Scribe notes for Ken Ribet’s Math 274

January 17, 1996

Scribe: Lawren Smithline, <lawren@math>

Here are some topics to be discussed in this course:

Galois representations and modular forms,

Hecke algebras,

modular curves, and Jacobians (Abelian varieties). This lecture is a brief overview of some connection between these concepts, and also an exercise in name-dropping.

We can describe an an elliptic curve, or the Jacobian of a higher genus curve, or abelian variety using a lattice. For E, the lattice is L = H_{1}(E(C), Z) ↪ C, by the map γ ↦ ∫_{γ} ω.

Weil considered curves over a finite field, k of characteristic p. There is an algebraic definition of L/nL for n ≥ 1, gcd(n, p) = 1.

E[n] = \{P ∈ E(k̄) : nP = 0\} = \frac{1}{n}L/L = L/nL.

For example, let n = ℓ^{ν} for ν ≥ 1, and ℓ a prime different from p.

Weil further considered the limit

E[ℓ^{∞}] = ⋃_{ν=1}^{∞} E[ℓ^{ν}].

Tate oberserved there is a map E[ℓ^{ν}] →^{ℓ} E[ℓ^{ν−1}], and so of course this inverse limit,

lim_{←} E[ℓ^{ν}] = T_{ℓ}(E)

is called the Tate module.

As E[n] is free of rank 2 over Z/nZ, so is T_{ℓ}(E) free of rank 2 over Z_{ℓ}. Also, V_{l}(e) = T_{ℓ}(E) ⊗ Q_{ℓ} is a 2 dimensional vector space over Q_{ℓ}. This is the first example of ℓ-adic étale cohomology. For topological space X, X ↦ H_{ét}^{i}(X/k̄, Q_{ℓ}).

Here are the names of some cool folks: Taniyama Shimura Mumford Tate.

Now, elliptic curve E/Q gets an action of G = Gal(Q̄/Q), and so does E[n]. I.e. σ(P + Q) = σ(P) + σ(Q). So we have a homomorphism ρ : G → Aut(E[n]) = GL_{2}(Z/nZ). Since we have an exact sequence

1 → ker ρ → G → im ρ → 0,

we get a tower of fields

Q̄ → K → Q,

and Gal(K/Q) = im ρ ⊂ GL_{2}(Z/nZ).

And now for something completely different. We can also get to these Galois representations via modular forms. Let k be the weight, such as 2. Let N be the level. The complex vector space S_{k}(N) is the set of cusp forms on Γ_{1}(N), a finite dimensional vector space, namely, the set of holomorphic functions f on ℌ such that

f((az + b)/(cz + d)) = (cz + d)^{k}f(z)

for

( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{1}(N), a, d ≡ 1, c ≡ 0(N).

Such an f has a power series (or Fourier series) expansion in q = exp(2πiz):

f(z) = ∑_{1}^{∞} c_{n}q^{n}.

Here is a famous example observed by Ramanujan, and proved by Mordell using (his) Hecke operators:

q ∏_{1}^{∞}(1 − q^{n})^{24} = ∑_{1}^{∞} τ(n)q^{n},

for Ramanujan’s τ function. Now, τ(n)τ(m) = τ(nm) for gcd(n, m) = 1. Also, there is a recurrence for prime powers. Amusingly, the normalized basis element of S_{12}(1) is

Δ = ∑_{1}^{∞} τ(n) exp(2πinz).

Even more amusingly,

τ(n) ≡ ∑_{d|n} d^{11} (691).

Experience and Shimura have shown that there exist f ∈ S_{k}(N) such that

T_{n}(f) = c_{n} · f

for all n ≥ 1 for some scalars c_{n}, and

f = ∑ c_{n}q^{n}

for the same c_{n}, and that these c_{n} are algebraic integers in a finitely generated number field. That is, [Q(c_{n} : n ≥ 1) : Q] is finite.

How can we study and interpret this? We start with the Hecke ring,

Q[T_{n}] ⊆ End(S_{k}(N)).

Serre in 1968 said there should be Galois representations attached to forms of arbitrary weight. Deligne constructed them. In a broad stroke, one can say that we get between Galois representations and modular forms via Frobenius elements.

Next time, we continue with the semihistorical overview.

January 19, 1996

Scribe: William Stein, <was@math>

Modular Representations and Modular Curves Arithmetic of Modular Forms

Suppose f = ∑_{n=1}^{∞} a_{n}q^{n} is a cusp form in S_{k}(N) which is an eigenform for the Hecke operators. The Mellin transform associates to f the L-function L(f, z) = ∑_{n=1}^{∞} n^{−s}a_{n}. Let K = Q(a_{1}, a_{2}, …). One can show that the a_{n} are algebraic integers and K is a number field. When k = 2, f is associated to f an abelian variety A_{f} over Q of dimension [K : Q], and A_{f} has a K action. (See Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Theorem 7.14.)

Example 1 (Modular Elliptic Curves) If a_{n} ∈ Q for all n, then K = Q and [K : Q] = 1. In this case, A_{f} is a one dimensional abelian variety, which is an elliptic curve, since it has nonzero genus. An elliptic curve arising in this way is called modular.

Definition 1 Elliptic curves E_{1} and E_{2} are isogenous if there is a morphism E_{1} → E_{2} of algebraic groups, which has a finite kernel.

The following conjecture motivates much of the theory.

Conjecture 1 Every elliptic curve over Q is modular, that is, isogenous to a curve constructed in the above way.

For k ≥ 2, Serre and Deligne found a way to associate to f a family of ℓ-adic representations. Let ℓ be a prime number and K be as above. It is well known that

K ⊗_{Q} Q_{ℓ} ≅ ∏_{λ|ℓ} K_{λ}.

One can associate to f a family of representations

ρ_{ℓ,f} : G = Gal(\overline{Q}/Q) → GL(K ⊗_{Q} Q_{ℓ})

unramified at all primes p |̸ℓN. By unramified we mean that for all primes P lying over p, the inertia group of the decomposition group at P is contained in the kernel of ρ. (The decomposition group D_{P} at P is the set of those g ∈ G which fix P and the inertia group is the kernel of the map D_{P} → Gal(O/P), where O is the ring of all algebraic integers.)

Now I_{P} ⊂ D_{P} ⊂ Gal(\overline{Q}/Q) and D_{P} /I_{P} is cyclic, since it is isomorphic to a subgroup of the galois group of a finite extension of finite fields. So D_{P} /I_{P} is generated by a Frobenious automorphism frob_{p} lying over p. We have

tr(ρ_{ℓ,f}(frob_{p})) = a_{p} ∈ K ⊂ K ⊗ Q_{ℓ}

and

det(ρ_{ℓ}) = χ_{ℓ}^{k−1}ε,  (0.1)

where χ_{ℓ} is the ℓth cyclotomic character and ε is a Dirichlet character.

Characters

Let f ∈ S_{k}(N). For all ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(Z) with c ≅ 0 mod N we have

f (\frac{az + b}{cz + d}) = (cz + d)^{k}ε(d)f(z),

where ε : (Z/nZ)^{∗} → C^{∗} is a Dirichlet character mod N. If f is an eigenform for the diamond bracket operator < d >, (so that f| < d >= ε(d)f) then ε actually takes values in K.

Let ϕ_{n} be the mod n cyclotomic character. The map ϕ_{n} : G → (Z/nZ)^{∗} takes g ∈ G to the automorphism induced by g on the nth cyclotomic extension Q(μ_{n}) of Q, where we identify Gal(Q(μ_{n})/Q) with (Z/nZ)^{∗}. The ε appearing in (0.1) is really the composition

G ⟶^{ϕ_{n}} (Z/nZ)^{∗} ⟶^{ε} C^{∗}.

For each positive integer ν we consider the ℓ^{ν}th cyclotomic character on G,

ϕ_{ℓ^{ν}} : G → (Z/ℓ^{ν}Z)^{∗}.

Putting these together give a map

ϕ_{ℓ^{∞}} = lim_{ν^{⟵}} ϕ_{ℓ^{ν}} : G ⟶^{χ_{ℓ}} Z_{ℓ}^{∗}.

Parity Conditions

Let c ∈ Gal(\overline{Q}/Q) be complex conjugation. We have ϕ_{n}(c) = −1, so ε(c) = ε(−1) and χ_{ℓ}(c) = (−1)^{k−1}. Let

( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) .

For f ∈ S_{k}(N),

f(z) = (−1)^{k}ε(−1)f(z),

so (−1)^{k}ε(−1) = 1. Thus,

det(ρ_{ℓ}(c)) = ϵ(−1)(−1)^{k−1} = −1.

The det character is odd so the representation ρ_{ℓ} is odd.

Remark 1 (Vague Question) How can one recognize representations like ρ_{ℓ,f} “in nature”? Mazur and Fontaine have made relevant conjectures. The Shimura-Taniyama conjecture can be reformulated by saying that for any representation ρ_{ℓ,E} comming from an elliptic curve E there is f so that ρ_{ℓ,E} ≅ ρ_{ℓ,f}.

Conjectures of Serre (mod ℓ version)

Suppose f is a modular form, ℓ a rational prime, λ a prime lying over ℓ, and the representation

ρ_{λ,f} : G → GL_{2}(K_{λ})

(constructed by Serre-Deligne) is irreducible. Then ρ_{λ,f} is conjugate to a representation with image in GL_{2}(O_{λ}), where O_{λ} is the ring of integers of K_{λ}. Reducing mod ℓ gives a representation

\overline{ρ}_{λ,f} : G → GL_{2}(F_{λ})

which has a well-defined trace and det, i.e., the det and trace don’t depend on the choice of conjugate used to reduce mod λ. One knows from representation theory that if such a representation is semisimple then it is completely determined by its trace and det. Thus if \overline{ρ}_{λ,f} is irreducible it is unique in the sense that it doesn’t depend on the choice of conjugate.

We have the following conjecture of Serre which remains open.

Conjecture 2 (Serre) All irreducible representation of G over a finite field which are odd, i.e., det(σ(c)) = −1, c complex conjugation, are of the form \overline{ρ}_{λ,f} for some representation ρ_{λ,f} constructed as above.

Example 2 Let E/Q be an elliptic curve and let σ_{ℓ} : G → GL_{2}(F_{ℓ}) be the representation induced by the action of G on the ℓ-torsion of E. Then detσ_{ℓ} = ϕ_{ℓ} is odd and σ_{ℓ} is usually irreducible, so Serre’s conjecture would imply that σ_{ℓ} is modular. From this one can, under Serre’s conjecture, prove that E is modular.

Definition 2 Let σ : G → GL_{2}(F) (F is a finite field) be a represenation of the galois group G. The we say that the representions σ is modular if there is a modular form f, a prime λ, and an embedding F ↪ \overline{F}_{λ} such that σ ≅ \overline{ρ}_{λ,f} over \overline{F}_{λ}.

Wile’s Perspective

Suppose E/Q is an elliptic curve and ρ_{ℓ,E} : G → GL_{2}(Z_{ℓ}) the associated ℓ-adic representation on the Tate module T_{ℓ}. Then by reducing we obtain a mod ℓ representation

\overline{ρ}_{ℓ,E} = σ_{ℓ,E} : G → GL_{2}(F_{ℓ}).

If we can show this is modular for infinitely many ℓ then we will know that E is modular.

Theorem 1 (Langlands and Tunnel) If σ_{2,E} and σ_{3,E} are irreducible, then they are modular.

This is proved by using the fact that GL_{2}(F_{2}) and GL_{2}(F_{3}) are solvable so we may apply “base-change”.

Theorem 2 (Wiles) If ρ is an ℓ-adic representation which is irreducible and modular mod ℓ with ℓ > 2 and certain other reasonable hypothesis are satisfied, then ρ itself is modular.

Scribe: Lawren Smithline, <lawren@math> (Note. These were done about two months after the fact, since the assigned person didn’t. So they are terse. It was a pretty easy lecture.)

Today, we have a limited goal: to explain modular forms as functions on lattices – or elliptic curves. (See Serre’s Course in Arithmetic, or Katz’s paper in the Proceedings of the Antwerp Conference in the Springer LNM series.)

Let the level N = 1. Consider a weight k cusp form f. For τ ∈ ℌ, we have

f(aτ + b/cτ + d) = (cτ + d)^{k}f(τ).

So f(τ + 1) = f(τ). By the map τ ↦ q = exp(2πiτ), we map ℌ to the punctured disc {z : 0 < |z| < 1}. We abuse notation and think of f as a function on this disc. Since f is a cusp form, f extends to 0, and f(0) = 0. So we have a q-expansion

f = ∑_{n=1}^{∞} a_{n}q^{n}.

Lattices inside C

Let L = Zω_{1} ⊕ Zω_{2}. We may assume that ω_{1}/ω_{2} ∈ ℌ. Let ℜ be the set of lattices in C. SL_{2}Z acts on the left of M = {(ω_{1}, ω_{2}) : ω_{1}, ω_{2} ∈ ℌ} by multiplication of the column vector. This action fixes the lattice.

Here is the relation with elliptic curves. A lattice L determines a complex torus C/L. There is a Weierstrass ℘ function on this torus. Consider an elliptic curve E over C. There is a lattice given by the inclusion H_{1}(E(C), Z) ↪ C. Choose a nonzero ω ∈ H^{0}(E, Ω′_{E}). Then γ ∈ H_{1} maps to ∫_{γ} ω ∈ C.

So maybe we should think of ℜ as the set of pairs {(E, ω)}.

We have a map M/C^{×} → ℌ sending (ω_{1}, ω_{2}) to ω_{1}/ω_{2}. Now take the quotient on the left by SL_{2}Z:

ℜ/C^{times} = SL_{2}Z\M/C^{×} ⟶ SL_{2}Z\ℌ.

But this just is the space of elliptic curves over C. So f : ℌ → C which is a modular form and F : M → C satisfying F (λL) = λ^{−k}F (L) amount to the same thing by a simple calculation.

Hecke Operators

Let F be a function on lattices. Define the Hecke operator T_{n} as

T_{n}F (L) = ∑_{L′⊂L,(L:L′)=n} F (L)n^{k−1}.

The essential case on elliptic curves is for n = ℓ a prime. In this case, the L′ correspond to the ℓ + 1 subgroups of order ℓ of (Z/ℓZ)^{2}.

Scribe: William Stein, <was@math>

More On Hecke Operators

We consider modular forms f on Γ_{1}(1) = SL_{2}(Z), that is, holomorphic functions on ℌ ∪ {∞} which satisfy

f(τ) = f(\frac{aτ + b}{cτ + d})(cτ + d)^{−k}

for all ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(Z). Using a Fourier expansion we write

f(τ) = ∑_{n=0}^{∞} a_{n}e^{2πiτn},

and say f is a cusp form if a_{0} = 0. There is a correspondence between modular forms f and lattice functions F satisfying F (λL) = λ^{−k}F (L) given by F (Zτ + Z) = f(τ).

Explicit Description of Sublattices

The nth Hecke operator T_{n} of weight k is defined by

T_{n}(L) = n^{k−1} ∑_{L′⊂L, (L:L′)=n} L′.

What are the L′ explicitly? Note that L/L′ is a group of order n and

L′/nL ⊂ L/nL = (Z/nZ)^{2}.

Write L = Zω_{1} + Zω_{2}, let Y_{2} be the cyclic subgroup of L/L′ generated by ω_{2} and let d = #Y_{2}. Let Y_{1} = (L/L′)/Y_{2}. Y_{1} is generated by the image of ω_{1} so it is a cyclic group of order a = n/d. We want to exhibit a basis of L′. Let ω′_{2} = dω_{2} ∈ L′ and use the fact that Y_{1} is generated by ω_{1} to write aω_{1} = ω′_{1} +bω_{2} for b ∈ Z and ω′_{1} ∈ L′. Since b is only well-defined modulo d we may assume 0 ≤ b ≤ d − 1. Thus

( \begin{matrix}ω′_{1} \\ ω′_{2} \end{matrix} ) = ( \begin{matrix}a & b \\ 0 & d \end{matrix} ) ( \begin{matrix}ω_{1} \\ ω_{2} \end{matrix} )

and the change of basis matrix has determinent ad = n. Since

Zω′_{1} + Zω′_{2} ⊂ L′ ⊂ L = Zω_{1} + Zω_{2}

and (L : Zω′_{1}+Zω′_{2}) = n (since the change of basis matrix has determinent n) and (L : L′) = n we see that L′ = Zω′_{1} + Zω′_{2}.

Thus there is a one-to-one correspondence between sublattices L′ ⊂ L of index n and

matrices ( \begin{matrix}a & b \\ 0 & d \end{matrix} ) with ad = n and 0 ≤ b ≤ d − 1. In particular, when n = p is prime there

p + 1 of these. In general, the number of such sublattices equals the sum of the positive divisors of n.

Action of Hecke Operators on Modular Forms

Now assume f(τ) = ∑_{m=0}^{∞} c_{m}q^{m} is a modular form with corresponding lattice function F. How can we describe the action of the Hecke operator T_{n} on f(τ)? We have

T_{n}F (Zτ + Z) = n^{k−1} ∑_{{a,b,d \atop ab=n} ; 0≤b<d} F ((aτ + b)Z + dZ)

= n^{k−1} ∑ d^{−k}F (\frac{aτ + b}{d}Z + Z)

= n^{k−1} ∑ d^{−k}f(\frac{aτ + b}{d})

= n^{k−1} ∑_{a,d,b,m} d^{−k}c_{m}e^{2πi( \frac{aτ+b}{d})m}

= n^{k−1} ∑_{a,d,m} d^{1−k}c_{m}e^{\frac{2πiamτ}{d}} \frac{1}{d} ∑_{b=0}^{d−1}(e^{\frac{2πim}{d}})^{b}

= n^{k−1} ∑_{m′≥0^{ad=n}} d^{1−k}c_{dm′}e^{2πiam′τ}

= ∑_{ad=n, m′≥0} a^{k−1}c_{dm′}q^{am′}.

In the second to the last expression we let m = dm′, m′ ≥ 0, then used the fact that the sum \frac{1}{d} ∑_{b=0}^{d−1}(e^{\frac{2πim}{d}})^{b} is only nonzero if d|m.

Thus

T_{n}f(q) = ∑_{ad=n, m≥0} a^{k−1}c_{dm}q^{am}

and if μ ≥ 0 then the coefficient of q^{μ} is

∑_{a|n, a|μ} a^{k−1}c_{\frac{nμ}{a^{2}}} .

Remark 2 When k ≥ 1 the coefficients of q^{μ} for all μ belong to the Z-module generated by the c_{m}.

Remark 3 Setting μ = 0 gives the constant coefficient of T_{n}f which is

∑_{a|n} a^{k−1}c_{0} = σ_{k−1}(n)c_{0}.

Thus if f is a cusp form so is T_{n}f. (T_{n}f is holomorphic since its original definition is as a finite sum of holomorphic functions.)

Remark 4 Setting μ = 1 shows that the coefficient of q in T_{n}f is just c_{n}. As an immediate corollary we have the following important result.

Corollary 1 Suppose f is a cusp form for which T_{n}f has 0 as coefficient of q for all n ≥ 1, then f = 0.

Remark 5 When n = p is prime we get an interesting formula for the action of T_{p} on the q-expansion of f. One has

T_{p}f = ∑_{μ≥0} ∑_{a|n, a|μ} a^{k−1}c_{\frac{nμ}{a^{2}}} q^{μ}.

Since n = p is prime either a = 1 or a = p. When a = 1, c_{pμ} occurs in the coefficient of q^{μ} and when a = p, we can write μ = pλ and we get terms p^{k−1}c_{λ} in q^{λp}. Thus

T_{n}f = ∑_{μ≥0} c_{pμ}q^{μ} + p^{k−1} ∑_{λ≥0} c_{λ}q^{pλ}.

January 26, 1996

Scribe: Amod Agashe, <amod@math>

Following the notation of the last few lectures, let M_{k} denote the space of modular forms of weight k for SL_{2}(Z) and S_{k} denote the subspace of cusp forms.

Then we have:

Proposition 1 . M_{k} is a finite dimensional C-vector space and is generated by modular forms having the coefficients of their Fourier expansion in Q.

Sketch of Proof. (For details, refer Serre’s “A course in Arithmetic” or Lang’s “Introduction to modular forms”.) The key ingredient that goes into proving finite dimensionality is the following result, which can be obtained by contour integration:

Let f ∈ M_{k} and let D = {z ∈ C : Im(z) > 0, | z |≥ 1, | Re(z) |≤ 1/2} be the fundamental domain for SL_{2}(Z). Then

∑_{p∈D∪∞} \frac{1}{e_{p}} ord_{p}(f) = \frac{k}{12}

where

e_{p} = \frac{1}{2}#\{γ ∈ SL_{2}(Z) : γp = p\} = \frac{1}{2}#Aut(E_{p})

Here, the latter equality follows from the observation that the category of elliptic curves over C with isogenies is the same as the category of lattices in C upto homothety with maps being multiplication by elemets of C. One can show that the invertible maps that preserve the lattice Z ⊕ Zp are in one-to-one correspondence with the set {γ ∈ SL_{2}(Z) : γp = p} and hence the latter equality.

In particular,

e_{p} = \{ \begin{matrix}2 & if p = i \\ 3 & if p = ^{3} \sqrt{−1} \\ 1 & otherwise \end{matrix}

Using this formula and relating the dimensions of M_{k} and S_{k}, one can show that M_{k} is finite dimensional and also explicitly calculate its dimension.

To get a basis with Fourier coefficients in Q, first observe that M_{k} is generated by the set of Eisenstein series G_{k} for all k. As a function on the complex upper half plane, the Eisenstein series G_{k} for k ∈ Z and k > 1 is given by

G_{k}(τ) = ∑_{(m,n)∈Z^{2}\(0,0)} \frac{1}{(mτ + n)^{k}}

One can then show that

G_{k}(τ) = \frac{1}{2}ζ(1 − k) + ∑_{k=1}^{∞} σ_{k−1}(n)q^{n}

where q = e^{2πiτ}, ζ is the Riemann zeta function and

σ_{k}(n) = ∑_{d|n} d^{k}

There is a theorem due to Euler which states that ζ(1 − k) = −\frac{b_{k}}{k} where b_{k} are the Bernoulli numbers defined by the following power series expansion:

\frac{x}{e^{x} − 1} = ∑_{k=0}^{∞} \frac{b_{k}x^{k}}{k!}

The constant term of G_{k} is thus −b_{k}/2k, which is rational. Thus the Fourier expansion of G_{k} has rational coefficients and thus we have found a basis with rational coefficients.

Next, let V be a subspace of M_{k} which is stable under the action of all the Hecke operators T_{n}. For example, observing that T_{n}(G_{k}) = σ_{k−1}(n)G_{k}, we see that V = C(G_{k}) is one such subspace.

Let T = T(V) = C-algebra generated by the T_{n}’s inside End(V) = C-vector space generated by the T_{n}’s inside End(V). The latter equality of sets follows because the product of two Hecke operators can be expressed as a linear combination of finitely many Hecke operators.

For k > 0, we define a bilinear map T × V → C by

(T, f) ↦ a_{1}(f | T).

Proposition 2 . The induced maps T → Hom(V, C) and V → Hom(T, C) are isomorphisms.

Proof. We first show that the maps are injective. Injectivity of the second map:

f ∈ V ↦ 0

⇒ a_{1}(f | T) = 0 ∀T ∈ T ⇒ a_{1}(f | T_{n}) = 0 ∀n ⇒ a_{n}(f) = 0 ∀n ≥ 1 ⇒ f is constant ⇒ f = 0 if k > 0 Injectivity of the first map:

T ∈ T ↦ 0
⇒ a_{1}(f | T) = 0 ∀f ∈ V
⇒ a_{1}((f | T_{n}) | T) = 0 ∀f ∈ V
⇒ a_{1}((f | T) | T_{n}) = 0 ∀f ∈ V
⇒ a_{n}(f | T) = 0 ∀n > 0, ∀f ∈ V
⇒ f | T = 0 ∀k > 0, ∀f ∈ V
⇒ T = 0

In the fourth line, f is replaced by f | T_{n}. Next observe that V, being a subspace of M_{k}, is finite dimensional. Hence we have from the injectivities of both the above maps that each map is actually an isomorphism.

A map ϕ : M → N of T-modules is said to be T-equivariant if ϕ(T m) = T ϕ(m) ∀m ∈ M.

Proposition 3 . The isomorphisms T ≅ Hom(V, C) and V ≅ Hom(T, C) as defined above are T-equivariant.

Proof. Consider the first map. Here is the T-module structure on Hom(V, C). Given ψ ∈ Hom(V, C), i.e. ψ : V → C, define T ψ : V → C by (T ψ)(f) = ψ(f | T). Let β denote the map T → Hom(V, C). Then given T′ ∈ T and T ∈ T, we have to show that β(T (T′)) = T (β(T′)). Let f ∈ V. Now (β(T T′))(f) = a_{1}(f | T T′), while (T (βT′))(f) = β(T′)(f | T) = a_{1}((f | T) | T′) = a_{1}(f | T T′). Thus (β(T T′))(f) = (T (β(T′)))(f) ∀f ∈ V and we are done. Next consider the second map. We define the T-module structure on Hom(T, C). Given ϕ ∈ Hom(T, C), i.e. ϕ : T → C, define T ϕ : T → C by (T ϕ)(T′) = ϕ(T T′). Let α denote the map V → Hom(T, C). Then given f ∈ V and T ∈ T, we have to show that α(T (f)) = T (α(f)). Let T′ ∈ T. Now

(α(T f))(T′) = a_{1}((f | T) | T′) = a_{1}(f | T T′),

while

(T (αf))(T′) = α(f)(T T′) = a_{1}(f | T T′).

Thus

(α(T f))(T′) = (T (αf))(T′) ∀T′ ∈ T

and we are done.

Definition 3 . An element f of M_{k} is said to be an eigenform if it is an eigenfunction for all the Hecke operators. i.e. f | T_{n} = λ_{n}f for some λ_{n} ∈ C ∀n ≥ 1.

Let f is an eigenform with eigenvalues λ_{n}. Then a_{n}(f) = a_{1}(f | T_{n}) = a_{1}(λ_{n}f) = λ_{n}a_{1}(f) ∀n ≥ 1. Thus if a_{1}(f) = 0, then a_{n}(f) = 0 ∀n ≥ 1. If k > 0 then this implies f = 0. Hence if k > 0, then if f ≠ 0, we can normalize f to \frac{1}{a_{1}(f)}f.

Definition 4 . An eigenform f is said to be normalized if a_{1}(f) = 1.

If f is a normalized eigenform with eigenvalues λ_{n}, then a_{n}(f) = λ_{n} and f | T_{n} = λ_{n}f = a_{n}(f)f.

Again, let α denote the map V → Hom(T, C) induced by the bilinear pairing mentioned earlier. Then if f ∈ V, we have the map α(f) : T → C.

Proposition 4 . Let f be an eigenform. Then f is normalized ⇔ α(f) is a ring homomorphism.

Proof. If f = 0 then the statement is trivial. So assume f ≠ 0. Then as discussed above, a_{1}(f) ≠ 0. Also recall from the same discussion that if f ≠ 0 is an eigenform, then

f | T_{n} = \frac{a_{n}(f)}{a_{1}(f)}f.

For ease of notation, let α_{f} denote the map α(f). So

α_{f}(T_{n}T_{m})
= a_{1}(f | T_{n}T_{m})
= a_{1}((f | T_{n})T_{m})
= a_{m}(f | T_{n})
= a_{m}((a_{n}(f)/a_{1}(f))f)
= a_{m}(f)a_{n}(f)/a_{1}(f).

We have

α_{f}(T_{n})α_{f}(T_{m}) = a_{1}(f | T_{n})a_{1}(f | T_{m}) = a_{n}(f)a_{m}(f).

The following are equivalent:

α_{f} is a ring homomorphism,

α_{f}(T_{n}T_{m}) = α_{f}(T_{n})α_{f}(T_{m}) ∀T_{n}, T_{m}, a_{1}(f) = 1, and f is normalized.

The first implication follows because T is generated by the T_{n}’s.

January 29, 1996

Scribe: János Csirik, <janos@math>

Today we’ll consider questions of rationality and integrality. (References: Serre: A Course in Arithmetic and Lang: Introduction to Modular Forms.) Let S = S_{k} be the space of cusp forms of weight k. Let

S(Q) = S_{k} ∩ Q[[q]]

and

S(Q) ⊇ S(Z) = S_{k} ∩ Z[[q]].

The following fact is easy to prove using explicit formulæ: S_{k} has a C-basis consisting of forms with integral coefficients (see Victor Miller’s construction below).

Recall that for all even k ≥ 4, there is an Eisenstein series

G_{k} = \frac{−b_{k}}{2k} + ∑_{n=1}^{∞} (∑_{d|n} d^{k−1}) q^{n},

which is a modular form of weight k. Renormalize this to obtain

E_{k} = \frac{2k}{−b_{k}} · G_{k} = 1 + · · · .

The first few Bernoulli numbers of even positive index are b_{2} = 1/6, b_{4} = −1/30, b_{6} = 1/42, b_{8} = −1/30, b_{10} = 5/66, b_{12} = −691/2730. The fact that the first four of these have numerator 1 is closely related to the arithmetic of cyclotomic fields.

The modular forms E_{4} and E_{6} have q-expansions with constant terms equal to 1, and all coefficients in Z. The functions E_{4}^{a}E_{6}^{b} with 4a + 6b = k form a basis for M_{k}. It is easy to see that they are modular forms. From the formula that the (weighted) number of zeros of any modular form of weight k is k/12, we deduce that E_{4} has a simple zero at ρ, and E_{6} hasn’t got a zero at ρ. Hence E_{4}^{a}E_{6}^{b} has a simple zero of order a at ρ so these expressions are linearly independent over C. To show that they span M_{k}, consider the following modular form of weight 12:

Δ = (E_{4}^{3} − E_{6}^{2})/1728.

Here the coefficient of q is a simple number: 1. Δ has a simple zero at ∞. Since a cusp form of weight k has k/12 zeros, it follows that (since the weighting e_{∞} is 1), that Δ does not vanish anywhere on ℌ. Therefore S_{k+12} = Δ · M_{k}. Since E_{k}(i∞) = 1 ≠ 0, it follows that M_{k} = E_{k} · C ⊕ S_{k} = E_{k} · C ⊕ Δ · M_{k−12}. Hence dim M_{k} = dim M_{k−12} + 1. Again using the fact that f ∈ M_{k} has k/12 zeros, we quickly deduce that the dimensions of M_{0}, M_{2}, M_{4}, M_{6}, M_{8}, M_{10} are 1,0,1,1,1,1 respectively. (e.g., for k = 4 any modular form must have just a simple zero at ρ. So for any f ∈ M_{4}, it is the case that f(τ) − f(i∞)E_{4}(τ) vanishes at τ = i∞ and hence is identically zero. So E_{4} spans M_{4}.) Thus we have determined dim M_{k} for all k ≥ 0, and it is easy to see that this number is equal to the number of solutions to 4a + 6b = k for a, b ≥ 0. Hence the E_{4}^{a}E_{6}^{b} span M_{k} and therefore we have proved that they form a basis.

The following construction comes from the first page of Victor Miller’s thesis. Let d = dim_{C} S_{k}. Then there exist f_{1}, …, f_{d} ∈ S(Z) such that a_{i}(f_{j}) = δ_{ij} for 1 ≤ i, j ≤ d. To show this, recall that E_{4} ∈ M_{4} and E_{6} ∈ M_{6} have q-expansions with coefficients in Z. Δ ∈ S_{12} has constant coefficient 0, and the coefficient of q is 1. Also Δ ∈ S_{12}(Z), as can be seen for example from the formula

Δ = q ∏_{n=1}^{∞}(1 − q^{n})^{24}.

Now pick a, b ≥ 0 so that 14 ≥ 4a + 6b ≡ k (mod 12), with a = b = 0 when k ≡ 0 (mod 12). Note that then 12d + 6a + 4b = k by our previous result on the dimension of M_{k} (and the fact that the dimension of S_{k} is one less than that for k ≥ 12). Hence the functions

g_{j} = Δ^{j}E_{6}^{2(d−j)+a}E_{4}^{b}

for 1 ≤ j ≤ d will be cusp forms of weight k. By our previous remarks on the coefficients of Δ, E_{6}, E_{4}, we have g_{j} ∈ S_{k}(Z) and

a_{i}(g_{j}) = δ_{ij}

for i ≤ j. A straightforward elimination now yields the f_{1}, …, f_{d} with the stated properties. It is clear that these f_{1}, …, f_{d} are linearly independent over C, hence they form a basis of S_{k}.

If you take T_{1}, …, T_{d} ∈ T = T(S_{k}), they are also linearly independent: for given any linear relation

∑_{i=1}^{d} c_{i}T_{i} = 0,

apply this to f_{j} and look at the first coefficient

0 = a_{1} (f_{j} |∑_{i=1}^{d} c_{i}T_{i} ) = ∑_{i=1}^{d} c_{i}a_{i}(f_{j}) = ∑_{i=1}^{d} c_{i}δ_{ij} = c_{j},

hence the linear relation given in the first place was trivial, so T_{1}, …, T_{d} form a basis for T(S_{k}), since they are linearly independent, and dim_{C} T = dim_{C} V = d.

Let R = Z[… T_{n} …] ⊆ End(S_{k}). Claim 1

R = ⨁_{i=1}^{d} ZT_{i}.

Proof. Since the T_{i} form a basis of T, we have T_{n} = ∑_{i=1}^{d} c_{n_{i}}T_{i} with c_{n_{i}} ∈ C. We need to check that c_{n_{i}} ∈ Z. With the f_{j} as above, consider

a_{n}(f_{j}) = a_{1}(f_{j}|T_{n})
= a_{1} (f_{j} |∑_{i=1}^{d} c_{n_{i}}T_{i} )
= ∑_{i=1}^{d} c_{n_{i}}a_{1}(f_{j}|T_{i})
= ∑_{i=1}^{d} c_{n_{i}}a_{i}(f_{j})
= ∑_{i=1}^{d} c_{n_{i}}δ_{ij}  = c_{n_{j}}.

Hence c_{n_{i}} ∈ Z. ■

R is called the integral Hecke algebra. It is a finite Z-module of rank d. We still have (from the last lecture) a pairing

S(Z) × R → Z
(f, T) ↦ a_{1}(f|T).

Now S(Z) ↪ Hom(R, Z) ≅ Z^{d} by the argument given before. Therefore S(Z) is a free Z-module of finite rank. But it also contains the f_{i}, so S(Z) ≅ Z^{d}.

What is S(Z) as an R-module?

Exercise 1 The map S(Z) ↪ Hom(R, Z) is in fact an isomorphism of T-modules.

Hint. The cokernel is a torsion (in fact finite) group. So if we show it torsion free, we are done.

Theorem 3 The T_{n} are all diagonalizable on S_{k}.

S_{n} supports a Hermitian non-degenerate inner product, the Petersson inner product

(f, g) ↦ ⟨f, g⟩ ∈ C.

We have ⟨f, f⟩ ≥ 0, with equality iff f = 0. Furthermore

⟨f|T_{n}, g⟩ = ⟨f, g|T_{n}⟩,

i.e., T_{n} is self-adjoint with respect to the given inner product.

An operator T is normal if it commutes with T^{∗} (which denotes its Hermitian transpose). Normal operators are diagonalizable (for a proof, refer to Math H110). In our case, T_{n}^{∗} = T_{n}, so this fact applies. It is also true (same proof) that a commuting family of semi-simple (i.e., diagonalizable) operators is simultaneously diagonalizable.

Proof. Put together the above facts. ■ We can also prove that the eigenvalues are real. This depends on the following trick. For

f ≠ 0 consider

a_{n}⟨f, f⟩ = ⟨a_{n}f, f⟩ = ⟨f|T_{n}, f⟩ = ⟨f, f|T_{n}⟩ = ⟨f, a_{n}f⟩ = ā_{n}⟨f, f⟩.

a_{n} ∈ R now follows since f ≠ 0 implies ⟨f, f⟩ ≠ 0.

Exercise 2 The a_{n} are totally real algebraic integers.

Hint. The space S_{k} is stable under the action of Aut(C) “on the coefficients”. Given a cusp form

f = ∑_{n=1}^{∞} c_{n}q^{n}

and some σ ∈ Aut(C), define ^{σ}f = ∑_{n=1}^{∞} σ(c_{n})q^{n}. This function is in S_{k}, since S_{k} has a basis in S(Q), which is fixed by σ. Then f is an eigenform iff ^{σ}f is an eigenform.

We’ll use a lame definition of the Petersson inner product for this section. Let z = x + iy ∈ ℌ. Then we have a volume form y^{−2}dx dy which is invariant under GL_{2}^{+}(R) (the subgroup of the general linear group of matrices of positive determinant). To prove this, note that dz ∧ dz̄ = (dx + i dy) ∧ (dx − i dy) = −2i(dx ∧ dy), and hence

dx dy = dx ∧ dy = \frac{−1}{2i}dz ∧ dz̄

Then for any α = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL_{2}^{+}(R) we can consider the usual action

α : z ↦ \frac{az + b}{cz + d} = \frac{(az + b)(cz̄ + d)}{|cz + d|^{2}}

where the imaginary part of the result is |cz + d|^{−2}(ad − bc)Im z = y|cz + d|^{−2}det(α). As for differentials,

d (\frac{az + b}{cz + d}) = \frac{a(cz + d) dz − (az + b)c dz}{(cz + d)^{2}} = \frac{ad − bc}{(cz + d)^{2}} dz

hence under application of α, dz ∧ dz̄ takes on a factor of

\frac{det(α)}{(cz + d)^{2}} \frac{det(α)}{(cz̄ + d)^{2}} = (\frac{det(α)}{|cz + d|^{2}} )^{2} .

This finally proves that the differential y^{−2}dx dy is invariant under the action of GL_{2}^{+}(R). The formula for the Petersson inner product is

⟨f, g⟩ = ∫_{SL_{2}(Z)\ℌ} (f(z)\overline{g(z)}y^{k}) \frac{dx dy}{y^{2}} .

This could be considered either an integral over the fundamental region or, noting that the integrand is invariant under SL_{2}(Z), an integral over the quotient space. One then checks that for f, g cusp forms, the integral will converge, since they go down exponentially as z tends to infinity. It is then clear that this inner product is Hermitian.

It is not immediately clear that the Hecke operators are self-adjoint.

January 31, 1996

Scribe: William Stein, <was@math>

Modular Curves 0.0.1 Cusp Forms

Recall that if N is a positive integer we define the congruence subgroups Γ(N) ⊂ Γ_{1}(N) ⊂ Γ_{0}(N) by

Γ_{0}(N) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(Z) : c ≡ 0 (mod N)\}
Γ_{1}(N) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(Z) : a ≡ d ≡ 1, c ≡ 0 (mod N)\}
Γ(N) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(Z) : ( \begin{matrix}a & b \\ c & d \end{matrix} ) ≡ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) (mod N)\}

Let Γ be one of the above subgroups. One can give a construction of the space S_{k}(Γ) of cusp forms of weight k for the action of Γ using the language of algebraic geometry. Let X_{Γ} = Γ\ℌ^{∗} be the upper half plane (union the cusps) modulo the action of Γ. Then X_{Γ} can be given the structure of Riemann surface. Furthermore, S_{2}(Γ) = H^{0}(X_{Γ}, Ω^{1}) where Ω^{1} is the sheaf of differential 1-forms on X_{Γ}. This works since an element of H^{0}(X_{Γ}, Ω^{1}) is a differential form f(z)dz, holomorphic on ℌ and the cusps, which is invariant with respect to

the action of Γ. If γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ then

d(γ(z))/dz = (cz + d)^{−2}

so

f(γ(z))d(γ(z)) = f(z)dz

iff f satisfies the modular condition

f(γ(z)) = (cz + d)^{2}f(z).

There is a similiar construction when k > 2.

0.0.2 Modular Curves

SL_{2}(Z)\ℌ parametrizes isomorphism classes of elliptic curves. The other congruence subgroups also give rise to similiar parametrizations. Γ_{0}(N)\ℌ parametrizes pairs (E, C) where E is an elliptic curve and C is a cyclic subgroup of order N, and Γ_{1}(N)\ℌ parametrizes pairs (E, P) where E is an elliptic curve and P is a point of exact order N. Note that one can also give a point of exact order N by giving an injection Z/NZ ↪ E[N] or equivalently an injection μ_{N} ↪ E[N] where μ_{N} denotes the Nth roots of unity. Γ(N)\ℌ parametrizes pairs (E, {α, β}) where {α, β} is a basis for E[N] ≅ (Z/NZ)^{2}.

The above quotients spaces are moduli spaces for the moduli problem of determining equivalence classes of pairs (E+ extra structure).

0.0.3 Classifying Γ(N)-structures

Definition 5 Let S be an arbitrary scheme. An elliptic curve E/S is a proper smooth curve

\begin{matrix}E \\ | \\ S \end{matrix}

with geometrically connected fibers all of genus one, together with a section “0”.

Loosely speaking, proper generalizes the notion of projective, and smooth generalizes nonsingularity. See Chapter III, section 10 of Hartshorne’s Algebraic Geometry for the precise definitions.

Definition 6 Let S be any scheme and E/S an elliptic curve. A Γ(N)-structure on E/S is a group homomorphism

φ : (Z/NZ)^{2} → E[N](S)

whose image “generates” E[N](S).

See Katz and Mazur, Arithmetic Moduli of Elliptic Curves, 1985, Princeton University Press, especially chapter 3.

Define a functor from the category of Q-schemes to the category of sets by sending a scheme S to the set of isomorphism classes of pairs

(E, Γ(N)-structure)

where E is an elliptic curve defined over S and isomorphisms (preserving the Γ(N)-structure) are taken over S. An isomorphism preserves the Γ(N)-structure if it takes the two distinguished generators to the two distinguished generators in the image (in the correct order).

Theorem 4 For N ≥ 4 the functor defined above is representable and the object representing it is the modular curve X(N) corresponding to Γ(N).

What this means is that given a Q-scheme S, the set X(S) = Mor_{Q-schemes}(S, X) is isomorphic to the image of the functor’s value on S.

There is a natural way to map a pair (E, Γ(N)-structure) to an Nth root of unity. If P, Q are the distinguished basis of E[N] we send the pair (E, Γ(N)-structure) to

e_{N}(P, Q) ∈ μ_{N}

where

e_{N} : E[N] × E[N] → μ_{N}

is the Weil pairing. For the definition of this pairing see chapter III, section 8 of Silverman’s The Arithmetic of Elliptic Curves. The Weil pairing is bilinear, alternating, non-degenerate, galois invariant, and maps surjectively onto μ_{N}.

February 2, 1996

Scribe: Lawren Smithline, <lawren@math> (Note. These were done about a month after the fact, since the assigned person dropped the course. So they are terse.)

Earlier, we looked at V. Miller’s construction for eigenforms. (See, for instance, Lang X §4 in the course references. This is a special miracle for SL_{2}(Z).) Over Z, T ≅ Z^{d} and S(Z) are dual, where d = dim S_{k}(C).

Here is Shimura’s explanation (Lang III §5, VIII). The Hecke operator T_{n} maps S_{k} to itself. Let A ⊂ C be a subring and

T_{A} = A[T_{n} : n > 0] ⊆ End_{C}S_{k}.

Denote by T, T_{Z}. There is a natural tensor product T_{A} ⊗_{A} C ↠ T_{C}.

There is also a complex conjugation automorphism of S_{k} by f ↦ \overline{f(−τ̄)}. This map is conjugate linear. The map τ ↦ exp(2πiτ) becomes τ ↦ \overline{exp(−2πiτ̄)} = exp(2πiτ). Say f = ∑ a_{n}q^{n}. Its conjugate g = ∑ ā_{n}q^{n}. If you know S_{k}(C) = C ⊗_{Q} S_{k}(Q), then you know that modular forms can be conjugated in this sense.

There is an isomorphism T_{R} ⊗_{R} C ⟶^{∼} T_{C} since the map is surjective and the complex dimensions on each side are equal.

Shimura (1959) exhibited the (Eichler-)Shimura isomorphism

S_{k}(C) ≅ H^{1}(X_{Γ}, R).

We have S_{k}(C) = H^{0}(X_{Γ}, Ω^{1}) and a map H_{1}(X_{Γ}, Z) × S_{k}(C) → C. Now, H_{1}(X, Z) ≅ Z^{2d} embeds in Hom_{C}(S_{k}(C), C) ≅ C^{2d} as a lattice.

So we have S_{k}(C) → Hom(H_{1}(X, Z), C)) and S_{k}(C) →^{∼} Hom(H_{1}(X, Z), R)) as real vector spaces. By the Shimura isomorphism, this is isomorphic to H^{1}(X, R) ∼ H_{p}^{1}(Γ, R), the parabolic cohomology of Γ.

So S_{k}(C) ≅ H_{p}^{1}(Γ, V_{k}), for a certain d − 1 dimensional subspace V_{k}. (Let W = R ⊕ R. Γ acts by linear fractional transformation. Let V_{k} = Sym^{k−2}W. There is a lattice in S_{k}(C) corresponding to H_{p}^{1}(Γ, Sym^{k−2}Z^{2})) We have an action of Γ by

f · γ ↦ ∫_{τ_{0}}^{γ(τ_{0})} f(τ)τ^{k−1}dτ.

Recall T = T_{Z} is a set of endomorphisms of a lattice L, and T has finite rank over Z. We have the inclusion S_{k}(Z) ↪ S_{k}(C), or equivalently,

Hom_{Z}(T, Z) ↪ Hom_{Z}(T, C) = Hom(T_{C}, C) = S_{k}(C) = S_{k}(Z) ⊗_{Z} C.

Here is a nifty inner product (the Petersson innner product) on S_{k}(C). For f, g ∈ S_{k}(C), let

⟨f, g⟩ = ∫_{Γ\ℌ} f(τ)g(τ)y^{k} \frac{dx dy}{y^{2}} .

The Hecke operators are self-adjoint for (p, N) = 1:

⟨f|T_{p}, g⟩ = ⟨f, g|T_{p}⟩.

Indeed, for α ∈ GL_{2}^{+}(R),

⟨f|α, g|α⟩ = ⟨f, g⟩.

February 5, 1996

Scribe: Shuzo Takahashi, <shuzo@math>

We have studied actions of T on S_{k}(C), S_{k}(Q) and S_{k}(Z) where Γ = SL_{2}(Z). What we know so far is

S_{k}(Z) ≃ Hom_{Z}(T, Z).

Also we have studied the Peterson product. It is Hermitian, i.e.,

⟨f|T_{n}, g⟩ = ⟨f, g|T_{n}⟩

for T_{n} ∈ T and for f, g ∈ S_{k}(C). Note: T_{n} defined on S_{k}(C) preserves S_{k}(Z).

Today we study when Γ = Γ_{1}(N) or Γ_{0}(N) for N ≥ 1.

1. The Diamond Operator and the Decomposition of S_{k}(Γ_{1}(N))

Theorem 5 Γ_{1}(N) is a normal subgroup of Γ_{0}(N) and we have Γ_{0}(N)/Γ_{1}(N) ≃ (Z/nZ)^{∗}.

Definition 7 The diamond operator <> is defined as follows: for ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(N), the

map on S_{k}(Γ_{1}(N))

f → f| ( \begin{matrix}a & b \\ c & d \end{matrix} )

defines an endmorphism < d > of S_{k}(Γ_{1}(N)) which depends only on d mod N. Thus we get an action <> of (Z/nZ)^{∗} on S_{k}(Γ_{1}(N)).

Since (Z/nZ)^{∗} is a finite group, we have the following decomposition theorem:

Theorem 6

S_{k}(Γ_{1}(N)) = ⨁_{ϵ} S_{k}(Γ_{0}(N), ϵ)

where ϵ runs over the set of characters (Z/nZ)^{∗} → C^{∗} and S_{k}(Γ_{0}(N), ϵ) is defined as:

S_{k}(Γ_{0}(N), ϵ) = {f ∈ S_{k}(Γ_{1}(N)) : f| < d >= ϵ(d)f}.

We have S_{k}(Γ_{0}(N), ϵ) = 0 unless ϵ(−1) = (−1)^{k}.

2. The Hecke Operators on S_{k}(Γ_{1}(N))

For n ≥ 1, we have the operation on S_{k}(Γ_{1}(N)) of the nth Hecke operator T_{n}. The following are basic properties:

Theorem 7 (1) T_{n}’s commute each other and with < d >.

(2) T_{n}’s preserve S_{k}(Γ_{0}(N), ϵ).

(3) if (n, N) = 1, then ⟨f|T_{n}, g⟩ = ⟨f, g|< n >^{−1}T_{n}⟩.

(4) ⟨f| < d >, g⟩ = ⟨f, g|< d >^{−1}⟩.

(5) if (n, N) = 1, then T_{n} is diagonalizable.

(6) if (n, N) ≠ 1, then T_{n} is not diagonalizable.

The action of T_{n} is described in the following theorem:

Theorem 8 Let f = ∑_{n=1}^{∞} a_{n}q^{n} ∈ S_{k}(Γ_{0}(N), ϵ). Then

(1) f|T_{p} = ∑_{n=1}^{∞} a_{pn}q^{n} + p^{k−1}ϵ(p) ∑_{n=1}^{∞} a_{n}q^{pn}. (Note: when p|N, ϵ(p) is considered to be 0 and U_{p} is used instead of T_{p} which is called Atkin-Lehner operator.)

(2) if (n, m) = 1, then T_{nm} = T_{n}T_{m}.

(3) if p |̸N, then T_{p^{l}} = T_{p^{l−1}}T_{p} − p^{k−1} < p > T_{p^{l−2}}.

(4) if p|N, then T_{p^{l}} = (T_{p})^{l}.

The last formula in the theorem can be proved by comparing the coefficients of q^{p^{l−1}} in both sides of the following formal identity:

(∑_{n=1}^{∞} T_{n}q^{n}) |T_{p} = ∑_{n=1}^{∞} T_{pn}q^{n} + p^{k−1} < p > ∑_{n=1}^{∞} T_{n}q^{pn}.

For example, the coefficient of q^{p} in the LHS is T_{p}T_{p}. On the other hand, the coefficient of q^{p} in the RHS is T_{p^{2}} + p^{k−1} < p > T_{1}. Thus, we have

T_{p^{2}} = (T_{p})^{2} − p^{k−1} < p > Id

where < p > should be considered to be a null map if p|N.

3. The Old Forms

Suppose M|N. Let f ∈ S_{k}(Γ_{1}(M)). Then for d such that d|\frac{N}{M}, f(dτ) ∈ S_{k}(Γ_{1}(N)). Thus we have a map

ϕ_{M} : ⨁_{d| \frac{N}{M}} S_{k}(Γ_{1}(M)) → S_{k}(Γ_{1}(N)).

The old part of S_{k}(Γ_{1}(N)) is defined as the subspace generated by the images of ϕ_{M} for M|N, M ≠ N.

Example 3 ϕ_{M} is not injective. Consider the case that k = 12, M = p and N = p^{2}. S_{k}(Γ_{1}(p)) contains Δ(τ) and Δ(pτ). But ϕ_{p} maps both of them to Δ(pτ) in S_{k}(Γ_{1}(p^{2})).

Theorem 9 Suppose p ∤ N. Consider f, g ∈ S_{k}(Γ_{1}(N)). Then f and g(pτ) are both in S_{k}(Γ_{1}(Np)). Then we have

f|U_{p} = (f|T_{p}) − p^{k−1}ϵ(p)(f(pτ))

and

g(pτ)|U_{p} = g(τ)

where f|T_{p} is considered in S_{k}(Γ_{1}(N)).

Proof. Let f = ∑_{n=1}^{∞} a_{n}q^{n}. Then, considering in S_{k}(Γ_{1}(N)), we have

f|T_{p} = ∑_{n=1}^{∞} a_{pn}q^{n} + p^{k−1}ϵ(p) ∑_{n=1}^{∞} a_{n}q^{pn}.

Also, considering in S_{k}(Γ_{1}(Np)), we have

f|U_{p} = ∑_{n=1}^{∞} a_{pn}q^{n}.

Thus, we have

f|U_{p} = (f|T_{p}) − p^{k−1}ϵ(p)(f(pτ)).

Now, let g = ∑_{n=1}^{∞} b_{n}q^{n}. Then

g(pτ)|U_{p} = (∑_{n=1}^{∞} b_{n/p}q^{n}) |U_{p} = g(τ)

where b_{n/p} = 0 unless p|n.

February 7, 1996

Scribe: Amod Agashe, <amod@math>

We are in the process of showing that the Hecke operators T_{p} acting on the space of cusp forms S_{k}(Γ_{1}(N)) are not necessarily semisimple if p | N.

Recall from last time that if M | N then for every divisor d of M/N, we had a map S_{k}(Γ_{1}(M)) → S_{k}(Γ_{1}(N)) given by f(τ) ↦ f(dτ).

Note that the various f(dτ)’s are linearly independent over C, because the Fourier expansion of f(dτ) starts with q^{d}.

Let f be an eigenfuntion for all the Hecke operators T_{n} in S_{k}(Γ_{1}(M)). Let p be a prime not dividing M. So f | T_{p} = af where a = a_{p}(f) and f |< p >= ϵ(p)f where ϵ(p) is the character associated to the modular form f. Note that one can prove that if f is an eigenfunction for the T_{n}’s then it is an eigenfunction for the diamond operators also (or alternatively, make it part of the definition of eigenform). Let N = p^{α}M with α ≥ 1. We will look at the action of the p^{th} Hecke operator U_{p} in S_{k}(Γ_{1}(N)) on the images of f under the maps described above. Let f_{i}(τ) = f(p^{i}τ) for 0 ≤ i ≤ α. As we showed earlier,

f | T_{p} = ∑ a_{np}q^{n} + ϵ(p)p^{k−1} ∑ a_{n}q^{pn}.

So

af = f_{0} | U_{p} + ϵ(p)p^{k−1}f_{1}.

Thus,

f_{0} | U_{p} = af_{0} − ϵ(p)p^{k−1}f_{1}.

From last time, we have f_{1} | U_{p} = f_{0}. In fact, in general, one can see easily that f_{i} | U_{p} = f_{i−1} for i ≥ 1.

So U_{p} preserves the 2-dimensional space spanned by f_{0} and f_{1}. The matrix of U_{p} (acting on the right) with this basis is given (from the equations above) by:

( \begin{matrix}a & 1 \\ −ϵ(p)p^{k−1} & 0 \end{matrix} )

The characteristic polynomial of this matrix is x^{2} − ax + p^{k−1}ϵ(p).

There is the following striking coincidence: Let E be the number field generated over Q by the coefficients of the Fourier series expansion of f and let λ be a prime ideal of O_{E} lying over some rational prime l. Then we have a Galois representation

ρ_{λ} : Gal(\overline{Q}/Q) → GL_{2}(E_{λ})

If p |̸Nl then ρ_{λ} is unramified and also det ρ_{λ}(F rob_{p}) = ϵ(p)p^{k−1} and tr ρ_{λ}(F rob_{p}) = a_{p}(f) = a. Thus the characteristic polynomial of ρ_{λ}(F rob_{p}) is x^{2} − ax + p^{k−1}ϵ(p), the same as that of the matrix of U_{p}!

A question one can ask is: Is U_{p} semisimple on the space spanned by f_{0} and f_{1}? The answer is yes if the eigenvalues of U_{p} are different.

Now, the eigenvalues are the same iff the discriminant of the characteristic polynomial is zero i.e. a^{2} = 4ϵ(p)p^{k−1} i.e. a = 2p^{\frac{k−1}{2}} ζ where ζ is some square root of ϵ(p). Here is a curious fact: the Ramanujan-Petersson conjecture proved by Deligne says |a| ≤ 2p^{\frac{k−1}{2}} ; thus the above equality is allowed by it, so we do not get any conclusion about the semisimplicity of U_{p}.

Let us now specialize to k = 2. Weil has shown that ρ_{λ}(F rob_{p}) is semisimple. Thus if the eigenvalues of U_{p} are equal, then ρ_{λ}(F rob_{p}) is a scalar. Edixhoven proved that it is not. So the eigenvalues of U_{p} are different and hence U_{p} is semisimple in this case. So this example (for k=2) does not give us an example of U_{p} being not semisimple.

There is the following example given by Shimura which shows that the Hecke operator U_{p} need not be semisimple. Let W denote the space spanned by f_{0}, f_{1} and V denote the space spanned by f_{0}, f_{1}, f_{2}, f_{3}. U_{p} preserves both spaces W and V, so it acts on V/W. The action is given by \overline{f_{2}} ↦ \overline{f_{1}} = 0 and \overline{f_{3}} ↦ \overline{f_{2}} where the bar denotes the image in V/W. Thus the matrix of U_{p} on the space V/W is

( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} )

which is nilpotent, and in particular not semisimple. If U_{p} were semisimple on V then it would be semisimple on V/W also; but we have just shown that it is not. Thus U_{p} is not semisimple on V, and hence not on S_{2}(Γ_{1}(M)) (because V is invariant under U_{p}).

We next discuss the structure of the C-algebra T = T_{C} generated by the Hecke and diamond operators and the structure of S_{k}(Γ_{1}(N)) as a T-module.

First we consider the case of level 1 i.e. N = 1. Then Γ_{1}(1) = SL_{2}(Z). All the T_{n}’s are diagonalizable. S_{k} = S_{k}(Γ_{1}(N)) has a basis of f_{1}, ...., f_{d} of normalized eigenforms where d = dim(S_{k}). Thus S_{k} ≅ C^{d} as a C-vector space. Then we have the C-algebra homomorphism T → C^{d} given by T ↦ (λ_{1}, ...., λ_{d}) where f_{i} | T = λ_{i}f_{i}. It is injective because if the image of T is zero, then it kills all f_{i} i.e. all of S_{k} i.e. it is the zero operator. The map is surjective because T has dimension d. Thus as a C-algbebra, T ≅ C^{d}. Next, we claim that the modular form v = f_{1} + ... + f_{d} generates S_{k} as a T-module. This follows because under the map S_{k} ≅ C^{d}, v ↦ (1, ...., 1) and our statement is just the trivial fact that (1, ...., 1) generates C^{d} as a C^{d}-module (acting component-wise).

Thus S_{k} is free of rank 1 as a T-module. We already know that S_{k} ≅ Hom(T, C) as T-modules. Thus T ≅ Hom(T, C) as T-modules. In fact the isomorphism is canonical since the f_{i}’s are normalized. We remark that v in fact lies in S_{k}(Q).

Next, we deal with the general case where the level is not necessarily 1.

First we need to talk about newforms. Recall the maps S_{k}(Γ_{1}(M)) → S_{k}(Γ_{1}(N)) for every divisor d of M/N mentioned at the beginning of this lecture. The old part of S_{k}(Γ_{1}(N)) is defined as the space generated by all the images of S_{k}(Γ_{1}(M)) for all M | N, M ≠ N under these maps. The new part of S_{k}(Γ_{1}(N)) can be defined in two different ways. Firstly we can define it as the orthogonal complement of the old part with respect to the Petersson inner product. There is also an algbraic definition as follows. There are certain maps going the other way: S_{k}(Γ_{1}(N)) → S_{k}(Γ_{1}(M)) for M | N, M ≠ N. The new part is the space killed under all these maps. The space of newforms, denoted S_{k}(Γ_{1}(N))_{new} is like S_{k}(Γ(1)) in the sense that all the T_{n}’s (including U_{p}) are semisimple and there is a basis consisting of newforms. A form of level N is said to be new of level N if it is in S_{k}(Γ_{1}(N))_{new}.

Next, one can show that the map ⨁_{M|N,M≤N} S_{k}(Γ_{1}(M))_{new} → S_{k}(Γ_{1}(N)) given by f(τ) ↦ f(dτ) for d | \frac{N}{M} is injective (See W.-C. W.Li, Newforms and functional equations, Math. Annalen, 212(1975), 285-315). Note that an eigenform in one of the subspaces of the source need not be an eigenfuntion for all the operators in the image. If f is a newform, then let M_{f} denote its level (i.e. f is new of level M_{f}). Let S be the set of newforms of weight k and some level dividing N. Let

v = ∑_{f∈S} f(\frac{N}{M_{f}} τ).

Then one can show that S_{k}(Γ_{1}(N)) is free of rank 1 over T_{C} with v as the basis element. Also one can show that v has rational coefficients.

February 9, 1996

Scribe: János Csirik, <janos@math>

Final comments about Hecke algebras

Recall that for the case Γ = SL(2, Z), if we set f_{1}, …, f_{d} to be the normalized eigenforms (newforms of level 1), then they have possibly complex coefficients but in any case {f_{i}} is finite and stable under automorphisms of C and all the coefficients a_{n}(f_{i}) lie in some number field. Furthermore this field is totally real: to show this we used that since the set is stable under conjugation, it suffices to show that all the a_{n}(f_{i}) are real, which followed by remarking that they are eigenvalues of the operators T_{n} which are self-adjoint with respect to the Petersson inner product.

More generally, let f ∈ S(Γ_{1}(N)) be a normalized eigenform of character ε. Then a_{n} = ε(n)\overline{a_{n}}.

Note that the algebra T_{Q} generated by the T_{i} over Q contains the diamond bracket operators: the formula relating T_{p^{2}} and (T_{p})^{2} tells us that the difference is p^{k−1}ε(p), so ε(p) ∈ T_{Q}. Using Dirichlet’s Theorem on primes in arithmetic progressions, for any d relatively prime to N we can find a prime p ≡ d (mod N), so ε(d) = ε(p) ∈ T_{Q}.

If the space of modular forms has dimension 1, then it is spanned by a (normalized) eigenform with rational coefficients, so the eigenvalues are all in Q.

The next simplest example is k = 24, which is the smallest weight such that the dimension is more than one. There are two eigenforms, which are conjugate to each other. If f = ∑ a_{n}q^{n} is an eigenform, then Q(… a_{n} …) = Q(\sqrt{144169}). In fact S_{24} is spanned by Δ^{2} and Δ^{2}|T_{2}. (Note that Δ^{2} is definitely not an eigenform since its q-coefficient is 0.) The action of T_{2} on S_{24} with respect to this basis is described by a two by two matrix of trace 1080 and determinant −2^{10}3^{2}2221. For high k, eigenforms tend to form a single orbit; however, no proof is known for this.

For every newform f, let E_{f} be the number field generated by its coefficients. Let Σ be a set of representatives for f’s modulo Gal(Q̄/Q). Define

T_{Q} → E_{f}
T ↦ λ_{T} ,

with f|T = λ_{T} f. Thus T_{n} ↦ a_{n} so this map is surjective.

In fact the induced

T_{Q} → ∏_{f∈Σ} E_{f}

is an isomorphism of Q-algebras: it is injective since if T dies on the image then it acts as zero on f ∈ Σ and so on all of f, by the rationality of Hecke operators: ^{σ}(g|T) = ^{σ}g|T. (And if an operator acts as zero on everything, then it is zero.)

Here we used the fact that there were no oldforms around.

For example, consider S_{2}(Γ(N)) for N prime. Then S_{2}(Γ(1)) is empty, hence we get an isomorphism

T_{Q} ≅ E_{1} × … × E_{t},

with the right hand side a product of totally real number fields. t > 1 is possible, e.g. for N = 37, T_{Q} = Q × Q.

In general, oldforms complicate the situation.

Final comments about Hecke algebras

We’ll only treat the case k = 2. Then (for Γ a congruence subgroup),

S_{2}(Γ) = H^{0}(X_{Γ}, Ω^{1})

where X_{Γ} = Γ \ ℌ ∪ Γ \ P^{1}(Q), with Γ \ P^{1}(Q) being the set of cusps that we need to adjoin to make it a compact Riemann surface.

Therefore

dim S_{2}(Γ) = g(X_{Γ}).

Example. SL(2, Z) \ P^{1}(Q) has just one point. The proof involves the Euclidean

algorithm: any element of P^{1}(Q) can be written as ( \begin{matrix}x \\ y \end{matrix} ) with x and y relatively prime

integers. By the Euclidean Algorithm, we can find a and b integers such that ax + by = 1. Then

( \begin{matrix}a & b \\ −y & x \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} ) = ( \begin{matrix}1 \\ 0 \end{matrix} )

To calculate g(X_{Γ}), use the following covering (recall that X_{Γ} is the compactification of Γ \ ℌ we obtain by adjoining the cusps)

X_{Γ} → X_{Γ(1)},

keeping in mind the isomorphism

j : X_{Γ(1)} → P^{1}(C)
(i, ρ, ∞) ↦ (1728, 0, ∞)

The only ramification in our covering occurs above the points 0, 1728, ∞.

Example. Let Γ = Γ_{0}(N). The degree of the covering is (P SL(2, Z) : Γ_{0}(N)/ ± 1) which is the number of cyclic subgroups of order N in SL(2, Z/NZ). We have a covering Y_{0}(N) → Y_{Γ(1)} where Y_{0}(N) parametrizes elliptic curves E with a cyclic subgroup of order N, C ⊆ E[N] ≅ Z/NZ × Z/nZ up to isomorphism; and Y_{Γ(1)} parametrises elliptic curves.

The isomorphism (E, C_{1}) ≅ (E, C_{2}) is an automorphism α of E with α : C_{1} ↦ C_{2}. Usually α = ±1, unless j = 1728, 0, ∞.

If we understand ramification, we can use the Riemann-Hurwitz formula. The following mnemonic way of thinking about it is due to N. Katz. The Euler characteristic (alternating sum of the dimensions of cohomology groups) should be thought of as totally additive:

χ(A ∐ B) = χ(A) + χ(B).

If X is a Riemann surface of genus G, χ(X) = 2 − 2G. A single point has Euler characteristic χ(P) = 1. Hence

χ(X \ {P_{1}, …, P_{n}}) = 2 − 2G − n.

Therefore

χ(X_{Γ(1)} \ {0, 1728, ∞}) = −1

and

χ(X_{Γ} \ {points over 1728, 0, ∞}) = 2 − 2g − n

if n points lie over 0, 1728, ∞. If the covering map X_{Γ} → X_{Γ(1)} has degree d, then we can think of the top space as d copies of the bottom space, so

d · (−1) = 2 − 2g − n

and therefore

2g − 2 = d − n = d − n_{0} − n_{1728} − n_{∞}.

Example. Let Γ = Γ(N). What happens over j = 0? This corresponds to an elliptic curve E with an automorphism α of order three. Let N > 3.

For any (E, P, Q), we have (E, αP, αQ) and (E, α^{2}P, α^{2}Q) which are isomorphic to it and hence they are the same point on X_{Γ}. So E only has one-third the usual number of points lying over it, n_{0} = d/3. (Except if the above three points are equal, i.e., α fixes E[N]. This can’t happen since N > 3.)

Similarly we get n_{1728} = d/2.

To determine the degree d, fix E and count the points lying over it: these are all of the form (C/Z ⊕ τZ, 1/n, τ/N) with Weil pairing e^{2πi/N} and all such occur, so we need to count the number of P, Q ∈ E[N] which form a basis of E[N] and e_{N}⟨P, Q⟩ = e^{2πi/N}. This gives us the order of SL(2, Z/nZ). However, (since N ≠ 2) we have to take into account that (E, P, Q) ≅ (E, −P, −Q) but they are not equal so the degree of the covering is #(SL(2, Z/NZ))/2.

We also have d = (P SL(2, Z) : Γ(N)/(Γ(N)∩±1)) since Γ(N)\SL(2, Z) ≅ SL(2, Z/NZ). So we have established that

2g − 2 = d − n_{0} − n_{1728} − n_{∞} = d/6 − n_{∞}.

To determine n_{∞}, note that SL(2, Z) acts on P^{1}(Q) ∋ ∞ = ( \begin{matrix}1 \\ 0 \end{matrix} ). The stabilizer Γ(N)_{∞}

of ( \begin{matrix}1 \\ 0 \end{matrix} ) is U = ± ( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} ) So the index ((SL_{2}(Z)/ ± 1)_{∞} : Γ(N)_{∞}) = N is the ramification

degree of a point over ∞, so n_{∞} = N/d.

Hence 2g(X(N)) − 2 = d/6 − d/N.

February 12, 1996

Scribe: János Csirik, <janos@math>

Plug: A useful reference for the next lecture is Andrew Ogg: Rational points on certain elliptic modular curves (1972).

From last lecture’s results on easily deduces that

g(X(N)) = 1 + \frac{d}{12N} (N − 6).

Example. Let N = 5, 7. If N is prime then the degree of the covering is \frac{(N^{2}−1)(N^{2}−N)}{N−1}. Therefore N = 5 gives d = 60 and g = 0 (the Galois group of the covering in this case is A_{5}.) Similarly, N = 7 yields g = 3 and d = 168. (Remark: For p ≥ 5 prime, the group SL(2, Z/pZ)/ ± 1 = L_{2}(p) is simple.)

Example. What is the genus of X_{0}(N), for N a prime?

It is an exercise to show that there are two cusps: ( \begin{matrix}1 \\ 0 \end{matrix} ) = ∞ and ( \begin{matrix}0 \\ 1 \end{matrix} ) = 0. The

covering X_{0}(N) → X(1) is easily seen to have degree N + 1, since an elliptic curve has N + 1 cyclic subgroups of order N. Here ∞ is unramified and 0 has ramification index N. Hence

2g − 2 = N + 1 − n_{∞} − n_{1728} − n_{0}

with n_{∞} = 2, n_{1728} approximately d/2 and n_{0} approximately d/3.

To calculate n_{0}, we need the number of isomorphism classes (E, C) with E fixed with End(E) = Z[μ_{6}]. μ_{6} acts on the set of C’s. Since ±1 is acting trivially, we really have an action of μ_{3} with (E, C) ≅ (E, ζC) ≅ (E, ζ^{2}C) with ζ some third root of unity.

If we consider E = C/O with O = Z[(−1+i\sqrt{3})/2], then E[N] = O/NO as an O-module. In fact

O/NO = \{ \begin{matrix}F_{N} ⊕ F_{N} & if N splits in O \\ F_{N^{2}} & if N does not split in O. \end{matrix}

The above covers all possibilities, because N > 3 can’t ramify in O. By Kummer’s theorem, N splits in O iff (\frac{−3}{N}) = 1, and there are exactly two O-stable submodules of O/NO. In the

second case, which happens iff (\frac{−3}{N}) = −1, O/NO has no O-stable submodules. Therefore

n_{0} = \{ \begin{matrix}\frac{N−1}{3} + 2 & if N splits in O, i.e., N ≡ 1 (mod 3). \\ \frac{N+1}{3} & if N does not split in O, i.e., N ≡ 2 (mod 3) \end{matrix}

In the first case, note that the two O-stable submodules have to be counted separately. Similarly we obtain

n_{1728} = \{ \begin{matrix}\frac{N−1}{2} + 2 & if N ≡ 1 (mod 4) \\ \frac{N+1}{2} & if N ≡ 3 (mod 4) \end{matrix}

depending on whetherN splits in Q(i).

Examples. For N = 37 the genus is 2. Using the formula, we get 2g − 2 = 36 − (2 + 18) − 14 = 2 (so it works).

Using the formula we can also obtain g(X_{0}(13)) = 0 and g(X_{0}(11)) = 1.

It is therefore clear that the genus of X_{0}(N) is approximately N/12. In the article by Serre in the Lecture Notes in Mathematics 349 (Antwerp), we find the following table. Write N = 12a + b with 0 ≤ b ≤ 11. Then

\begin{matrix}b & 1 & 5 & 7 & 11 \\ g & a − 1 & a & a & a + 1. \end{matrix}

Hence

1 + g(X_{0}(p)) = dim M_{p+1}(SL_{2}(Z)).

It was Serre’s idea to think of “modular forms mod p”, for some congruence subgroup

Γ ∋ ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ), like Γ_{0}(N) or Γ_{1}(N). We could use our moduli theoretic interpretations, but

instead we’ll define

M_{k}(Γ, F_{p}) ⊆ F_{p}[[q]].

By Shimura’s cohomology trick, we know that M_{k}(Γ, Z) is a lattice in M_{k}(Γ, C). Hence we can set

M_{k}(Γ, F_{p}) = M_{k}(Γ, Z) ⊗_{Z} F_{p}.

Then Serre’s equality states that for a prime p,

M_{p+1}(SL_{2}(Z), F_{p}) = M_{2}(Γ_{0}(p), F_{p})

in F_{p}[[q]]. The philosophy is mod p forms with p in the level can be taken to mod p formswith no p in the level, but of a higher weight. So for example M_{k}(Γ_{1}(p^{α}N), F_{p}) is a subset of M_{?}(Γ_{1}(N), F_{p}) of forms of some higher level.

Finally, consider the map from the right hand side to the left hand side in Serre’s equality. Recall that

G_{k} = \frac{−B_{k}}{2k} + ∑_{n=1}^{∞} σ_{k}(n)q^{n} ∈ M_{k}(SL_{2}(Z)).

By Kummer, ord_{p}(B_{p−1}) = −1, so

E_{p−1} = 1 + \frac{−2(p − 1)}{B_{p−1}} ∑ σ_{p−1}(n)q^{n} ≡ 1 (mod p).

Hence we got the map from the right to the left: multiply by E_{p−1} to get to M_{p+1}(Γ_{0}(p), F_{p}). Then take the trace to get to M_{p+1}(SL_{2}(Z), F_{p}). The trace map is dual to the inclusion and is expressed by

tr(f) = ∑_{i=1}^{p+1} f|γ_{i} γ_{i} ∈ Γ_{0}(p) \ SL_{2}(Z).

February 14 and 16, 1996

Scribe: Jessica Polito, <polito@math>

Our goal, for these two days, is to define the modular curves X(N) over Q(μ_{N}), and X_{1}(N), and X_{0}(N) over Q. These notes will spell out the construction of X(N), with some discussion of the construction of the other two types of curves. The idea comes from Shimura.

Let Q(t) be the function field of P^{1}/Q, and pick an elliptic curve E_{j}/Q(t) with j-invariant t:

E_{j} : y^{2} = 4x^{3} − \frac{27t}{t − 1728}x − \frac{27t}{t − 1728}

(Note that the general formula for the j-invariant of a curve y^{2} = 4x^{3}−g_{2}x−g_{3} is j = \frac{1728g_{2}^{3}}{g_{2}^{3}−27g_{3}^{2}} , and that j(E) determines the isomorphism class of the given curve E over the algebraic closure of the field of definition.)

By substituting in a given value j for t, we would get a formula for an elliptic curve over Q(j) with j-invariant j; for j = 0 or 1728, we could pick a diferent formula for E_{j} which would give an isomorphic curve over Q(t), for which that substitution would make sense.

Notice that, in general, if we have an elliptic curve E/K, with K some field of charictaristic prime to N, then we can consider E[N](\overline{K}), the set of all N-torsion point of E defined over \overline{K}, which is isomorphic to (Z/NZ)^{2}. Then we let K(E[N]) be the smallest extension of K over which all the points of E[N] are defined. Notice also that E/K and N together define a representation of the Galois groups Gal(\overline{K}/K) into the automorphisms of the N-torsion points of E, as they are defined by polymial equations with coefficients in E. We get

ρ_{E,N} : Gal(\overline{K}/K) → Aut(E[N]) ≅ GL_{2}(Z/NZ)

with Gal(\overline{K}/K(E[N])) = ker(ρ_{E,N}). Unsurprisingly, we will frequently leave out the N, writing simply ρ_{E}.

Now, to return to our construction, where K = Q(t). We will show that Q(t)(E[N]) is the function field of a curve defined over Q(μ_{N}) (so Q(t)(E[N]) ∩ \overline{Q} = Q(μ_{N})), and that this curve corresponds to X(N).

We will actually do this, not by looking at ρ_{E}, but instead at \overline{ρ_{E}}, where \overline{ρ_{E}} is given by reducing the image of ρ_{E} mod ±1:

Gal(\overline{K}/K) \overline{ρ_{E}_{}}GL_{2}(Z/NZ)
[figure: \overline{ρ_{E}}]
GL_{2}(Z/NZ)/{±1}^{}

Note: ρ_{E} is surjective iff \overline{ρ_{E}} is.

Proof. If \overline{ρ_{E}} is surjective, then

± ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ∈ im(ρ_{E}).

By squaring, we see that

( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) ∈ im(ρ_{E}),

and thus −1 ∈ im(ρ_{E}), so ρ_{E} is surjective.

(Obviously, if ρ_{E} is surjective, so is \overline{ρ_{E}}.) ■ Why will we use \overline{ρ_{E}} instead of ρ_{E}? The kernal of ρ_{E} is the field generated over K by the

X and Y coordinates of the non-zero points of E_{j}[N], whereas the kernal of \overline{ρ_{E}} is the field generated by just the X coordinates of those points. This follows because X(P) = X(−P), while Y (P) = −Y (−P), as then the X but not the Y coordinates are fixed by ±1. (We are assuming that we have a Weierstrass model for E_{j} here.)

Remark 6 Let E_{1} and E_{2} be elliptic curves over K with equal j invariants. (In other words, E_{1} and E_{2} are isomorphic over \overline{K}.) Then ρ_{E_{1}} is surjective iff ρ_{E_{2}} is. (We’ll assume that either N ≠ 2, 3 or E does not have complex multiplication.)

Proof. Assume ρ_{E_{1}} is surjective. Then E_{1} does not have complex multiplication over \overline{K}. (If E has CM, then the image of ρ_{E} is abelian for some quadratic extension of K, but GL_{2}(Z/NZ) has no abelian subgroup of index 2, as N > 3.)

Then Aut(E_{1}) = \{±1\} Pick an isomorphism ϕ : E_{1} ≅ E_{2} over \overline{K}. Then, for any σ ∈ Gal(\overline{K}/K), we have^{σ}ϕ : ^{σ}E_{1} ≅ ^{σ}E_{2}, and^{σ}E_{i} = E_{i}. Thus^{σ}ϕ = ±ϕ for all σ ∈ Gal(\overline{K}/K). So ϕ : E_{1}[N] ≅ E_{2}[N] does not quite give us an isomorphism of representations ρ_{E_{1}} and ρ_{E_{2}}, but it does give us an isomorphism \overline{ρ_{E_{1}}} ≅ \overline{ρ_{E_{2}}}. We then have that \overline{ρ_{E_{2}}}, and thus ρ_{E_{2}}, is surjective. ■

From here on, let K = C(t), E_{j} and ρ_{E}, \overline{ρ_{E}} as above. We have

ρ_{E} : Gal(\overline{K}/K) → GL_{2}(Z/NZ),

so the determinant of ρ_{E} is a cyclotomic character. As the Nth roots of unity are all already in K, this means that the determinant of ρ_{E} is trivial, so

ρ_{E} : Gal(\overline{K}/K) → SL_{2}(Z/NZ).

Our goal is to prove that ρ_{E} is surjective onto SL_{2}(Z/NZ).

(For any field F of characteristic not dividing p, there is an algebraic proof by Igusa that a “generic elliptic curve” (such as E_{j} here), has the maximal possible Galois action on its division points.)

We will now introduce another way of looking at K. It is equal to C(j) = F_{1}, the field of modular functions of level 1. (A modular function of level N is a Γ(N)-invariant function on ℌ which is meromorphic on ℌ and at the cusps.) The set of all modular functions for SL_{2}(Z) are generated over C by a single function j. Similarly, F_{N} is the field of modular functions of level N. We then have the tower of field extensions

\begin{aligned}\overline{K} & = & \overline{F_{1}} \\ & & | \\ & & F_{N} \\ & & | \\ K & = & F_{1} \end{aligned}

SL_{2}(Z) acts on F_{N}, via the map f(τ) → f (\frac{aτ+b}{cτ+d}), and fixes F_{1}. By reviewing definitions, we see that SL_{2}(Z)/Γ(N) = SL_{2}(Z/NZ) acts on F_{N}, and that −1 acts trivially. We will show that SL_{2}(Z/NZ)/{±1} is that Galois group of F_{N}/F_{1}, by exhibiting a set of functions on whose permutations this group acts faithfully.

So, the entire picture so far is:

\overline{F_{N}}
[figure]
K(E_{j}[N])/{±1}  F_{N}
[figure]
F_{1} = K

(Recall that K(E_{j}[N])/\{±1\} is the fiexed field of the kernal of \overline{ρ_{E}}.) We will show that K(E[N]) = F_{N}, at the same time that we show that the Galois group is what we want it to be. To do this, we’ll use the Weierstrass ℘-function attached to a given lattice L of C (and thus to the corresponding elliptic curve E/C). So we will pick a specific τ ∈ ℌ. (Then sending t to j(τ) gives us a map from E_{j}_{t} to E_{τ}, the elliptic curve over C with j-invariant j(τ), which is also given by the lattice Z + Zτ.) To the lattice L = Z + Zτ we attach ℘z, which is a modular form of weight 2 satisying the differential equation

℘′(z)^{2} = ℘(z)^{2} − 4g_{2}(τ)℘(z)g_{3}(τ),

which, replacing ℘ by X and ℘′ by Y, is a Weierstrass equation for E_{τ}. Thus we have ℘(\frac{rτ+s}{N}, Z+Zτ), with r, s ∈ Z and not both divisible by N, running through the X-coordinates of points in E_{τ}[N]. This form is of weight 2; we need to find modular functions, so let

f_{(r,s)}(τ) : τ ↦ \frac{g_{2}(τ)}{g_{3}(τ)}℘ (\frac{rτ + s}{N}, Z + Zτ ) .

(The f are certainly meromorphic on ℌ and at the cusps, as g_{2}, g_{3} and ℘ all are.) SL_{2}(Z) acts on these functions f by α : f_{(r,s)}(ατ) = f_{(r,s)α}(τ), where α simply multiplies the row vector (r, s) on the right. Then we can see that {f_{(r,s)}} is invariant under the action of Γ[N], so they are permuted by SL_{2}(Z/NZ). In fact, as ℘ is an even function (X(P) = X(−P) for any point on the eliptic curve), f_{(r,s)} is also fixed by ±1, so they are permuted by SL_{2}(Z/NZ)/{±1}.

We have defined f_{r,s} for r, s ∈ Z,, not both divisible by N. In fact, if (r, s) = (r′, s′) (mod N), then f_{(r,s)} = f_{(r′,s′)}, so we only need to consider (r, s) ∈ Z/NZ^{2}/{±1}\(0, 0) As f_{(r,s)}(τ) runs over all of the X-coordinates of points in E_{τ}[N] which are not zero for these (r, s), we must have these f_{(r,s)} all distinct. (There are N^{2} − 1 non-zero points on E[N], and X(P) = X(Q) iff P = −Q, so there are exactly as many pairs (r, s) as there are X coordinates of points in E[N].)

Thus we have shown that the non-zero points of E_{j}[N] have X-coordinates (in \overline{F_{N}}) given by the f_{(r,s)} ∈ F_{N}, and thus F_{N} = K(E_{j}[N]).

(As a corollary, we have shown that the Galois group of F_{N} over F_{1} is SL_{2}(Z/NZ)/{±1}, as this group acts faithfullly on the f_{(r,s)}.)

Thus, finally, we have shown that for any E/C(t) with j-invariant t, the associated representation ρ_{E} has image SL_{2}(Z/NZ).

We know that, if we let K = Q(μ_{N})(t), E an elliptic curve with j-invariant t, then ρ_{E} has image contained in SL_{2}(Z/NZ). (This doesn’t work for K = Q(t), as we need to have μ_{n} ⊂ K in order to have the image of ρ_{E} contained in SL_{2}(Z/NZ).) Furthermore, if we replace C(t) by Q(μ_{N})(t), the image of ρ_{E} can only get larger. In the following diagram

C(t)(E[N])
[figure: G′]
C(t)  Q(μ_{N})(t)(E[N])
[figure: G]
Q(μ_{N})(t)

G and G′ are the Galois groups (and also the images of ρ_{E} over Q(μ_{N})(t) and C(t) respectively). Basic Galois theory tells us that G′ ⊂ G, but we also know that G′ = SL_{2}(Z/NZ) and G ⊂ SL_{2}(Z/NZ), so G = G′ = SL_{2}(Z/NZ).

Next, let us consider E/Q(t), as always with j-invariant t. Then ρ_{E} : Gal(Q(t)(E[N]) → GL_{2}(Z/NZ). The determinant of ρ_{E} is again a cyclotomic character, but is not trivial this time; indeed, it is surjective and thus contains a set of coset representatives for SL_{2}(Z/NZ) in GL_{2}(Z/NZ). We see that ρ_{E} is again surjective, although onto GL_{2}(Z/NZ) this time.

Notice that, in this case, Q(t)(E[N]) contains μ_{N}, so we now have the diagram

\overline{Q}  Q(t)(E[N])

Q(μ_{N})  Q(μ_{N})(t)
[figure]
Q(t)
[figure]

Q

Basic Galois theory now tells us that \overline{Q} ∩ Q(t)(E[N]) = Q(μ_{N}):

Proof. Let F be the intersection of these two fields. It clearly contains Q(μ_{N}). If F is bigger than Q(μ_{N}), then F (t) is bigger than Q(μ_{N})(t), and thus

Gal(Q(t)(E[N])/F (t)) ⊊ SL_{2}(Z/NZ).

But we know (as above) that

SL_{2}(Z/NZ) = Gal(C(t)(E[N]/C(t)) ⊂ Gal(Q(t)(E[N])/F (t))

so we must have equality above, and thus F = Q(μ_{N}).

Recall that C(t)(E[N])/C(t) is the function field for X(N). We have shown that Q(t)(E[N]) is a function field for the corresponding curve defined over Q(μ_{N}), so we have defined X(N) over Q(μ_{N}).

We can similarly define X_{1}(N) and X_{0}(N) over Q. Let L = K(E[N]) (where Q(t) = K). Then Gal(L/K) = GL_{2}(Z/NZ). Consider the subgroup H of this Galois group consisting of

all matrices of the form ( \begin{matrix}∗ & ∗ \\ 0 & ∗ \end{matrix} ). Then L^{H} ∩ \overline{Q} = Q(μ_{N})^{H} = Q. As L^{H} is an extension of

Q of transcendence degree 1, it is the function field of a curve defined over Q. It turns out

to be the curve X_{0}(N). To get X_{1}(N), we would use the subgroup H = ( \begin{matrix}∗ & ∗ \\ 0 & 1 \end{matrix} ).

21 February, 1996

Scribe: David M Jones, <dmjones@math>

First of all, we start with a correction from last time. Let L/K be a Galois extension. Let E_{/L} be an elliptic curve. Suppose that j(E) ∈ K and for all g ∈ Gal(L/K), ^{g}E ≅_{/L} E.

Last time it was stated that we could conclude that there exists E_{0}_{/K} such that E ≅_{/L} E_{0}. This is not correct. We may only conclude that E ≅_{/L̄} E_{0}. The point is that L = K̄.

Let E_{/K} be an elliptic curve. Let g ∈ Gal(L/K) for some extension L/K. Then let

λ_{g} : ^{g}E →^{∼}^{λ_{g}} E be a family of isomorphisms. Now, λ_{gh} and λ_{g} ∘^{g}λ_{h} are both isomorphisms from ghE to E so they differ by an element of Aut(E) = {±1}.

Definition 8 Let g, h ∈ Gal(L/K). We define c(g, h) ∈ Aut(E) by the relation c(g, h)λ_{gh} = λ_{g} ∘^{g}λ_{h}.

Claim 2 c(g, h) is a 2-cocyle.

Proof. First, let’s rewrite c(g, h) = λ_{g} ∘^{g}λ_{h} ∘ λ_{gh}^{−1}. Then in order to prove that c(g, h) is a 2-cocycle, we must show the following (see Serre’s “Local Fields”, p.113):

gc(g′, g′′) · c(g, g′g′′) = c(g, g′) · c(gg′, g′′)

Then c(g, g′)·c(gg′, g′′) = λ_{g} ∘^{g}λ_{g′} ∘λ_{gg′}^{−1} ∘λ_{gg′} ∘^{gg′}λ_{g′′} ∘λ_{gg′g′′}^{−1} = λ_{g} ∘^{g}λ_{g′} ∘^{gg′}λ_{g′′} ∘λ_{gg′g′′}^{−1} = λ_{g} ∘^{g}(λ_{g′} ∘^{g′}λ_{g′′}) ∘ λ_{gg′g′′}^{−1} = λ_{g} ∘^{g}(c(g′, g′′) · λ_{g′g′′}) ∘ λ_{gg′g′′}^{−1} = ^{g}c(g′, g′′) · λ_{g} ∘^{g}λ_{g′g′′} ∘ λ_{gg′g′′}^{−1} = gc(g′, g′′) · c(g, g′g′′). ■

We want this 2-cocyle c(g, h) to be trivial.

Lemma 1 Let G = Gal(L/K). If L = K̄, then H^{2}(G, {±1}) → H^{2}(G, L^{∗}) is injective.

Proof. Consider the exact sequence

0 → {±1} → K̄^{∗} →^{•^{2}} K̄^{∗} → 0.

Looking at a piece of the long exact cohomology sequence we get

... → H^{1}(G, K̄^{∗}) → H^{2}(G, {±1}) → H^{2}(G, K̄^{∗})[2] → 0.

Now H^{1}(G, K̄^{∗}) = 0 by Hilbert Theorem 90 so H^{2}(G, {±1}) is isomorphic to the 2-torsion in H^{2}(G, K̄^{∗}). So H^{2}(G, {±1}) certainly injects into H^{2}(G, K̄^{∗}). Since L = K̄, the result follows. ■

In order to show that the class of c(g, h) in H^{2}(G, {±1}) is trivial, by the above lemma, it is enough to show that the class of c(g, h) in H^{2}(G, K̄^{∗}) is trivial. To do this we will use

Differentials: Pick a non-zero differential ω ≠ 0 on E. That is, ω ∈ H^{0}(E, Ω^{1}). Let g ∈ Gal(L/K). Then λ_{g} : ^{g}E→^{∼}E so the pullback λ_{g}^{∗} : H^{0}(E, Ω^{1})→^{∼}H^{0}(^{g}E, Ω^{1}). Thus, λ_{g}^{∗}ω ∈ H^{0}(^{g}E, Ω^{1}), which has basis ^{g}ω so we must have λ_{g}^{∗}ω = a_{g} ·^{g}ω for some a_{g} ∈ L^{∗}.

Note 1 What is ^{g}ω? Let E_{/L} be an elliptic curve and let L ↪ M be an injection of fields. Then H^{0}(E, Ω^{1}) ⊗_{L} M = H^{0}(E_{/M}, Ω^{1}). In our case, M = L and the map L ↪ M is given by the action of g ∈ Gal(L/K). So we get H^{0}(E, Ω^{1}) ⊗_{L} L = H^{0}(^{g}E, Ω^{1}). The element of H^{0}(^{g}E, Ω^{1}) which corresponds to ω ⊗ 1 under this map is ^{g}ω.

Lemma 2 Show that c(g, h) = (^{g}a_{h} · a_{g})/a_{gh}.

Proof. First, consider the following:

a_{g} ·^{g}a_{h} · a_{gh}^{−1}ω = a_{g} ·^{g}a_{h} · ((λ_{gh}^{∗})^{−1}^{gh}ω) = a_{g} · (λ_{gh}^{∗})^{−1} ∘^{g}(a_{h} ·^{h}ω) = a_{g} · (λ_{gh}^{∗})^{−1} ∘^{g}(λ_{h}^{∗}ω) = (λ_{gh}^{∗})^{−1} ∘^{g}(λ_{h}^{∗}) ∘ (a_{g}^{g}ω) = (λ_{gh}^{∗})^{−1} ∘^{g}(λ_{h}^{∗}) ∘ λ_{g}^{∗}ω = ((λ_{gh})^{−1} ∘^{g}(λ_{h}) ∘ λ_{g})^{∗}ω.

But c(g, h) = λ_{g} ∘^{g}λ_{h} ∘ λ_{gh}^{−1} is multiplication by a constant (±1) so the dual is multiplication by the same constant. Thus, we have a_{g} ·^{g}a_{h} · a_{gh}^{−1}ω = c(g, h)ω and as ω ≠ 0, we must have c(g, h) = (a_{g} ·^{g}a_{h})/a_{gh}. ■

This shows that c(g, h) is a coboundary, and thus its class is trivial in H^{2}(G, K̄^{∗}).

Let N > 3 be an integer. Recall that we wrote down an elliptic curve E_{/Q(j)} whose j-invariant was j(E) = j.

Let F_{N} be the field of modular functions of level N. That is, F_{N} = Q(j)(E[N]/{±1}). This field lies over F_{1} = Q(j). We have Gal(F_{N}/F_{1}) = GL_{2}(Z/NZ)/{±1}.

Now define F = ∪_{N}F_{N}. This field, which is the compositum of all the of the F_{N}’s, corresponds to a projective system of modular curves.

Definition 9 The finite adele ring of Q, denoted A_{f}, is

{(x_{p}) ∈ ∏_{p} Q_{p} : x_{p} ∈ Z_{p} for almost all p}

We may also describe A_{f} as Q̂ = Ẑ ⊗ Q.

Notice that GL_{2}(Ẑ) ⊂ GL_{2}(A_{f}).

Claim 3 The group GL_{2}(A_{f}) acts on F.

First, recall that F = Q(f_{(r,s)} : (r, s) ∈ (Z/NZ)^{2}\(0, 0), N ≥ 1). GL_{2}(Z/NZ) acts on the set of f_{(r,s)} via

( \begin{matrix}a & b \\ c & d \end{matrix} ) : f_{(r,s)} ↦ f_{(ar+cs,br+ds)}

Consider the object lim_{⟵} X(N). It is not a variety but does exist in some appropriate category. We have the following two notions of “points” on lim_{⟵} X(N), which are contra-dual to each other:

1. Pairs (E, ι) where E is an elliptic curve and ι : (Q/Z)^{2}→^{∼}E_{tors}

2. Pairs (E, α) where E is an elliptic curve and α : Ẑ^{2}→^{∼}T a(E), where T a(E) =

lim_{⟵} E[N] = ∏_{p} T a_{p}(E)

We will now do a warm-up to show that GL_{2}(A_{f}) acts on F.

Let g = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL_{2}^{+}(Q). Look at pairs (E, α) with E_{/C} and α : Z^{2}→^{∼}H_{1}(E(C), Z).

Note that we have

Z^{2} →^{∼} H_{1}(E(C), Z)
⋂ |  ⋂ |
Q^{2} →^{∼} H_{1}(E, Q) = H_{1}(E(C), Z) ⊗ Q

We denote this bottom map α : Q^{2}→^{∼}H_{1}(E, Q) as well.

Then we find

α ∘ g : Q^{2} →^{∼} H_{1}(E, Q)
⋃  ⋃
Z^{2} →^{∼}  L′

where L′ is defined as the image of Z^{2} under α ∘ g.

As we have a new lattice L′, it determines a new elliptic curve E′. We also get a map

λ : L′→^{∼}H_{1}(E′, Z).

So g : (E, α) ↦ (E′, α′) where α′ is the composed map λ ∘ α ∘ g : Z^{2}→^{∼}L′→^{∼}H_{1}(E′, Z).

Let τ ∈ ℌ, E = E_{τ}, L_{τ} = Z + Zτ ≅ Z^{2}. Let us denote the inverse to this latter isomorphism by α_{τ} : Z^{2}→^{∼}Z + Zτ. Under α_{τ}, (1, 0) ↦ τ and (0, 1) ↦ 1.

Claim 4 Check that g : (E_{τ}, α_{τ}) ↦ (E_{τ′}, α_{τ′}) where τ′ = \frac{aτ+b}{cτ+d} = gτ.

Proof. First of all, α_{τ′} takes (1, 0) to τ′ = \frac{aτ+b}{cτ+d} and takes (0, 1) to 1.

Now we look at λ ∘ α ∘ g. GL_{2}^{+}(Q) acts on the right on Z^{2} so g sends (1, 0) to

(1, 0) · ( \begin{matrix}a & b \\ c & d \end{matrix} ) = (a, b)

and sends (0, 1) to (c, d). Next α sends (a, b) to aτ + b = (cτ + d)τ′ and sends (c, d) to cτ + d. Now λ just scales the result so that it is of the form αsomething. That is, (0, 1) gets sent to 1. This is dividing by cτ + d and we see that under λ ∘ α ∘ g, (0, 1) is sent to 1 and (1, 0) is sent to τ′. ■

In the end we want to show that GL_{2}(A_{f}) acts on the left on functions and on the left on curves.

23 February, 1996

Scribe: David M Jones, <dmjones@math>

Today we start of with some philosophy about where we are headed in this course and why we are headed there.

We defined modular forms over C early on in the course. We also discussed a trick by which we could define modular forms over any ring R. We also defined the Hecke operators and algebras for these modular forms over C.

Recently, we defined, following Shimura, modular curves and modular forms over more “arithmetic” rings (such as Q). The construction of the modular curves over these rings was done on 2/14 and 2/16 and then the modular form were realized at differentials on the modular curves; elements of H^{0}(X, Ω^{1}) where X is the modular curve.

Now we are setting out to define the Hecke operators on these modular forms – to define the Hecke operators and Hecke algebras in a more algebraic (or “arithmetic”) way.

Recall: For N′|N, we had a tower:

\begin{matrix}X(N) & F_{N} \\ ↓ & ↑ \\ X(N′) & F_{N′} \\ ↓ & ↑ \end{matrix}

And we also have F = lim_{⟶} F_{N}. That is, F is the compositum of the fields F_{N}.

Intuitively, F, which is a field, corresponds to the “pro-curve” lim_{⟵} X(N), which is not an algebraic curve.

We were discussing the operation of GL_{2}(A_{f}), where A_{f} is the ring of finite adeles of Q. A_{f} = Ẑ ⊗ Q and A_{f} contains Ẑ as a subring.

Consider a pair (E, α), where E_{/Q} is an elliptic curve and

α : Ẑ^{2}→^{∼} ∏_{p} T a_{p}(E)

We refer to ∏_{p} T a_{p}(E) as T a(E), the Tate module of E.

Now, T a(E) is free of rank 2 over Ẑ^{2}. Let V (E) = T a(E) ⊗_{Z} Q = ∏_{p} V_{p}(E) where V_{p}(E) = T a_{P} (E) ⊗_{Z_{p}} Q_{p}.

Since α maps Ẑ^{2} to T a(E), we may extend by Q to get a map (also called α) Q̂^{2}→^{∼}V (E). If g ∈ GL_{2}(A_{f}), then g : Q̂^{2} → Q̂^{2}. So by composing α and g, we get a new map

Q̂^{2}→^{∼}V (E).

Look at the subring Ẑ^{2} in Q̂^{2}, and take its image under α ∘ g, and we get a lattice T′ in V (E) which is different from T a(E). The following picture illustrates the idea:

T a(E)

⋂ |
Q̂^{2} →^{∼}^{α∘g} V (E)
⋃ |  ⋃ |
Ẑ^{2} →^{∼} T′

Our goal, as stated earlier, is to make the Hecke operators T_{n} and < d > into objects defined over Q.

The key idea is that X_{1}(N) and X_{0}(N) have Hecke operators acing as correspondences.

Definition 10 A correspondence between two curves X and Y is a third curve C and two maps α : C → X and β : C → Y.

Given a correspondence (C, α, β) of curves X and Y, we can form the maps

α^{∗} : H^{0}(X, Ω^{1}) → H^{0}(C, Ω^{1})

and

β_{∗} : H^{0}(C, Ω^{1}) → H^{0}(Y, Ω^{1}),

this latter map β_{∗} being a trace map of sorts.

α^{∗} is the standard pull-back map. β_{∗} can be defined using Serre duality. Recall that H^{0}(C, Ω^{1})^{∗} = H^{1}(C, O_{C}) and H^{0}(Y, Ω^{1})^{∗} = H^{1}(Y, O_{Y}). So the dual of β_{∗} (which may be described as (β_{∗})^{∗} if one likes) may be defined via the pull-back

β^{∗} : H^{1}(Y, O_{Y}) → H^{1}(C, O_{C}).

We will make this explicit for T_{p}, p not dividing N.

In order to define T_{p} (for p not dividing N) as a correspondence from X_{0}(N) to itself, we will say what C is and what α and β are.

Let C = X_{0}(pN). View

X_{0}(N) = {pairs (E, C)|C ≅ Z/NZ}
X_{0}(pN) = {triples (E, C, D)|C ≅ Z/NZ, D ≅ Z/pZ}

Then we can consider the maps α and β as given by:

α : (E, C, D) ↦ (E, C)
β : (E, C, D) ↦ (E/D, (C + D)/D)

One must also describe what happens at the cusps but we will gloss over that. We may also describe this complex analitically are follows: Γ_{0}(pN) ⊆ Γ_{0}(N) so Γ_{0}(pN)\ℌ → Γ_{0}(N)\ℌ, which corresponds to α : X_{0}(pN) → X_{0}(N).

Also, conjugation by ( \begin{matrix}p & 0 \\ 0 & 1 \end{matrix} ) maps Γ_{0}(pN) to Γ_{0}(N) since

( \begin{matrix}p & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & b \\ pNc & d \end{matrix} ) ( \begin{matrix}p & 0 \\ 0 & 1 \end{matrix} )^{−1} = ( \begin{matrix}a & pb \\ Nc & d \end{matrix} )

So we get Γ_{0}(pN)\ℌ → ( \begin{matrix}p & 0 \\ 0 & 1 \end{matrix} ) Γ_{0}(pN) ( \begin{matrix}p & 0 \\ 0 & 1 \end{matrix} )^{−1} \ℌ → Γ_{0}(N)\ℌ and this is

β : X_{0}(pN) → X_{0}(N).

Now we consider pulling back differentials via these maps. As always, recall that

H^{0}(X_{0}(N), Ω^{1}) = S_{2}(Γ_{0}(pN))

so as we have α, β : X_{0}(pN) → X_{0}(N), we get α^{∗}, β^{∗} : H^{0}(X_{0}(N), Ω^{1}) → H^{0}(X_{0}(pN), Ω^{1}) and thus α^{∗}, β^{∗} : S_{2}(Γ_{0}(N)) → S_{2}(Γ_{0}(pN)). These are given by α^{∗}(f) = f and, if f = ∑_{n=1}^{∞} a_{n}q^{n}, then β^{∗}(f) = p ∑_{n=1}^{∞} a_{n}q^{pn}.

Exercise 3 Check that β^{∗}(f) = p ∑_{n=1}^{∞} a_{n}q^{pn} rather than β^{∗}(f) = ∑_{n=1}^{∞} a_{n}q^{pn}.

Exercise 4 Show that H^{0}(X_{0}(N), Ω^{1}) → H^{0}(X_{0}(N), Ω^{1}) given by β_{∗} ∘ α^{∗} is the same as that given by α_{∗} ∘ β^{∗}, which both equal T_{p}. One must show that they both give

T_{p} : ∑_{n=1}^{∞} a_{n}q^{n} ↦ ∑_{n=1}^{∞} a_{pn}q^{n} + p ∑_{n=1}^{∞} a_{n}q^{pn}

Another way to look at this is as follows:

Let φ : X → Y. Then let Γ = the graph of φ ⊂ X × Y. So we get

Γ ↪ X × Y →^{α} X
β ↓
Y

as correspondences. We have α : Γ → X and β : Γ → Y.

Γ →^{α} X has degree d ≥ 1. So β(α^{−1}(x)) ∈ Div(Y), and deg(β(α^{−1}(x))) = d. So we have, for each n, Div^{n}(X) → Div^{dn}(Y). For n = 0, we get Div^{0}(X) → Div^{0}(Y).

Applying this to our situation, we have T_{p}

X_{0}(pN) →^{β} X_{0}(N)
α ↓
X_{0}(N)

And we get

(E, C) ↦^{α^{∗}} ∑_{D⊆E[p]}(E, C, D) ↦^{β} ∑_{D⊆E[p]}(E/D, (C + D)/D)

What happens on Jacobians?

Given a curve X, we get Jac(X) = J(X) which is an abelian variety. If genus(X) = g, then dim(J(X)) = g.

Over C, J(X) = H^{0}(X, Ω^{1})^{∗}/H_{1}(X, Z). J(X) is covariantly associated to X. This is the Albanese construction of J(X).

There is another construction of J(X) which is contravariant.

If X = Y, you need to be careful about which construction you use so that correspondences from X to X give the right morphisms J(X) to J(X).

February 26, 28 and March 1, 1996

Scribe: Shuzo Takahashi, <shuzo@math> The topics covered in this week’s lectures are

1. Generality on Correspondences

2. Hecke Operators on J_{0}(N), H^{0}(X_{0}(N), Ω^{1}), and Cot(J_{0}(N))

3. Reduction of J_{0}(N) mod p

4. The Eichler-Shimura Relation

5. Galois Action on Tate Module

6. Shimura Construction of an Abelian Variety Associated to a New Form 7. Galois Representations Associated to New Forms of Weight 2

1. Generality on Correspondences

Before considering Hecke operators on J_{0}(N), H^{0}(X_{0}(N), Ω^{1}), and Cot(J_{0}(N)), let’s consider correspondences in a general setting. Consider a correspondence

Γ

[figure: α; β]
X  Y

This induces naturally a map

α_{∗} ∘ β^{∗} : H^{0}(Y, Ω^{1}) → H^{0}(X, Ω^{1}).

Since X → J(X) induces Cot(J(X)) ≃ H^{0}(X, Ω^{1}), the above map induces

Cot(J(Y)) → Cot(J(X)).

Also, the correspondence induces

β_{∗} ∘ α^{∗} : J(X) → J(Y)

by the Albanese functoriality of J(.). Moreover , if we have ψ : J(X) → J(Y), then we have ψ^{∨} : J(Y)^{∨} → J(X)^{∨}. However since we have the canonical duality J(X) ≅ J(X)^{∨}, ψ induces J(Y) → J(X). We have (β_{∗} ∘ α^{∗})^{∨} = α_{∗} ∘ β^{∗}.

2. T_{p} on J_{0}(N), H^{0}(X_{0}(N), Ω^{1}), and Cot(J_{0}(N))

Assume p |̸N. Consider a correspondence

X_{0}(pN)

[figure: α; β]
X_{0}(N)  X_{0}(N)

where α and β are degeneracy maps (Consider a point of X_{0}(pN) as (E, Z/pZ), regarding a group of order N as a part of E.). This correspondence is by definition T_{p}. T_{p} naturally induces

T_{p}^{∗} = α_{∗} ∘ β^{∗} : H^{0}(X_{0}(N), Ω^{1}) → H^{0}(X_{0}(N), Ω^{1}).

Identifying H^{0}(X_{0}(N), Ω^{1}) and S_{2}(Γ_{0}(N)), this T_{p}^{∗} is the same as the usual T_{p}.

H^{0}(X_{0}(N), Ω^{1}) ≃ S_{2}(Γ_{0}(N))
T_{p}^{∗} ↓  ↓ T_{p}
H^{0}(X_{0}(N), Ω^{1}) ≃ S_{2}(Γ_{0}(N))

Now consider the Jacobian of X_{0}(N) = J_{0}(N) which is an abelian variety over Q. α : X_{0}(pN) → X_{0}(N) induces α^{∗} : J_{0}(N) → J_{0}(pN) by the Picard functoriality and α_{∗} : J_{0}(pN) → J_{0}(N) by the Albanese functoriality.

Theorem 10 The action of α_{∗} ∘ α^{∗} on J_{0}(N) is multiplication by p + 1.

Theorem 11 We have β_{∗} ∘ α^{∗} = α_{∗} ∘ β^{∗}. Let T_{p} = β_{∗} ∘ α^{∗}. Then T_{p} is an endmorphism of J_{0}(N). (T_{p} is called ξ_{p} in Shimura’s book.)

ξ_{p} induces

ξ_{p}^{∗} : Cot(J_{0}(N)) → Cot(J_{0}(N))

Next, consider θ : X_{0}(N) → J_{0}(N), mapping x to the class of (x) − (∞). θ induces

θ^{∗} : Cot(J_{0}(N)) ≃ H^{0}(X_{0}(N), Ω^{1})

Identifying Cot(J_{0}(N)) and H^{0}(X_{0}(N), Ω^{1}), ξ_{p}^{∗} is the same as T_{p}^{∗}.

Cot(J_{0}(N)) ≃ H^{0}(X_{0}(N), Ω^{1})
ξ_{p}^{∗} ↓  ↓ T_{p}^{∗}
Cot(J_{0}(N)) ≃ H^{0}(X_{0}(N), Ω^{1})

So far we have considered the case p |̸N. If p|N, we need to choose between β_{∗} ∘ α^{∗} and α_{∗}∘β^{∗}. The usual choice ξ_{p}^{∗} induces the usual T_{p} on S_{2}(Γ_{0}(N)). Now we have ξ_{p} ∈ End(J_{0}(N)) for every p. T = Z[⋯ T_{p} ⋯] ⊆ End(J_{0}(N)) is the same ring as T_{Z} ⊆ End(S_{2}(Γ_{0}(N))).

3. Reduction of J_{0}(N) mod p

Igusa showed that for all p |̸N, J_{0}(N) over Q ⊃ Z[1/N] has a good reduction mod p considering X_{0}(N) over Z[1/N].

On the other hand, X_{0}(pN)/F_{p} is obtained by attaching the two copies of X_{0}(N)/F_{p} at their supersingular points (those which arise from supersingular elliptic curves over \overline{F}_{p}) a supersingular point x on the first copy being identified with its Frobenius transform x^{(p)} on the second.

4. The Eichler-Shimura Relation

Suppose p |̸N. J_{0}(N) has a good reduction mod p. We have

T_{p} ∈ End(J_{0}(N)) ↪ End(J_{0}(N)/F_{p}).

Frobenius morphism F : X_{0}(N)/F_{p} → X_{0}(N)/F_{p} of degree p induces

frob : J_{0}(N)/F_{p} → J_{0}(N)/F_{p}

by covariant functoriality, and

ver : J_{0}(N)/F_{p} → J_{0}(N)/F_{p}

by contravariant functoriality. Since the degree of F is p, we have

Theorem 12 ver ∘ frob = frob ∘ ver = p

The Eichler-Shimura relation is

Theorem 13 T_{p} = frob + ver.

Note. If we use X_{1}(N), then we have T_{p} = frob + ⟨p⟩ver. Idea of Proof. First, by the work of Deligne-Rapport, the reduction of X_{0}(pN) mod p can be considered as two copies of X_{0}(N)/F_{p}. Consider points on X_{0}(pN) as isogenies (E → E′). One copy of X_{0}(N)/F_{p} is obtained by the map

r : E ↦ (E⟶^{frob}E^{(p)}).

The other copy is obtained by the map

s : E ↦ (E^{(p)}⟶^{ver}E).

Through these maps, we have

X_{0}(pN)/F_{p} ≅ \frac{X_{0}(N)/F_{p} ∐ X_{0}(N)/F_{p}}{identifying supersingular points of X_{0}(N)/F_{p} via frob}

Consider α : (E → E′) ↦ E and β : (E → E′) ↦ E′. Then we have

X_{0}(N)  X_{0}(N)
[figure: r; s]
id  X_{0}(pN)  id
[figure: ↓; α; β; ↓]
X_{0}(N)  X_{0}(N)

where α ∘ r = id = β ∘ s and α ∘ s = F = β ∘ r. Thus the correspondence T_{p} can be defined by two correspondences as follows:

X_{0}(pN)  X_{0}(N)  X_{0}(N)
[figure: α; β; =; id; F; +; F; id]
X_{0}(N)  X_{0}(N)  X_{0}(N)  X_{0}(N)  X_{0}(N)  X_{0}(N)

The first diagram of the right hand side is the graph of the Frobenius F : X_{0}(N) → X_{0}(N) and the second one is its dual. From this, one can deduce the Eichler-Shimura relation.

5. Galois Action on Tate Module

Look at T ate_{l}(J_{0}(N)) = lim_{⟵} J_{0}(N)[l^{ν}] ⊗_{Z_{l}} Q_{l} (l ≠ p). Let’s call this X. This is the same in char p as in char 0.

Theorem 14 We have operators T_{p}, frob_{p}, and ver_{p} on X satisfying the following: (a) T_{p} = frob_{p} + ver_{p},

(b) frob_{p}ver_{p} = p,

(c) T_{p} = frob_{p} + pfrob_{p}^{−1},

(d) frob_{p}^{2} − T_{p}frob_{p} + p = 0. Moreover X has actions of

(1) Gal(\overline{Q}/Q),

(2) End_{Q}J_{0}(N) ⊗_{Z} Q_{l},

(3) End_{F_{p}}J_{0}(N) ⊗_{Z} Q_{l},

(4) Gal(\overline{F_{p}}/F_{p}).

(1) and (2) commute, and (3) and (4) commute. (4) is a special case of (1) because we have a map D_{p} ↠ Gal(\overline{F_{p}}/F_{p}) where D_{p} ⊂ Gal(\overline{Q}/Q) is a decomposition group.

Point. There is a Frobenius element φ_{p} ∈ Gal(\overline{F_{p}}/F_{p}) and a Frobenius endmorphism frob_{p} ∈ End_{F_{p}}J_{0}(N) ⊗_{Z} Q_{l}. On Tate module X, frob_{p} acts as φ_{p}, and we have

φ_{p}^{2} − T_{p}φ_{p} + p = 0

which reminds Cayley-Hamilton theorem with detφ_{p} = p and trφ_{p} = T_{p}. Also, we have

T_{p} = φ_{p} + pφ_{p}^{−1}.

6. Shimura Construction of an Abelian Variety Associated to a New Form

Consider T_{0} = Z[⋯ T_{n} ⋯] where (n, N) = 1. We have

T_{0} ⊗ Q ≃ ∏_{f} E_{f}

where f runs over the set of new forms of level dividing N modulo conjugation by Gal(\overline{Q}/Q). Also, we have

T ⊗ Q ≃ ∏_{f} A_{f}

where A_{f} is an E_{f} algebra.

Exercise 5 Show dim_{E_{f}} A_{f} = # of divisors of \frac{N}{N(f)} where N(f) is the level of f.

Example 4 Suppose N = N(f)q where q |̸N(f). Then we have

A_{f} = E_{f}[U]/(U^{2} − a_{q}U + q).

Exercise 6 Write A_{f} in general as

E_{f}[⋯ U_{q} ⋯]/(relations)

where q|\frac{N}{N(f)}.

More explanation about the above example. A_{f} is the algebra generated by the T_{p} acting on the C vector space generated by the forms

f(dτ) where d|\frac{N}{N(f)}. T_{p} acts as the scalar a_{p} if p |̸N. If q|N, T_{q} = U_{q} acts as ( \begin{matrix}a_{q} & 1 \\ −q & 0 \end{matrix} ).

An abelian subvariety A_{f} of J_{0}(N). Let π_{f} be the projection of T_{0} ⊗Q ≃ ∏_{f} E_{f} over E_{f}. Then define A_{f} = π_{f}(J_{0}(N)) (π_{f} is not exactly an endmorphism but is in End(J_{0}(N)) ⊗ Q). In the category of abelian varieties up to isogeny, ∏ A_{f} acts on J_{0}(N) and pulls apart J_{0}(N) into ∏_{f} A_{f}. We have ∑_{f} A_{f} = J_{0}(N) and ∏_{f} A_{f} → J_{0}(N) is an isogeny.

Also, J_{0}(N)/C can be considered as

J_{0}(N) = Hom(S_{2}(Γ_{0}(N)), C)/H_{1}(X_{0}(N), Z)

(call it V/L). Then V = ⨁ V_{f} and A_{f} = V_{f}/(V_{f} ∩L) where the basis of V_{f} consists of f(dτ)’s as well as conjugates of them. Special case when the level of f is N. Now, let’s consider a special case where f is a new form (N(f) = N). Then we have A_{f} = E_{f}. Also d = dim A_{f} = dim_{Q} A_{f} = [E_{f} : Q]. Fix f and let A = A_{f} and E = E_{f}. Let T ate_{l} be the Q_{l}-adic Tate module of A_{f} ≅ Q_{l}^{2d}. Action of E ⊗_{Q} Q_{l} on T ate_{l} is ∏_{λ|l} E_{λ}. Thus T ate_{l} = ∏_{λ|l} T ate_{λ} where T ate_{λ} is an E_{λ} vector space. We have

Lemma 3 dim_{E_{λ}} T ate_{λ} = 2 for all λ|l. That is, T ate_{l} is free of rank 2 over E ⊗ Q_{l}.

Proof. A_{f} = V_{f}/L_{f} where L_{f} = V_{f} ∩L. L_{f} ⊗Q affords an action of E_{f} (a field). So L_{f} ⊗Q is a vector space over E_{f} of dim = 2. On the other hand, T ate_{l} is (L_{f} ⊗ Q) ⊗_{Q} Q_{l}.

7. Galois Representations Associated to New Forms of Weight 2

Gal(\overline{Q}/Q) acts on T ate_{l} which is compatible with E ⊗ Q_{l}. That is, we have

Gal(\overline{Q}/Q)→^{ρ_{l}}Aut_{E⊗Q_{l}}T ate_{l} = GL(2, E ⊗ Q_{l}) = ∏_{λ} GL(2, E_{λ}).

Then we have

ρ_{l} = ⨁ ρ_{λ}

where

ρ_{λ} : Gal(\overline{Q}/Q) → Aut_{E_{λ}}T ate_{λ}

If p |̸lN, then ρ_{λ} is unramified at p and ρ_{λ}(φ_{p}) has a well defined (trace, det) characteristic polynomial Φ(X) given as

Theorem 15

Φ(X) = X^{2} − a_{p}X + p

where a_{p} is the image of T_{p} in E_{f}.

Idea of Proof. By Cayley-Hamilton, we have

(1) φ_{p}^{2} − (trρ_{λ}(φ_{p}))φ_{p} + detρ_{λ}(φ_{p}) = 0 By Eichler-Shimura, we have

(2) φ_{p}^{2} − a_{p}φ_{p} + p = 0 Thus it suffices to prove (detρ_{λ})(φ_{p}) = p which is equivalent to detρ_{λ} = χ_{l} by Cebotarev density theorem where χ_{l} : Gal(\overline{Q}/Q) → Z_{l}^{∗} ⊆ E_{λ}^{∗} is the l-adic cyclotomic character. A complete proof of this theorem will be given next week.

March 4, 1996

Scribe: János Csirik, <janos@math>

Last time we had some weight 2 eigenform f on Γ_{0}(N) of exact level N. (Γ_{1}(N) would give an interesting variant which we shall consider briefly at the end of the lecture.) We found some abelian subvariety A_{f} ⊆ J_{0}(N), it has an action on it by T ⊗ Q which acts through the quotient E_{f} = Q(… a_{n}(f) …), and [E_{f} : Q] = dim A_{f}. E_{f} is a totally real algebraic number field.

We have discovered that Ta_{ℓ}(A_{f}) is free of rank 2 over the ℓ-adic completion E_{f} ⊗ Q_{ℓ} = ∏_{λ|l} E_{λ}, and the Tate module breaks apart thus:

Ta_{ℓ} = ∏_{λ} Ta_{λ}.

We also have

ρ_{λ} : Gal(\overline{Q}/Q) → Aut_{E_{λ}}Ta_{λ} = GL(2, E_{λ}).

For p |̸ ℓN, ρ_{λ}(frob_{p}) satisfies

X^{2} − a_{p}X + p.

(Note that a_{p} is well-defined as E_{f} ↪ C via f: a_{p} is a certain eigenvalue ∈ C.) By the Cayley-Hamilton theorem, ρ_{λ}(frob_{p}) also satisfies

X^{2} − tr ρ_{λ}(frob_{p}) + det ρ_{λ}(frob_{p}) = 0.

Theorem 16 (1) tr ρ_{λ}(frob_{p}) = a_{p}.

(2) det ρ_{λ}(frob_{p}) = p.

Note that (2) implies (1) by the Cayley-Hamilton theorem: use that ρ_{λ}(frob_{p}) is an invertible matrix and subtract the above two polynomials that it is known to satisfy.

To prove (2), we need the following “true theorem”:

Theorem 17 Using χ_{ℓ} to denote the ℓ-adic cyclotomic character,

det ρ_{λ} = χ_{ℓ}.

Once this theorem is proved, we can substitute frob_{p} (on both sides) and get (2) above.

Before tackling the true theorem, we need to talk about a generalization of the Weil pairing that we’ll need in the proof.

We have an isogeny

A_{f} ↪ J_{0}(N)→^{∼}J_{0}(N)^{∨} → A_{f}^{∨}.

This is a polarisation.

Generally, for a polarisation ϕ_{L} of an abelian variety A defined by the line bundle L, we need

ϕ_{L} : A → Pic^{0} A = A^{∨}
a ↦ T_{a}^{∗}L ⊗ L^{−1},

where T_{a} stands for translation by a. By the theorem of the square, this is a homomorphism. We also require that ϕ_{L} be an isogeny (L needs to be ample).

The Weil pairing has a natural generalization to A = A_{f}: For each n, it is a natural map

A[n] × A^{∨}[n] → μ_{n}

which is a bilinear perfect pairing. Given ϕ : A → A^{∨}, we get

A[n] × A[n] → μ_{n}

which is an alternating, but not perfect, pairing. We can make it perfect by setting n = ℓ^{r} and taking lim_{r^{⟵}}:

(., .) : Ta_{ℓ}A × Ta_{ℓ}A → Q_{ℓ}(1) = (lim_{r^{⟵}} μ_{ℓ^{r}}) ⊗_{Z_{ℓ}} Q_{ℓ}.

In fact for a map of abelian varieties t : A → B, we have for β ∈ B^{∨} and a ∈ A,

(ta, β)_{B} = (a, t^{∨}β)_{A}.

Now take A = B and recall our isogeny ϕ : A → A^{∨} above. Then taking t ∈ T_{0} ⊗ Q, we get

(ta, a′) = (a, ta′).

To put this into context, note that since ϕ is an isogeny, it has an inverse if we throw in denominators, so we can take t ∈ End A into t^{∗} ∈ End A ⊗ Q. This is called the Rosati involution associated to the polarisation ϕ. It follows easily that in this case

(ta, a′) = (a, t^{∗}a′).

Looking at J_{0}(N) and its natural polarisation, if t ∈ T comes from a correspondence then t^{∗} comes from its dual.

In our case, ϕ = w is an involution, so t^{∗} = wtw. We have

T_{0} ⊗ Q ⊆ T ⊗ Q ⊆ End(J_{0}(N))

and ^{∗} acts as the identity on T_{0} ⊗ Q. In general, wtw is not always in T ⊗ Q: if t = T_{p} for p|N, t^{∗} does not commute with t. Note that

dim_{Q} T ⊗ Q = dim_{C} S_{2}(Γ_{0}(N))

and

dim_{Q} T_{0} ⊗ Q = ∑_{d|N} dim_{C} S_{2}(Γ_{0}(d))^{new}.

Let’s try to prove our determinant assertion now. Restrict

(., .) : Ta_{ℓ}A × Ta_{ℓ}A → Q_{ℓ}(1)

to

⟨., .⟩ : Ta_{λ}A × Ta_{λ}A → Q_{ℓ}(1).

This is an alternating pairing. Since the projectors come from elements of E, which are their own inverses with respect to the pairing, this pairing is also non-degenerate.

The pairing in Galois-invariant so for σ ∈ Gal(\overline{Q}/Q) and x, y ∈ Ta_{λ}, ⟨σx, σy⟩ = ^{σ}⟨x, y⟩, and since σ acts on Q_{ℓ}(1) by multiplication by χ_{ℓ}(σ), we get

σ⟨x, y⟩ = χ_{ℓ}(σ)⟨x, y⟩.

Fancy proof. ⟨tx, y⟩ = ⟨x, ty⟩ for t ∈ E_{λ}, so ⟨., .⟩ factors through D = ⋀_{E_{λ}}^{2} Ta_{λ}:

⟨., .⟩ : D → Q_{ℓ}(1).

The left hand side is one dimensional over E_{λ}. So we get

⟨., .⟩ ∈ Hom_{Q_{ℓ}}(D, Q_{ℓ}(1)) = Hom_{Q_{ℓ}}(D, E_{λ}(1)),

where the equality is through composition with the trace. So we got D ≅ E_{λ}(1), the left hand side the determinant of our representation, the right hand side acted on by χ_{ℓ}. So we proved

det ρ_{λ} = χ_{ℓ}.

Non-fancy proof. Let σ ∈ Gal(\overline{Q}/Q), and take E_{λ}-linearly independent vectors x, y ∈ Ta_{ℓ}. Set

σx = a(σ)x + b(σ)y
σy = c(σ)x + d(σ)y.

We want to show that

δ = δ(σ) = a(σ)d(σ) − b(σ)c(σ) − χ_{ℓ}(σ) = 0.

Observe that

⟨ax, dy⟩ = ⟨adx, y⟩,

and

⟨by, cx⟩ = ⟨bcy, x⟩ = −⟨x, bcy⟩ = −⟨bcx, y⟩.

Then since

⟨bdy, y⟩ = ⟨by, dy⟩ = −⟨dy, by⟩ = −⟨bdy, y⟩,

if follows that

⟨by, dy⟩ = 0,

and similarly

⟨ax, cx⟩ = 0.

Therefore

χ(σ)⟨x, y⟩ = ⟨σx, σy⟩ = ⟨ax + by, cx + dy⟩ = ⟨(ad − bc)x, y⟩,

So ⟨δx, y⟩ = 0 for any choice of linearly independent vectors x, y. So we proved once again that

δ = 0.

The case of Γ_{1}(N)

For a newform f, we have f|T_{n} = a_{n}(f) for all n ≥ 1, and f|⟨d⟩ = ε(d)f for d ∈ (Z/NZ)^{∗}. Repeating the construction we get an abelian variety A_{f} with an action of a number field E_{f}. However in this case E_{f} is not necessarily real: but if it isn’t then it is equipped with a canonical complex conjugation: \overline{a_{n}} = ε(n)^{−1}a_{n}. Also we have ε(d) ∈ E_{f}^{×}.

In this case, we have

Theorem 18 (1) tr ρ_{λ}(frob_{p}) = a_{p}.

(2) det ρ_{λ}(frob_{p}) = ε(p)p.

This can be proved from the following:

Theorem 19 ε can be considered representation of the Galois group. Still using χ_{ℓ} to denote the ℓ-adic cyclotomic character,

det ρ_{λ} = εχ_{ℓ}.

Signs. There is less gratuitous self-duality going on here than in the case of Γ_{0}(N), so we shall have two occasions to make choices.

For every p, T_{p} is defined as a correspondence on X_{1}(N), with the same α and β that we have used before. Do we want to do (T_{p})^{∗} or (T_{p})_{∗}?

We have another choice in Eichler-Shimura. What does X_{1}(N) classify? Either an elliptic curve E and a point P of order N (which amount to a pair (E, Z/nZ ↪ E)); Or a pair (E, μ_{n} ↪ E). The second one is more suitable for us (and for use with the Tate curve), and it forces us to use the Picard functoriality of the Jacobians. Therefore we shall make T_{p} : J_{1}(N) → J_{1}(N) so that (T_{p}^{∨})^{∗} on H^{0}(X_{1}(N), Ω^{1}) = S_{2}(Γ_{1}(N)) is the classical T_{p}. (Here regard T_{p}^{∨} as an element of End(J_{1}(N)) by the autoduality of J_{1}(N).)

6 March, 1996

Scribe: David M Jones, <dmjones@math>

Today we will discuss the process of reducing a λ-adic representation mod λ.

Recall: Let f = ∑_{n=1}^{∞} a_{n}q^{n} be a weight 2 newform on Γ_{0}(N). That is, f has trivial Nebentypus and is new of level N. We looked at the field E = Q(...a_{n}...). For each prime λ of E lying above a fixed rational prime ℓ, let E_{λ} be the completion of E at the valuation corresponding to λ. Then we created a representation

ρ_{λ} : Gal(Q̄/Q) → GL_{2}(E_{λ})

such that detρ_{λ} = χ_{ℓ}, the ℓ-adic cyclotomic character, and trρ_{λ}(frob_{p}) = a_{p} for p ∤ ℓN.

Lemma 4 ρ_{λ} is equivalent to a representation with values in GL_{2}(O), where O = the ring of integers in E_{λ}.

Proof. An equivalent statement to the lemma is: there is an O-lattice L in V (the vector space on which Gal(Q̄/Q) acts via ρ_{λ}) which is Gal(Q̄/Q)-stable.

Pick any lattice L_{0} in V. The set of lattices gL_{0} with g ∈ Gal(Q̄/Q) is finite. This is because Gal(Q̄/Q) is compact and L_{0} is discrete. So we take L = ∑_{g} gL_{0}, which is stable under Galois. ■

So we are led to ask the following Question: Can we define ρ̄_{λ} = ρ_{λ} mod λ by taking an equivalent representation as in the

lemma and reducing it mod λ?

More specifically, let F = O/λ, a finite field. Then the map O → F induces GL_{2}(O) → GL_{2}(F). Can we define ρ̄_{λ} by the composition

σ : Gal(Q̄/Q) → GL_{2}(O) → GL_{2}(F)?

The answer is no, because this map depends on the choice of L in the lemma. So we do the following:

Definition 11 ρ̄_{λ} = (σ : Gal(Q̄/Q) → GL_{2}(F))^{ss}, the semisimplification of the map described above. That is,

1. If σ is irreducible, ρ̄_{λ} = σ.

2. If σ ∼ ( \begin{matrix}α & ∗ \\ 0 & β \end{matrix} ), then σ^{ss} = α ⊕ β.

Now we list the key properties of ρ̄_{λ}:

(a) ρ̄_{λ} is semisimple.

(b) detρ̄_{λ} = (χ_{ℓ} mod ℓ).

(c) trρ̄_{λ}(frob_{p}) = (a_{p} mod λ) ∈ F, ∀p ∤ ℓN.

Theorem 20 (Brauer-Nesbitt Theorem) Suppose ρ, ρ′ are two representations as above, both semisimple, with the same traces and determinants, then ρ ≅ ρ′.

So ρ̄_{λ} does not depend on the lattice L; it is well-defined. Now let’s view this situation in a new way: T_{n} ∈ End(S_{2}(Γ_{0}(N))), T = Z[...T_{n}...] ⊆

End(S_{2}(Γ_{0}(N)))

Take 𝔪 ⊆ T, a maximal ideal, and T/𝔪 = k = F_{ℓ^{ν}}.

Proposition 5 There exists a (unique) semisimple representation

ρ_{𝔪} : Gal(Q̄/Q) → GL_{2}(k)

such that:

1. ρ_{𝔪} is unramified outside ℓN.

2. trρ_{𝔪}(frob_{p}) = T_{p}∀p |̸ℓN.

3. detρ_{𝔪}(frob_{p}) = p∀p ∤ ℓN.

Proof. First of all, it is enough to prove this for T_{0} = Z[...T_{n}...] ⊆ T, where T_{0} is formed using only those T_{n} with (n, N) = 1. Then T_{0}/𝔪_{0} = k_{0} ⊆ k.

T_{0} ⊗ Q = E_{1} × ... × E_{t}
⋃ |  ⋃ |
T_{0} ⊆ O_{1} × ... × O_{t} = O
⋃ |  ⋃ |
𝔪_{0} =  λ

We can make ρ̄_{λ} : Gal(Q̄/Q) → GL_{2}(O/λ). This has the required properties of ρ_{𝔪} except that it takes values in GL_{2}(O/λ) ⊇ GL_{2}(k_{0}). O/λ being bigger than k_{0} is a potential obstruction to defining ρ_{𝔪} = ρ̄_{λ}.

But, the characteristic polynomials of ρ̄_{λ}(g) have coefficients in k_{0}, and since the Brauer group of a finite field is trivial, there is a model for ρ̄_{λ} over k_{0}. We take ρ_{𝔪} to be this model. ■

Let us digress in order to explain in a different way what we’ve just done.

Recall: T_{0} ⊗ Q = E_{1} × ... × E_{t} and each factor E_{i} corresponds to a newform of level M for some M|N. Suppose E_{1} corresponds to a newform of level N.

We get J_{0}(N) ∼ A_{1} × ... × A_{t}, where we have dim A_{1} = [E_{1} : Q]. Also, T a_{ℓ}A_{1} is free of rank 2 over E_{1} ⊗_{Q} Q_{ℓ}.

But if E_{i} corresponds to a form which is new of level M, M|N, M ≠ N, then this doesn’t work. We must go back to Γ_{0}(M).

Question: Let A = A_{1} ⊆ J_{0}(N). Then we have T → End(A_{1}). Let T̄ be the image of this map. So T̄ ⊆ O_{1}, but T̄ may not be integrally closed.

But there exists a Z_{ℓ}-adic Tate module: H_{ℓ}(A) = ∏_{𝔪} H_{𝔪}(A) which is isomorphic as a group to Z_{ℓ}^{2 dim A}. H_{ℓ}(A) has an action of T̄ ⊗ Z_{ℓ} = ∏_{𝔪} T̄_{𝔪}. For 𝔪|ℓ, we have H_{𝔪} free of rank 2 over T̄_{𝔪}.

Let J = J_{0}(N), T ⊆ EndJ. Let 𝔪 ⊆ T be a maximal ideal. Then define J[𝔪] = {t ∈ J(Q̄)|xt = 0∀x ∈ 𝔪}. Then J[𝔪] ⊆ J[ℓ], and J[ℓ] is isomorphic as a group to (Z/ℓZ)^{2g}. J[ℓ] is a T/ℓT-module.

But J[𝔪] is “better” because it is a T/𝔪-module and T/𝔪 is a field, so J[𝔪] is a vector space. J[𝔪] is a representation of k[G] = (T/𝔪)[Gal(Q̄/Q)].

Naive Idea: J[𝔪] is a model for ρ_{𝔪}, at least when ρ_{𝔪} is irreducible.

Theorem 21 If ℓ ≠ 2 and ℓ ∤ N, then J[𝔪] is a model for ρ_{𝔪}.

Note: J[𝔪] may indeed be a model for ρ_{𝔪} in other cases.

Let V = k ⊕ k with the ρ_{𝔪}-action of Gal(Q̄/Q).

Fact: dim_{T/𝔪} H^{0}(X_{0}(N)_{/F_{ℓ}}, Ω^{1})[𝔪] ≤ 1.

8 March, 1996

Scribe: David M Jones, <dmjones@math>

T = Z[...T_{n}...] ⊆ EndJ_{0}(N). Let 𝔪 ⊂ T be a maximal ideal with 𝔪|ℓ. As we showed last time, there exists a representation

ρ_{𝔪} : Gal(Q̄/Q) → GL_{2}(T/𝔪)

which is semisimple.

Definition 12 Let R be a Z_{ℓ}-algebra. Then we say that R is Gorenstein if Hom_{Z_{ℓ}}(R, Z_{ℓ}) ≅ R.

Goal: To prove that if 𝔪 ∤ 2N and ρ_{𝔪} is irreducible, then T_{𝔪} is Gorenstein.

Theorem 22 If J = J_{0}(N), J[𝔪] ⊆ J[ℓ], then dim_{T/𝔪} J[𝔪] = 2. That is, ρ_{𝔪} arrises from J[𝔪].

Remarks: Let M = Hom(T, Z) = (S_{2}(Γ_{0}(N)), Z), a T-module. Saying that M_{𝔪} is Gorenstein amounts to M_{𝔪} being free of rank 1 over T_{𝔪}.

Let W = J[𝔪]. We will compare W and V, where V is the Galois module given by ρ_{𝔪} : Gal(Q̄/Q) → Aut_{T/𝔪}V.

Let W^{ss} be the direct sum of the Jordan-Holder quotients of W.

Theorem 23 (Mazur) W^{ss} is isomoprhic as a group to V × ... × V, with t copies of V. (This is true ∀N as long as ρ is irreducible.)

Remark: A variant of this is the following

Definition 13 A representation ρ_{𝔪} : Gal(Q̄/Q) → GL_{2}(T/𝔪) is absolutely irreducible if the composition Gal(Q̄/Q) → GL_{2}(T/𝔪) ↪ GL_{2}(\overline{T/𝔪}) is irreducible, where \overline{T/𝔪} is the algebraic closure of T/𝔪.

Theorem 24 (Boston-Lenstra-Ribet) If ρ_{𝔪} is absolutely irreducible, then W is isomorphic as a group to V × ... × V.

Proof. (Of Mazur’s theorem) Suppose d = dim W. Consider

Hom_{F−ℓ}(W, μ_{ℓ}) = Hom_{T/𝔪}(W, T/𝔪(1)) = W^{∗}.

We will show that

W^{ss} ⊕ (W^{∗})^{ss} = V × ... × V

where the product has 2d copies of V.

We will show that the two representations above have the same characteristic polynomials. Then the Brauer-Nesbitt theorem will prove that they are isomorphic.

Look at frob_{p}. It has the following characteristic polynomial on V: X^{2} − T_{p}X + p = (X − r)(X − pr^{−1}) ∈ (T/𝔪)[X]. So the characteristic polynomial of frob_{p} on V^{d} is (X − r)^{d}(X − pr^{−1})^{d}.

And the characteristic polynomial of frob_{p} on W is (X − α_{1}) · ... · (X − α_{d}), where α_{i} is r or pr^{−1}. The α_{i}’s are of this form because frob_{p}^{2} − T_{p}frob_{p} + p = 0. So the characteristic polynomial of frob_{p} on W ⊕ W^{∗} is (X − α_{1})(X − pα_{1}^{−1}) · ... · (X − α_{d})(X − pα_{d}^{−1}) = (X − r)^{d}(X − pr^{−1})^{d}. ■

Theorem 25 J[𝔪] ≠ 0. ie. t ≥ 1.

Proof. Suppose J[𝔪] = 0. Then J[𝔪^{i}] = 0∀i ≥ 1. This follows from Nakayama’s lemma. So now look at the ℓ-divisible group J_{𝔪} = ∪_{i}J[𝔪^{i}]. So in order to show J[𝔪] ≠ 0, equivalently, we will show that J_{𝔪} ≠ 0. From the discussion to follow, we will see that J_{𝔪} is dual to T a_{𝔪}^{∗}J. But T a_{𝔪}^{∗}J is non-zero by the theorem of Weil stated below, and thus J_{𝔪} is non-zero; hence, J[𝔪] is non-zero. ■

Theorem 26 (Weil) T ⊗ Z_{ℓ} ⊆ EndJ ⊗ Z_{ℓ} injects into End(T a_{ℓ}J).

Corollary 2 T a_{ℓ}J ≠ 0.

Proof. Since T ⊗ Z_{ℓ} ≠ 0 and T ⊗ Z_{ℓ} injects into End(T a_{ℓ}^{∗}J), thus End(T a_{ℓ}^{∗}J) ≠ 0 and so we must have T a_{ℓ}^{∗}J ≠ 0. ■

Now we will discuss the above duality between J_{𝔪} and T a_{𝔪}^{∗}J.

T a_{ℓ}J = lim_{⟵} J[ℓ^{i}], which is isomorphic as a group to Z_{ℓ}^{2 dim J}. On the other hand, J_{ℓ} = ∪_{i=1}^{∞}J[ℓ^{i}] which is isomorphic as a group to (Q_{ℓ}/Z_{ℓ})^{2 dim J}.

Definition 14 The covariant Tate module T a_{ℓ}J is Hom(Q_{ℓ}/Z_{ℓ}, J_{ℓ}).

Definition 15 The contravariant Tate module T a_{ℓ}^{∗}J is Hom(J_{ℓ}, Q_{ℓ}/Z_{ℓ}).

The contravariant Tate module is the Pntryajin dual of J_{ℓ}.

Note that Hom(J_{ℓ}, Q_{ℓ}/Z_{ℓ}) = Hom_{Z_{ℓ}}(T a_{ℓ}J, Z_{ℓ}) so T a_{ℓ}^{∗}J = Hom(T a_{ℓ}J, Z_{ℓ}).

Now, we have the Weil pairing, J[ℓ^{i}] × J[ℓ^{i}] → μ_{ℓ^{i}}. Passing to the limit induces a pairing <, >: T a_{ℓ}J × T a_{ℓ}J → Z_{ℓ}(1) = lim_{⟵} μ_{ℓ^{i}}. This pairing is perfect and gives us that

T a_{ℓ}J→^{∼}Hom(T a_{ℓ}J, Z_{ℓ}(1)) = (T a_{ℓ}^{∗}J)(1).

Now if t ∈ T, then ⟨tx, y⟩ = ⟨x, t^{∨}y⟩ where t^{∨} = wtw and w = w_{N} ∈ EndJ_{0}(N) is the Atkin-Lehner involution. So this pairing is not T-compatible.

So we define a new pairing, which is T-compatible, by

[ , ] = ⟨x, wy⟩.

Check for T-compatibility: [tx, y] = ⟨tx, wy⟩ = ⟨x, t^{∨}wy⟩ = ⟨x, wty⟩ = [x, ty].

As a reminder: w is an involution on X_{0}(N) given by (E, C) ↦ (E/C, E[N]/C).

[ , ] defines an isomorphism of T ⊗ Z_{ℓ}-modules:

T a_{ℓ}J→^{∼}T a_{ℓ}^{∗}J(1) = Hom_{Z_{ℓ}}(T a_{ℓ}J, Z_{ℓ}(1))

which is isomorphic as a group to Hom(T a_{ℓ}J, Z_{ℓ}) = T a_{ℓ}^{∗}J.

1

March 11, 1996

Scribe: János Csirik, <janos@math>

Our basic situation is that we have J_{0}(N) (or J_{1}(N)) and a maximal ideal 𝔪 ⊆ T ⊆ End(J_{0}(N)) lying over ℓ. We have T ⊗_{Z} Z_{ℓ} = ∏_{𝔪|ℓ} T_{𝔪}, where T_{𝔪} is the 𝔪-adic completion of T. This ring acts on Ta_{ℓ}, breaking it up as ∏_{𝔪|ℓ} Ta_{𝔪}. The last time we have confirmed that Ta_{𝔪} ≠ 0, and that Ta_{ℓ} is T ⊗ Z_{ℓ}-autodual, hence each T_{𝔪} is Z_{ℓ}-autodual.

On a finite level, take J[𝔪] = W, a representation of Gal(\overline{Q}/Q) over the field T/𝔪. Also

V = 2-dimensional representation of Gal(\overline{Q}/Q) over T_{𝔪}

which gives the representation ρ_{𝔪} (which is given by its trace, determinant and the fact that it is semisimple.)

Now assume that V is irreducible. Then

W^{ss} ≅ V × ⋯ × V

(with t copies of V on the right hand side). This was proved in the last lecture using an argument of Mazur. We want to show that t = 1.

We already know that t > 0, since by Nakayama’s Lemma, W = 0 implies Ta_{𝔪} = 0 which we know is false.

How is one supposed to think about J[ℓ]? Think of it as J[ℓ] = Ta_{ℓ}/ℓTa_{ℓ}. Since Ta_{ℓ}^{∗} is the isomorphic linear dual of Ta_{ℓ}, you can think of J[ℓ] as Ta_{ℓ}^{∗}/ℓTa_{ℓ}^{∗}. Then

J[𝔪] = Ta_{𝔪}^{∗}/𝔪Ta_{𝔪}^{∗}

by Pontrjagin duality.

More on Pontrjagin duality: J_{ℓ} = J[ℓ^{∞}] is Pontrjagin dual to Ta_{ℓ}^{∗} = Hom_{Z_{ℓ}}(Ta_{ℓ}, Z_{ℓ}) = Hom_{Z_{ℓ}}(J[ℓ^{∞}], Q_{ℓ}/Z_{ℓ}).

In general, Pontrjagin duality assigns to each discrete abelian group M a compact abelian group M^{∗} = Hom(M, Q/Z); and to each compact abelian group N a discrete Hom(N, Q/Z). Therefore the groups Z and Q/Z correspond to one another.

We are using a kind of local version under which the group Q_{ℓ}/Z_{ℓ} is sent to Z_{ℓ} and the group Z_{ℓ} is sent to Q_{ℓ}/Z_{ℓ}.

Given the group

J[ℓ^{∞}] = Hom(Ta_{ℓ}^{∗}, Q_{ℓ}/Z_{ℓ}),

it has the subgroup

J[𝔪^{∞}] = Hom(Ta_{𝔪}^{∗}, Q_{ℓ}/Z_{ℓ}),

which in turn contains

J[m] = Hom_{Z/ℓZ}(Ta_{𝔪}^{∗}/𝔪Ta_{𝔪}^{∗}, Q_{ℓ}/Z_{ℓ}).

Since the part of Q_{ℓ}/Z_{ℓ} killed by ℓ is just Z/ℓZ, this can be written as

J[m] = Hom_{Z/ℓZ}(Ta_{𝔪}^{∗}/𝔪Ta_{𝔪}^{∗}, Z/ℓZ).

Now if J[m] is zero, then so must be Ta_{𝔪}^{∗}/𝔪Ta_{𝔪}^{∗}, but then by Nakayama’s Lemma Ta_{𝔪}^{∗} should be zero, which it isn’t.

We have the following to goals:

First goal. To show that t = 1, i.e., that J[𝔪] is generated by two elements over T/𝔪. (Since J[𝔪] is killed by 𝔪, this is equivalent to showing that J[𝔪] is generated over T by two elements.)

Second goal. To show that T_{𝔪} is Gorenstein, i.e., that

T_{𝔪} ≅ Hom_{Z_{ℓ}}(T_{𝔪}, Z_{ℓ}).

(Why would we want this? Well, recall that S(Z) ≅ Hom_{Z}(T, Z) as T-modules. Our goal just says that S(Z) is locally free at 𝔪.)

We are assuming that ρ_{𝔪} is irreducible, and that ℓ |̸2N. ℓ |̸N means that J has good reduction at ℓ, and 2 |̸N means that we can apply Raynaud’s results on group schemes.

Postpone proving the first goal now, and show instead why the first goal implies the second one.

By Nakayama’s Lemma, Ta_{𝔪} is generated by 2 elements as a module over T_{𝔪}. We need that Ta_{𝔪}^{∗}/𝔪Ta_{𝔪}^{∗} is also generated by 2 elements: but for T/𝔪 vector spaces R, we have

Hom_{Z/ℓZ}(R, Z/ℓZ) ≅ Hom_{T/𝔪}(R, Z/ℓZ)

by considering the trace. So we can find a map

T_{𝔪} × T_{𝔪} ↠ Ta_{𝔪}.

Claim 5

2rank_{Z_{ℓ}}T_{𝔪} = rank_{Z_{ℓ}}Ta_{𝔪}.

Recall that rank_{Z_{ℓ}}Ta_{ℓ} = 2 dim J = 2 dim T = 2rank_{Z_{ℓ}}T ⊗ Z_{ℓ}. This makes the claim pausible, but isn’t a proof. The proof is postponed for now.

Accepting the Claim, it follows by basic graduate algebra (Math 250) that T_{𝔪} × T_{𝔪} ≅ Ta_{𝔪}. Recall that Ta_{𝔪} is auto-dual in the following sense:

Ta_{𝔪} ≅ Hom_{Z_{ℓ}}(Ta_{𝔪}, Z_{ℓ}).

Hence T_{𝔪} × T_{𝔪} is self-dual in the same sense. Does it follow that T_{𝔪} is self-dual?

Yes. To do this, use a stupid thing: call two copies of T_{𝔪} F_{1} = T_{𝔪} and F_{2} = T_{𝔪}. Then we have

F_{1}^{∗} × F_{2}^{∗}→^{∼}F_{1} × F_{2},

where ^{∗} denotes taking linear duals (Hom of the thing to Z_{ℓ}). Then F_{1}^{∗} × F_{2}^{∗} is free, so F_{1}^{∗} is projective, so it’s free (over a local ring, all projective modules are).

Now going back to our claim, note that rank_{Z_{ℓ}}T_{𝔪} = dim_{Q_{ℓ}}(T_{𝔪}⊗_{Z_{ℓ}}Q_{ℓ}). Also rank_{Z_{ℓ}}Ta_{𝔪} = dim_{Q_{ℓ}}(Ta_{𝔪} ⊗_{Z_{ℓ}} Q_{ℓ}).

Ta_{ℓ} is gotten by considering the ℓ-power torsion points in the Jacobian, or else it is just those complex points in J(C). But using the Albanese covariant interpretation, (and recalling that H^{1}(X_{0}(N), C) = S_{2}(Γ_{0}(N), C)),

J(C) = \frac{Hom_{C}(S_{2}(Γ_{0}(N), C), C)}{H_{1}(X_{0}(N), Z)}.

The lattice L = H_{1}(X_{0}(N), Z) is a T-module, and Ta_{ℓ} = L ⊗_{Z} Z_{ℓ}.

Instead, we’ll look at

L ⊗_{Z} R = Hom_{C}(S_{2}(Γ_{0}(N), C), C) = Hom_{R}(S_{2}(Γ_{0}(N), R), C)
= Hom_{R}(S_{2}(Γ_{0}(N), R), R) ⊗_{R} C = (T ⊗_{Z} R) ⊗_{R} C,

hence L ⊗_{Z} R = (T ⊗_{Z} R) ⊗_{R} C, so L ⊗_{Z} R is free of rank 2 over T ⊗_{Z} R, therefore L ⊗_{Z} C is free of rank 2 over T ⊗_{Z} C.

Now choose an embedding Q_{ℓ} ↪ C. Ta_{ℓ} = ∏_{𝔪} Ta_{𝔪} is a module over T ⊗ Z_{ℓ} = ∏_{𝔪} T_{𝔪}. Consider Ta_{ℓ} ⊗ ZQ_{ℓ} = ∏_{𝔪}(Ta_{𝔪} ⊗_{Z_{ℓ}} Q_{ℓ}.

∏_{𝔪}(Ta_{𝔪} ⊗_{Z_{ℓ}} Q_{ℓ}) = Ta_{ℓ} ⊗_{Z_{ℓ}} C = L ⊗_{Z} C is free of rank 2 over T ⊗_{Z} C = (T ⊗_{Z} Z_{ℓ}) ⊗_{Z_{ℓ}} C = ∏_{𝔪}(Ta_{𝔪} ⊗_{Z_{ℓ}} C). Hence Ta_{𝔪} ⊗_{Z_{ℓ}} C is free of rank 2 over T_{𝔪} ⊗_{Z_{ℓ}} C, therefore dim_{C} Ta_{𝔪} ⊗_{Z_{ℓ}} C = 2 dim_{C} T_{𝔪} ⊗_{Z_{ℓ}} C, so dim_{Q_{ℓ}} Ta_{𝔪} ⊗_{Z_{ℓ}} Q_{ℓ} = 2 dim_{Q_{ℓ}} T_{𝔪} ⊗_{Z_{ℓ}} Q_{ℓ}, as the dimension of a vector space does not change if we tensor up to a bigger field.

This settles our claim that the second goal implies the first one. The first goal shall be proved in the next lecture.

March 13, 1996

Scribe: Amod Agashe, <amod@math>

Let S be a scheme. Then a group scheme over S is a group object in the category of S-schemes i.e. it is an S-scheme G with S-morhphisms m : G×G → G, i : G → G, e : S → G satisfying the usual group axioms (See [1] for details).

We will be interested in finite flat group schemes over F_{p} where p is a prime and F_{p} is the finite field with p elements. Let A be a F_{p}-vector space and let G = Spec(A). Then the order of G, #G = dim_{F_{p}}A. If R is an F_{p}-algebra, then G(R) = Mor(SpecR, G) is the group of R-valued points of G.

Here are some examples: 1)The additive group scheme Γ_{a} = Spec(F_{p}[T]).

2) The multiplicative group scheme Γ_{m} = Spec(F_{p}[T, T^{−1}]).

3) The pth roots of unity μ_{p} is the kernel of the p-power map on Γ_{m}.

4) The group α_{p} is the kernel of the Frobenius map on Γ_{a}.

5) Let E/F_{p} be an elliptic curve thought of as a group scheme. Then E[p] is the kernel of the multiplication by p map on E. It has order p^{2} and

E[p](\overline{F_{p}}) = \{ \begin{matrix}1 & if E is supersingular \\ p & if E is ordinary \end{matrix}

The primary reference for what follows is [2]. Let J = J_{0}(N) be the Jacobian of X_{0}(N). Then J is defined over Q and has good reduction at primes not dividing N. Let l be a prime not dividing N. Then J[l] extends to a finite flat group scheme over Z[\frac{1}{N}] (Ref: [3, exp. IX]). This in turn gives rise to a group scheme over F_{l} which can be thought of as J_{/F_{l}}[l].

Let V be the two dimensional vector space over T/𝔪 which gives rise to the representation ρ_{𝔪} constructed last week. We want to show that this is isomorphic to the naturally defined Galois representation W = J[𝔪]. We have 0 ⊂ V ⊂ W ⊂ J[l]. We make the following assumptions: l |̸N, l ≠ 2 and V is irreducible. This is because we will be invoking theorems of Raynaud.

Let \underline{J} denote J considered as a group scheme. Then \underline{J}[l] is a finite flat group scheme over Z_{l}; in fact, it is the Spec of a free finite rank module over Z_{l} of rank l^{2g}. Raynaud showed (See [4] or [1]) that if l ≠ 2 (so that the absolute ramification index ρ = 1 < l − 1), almost everything about the group scheme \underline{J}[l] can be seen in terms of its Galois module J[l](\overline{Q_{l}}). Let \underline{V} and \underline{W} be the group schemes given by the Zariski closure of V and W in \underline{J}[l]_{/Z_{l}} respectively. Then we have:

\underline{V} ⊂ \underline{W} ⊂ \underline{J}[l].

Our goal is to show that V ↪ W is an isomorphism. Raynaud showed that the category of finite flat group schemes over Z_{l} is an abelian category, so it makes sense to talk about the quotient \underline{Q} = \underline{W} /\underline{V}, which is again a finite flat group scheme over Z_{l}. Note that \underline{Q}_{/Q_{l}} corresponds to W/V. Thus, V = W ⇔ \underline{Q}_{/Q_{l}} = 0. Since \underline{Q} is flat, it has the same rank over Q_{l} as over F_{l}. Passing to characteristic l yields the following exact sequence:

0 → \underline{V}_{/F_{l}} → \underline{W}_{/F_{l}} → \underline{Q}_{/F_{l}} → 0.

Again, since the ranks remain constant,

V ↪ W is an isomorphism ⇔ \underline{V}_{/F_{l}} ↪ \underline{W}_{/F_{l}} is an isomorphism ⇔ \underline{Q}_{F_{l}} = 0.

Following Raynaud, \underline{V}, \underline{W} and \underline{Q} inherit the action of k = T/𝔪 (i.e. they are k-vector space schemes). We will show that \underline{Q}_{F_{l}} = 0. For that, we need Dieudonné module theory.

Let G/k be a finite k-vector scheme where k is a finite field of order q. Suppose the order of G is q^{n}. Then the Dieudonné module functor D (See [5]) is a contravariant functor which maps G to a k-vector space D(G) of dimension n, called the Dieudonné module of G. Let F rob : G → G be the morphism on G induced by the pth power map on the underlying sets and let V er be the dual of F rob. Then the maps ϕ and ν induced on D(G) by F rob and V er respectively satisfy ϕ ∘ ν = ν ∘ ϕ = 0. D is a fully faithful functor i.e. we can tell everything about the group scheme from its Dieudonné module.

For example, let k = F_{l} = F_{p}. Let G = μ_{p}, α_{p} or Z/pZ. Then in all cases, dim_{k}(G) = 1. For α_{p}, ϕ = ν = 0; for μ_{p}, ϕ = 0, ν = id and for Z/pZ, ϕ = id, ν = 0.

Let G^{∨} = Hom(G, μ_{p}) denote the Cartier dual of G. Then D(G^{∨}) = Hom_{k}(D(G), k) with ϕ and ν interchanged.

Here is a sophisticated example of the above theory: Let A be an abelian variety over F_{l} and G = A[l]. Then, G is an F_{l}-vector space scheme of order l^{2g} and thus D(G) has dimension 2g. Furthermore, D(G) = H_{DR}^{1}(A_{/F_{l}}). Consider the Hodge filtration:

0 → H^{0}(A, Ω^{1}) → H_{DR}^{1}(A_{/F_{l}}) → H^{1}(A, O_{A}) → 0.

Now H^{1}(A, O_{A}) = T an(A_{F_{l}}^{∨}) = Hom(H^{0}(A^{∨}, Ω^{1}), F_{l}) and it corresponds to the subgroup D(A_{F_{l}}[V er]) of D(A_{/F_{l}}) where V er is the dual to F rob.

Consider the exact sequence

0 → W_{/F_{l}} → J_{/F_{l}}[l] →^{𝔪} J_{/F_{l}}[l].

Since D is an exact functor, applying D to the above, we get the exact sequence

D(J_{/F_{l}}[l]) →^{𝔪} D(J_{/F_{l}}[l]) → D(W_{/F_{l}}) → 0.

In other words, D(W_{/F_{l}}) = D(J_{/F_{l}}[l])/𝔪D(J_{/F_{l}}[l]).

Next time, following Fontaine, we shall consider D(W_{/F_{l}}[V er]) = H^{1}(J, O)/𝔪H^{1}(J, O). The latter is dual (in the sense of Hom of a vector space into F_{l}) to H^{0}(J^{∨}, Ω^{1})[𝔪] = H^{0}(X_{0}(N)_{/F_{l}}, Ω^{1})[𝔪]. For the duality above, we need to use J^{∨} = Alb(X_{0}(N)) i.e. J = P ic^{0}(X_{0}(N)). We will be using the q-expansion principle to show that D(W_{/F_{l}}[V er]) is small.

References:

[1] S. Shatz, Group Schemes, Formal Groups, and p-Divisible Groups. In Arithmetic Geometry, G.Cornell and J.Silverman, eds., Springer-Verlag, New York, 1986, pg. 29-78.

[2] B.Mazur, Modular Curves and the Eisenstein Ideal, Publ.Math. I.H.E.S., 47 (1977), 33-186.

[3] Séminaire de Géométrie algébrique du Bois-Marie, 67-69. P.Deligne and N.Katz, Lecture Notes in Mathematics, nos. 288,340, Berlin-Heidelberg-New York, Springer, 1972, 1973.

[4] M. Raynaud, Schémas en groupes de type (p,p,...,p), Bull. Soc. Math. Fr., 102 (1974), 241-280.

[5] T.Oda, The first De Rham cohomology group and Dieudonné modules, Ann. scient. É.Norm.Sup., 4^{e} séie, t. 2 (1969), 63-135.

April 1, 1996

Scribe: János Csirik, <janos@math>

Local properties of ρ_{λ}

Summary of all we know, without proofs: ρ_{λ} is associated to f, a newform of weight 2, level N, character ε : (Z/nZ)^{∗} → C^{∗} (i.e., we don’t necessarily only consider Γ_{0}(N) any more), conductor of ε dividing N (not necessarily equal to N, could be trivial). Let λ|ℓ, and look locally at p (with p ≠ ℓ).

The real story is that f gives rise to some irreducible representation of GL(2, A), which corresponds to a family of representations (π_{v}), with π_{v} being a representation of GL(2, Q_{v}). Here v stood for any valuation of Q, but for the archimedean prime in the weight 2 case we always get a well-known discrete series representation on GL(2, R), so we really only care about the representations π_{p} on GL(2, Q_{p}).

It is a theoretical result of Carayol (building on the work of Deligne and Langlands) that ρ_{λ}|_{D_{p}} up to isomorphism only depends on π_{p}.

In our case, we are most interested in the case where p^{2} |̸N. A reference is R. P. Langlands: Modular Forms and ℓ-adic representations, pp. 361–500 in Antwerp II.

For a warmup, consider the case of p |̸N. Then we have the following “characteristic polynomial” floating about: X^{2} − a_{p}X + pε(p). Factor it as (X − r)(X − s) with r, s ∈ C. From the Riemann Hypothesis for abelian varieties (proved by André Weil) it follows that |r| = |s| = \sqrt{p}.

ρ_{λ}|_{D_{p}} is an unramified representation so I_{p} is killed by it, so frob_{p} makes sense up to conjugacy. So ρ_{λ}(frob_{p}) has a well-defined characteristic polynomial X^{2}−a_{p}X+pε(p) ∈ E[X], where E is the field generated over Q by the coefficients of our newform f. So since p does

not divide N, ρ_{λ}(frob_{p}) is semisimple, so it follows that ρ_{λ}(frob_{p}) is conjugate to ( \begin{matrix}r & 0 \\ 0 & s \end{matrix} ).

What is π_{p} supposed to be? Let

α, β : Q_{p}^{∗} → C^{∗}

be Grössencharacters (i.e., charaters with values that don’t necessarily have absolute value 1), such that

1. α, β are unramified: α|_{Z_{p}^{∗}} = β|_{Z_{p}^{∗}} = 1 (here we used 1 to denote the unit character). Note that this condition is called being unramified because by local class field theory, Z_{p}^{∗} corresponds to I_{p} in Gal(\overline{Q}_{p}/Q_{p});

2. α(p^{−1}) = r and β(p^{−1}) = s.

By the relation between Grössencharacters and characters of Gal(\overline{Q}/Q) which is explained in Weil (1950), or J.-P. Serre and J. Tate, Good reduction of abelian varieties, Ann. of Math., 2, 1968, pp. 492–517, or Serre (1972). α and β correspond to characters

α_{λ}, β_{λ} : Gal(\overline{Q}/Q) → \overline{E_{λ}}^{∗},

which satisfy

1. α_{λ}, β_{λ} are unramified;

2. α_{λ}(frob_{p}) = r, β_{λ}(frob_{p}) = s.

Then ρ_{λ} = α_{λ} ⊕ β_{λ}.

Define a character of the Borel subgroup B = ( \begin{matrix}∗ & ∗ \\ 0 & ∗ \end{matrix} ) ∈ GL(2, Q_{p}),

θ : ( \begin{matrix}x & y \\ 0 & z \end{matrix} ) ↦ α(x)β(z).

This is a character θ : B → C^{∗}. Then π_{p} = Ind_{B}^{GL(2,Q_{p})}θ is irreducible since r/s is neither p nor p^{−1}. This is called the (unramified) principal series representation attached to α and β. (Unramified in this case since α and β are unramified.) This is denoted

π_{p} = PS(α, β).

Here is the end of the warmup.

Now look at the case where p exactly divides N. There are two very different cases, according as ε is ramified or unramified at p. (It is ramified iff p divides the conductor of ε.)

The case of ε ramified at p. Here π_{p} = PS(α, β), and ρ_{λ} = α_{λ} ⊕ β_{λ}. α (say) is unramified and β is ramified (with conductor = p), i.e. β|_{Z_{p}^{∗}} is nontrivial. But it is not too bad since β|_{1+pZ_{p}^{∗}} = 1 for p ≠ 2.

To see what’s going on (what is α and β?), think about L-functions. Let f = ∑ a_{n}q^{n} and fix attention to a_{p} (which is non-zero by the analytic point of view). By definition,

L(f, s) = ∑_{n=1}^{∞} a_{n}/n^{s} = ∏_{p|̸N}(1 − a_{p}p^{−s} + pε(p)p^{−2s})^{−1} ∏_{p|N}(1 − a_{p}p^{−s})^{−1},

the L-factor corresponding to p being (1 − a_{p}p^{−s})^{−1}.

Now V_{λ} has a one-dimensional quotient on which frob_{p} acts by multiplication by the number a_{p}. So the unramified character α can only be the one with α(p^{−1}) = a_{p}. But we know that α_{λ}β_{λ} = detρ_{λ} = εχ_{ℓ} (with ε corresponding to the Grössencharacter taking p^{−1} to p). Hence we must have β(p^{−1}) = ε(p)pa_{p}^{−1}.

Now use the formula \overline{f} = ∑ \overline{a_{n}}q^{n} = f ⊗ ε^{−1}. For (n, N) = 1, we have a_{n} = \overline{a_{n}}ε(n), so in particular this holds for almost all primes, so by the Chebotarev density theorem ρ_{λ,f} = ρ_{λ,\overline{f}} ⊗ ε, hence if we let α_{λ}, β_{λ} to correspond to f and δ_{λ}, γ_{λ} to correspond to \overline{f}, we have

( \begin{matrix}α_{λ} & 0 \\ 0 & β_{λ} \end{matrix} ) = ( \begin{matrix}γ_{λ}ε & 0 \\ 0 & δ_{λ}ε \end{matrix} ) ,

so up to permutation we must have a_{λ} = γ_{λ}ε and b_{λ} = δ_{λ}ε (*). Now exactly one of γ_{λ} and δ_{λ} is ramified (exactly as in the case of α and β). Hence by the (*) above, δ_{λ} is unramified while γ_{λ} is ramified. So using det = χ_{ℓ}ε, we get

β_{λ} = δ_{λ}ε = εχ_{ℓ}α_{λ}^{−1}.

Looking at the Frobenii, \overline{a_{p}} = pa_{p}^{−1}, hence \overline{a_{p}}a_{p} = p, so

|a_{p}| = \sqrt{p}.

The case of ε unramified at p. So here we can talk of ε(p), and it is going to be a root of 1 in E^{∗}. It is supposed to remind you of the Tate curve (elliptic curve with multiplicative reduction at p). Now

ρ_{λ}|_{D_{p}} = ( \begin{matrix}α_{λ} & ∗ \\ 0 & β_{λ} \end{matrix} ) .

Here α_{λ} = χ_{ℓ}β_{λ} and α_{λ} = β_{λ} = det = εχ_{ℓ}, so plugging in we get β_{λ}^{2} = ε. So β_{λ} is unramified,

β_{λ}(frob_{p}) = a_{p} α_{λ}(frob_{p}) = pa_{p}.

ρ_{λ}|_{D_{p}} is an indecomposable representation: the ∗ in the above formula is non-zero. Since p ≠ ℓ, this property characterizes it. We can say that

ρ_{λ} = β_{λ} ⊗ ( \begin{matrix}χ_{ℓ} & ∗ \\ 0 & 1 \end{matrix} ) ,

where the second term in the tensor product is the unique indecomposable λ-adic representation of Gal(\overline{Q}_{p}/Q_{p}) with semisimplification 1 ⊕ χ_{ℓ}. Correspondingly

π_{p} = β ⊗ st,

where st is a pun for both the “standard representation” and the “Steinberg representation”. Morally, a PS representation attached to (1, ||.||) is not irreducible, but it has a unique irreducible quotient.

Remarks. A special representation is any abelian character tensored with the Steinberg representation.

Langlands invented a dictionary between representations of GL(2, Q_{p}) and representations of Gal(\overline{Q}_{p}/Q_{p}), which goes through complex representations of a smaller group (the Weil group).

0.1 The weight and fundamental characters

Let

ρ : Gal(\overline{Q}/Q) → GL(2, F_{ℓ^{ν}})

be an irreducible modular mod ℓ representation. Let D = D_{p} = Gal(\overline{Q}_{p}/Q_{p}) be the decomposition group at p. Today we will see how studying the local properties of ρ help us to understand the following theorem.

Theorem 27 If ℓ > 2 and ρ is modular then ρ is modular of a certain weight k(ρ) and level N(ρ).

Since ρ is modular, ρ ∼ \overline{ρ}_{λ,f} for some modular form f. The theorem asserts that we can actually take f ∈ S_{k(ρ)}(Γ_{1}(N(ρ))). As we have seen before N(ρ) is the prime to ℓ Artin conductor of ρ.

Today we attack the question of what k(ρ) should be. To do this we restrict to the inertia group I_{p}. Before doing this we introduce a certain pair of characters. First we prove a lemma. Let σ be the semisimplification of ρ|_{D}. Thus σ is either a direct sum of two characters or ρ|_{D} depending on whether or not ρ|_{D} is irreducible.

Lemma 5 The representation σ is tame, i.e, σ is trivial on the ℓ-Sylow subgroup of I_{ℓ}.

Proof. Let P be the ℓ-Sylow subgroup of I_{ℓ}. Since I_{ℓ} is normal in D and P is unique it follows that P is normal in D. Let W = F_{ℓ^{ν}} × F_{ℓ^{ν}} be the representation space of σ. Then

W^{P} = {w ∈ W : σ(τ)w = w for all τ ∈ P }

is a subspace of W invariant under the action of D. To see this let α ∈ D and suppose w ∈ W^{P}. Since P is normal in D, α^{−1}τα = τ′ for some τ′ ∈ P. Therefore

σ(α)^{−1}σ(τ)σ(α)w = σ(τ′)w = w

so σ(τ)σ(α)w = σ(α)w hence σ(α)w ∈ W^{P}.

But W^{P} ≠ 0. To see this write W as a disjoint union of its orbits under the action of P. Since P is an ℓ-Sylow group and W is finite we see that the size of each orbit is either 1 or a positive power of ℓ. Now {0} is a singleton orbit, W has ℓ-power order, and all non-singleton orbits have order a positive power of ℓ so there must be at least ℓ − 1 other singleton orbits. Each of these other singleton orbits gives a nonzero element of W^{P}.

If W^{P} = W then P acts trivially so we are done. If W^{P} ≠ W then W^{P} is a one dimensional subspace invariant under D so by semisimplicity σ is a diagonal representation. Suppose τ ∈ P then τ has order ℓ^{n} for some n. Write

σ(τ) = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )

then α^{ℓ^{n}} = 1 and β^{ℓ^{n}} = 1. Since α, β ∈ F_{ℓ^{ν}} it follows that α = β = 1. ■

The lemma implies σ|_{I} factors through the tame quotient I_{t} = I/P. We now describe I_{t} explicitly. Let Q_{p}^{tame} be the maximal tame extension of Q_{p} and let K = Q_{p}^{ur} be the maximal unramified extension. Since P is the part of inertia fixing Q_{p}^{tame},

I_{t} = I/P = Gal(Q_{p}^{tame}/Q_{p}^{ur}).

For each n prime to p there is a tower of fields

Q_{p}^{tame}
|
K(p^{\frac{1}{n}})
|
K = Q_{p}^{ur}
|
Q_{p}

By Kummer theory Gal(K(p^{\frac{1}{n}})/K) = μ_{n}(K) where μ_{n}(K) denotes the group of nth roots of unity in K. Thus for each n prime to p we obtain by restriction a map I_{t} → μ_{n}(K). They are compatible so passing to the limit we obtain a map

I_{t} → lim_{←n} μ_{n}(K) = ∏_{r≠p} lim_{←a} μ_{r^{a}}(K) = ∏_{r≠p} Z_{r}(1).

Viewed more cleverly mod p we obtain a map

I_{t} → lim_{←n} μ_{n}(\overline{F}_{p}) = lim_{←n} F_{p^{n}}^{∗}.

We now fix attention on n = p^{i} −1. The point is, that for every such i we have a standard map

I_{t} → F_{p^{i}}^{∗}.

We call this map the fundamental character of level i. This is unnatural because we had to take an embedding F_{ℓ^{ν}} ↪ \overline{F}_{ℓ}. Instead Serre begins with some disembodied field F of order p^{i}. There are i different maps F_{p^{i}} → F corresponding to the i automorphisms of F_{p^{i}}. Restricting these maps to F_{p^{i}}^{∗} and composing with the fundamental character I_{t} → F_{p^{i}}^{∗} defined above gives Serre’s fundamental characters of I_{t} with values in F. We have thus defined some characters on I_{t} which can be viewed as standard.

Recall that we have a mod ℓ^{ν} representation ρ which gives rise to

σ : I_{t} → GL(2, F_{ℓ^{ν}}).

Since the elements of I_{t} have order prime to the characteristic ℓ this representation is semisimple so it can be diagonalized upon passing to an algebraic closure of F_{ℓ^{ν}}. Thus σ corresponds to a pair of characters

α, β : I_{t} → F_{ℓ^{2ν}}^{∗}.

These characters have some stability properties since σ is the restriction of a homomorphism from the full decomposition group. Consider the tower of fields

K(p^{\frac{1}{n}}) ⊃ K = Q_{p}^{ur} ⊃ Q_{p}.

Let G = Gal(K(p^{\frac{1}{n}})/Q_{p}). Recall that Gal(K(p^{\frac{1}{n}})/K) = μ_{n}(K) and Gal(K/Q_{p}) is generated by Frob_{p}. What happens is that if h ∈ μ_{n}(K) and g ∈ G is such that g restricts to Frob_{p} then we have a conjugation formula: ghg^{−1} = h^{p}. Applying this reasoning to the σ situation with h ∈ I_{t} and g restricting to Frob_{p} we find that

σ(ghg^{−1}) = σ(h^{p}) = σ(h)^{p}.

Thus the pair of characters {α, β} is stable under pth powering, i.e., {α, β} = {α^{p}, β^{p}}. The point is that

σ(g)σ(h)σ(g^{−1}) = σ(ghg^{−1}) = σ(h)^{p}.

Thus the representation h ↦ σ(h)^{p} is isomorphic to h ↦ σ(h) via conjugation by σ(g). Thus α, β as a pair are the same as α^{p}, β^{p} as a pair since they came from isomorphic representations.

What does this mean? One possibility is that α = α^{p} and β = β^{p}. This means α, β take values in F_{p}^{∗}. The other possibility is that α^{p} = β and β^{p} = α. Then α^{p^{2}} = α and β^{p^{2}} = β so α, β take values in F_{p^{2}}^{∗}. The first situation in which α and β take values in F_{p}^{∗} is called the level 1 situation. The second situation is called the level 2 situation.

We will play a carnival game, “guess your weight.” First we consider the level 2 case. Our strategy is to try and express α and β in terms of the two fundamental characters of level 2.

First some background. Before stating the general result we talk about a special case. Suppose E is an elliptic curve over Q and p is a prime (2 is allowed). Assume E has good supersingular reduction at p. Then there is a representation

Gal(\overline{Q}/Q) → Aut(E[p])

which may or may not be irreducible. It gives rise via restriction to two characters α, β : I_{t} → F_{p}^{∗}. Serre [Groupes de Galois attachés aux points d’ordre fini des courbes elliptiques sur un corps de nombres, 1972] proves that α, β are the two fundamental characters of level 1 and that I_{t} acts tamely and irreducibly. He also obtained a map

I_{t} → F_{p^{2}}^{∗} ⊂ GL(2, F_{p})

where F_{p^{2}}^{∗} sits inside GL(2, F_{p}) via the action of the multiplicative group of a field on itself after choice of a basis. The action of I_{t} on F_{p^{2}}^{∗} is through the fundamental character. This was the world’s first view of fundamental characters.

Serre next asked Fontaine the corresponding question for more general weight. The first published proof appears in [Edixhoven, The weight in Serre’s conjectures on modular forms, Invent. Math., 1992]. Suppose f is a newform of weight k such that 2 ≤ k ≤ p and that the level of f is prime to p = ℓ. Take \overline{ρ}_{f,λ} where E_{f} = Q(coefficients of f) and λ is a prime of O_{E_{f}} dividing p. Suppose we are in the supersingular case so a_{p} ≡ 0 mod λ. Semisimplifying and restricting as before gives a pair of characters α, β : I_{t} → F_{p^{ν}}^{∗}.

Theorem 28 Under the above hypothesis, {α, β} equals {ψ^{k−1}, (ψ′)^{k−1}} where ψ and ψ′ are the two fundamental characters of level 2, i.e., ψ = (ψ′)^{p} and ψ′ = ψ^{p}.

Edixhoven’s method is to reduce from weight k to weight 2. Given a modular form mod p on Γ_{1}(N) of weight k and character ε he shows that it is enough to look at a corresponding modular form on Γ_{1}(Np) of weight 2 and character εω^{k−2} where ω is a Teichmuller character on (Z/pZ)^{∗}. This is a method which helps us understand what happens at primes p where a certain abelian variety does not have good reduction but has potentially good reduction. In this direction Ogus suggests reading the paper [Faltings and Jordan, Crystalline cohomology and GL(2, Q), Israel J. Math. 90 (1995), no. 1-3, 1–66.]

What can we guess about the arbitrary situation? The situation is as follows. We begin with a representation ρ which gives rise to a representation σ which in turn gives rise to a pair of fundamental characters ψ^{a} and (ψ′)^{a} = ψ^{pa}. (Here a should be thought of as a number mod p^{2} − 1.) The condition that we are not in level 1 means that a is not divisible by p + 1 since

ψψ′ = ψ^{p+1} : I_{t} → F_{p}^{∗}

is the unique fundamental character of level 1 (namely the mod p cyclotomic character χ).

We next do a Euclidean division. Take 0 ≤ a < p^{2} − 1 and write a = qp + r. Then pa = qp^{2} + rp, but we are working mod p^{2} − 1 so this becomes pa = q + rp. What are the possible values for q and r? By the Euclidean algorithm 0 ≤ r ≤ p − 1 and 0 ≤ q ≤ p − 1. Now what are the constraints on r and q together? The main constraint is that r ≠ q, since if r = q then a is a multiple of p + 1. So we can assume that 0 ≤ r < q ≤ p − 1. We know that

{α, β} = {(ψψ′)^{r}(ψ′)^{q−r}, (ψψ′)^{r}ψ^{q−r}}.

For example, if α = ψ^{a} then

α = ψ^{qp+r} = (ψ′)^{q}ψ^{r} = (ψψ′)^{r}(ψ′)^{q−r}.

Since ψψ′ = χ we can view {α, β} as a pair of characters (ψ′)^{q−r} and ψ^{q−r} which has been multiplied as a pair by χ^{r}.

What would happen if for example r = 0? In this case

{α, β} = {(ψ′)^{k−1}, ψ^{k−1}}

where k = q + 1. So when r = 0 we guess the weight to be k(ρ) = q + 1.

What happens more generally? Suppose f is a modular form thought of mod p where λ|p. If f = ∑ a_{n}q^{n} gives rise to ρ then we might well ask what modular form gives rise to ρ ⊗ χ. In [Lecture Notes in Math, no. 601] we find that θf = ∑ na_{n}q^{n} is a form giving rise to ρ ⊗ χ. If k is the weight of f, then θf has weight k + p + 1. More generally, since χ^{r} appears in the pair

{ψ^{q−r}, (ψ′)^{q−r}} · χ^{r}

we obtain

k(ρ) = q − r + 1 + (p + 1)r = q + pr + 1.

But be careful, the weight k does not have to go up. In fact, it goes up p − 1 times then ends up where it started. This gives rise to the theory of theta cycles which we will study in the next lecture.