Dimension Formulas
3.7 More on elliptic points
This shows that ντ( ˜f) is even for all τ ∈ H, since ντ( ˜f) = 3νπ(τ)( ˜f) when π(τ) is an elliptic point x3,i, and the cusps are not in H. So there exists a functionf1 onHsuch thatf12= ˜f. Since ˜f is weight-2 invariant underΓ,f1
transforms by the rule (f1[γ]1) =χ(γ)f1for allγ∈Γ, whereχ:Γ −→ {±1} is a character. LetΓ={γ∈Γ :χ(γ) = 1}, a subgroup of index 1 or 2. If the index is 1 thenf1∈ A1(Γ).
If the index is 2 thenΓ =Γ∪Γαwheref[α]1=−f1. The surjection of Riemann surfacesπ:X(Γ)−→X(Γ) pulls back to an inclusion of their fields of meromorphic functions, π∗ :C(X(Γ))−→ C(X(Γ)), where the pullback of a function f ∈ C(X(Γ)) is the function π∗f = f ◦π ∈ C(X(Γ)). After identifying C(X(Γ)) with its image, the extension of function fields has the same degree degree as the original surjection, namely 2, and the Galois group Γ/Γ acts as {±1} on generators of the field extension. So there exists some f2 ∈C(X(Γ)) such thatf2◦α=−f2 and thus (f1f2)[α]1=f1f2, showing thatf1f2∈ A1(Γ).
Whichever off1 orf1f2 lies inA1(Γ), itskth power is a nonzero element ofAk(Γ) as desired.
Exercises
3.6.1.Establish the formula div(f) = 2 div(f) +kdiv(dτ) in the context of the section.
3.6.2.(a) Show that at a pointx=xi the integral part of νx(f) is 12νx(ω) + k/2, a sum of half-integers.
(b) Show that at a pointx=xithe integral part ofνx(f) is 12νx(ω) + (k− 1)/2, a sum of integers.
3.6.3.(a) Justify the calculation that deg(div(f))>2g−2 ifk≥3.
(b) Establish the formula for(div(f)−
ixi−
i(1/2)xi) whenk≥3.
3.6.4.Prove the remaining statements in Theorem 3.6.1.
3.6.5.Show that dim(S3(Γ1(7)) = 1, so Proposition 3.2.2(a) shows that this space is spanned by the functionϕ3(τ) =η(τ)3η(7τ)3.
conjugate isotropy subgroupΓ0(N)α(τ)=αγα−1. An elliptic pointΓ0(N)τ of the modular curveX0(N) = X(Γ0(N)) thus corresponds to a conjugacy class of isotropy subgroups ofΓ0(N),
Γ0(N)τ ←→ {αγα−1:α∈Γ0(N)} whereΓ0(N)τ =γ.
Conjugacy classes of elements are more convenient than conjugacy classes of subgroups, so it is tempting to replace the subgroupγby its generator γin the correspondence. But this is ill defined since each nontrivial isotropy subgroupγis also generated byγ−1. Instead each elliptic point corresponds to a pair of conjugacy classes,
Γ0(N)τ←→(
{αγα−1:α∈Γ0(N)},{αγ−1α−1:α∈Γ0(N)}) , where again Γ0(N)τ = γ. The inverse pair of generators γ and γ−1 are not conjugate inΓ0(N) (Exercise 3.7.1(a)), so the two conjugacy classes are distinct and in fact disjoint (Exercise 3.7.1(b)). Thus the number of period 2 elliptic points is half the number of conjugacy classes of order 4 elements, and similarly for period 3 and order 6.
To correct the factor of a half, introduce the group Γ0±(N) =
a b c d
∈GL2(Z) : a b
c d
≡ ∗ ∗
0∗
(modN)
(where GL2(Z) is the group of invertible 2-by-2 matrices with integer en- tries), similar toΓ0(N) but its elements can have determinant±1. Since the map det :Γ0±(N)−→ {±1} is a surjective homomorphism,Γ0(N) is normal in Γ0±(N) and [Γ0±(N) :Γ0(N)] = 2. Thus for anyγ ∈Γ0(N) the extended conjugacy class
{αγα−1:α∈Γ0±(N)}
lies inΓ0(N). This is the union of two conjugacy classes underΓ0(N) (Exer- cise 3.7.1(c)), so the number of period 2 (or period 3) elliptic points ofΓ0(N) is the number of extended conjugacy classes of order 4 (or order 6) elements ofΓ0(N). The extended conjugacy class of γ under Γ0±(N) is not in general the union of the conjugacy classes of γ and γ−1 under Γ0(N) (see Exer- cise 3.7.1(d)).
Counting the elliptic points ofΓ0(N) by counting these conjugacy classes is done in the same environment as the proof of Proposition 2.3.3. Consider elliptic pointsτ of period 3 with Γ0(N)τ =γ, γ6 =I. Letµ6 =e2πi/6 be the complex sixth root of unity and letA=Z[µ6], a principal ideal domain.
Each extended conjugacy class will correspond to an idealJ of A such that A/J∼=Z/NZas an Abelian group,
{αγα−1:α∈Γ0±(N)} ←→Jγ.
To construct the idealJγ fromγ, note that the latticeL=Z2is anA-module with the multiplicationγ :A×L−→Lgiven by
(a+bµ6)γl= (aI+bγ)l fora, b∈Zandl∈L.
Define L0(N) = {[xy] ∈ L : y ≡ 0 (modN)}, a subgroup of L such that the quotient L/L0(N) is isomorphic to Z/NZ. Then L0(N) is also an A- submodule of L since γ ∈ Γ0(N) (Exercise 3.7.2), making L/L0(N) an A- module. LetJγ be its annihilator,
Jγ = Ann(L/L0(N)) ={a+bµ6∈A: (a+bµ6)γL/L0(N) = 0}. ThusA/Jγ ∼=AγL/L0(N) =L/L0(N)∼=Z/NZas desired.
To see that the ideal Jγ depends only on the extended conjugacy class ofγ, letγ =αγα−1 with α∈Γ0±(N). Then γ is also an order-6 element of Γ0(N), andLis also anA-module associated toγ with multiplication
(a+bµ6)γ l= (aI+bγ)l.
LetLdenoteLas theA-module associated toγin this fashion, and similarly forL0(N). For anya, b∈Zandl∈Lcompute that
α((a+bµ6)γl) =α(aI+bγ)l= (aI+bγ)αl sinceαγ=γα
= (aI+bµ6)γαl.
This shows that multiplication byαgives anA-module isomorphismL−→∼ L, and sinceαL0(N)⊂L0(N) becauseα∈Γ0±(N) this induces an isomorphism of quotientsL/L0(N)−→∼ L/L0(N). Compute that for anyj=a+bµ6∈A,
j∈Jγ ⇐⇒ jγL/L0(N) = 0L/L0(N)
⇐⇒ α(jγL/L0(N)) = 0L/L0(N)
⇐⇒ jγα(L/L0(N)) = 0L/L0(N)
⇐⇒ jγL/L0(N) = 0L/L0(N)
⇐⇒ j∈Jγ,
showing that indeedJγ depends only on the extended conjugacy class.
Conversely to the map from conjugacy classes to ideals, start now from an idealJ ofAwithA/J∼=Z/NZ. By the structure theorem for modules over a principal ideal domain again, there exists a basis (u, v) ofAoverZsuch that (u, N v) is a basis ofJ overZ. Since (u, v) is a basis ofAoverZ,
(µ6u, µ6v) = (ua+vc, ub+vd) for some integersa, b, c, d, or
µ6(u, v) = (u, v)γJ, for someγJ =a b
c d
∈M2(Z).
So the matrixγJhas order 6 and has the same minimal polynomialX2−X+1 asµ6, showing thatγJ ∈SL2(Z) (Exercise 3.7.3(a)). And sinceu∈J andJ is anA-submodule ofA, the formulaµ6u=ua+vcshows thatc≡0 (modN), i.e.,γJ∈Γ0(N). Furthermore, for anyα∈Γ0±(N) consider theZ-basis ofA
(u, v) = (u, v)α.
A calculation (Exercise 3.7.3(b)) shows that (u, N v) is again aZ-basis ofJ. So now
µ6(u, v) =µ6(u, v)α= (u, v)γα= (u, v)α−1γJα,
and so the basis (u, v) ofAand itsΓ0±(N)-translates map back to aΓ0±(N)- conjugacy class of an order-6 elementγJ in Γ0(N). Finally, any twoZ-bases (u, v) and (u, v) ofAsuch that (u, N v) and (u, N v) areZ-bases ofJ satisfy the relation (u, v) = (u, v)αfor some α∈Γ0±(N) (Exercise 3.7.3(c)), so the conjugacy class is the same for any suchZ-basis ofA.
Thus we have a map from extended conjugacy classes to ideals and a map back from ideals to extended conjugacy classes. Suppose the class of γ ∈Γ0(N) maps to the ideal Jγ ofA. The proof of Proposition 2.3.3 shows thatγ=m−10−1
1 1
mfor somem=a b
c d
∈GL2(Z), and this computes out to
γ=
∗ ∗ a2+ac+c2 ab+ad+cd
, a2+ac+c2≡0 (modN).
Let (u, v) = (1, µ6)m, aZ-basis ofA. A calculation using the displayed descrip- tion ofγshows thatuannihilatesL/L0(N) (Exercise 3.7.4), asN vclearly does as well. Thus (u, N v) is aZ-basis of anA-submoduleJof Ann(L/L0(N)) =Jγ
with [A :J] =N = [A :Jγ], showing thatJ =Jγ. This ideal maps back to the extended conjugacy class of the matrix γ such that µ6(u, v) = (u, v)γ. Compute that
µ6(u, v) =µ6(1, µ6)m= (µ6,−1 +µ6)m
= (1, µ6)0−1
1 1
m= (u, v)m−10−1
1 1
m= (u, v)γ, so Jγ maps back to the class of γ as desired. On the other hand, suppose the ideal J of A such that A/J ∼= Z/NZ maps to the extended conjugacy class ofγ∈Γ0(N). This means thatµ6(u, v) = (u, v)γfor someZ-basis (u, v) ofAsuch that (u, N v) is aZ-basis ofJ. Note that (u, v) = (1, µ6)mfor some m=a b
c d
∈GL2(Z). As above,µ6(u, v) = (u, v)m−10−1
1 1
m, showing that γ=m−10−1
1 1
m. Again the displayed description of such aγ shows thatu annihilatesL/L0(N) and consequently J =Jγ. That is, the class of γ maps back toJ.
A virtually identical discussion with A=Z[i] applies to period 2 elliptic points (Exercise 3.7.5(a)), proving
Proposition 3.7.1.The period 2 elliptic points ofΓ0(N)are in bijective cor- respondence with the idealsJ ofZ[i]such that Z[i]/J∼=Z/NZ. The period 3 elliptic points of Γ0(N) are in bijective correspondence with the ideals J of Z[µ6] (whereµ6=e2πi/6) such thatZ[µ6]/J ∼=Z/NZ.
Counting the ideals gives
Corollary 3.7.2.The number of elliptic points forΓ0(N) is ε2(Γ0(N)) =
p|N(1 + −1
p
) if4N,
0 if4|N,
where(−1/p)is±1if p≡ ±1 (mod 4)and is0 ifp= 2, and ε3(Γ0(N)) =
p|N(1 + −3
p
) if9N,
0 if9|N,
where(−3/p)is±1if p≡ ±1 (mod 3)and is0 ifp= 3.
These formulas extend Exercise 2.3.7(c) and Exercise 3.1.4(b,c).
Proof. This is an application of beginning algebraic number theory; see for example Chapter 9 of [IR92] for the results to quote. For period 3, the ring A=Z[µ6] is a principal ideal domain and its maximal ideals are
• for each prime p≡1 (mod 3), two ideals Jp =a+bµ6 and Jp =a+ bµ6 such thatp=JpJp and the quotients A/Jpe and A/Jep are group- isomorphic toZ/peZfor alle∈N,
• for each primep≡ −1 (mod 3), the idealJp=psuch that the quotient A/Jpe is group-isomorphic to (Z/peZ)2 for alle∈N,
• forp= 3, the idealJ3=1+µ6such that3=J32and the quotientA/J3e is group-isomorphic to (Z/3e/2Z)2for evene∈Nand is group-isomorphic toZ/3(e+1)/2Z⊕Z/3(e−1)/2Zfor odde∈N.
The formula forε3(Γ0(N)) now follows from Proposition 3.7.1 and the Chi- nese Remainder Theorem. Counting the period 2 elliptic points is left as Ex- ercise 3.7.5(b), similarly citing the theory of the ringA=Z[i].
The elliptic points ofΓ0(N) can be written down easily now that they are counted. Consider the set of translates inH
1 0 n1
(µ3) : 0≤n < N . The corresponding isotropy subgroup generators [1 0n1]0−1
1 1
−1 0n1
are n −1
n2−n+ 1 1−n
: 0≤n < N .
The number of these that are elements of Γ0(N) is the number of solutions to the congruence x2 −x+ 1 ≡ 0 (modN), and this number is given by the formula for ε3(Γ0(N)) in Corollary 3.7.2 (Exercise 3.7.6(a)). The cosets {Γ0(N) [n1 01] : 0≤n < N} are distinct in the quotient spaceΓ0(N)\SL2(Z), though they do not constitute the entire quotient space, and the corresponding
orbits Γ0(N) [n1 01] (µ3) for such n such that n2 −n+ 1 ≡ 0 (modN) are distinct in X0(N) (Exercise 3.7.6(b)). Thus we have found all the period 3 elliptic points ofΓ0(N) (Exercise 3.7.6(c)),
Γ0(N) n+µ3
n2−n+ 1, n2−n+ 1≡0 (modN). (3.14) Similarly, the period 2 elliptic points are (Exercise 3.7.6(d))
Γ0(N) n+i
n2+ 1, n2+ 1≡0 (modN). (3.15) Exercises
3.7.1.(a) Show that if γgenerates a nontrivial isotropy subgroup in SL2(Z) thenγ andγ−1 are not conjugate in GL+2(Q). (Hints for this exercise are at the end of the book.)
(b) Show that any two conjugacy classes in a group are either equal or disjoint.
(c) Show that theΓ0±(N)-conjugacy class ofγ∈Γ0(N) is the union of the Γ0(N)-conjugacy classes ofγand1 0
0−1
γ1 0
0−1
. Show that ifγ has order 4 or 6 then this union is disjoint.
(d) Let γ = [1 11 2]0−1
1 0
[1 11 2]−1 = 3−2
5−3
, an order-4 element of Γ0(5).
Show thatγ is not conjugate to its inverse inΓ0±(5).
3.7.2.In the context of mapping a matrix conjugacy class to an ideal, show thatL0(N) is anA-submodule ofL.
3.7.3.(a) In the context of mapping an idealJ ofAsuch thatA/J ∼=Z/NZ back to a matrix conjugacy class, retain the notation (u, v) for aZ-basis ofA such that (u, N v) is a Z-basis of J. Show that the matrixγJ ∈M2(Z) such thatµ6(u, v) = (u, v)γJ lies in SL2(Z).
(b) For anyα∈Γ0±(N) consider theZ-basis (u, v) = (u, v)αofA. Show that (u, N v) is again aZ-basis ofJ.
(c) Show that any two such Z-bases (u, v) and (u, v) of A satisfy the relation (u, v) = (u, v)αfor someα∈Γ0±(N).
3.7.4.In the context of checking that the maps between conjugacy classes and ideals invert each other, let γ ∈ Γ0(N) of order 6 be given and define (u, v) = (1, µ6)m as in the section. Show that uγ L ⊂ L0(N), so that u annihilatesL/L0(N). (A hint for this exercise is at the end of the book.) 3.7.5.(a) Similarly to the methods of the section, check that the first half of Proposition 3.7.1 holds.
(b) Prove the first half of Corollary 3.7.2. (A hint for this exercise is at the end of the book.)
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).