d is also represented by [ac] whenever (y0+iN/d)a ≡ a+jc(mod N) for some i and j, or a ≡ y0a (mod gcd(c, N, aN/d)), or a≡y0a (mod gcd(d, N/d)). Also, ais relatively prime to dsince gcd(a, d)| gcd(a, c) = 1, soais relatively prime to gcd(d, N/d). Thus for each divisord ofN there areφ(gcd(d, N/d)) cusps, and the number of cusps of Γ0(N) is
ε∞(Γ0(N)) =
d|N
φ(gcd(d, N/d)).
The reader should use this discussion to find the cusps ofΓ0(12) with the help of Figure 3.2 (Exercise 3.8.5).
Let N be any positive integer and let k be odd. All cusps of Γ0(N) are regular since−I∈Γ0(N). To study the cusps ofΓ1(N), take anys∈Q∪{∞}
and letγ∈Γ1(N)s, so ifα∈SL2(Z) takes∞tosthenα−1γα∈(α−1Γ α)∞. If α−1γα takes the form −[10 1h] then γ has trace −2. But also, trace(γ) ≡ 2 (modN), so this can happen only ifN|4. Since−I∈Γ1(2), the only case where an irregular cusp might occur isΓ1(4). This argument shows that all cusps ofΓ(N),N = 4, are regular as well. In fact, the only irregular cusp in this context turns out to be the example we have already seen,s = 1/2 for Γ1(4) (Exercise 3.8.7).
Finally, the cusps of any congruence subgroup of SL2(Z) have a purely group-theoretic description. LetGbe an arbitrary group and letH1 and H2
be subgroups. Adouble coset of Gis a subset ofGof the form H1gH2. The space of double cosets is denotedH1\G/H2, that is,
H1\G/H2={H1gH2:g∈G}.
As with one-sided cosets, any two double cosets are disjoint or equal, so there is a disjoint set decomposition
G= 4
g∈R
H1gH2 whereRis a set of representatives.
A specific case of all this describes the cusps of any congruence subgroup.
Proposition 3.8.5.Let Γ be a congruence subgroup of SL2(Z) and letP be the parabolic subgroup of SL2(Z), P = {±1j
0 1
: j ∈ Z} = SL2(Z)∞, the fixing subgroup of∞. Then the map
Γ\SL2(Z)/P −→ {cusps ofΓ} given by
Γ αP →Γ α(∞) is a bijection. Specifically, the map isΓa b
c d
P →Γ(a/c).
Proof. The map is well defined since ifΓ αP =Γ αP thenα=γαδfor some γ ∈Γ and δ∈ P, so that Γ α(∞) =Γ γαδ(∞) =Γ α(∞), the last equality becauseγ∈Γ andδfixes∞.
The map is injective since the condition Γ α(∞) =Γ α(∞) is equivalent to α(∞) =γα(∞) for some γ ∈ Γ, or α−1γ−1α ∈ P for some γ ∈ Γ, or α ∈Γ αP, meaningΓ αP =Γ αP.
The map is clearly surjective.
Another proof, essentially identical, is to show first that SL2(Z)/P identi- fies withQ∪ {∞}, so that the double coset spaceΓ\SL2(Z)/P identifies with the cuspsΓ\(Q∪ {∞}).
In particular, the double coset spaceΓ(N)\SL2(Z)/P is naturally viewed as SL2(Z/NZ)/P whereP denotes the projected image ofP in SL2(Z/NZ), that is, P = (
±1j
0 1
:j∈Z/NZ)
. Thus double cosets Γ(N)αP with α ∈ SL2(Z) can be viewed as ordinary cosets αP withα∈SL2(Z/NZ). Since P acts from the right by
a b c d
→ ±
a b+ja c d+jc
,
the cosets are represented by pairs±[ac] where the vector is the left column of someα∈SL2(Z/NZ) as before. Similarly, whenΓ =Γ1(N) =
j
1j
0 1
Γ(N), the double coset space is naturally viewed as P+\SL2(Z/NZ)/P where P+=(1j
0 1
:j∈Z)
is the “positive” half ofPand again the overbar signifies reduction moduloN. Since1j
0 1
[ac∗∗] = [a+jcc ∗∗] this recovers the description of the cusps ofΓ1(N). The double coset decompositionΓ0(N)\SL2(Z)/P re- covers the cusps ofΓ0(N) in the same way. See Section 1.6 of [Shi73] for an elegant enumeration of the cusps ofΓ0(N) using double cosets.
In Chapter 5 of this book double cosets will define the Hecke operators mentioned in the preface. For future reference in Chapter 5 we now state that for any primep,
Γ1(N) 1 0
0p
Γ1(N) =
γ∈M2(Z) :γ≡ 1∗
0p
(modN), detγ=p . (3.16)
Supply details as necessary to the following argument (Exercise 3.8.8(a)).
One containment is clear. For the other containment, let L = Z2 and let L0 = L0(N) = {[xy] ∈ L : y ≡ 0 (modN)}. Then M2(Z) acts on L by left multiplication. Takeγ∈M2(Z) such thatγ≡1∗
0p
(modN) and detγ= p. Becauseγ∈M2(Z) andγ≡[∗ ∗0∗] (modN),γL0⊂L0. Because detγ=pis positive, [L:γL0] = [L:L0][L0:γL0] =N p. By the theory of Abelian groups there exists a basis{u, v} ofLsuch that det[u v] = 1 andγL0=muZ⊕nvZ where 0< m|nandmn=N p. The first column ofγis [a0γ] (modN), but it is alsoγe1(wheree1is the first standard basis vector), an element ofγL0and therefore [00] (modm). Because gcd(aγ, N) = 1, also gcd(m, N) = 1. Because pis prime it follows thatm= 1 and n=N p,
γL0=uZ⊕N pvZ.
The right side has unique supergroups of indexpandN insideL=uZ⊕vZ, so now
L0=uZ⊕N vZ, γL=uZ⊕pvZ, γL0=uZ⊕N pvZ.
Let γ1 = [u v]. The condition u ∈ L0 shows that γ1 ∈ Γ0(N). Let γ2 = 5γ11 0
0p
6−1
γ, an elementa b
c d
of GL+2(Q) with determinant 1 such that as desired,
γ=γ1 1 0
0p
γ2. (3.17)
The conditionγe1 ∈γL0 is au+cpv ∈uZ⊕N pvZ, or a∈Z andc ∈NZ.
The conditionγe2∈γLisub+pvd∈uZ⊕pvZ, orb, d∈Z. Thusγ2∈Γ0(N) as well. Becauseaγ ≡1 (modN), γe1 ≡e1 (modN) and thus (3.17) shows that au1 ≡1 (modN) using only thatγ1, γ2 ∈Γ0(N) whereγ2=a b
c d
. So ifγsatisfies (3.17) forγ1, γ2∈Γ0(N) such that eitherγ1 orγ2 lies in Γ1(N) then both do. To complete the argument it suffices to show that
Γ0(N) 1 0
0p
Γ0(N) =Γ1(N) 1 0
0p
Γ0(N).
For the nontrivial containment, sinceΓ1(N)\Γ0(N) is represented by matrices of the forma k
N d
∈SL2(Z), it suffices to show that for each such matrix there exists a matrixδ∈Γ1(N) such that a k
N d 1 0 0p
Γ0(N) =δ1 0
0p
Γ0(N), or a matrixδ ∈Γ1(N) such that
1 0 0p
−1
δ a k
N d 1 0 0p
∈Γ0(N).
If p | N then δ = dN+1−1
−dN 1
works. If p N then any δ = [N d∗ ∗] where d ≡ 1 (modN) and d ≡ −a(modp) will do. This completes the proof. Exercise 3.8.8(b) requests a similar description of the double coset Γ0(N)1 0
0p
Γ0(N).
Exercises
3.8.1.(a) In the proof of Lemma 3.8.1 show that the fraction (pa+qc)/(ra+tc) is in lowest terms. (A hint for this exercise is at the end of the book.)
(b) Complete the proof of Lemma 3.8.1.
3.8.2.(a) Letdbe a divisor ofN. Show that the number of valuesamoduloN such that gcd(a, d) = 1 is (N/d)φ(d).
(b) Show that
d|N(N/d)φ(d)φ(N/d) = N2
p|N(1−1/p2). (A hint for this exercise is at the end of the book.)
3.8.3.Make a table counting the cusps ofΓ(4), of Γ1(4), and ofΓ0(4).
3.8.4.Confirm the calculations that count the cusps ofΓ0(N).
3.8.5.Find the cusps ofΓ0(12).
3.8.6.Find the cusps ofΓ(10), ofΓ1(10), and ofΓ0(10).
3.8.7.Show that then when kis odd, all cusps ofΓ(4) are regular and only the cusps= 1/2 is irregular forΓ1(4). (A hint for this exercise is at the end of the book.)
3.8.8.(a) Supply details as necessary to the proof of (3.16) in the section.
(b) Analogously to (3.16), describe the double cosetΓ0(N)1 0
0p
Γ0(N) as a subset of M2(Z).