EULER SYSTEMS FOR HIGHER-WEIGHT MODULAR FORMS

AMNON BESSER

The Selmer and Tate-Šafarevič groups of a Galois representation

K-: A number field

G-: Gal(ℚ̄/K)

T -: A free ℤ_{p} module of finite rank with a continuous G action.

Let V = T ⊗ ℚ, A defined by the sequence

0 ⟶ T ⟶^{i} V ⟶^{pr} A ⟶ 0.

Assumption: There is a perfect pairing V × V → ℚ(1).

Let v be a place of K. The following definitions are due to Bloch and Kato

H_{f}^{1}(K_{v}, V) := Ker H^{1}(K_{v}, V) ⟶^{res} H^{1}(K_{v}^{ur}, V);
H_{f}^{1}(K_{v}, T) := i^{−1}H_{f}^{1}(K_{v}, V);
H_{f}^{1}(K_{v}, A) := Im H_{f}^{1}(K_{v}, V) ↪ H^{1}(K_{v}, V) ⟶^{pr} H^{1}(K_{v}, A),

This is when v ∤ p. Otherwise there is another definition.

\begin{matrix}Sel_{p}(T/K) := Ker H^{1}(K, A) ⟶ \\ ∏_{v} H^{1}(K_{v}, A)/H_{f}^{1}(K_{v}, A), \end{matrix}

\begin{matrix}H_{f}^{1}(K, V) := Ker H^{1}(K, V) ⟶ \\ ∏_{v} H^{1}(K_{v}, V)/H_{f}^{1}(K_{v}, V). \end{matrix}

�_{p}(T/K) = Sel_{p}(T/K)/ Im H_{f}^{1}(K, V)

Examples:

1. T = T_{p}(E) is the Tate module of a modular elliptic curve E. �_{p}(T/K) is the p part of

�(E/K) := Ker H^{1}(K, E) ⟶ ∏_{v} H^{1}(K_{v}, E)

Date: May 9, 1996.

K is a quadratic imaginary field, some restrictions. Kolyvagin: If ord_{s=1}L(E, s) ≤ 1 then �_{p}(T/K) is finite and 0 for almost every p.

2. Flach T = Symm^{2}(T_{p}(E))(−1) Then Sel_{p}(T/ℚ) is finite and 0 for almost all p.

In both cases there are certain bounds.

Results of Nekovář (Inventioness 92):

f-: New form of weight 2r for Γ_{0}(N).

f(q) = ∑ a_{n}q^{n}. K-: A quadratic imaginary field of discriminant D in which all primes dividing

N split T_{p}(f)-: The representation associated to f by Deligne - some part of the coho-

mology

H^{2r−1}(E^{2r−2}, ℤ_{p}).

E is the universal generalized elliptic curve over X_{0}(N) (desingularized). There is a projector Π_{f} : H^{2r−1}(E^{2r−2}, ℤ_{p}) → T_{p}(f).

T: = T_{p}(f)(r).

K_{n}-: The ring class field of conductor n over K.

There is a CM cycle associated to K - Take E_{0} with CM by O_{K}. On E_{0}^{2} there is a cycle Γ\sqrt{D} corresponding to the graph of \sqrt{D}

Consider C_{1} = Symm^{2r−2} Γ\sqrt{D} - Cycle of codim r − 1 on Symm^{2r−2} E_{0} and r on E^{2r−2}. It’s defined over K_{1} and cohomologically trivial.

We have an Abel-Jacobi map

\begin{matrix}α_{p} : CH^{r}(E^{2r−2} ⊗ K_{1})_{0} → \\ H^{1}(K_{1}, H^{2r−1}(E^{2r−2}, ℤ_{p}(r))) \end{matrix}

Define

y_{1} = Π_{f}(r)(α_{p}(C_{1})) ∈ H^{1}(K_{1}, T)

and

P (1) = cor_{K_{1},K} y_{1} ∈ H^{1}(K, T).

Theorem (Nekovář ): Assume p ∤ 2N. If P (1) is not torsion, then ℚ_{p} Im(α_{p}) = ℚ_{p}P (1) in H^{1}(K, T) and �_{p}(T) is finite.

There are some bounds there, but they are not best possible, so one can’t show that �_{p} is trivial for almost all p.

Theorem (B): For almost all p the following is true: if

red P (1) ∈ H^{1}(K, T/p) ≠ 0,

then �_{p}(T) = 0. Set of exceptional p’s can be explicitly determined.

EULER SYSTEMS FOR HIGHER-WEIGHT MODULAR FORMS 3

General program - Method of Euler systems Step 1 - One constructs classes y_{n} ∈ H^{1}(K_{n}, T) in a similar fashion to y_{1}. These are in H_{f} at all primes of good reduction (We don’t know how to treat places which divide both N and p hence the constraint p ∤ N. The y_{n} satisfy the relation

cor_{K_{n},K_{m}}(y_{n}) = a_{ℓ}y_{m}

when n = mℓ, ℓ is prime and ℓ ∤ n. Step 2 - Fix k and consider n which are products of distinct primes ℓ satisfying the following conditions

• ℓ ∤ NDp;

• ℓ is inert in K;

• p^{k} divides a_{ℓ} and ℓ + 1; Use the y_{n} to make classes P (n) ∈ H^{1}(K, A_{p^{k−M}}) for some fixed M - Use some D_{n} ∈ ℤ[Gal(K_{n}/K_{1})] such that

D_{n}y_{n} ∈ H^{1}(K_{n}, T/p^{k})^{Gal(K_{n}/K_{1})}

You can lift D_{n}y_{n} to z_{n} ∈ H^{1}(K_{1}, •) after reducing further. Take P (n) = cor_{K_{1},K} z_{n}.

Step 3 - Properties of the P (n):

1. P (n) belongs to the (−1)^{par(n)}ε_{L}-eigenspace of the complex conjugation τ acting on H^{1}(K, A_{p^{k}}). ε_{L} is the negative of the sign of the functional equation of L(f, s).

2. For a place v of K such that v ∤ Nnp, P (n) ∈ H_{f^{∗}}^{1}(K_{v}, A_{p^{k}}).

3. If n = m · ℓ and λ is the unique prime of K above ℓ. In H^{1}(•, A_{p^{k}}) we have:

P (m)_{λ} ≠ 0 ⟹ P (n)_{λ,sin} ≠ 0.

Step 4 - Local Tate duality There is a perfect pairing ⟨, ⟩ : H^{1}(K_{v}, A_{p^{k}}) × H^{1}(K_{v}, A_{p^{k}}) → ℤ/p^{k}. Let: H_{f}^{1}(K_{v}, A_{p^{k}}) be the preimage in H^{1}(K_{v}, A_{p^{k}}) of H_{f}^{1}(K_{v}, A). H_{f^{∗}}^{1}(K_{v}, A_{p^{k}}) be the annihilator of H_{f}^{1}(K_{v}, A_{p^{k}}) in H^{1}(K_{v}, A_{p^{k}}) under local Tate duality (group of dual finite classes).

H_{sin}^{1}(K_{v}, A_{p^{k}}) := H^{1}(K_{v}, A_{p^{k}})/H_{f^{∗}}^{1}(K_{v}, A_{p^{k}})

(Singular cohomology, this definition is due to Mazur). For v ∤ p a place of good reduction it turns out that H_{f}^{1} = H_{f^{∗}}^{1} = H_{ur}^{1} := Ker H^{1}(K_{v}, •) → H^{1}(K_{v}^{ur}, •).

\begin{matrix}Sel(K, A_{p^{k}}) := Ker H^{1}(K, A_{p^{k}}) ⟶ \\ ∏_{v} H^{1}(K_{v}, A_{p^{k}})/H_{f}^{1}(K_{v}, A_{p^{k}}). \end{matrix}

To prove the result we need to show that Sel(K, A_{p}) is generated by P (1) for almost all primes p. Step 5 - Proof of the main result (under simplified conditions), first steps We’ll need to assume P (n) ∈ H_{f^{∗}}^{1}(K_{v}, A_{p^{k}}) for all v ∤ n. It’s a problem only of v|Np. For v|p solved by Nekovář . Assume (for the moment), all coefficients or f are in ℚ. Then A_{p} is a ℤ/p v.s. of dim 2. Also assume f is not a CM form. Let L = K(A_{p}). Proposition: H^{i}(Gal(L/K), A_{p}) = 0 for all i, almost all p. This is a consequence of a result of Ribet that Gal(L/K) = Aut(A_{p}) for almost all p (only true if the coefficients are in ℚ but conclusion holds in general). The exact sequence

0 → H^{1}(Gal(L/K), A_{p}) → H^{1}(K, A_{p}) →
H^{1}(L, A_{p})^{Gal(L/K)} → H^{2}(Gal(L/K), A_{p})

Shows that

H^{1}(K, A_{p}) ≅ H^{1}(L, A_{p})^{Gal(L/K)}
≅ Hom_{Gal(L/K)}(Gal(ℚ̄/L), A_{p})

For any S ∈ H^{1}(K, A_{p}) a f.d.v.s. corresponds a field L_{S}/L cut out by S, and we have:

[ , ]_{S} : S × Gal(L_{S}/L) → A_{p}.

s.t.

x ∈ S and [x, Gal(L_{S}/L)]_{S} = 0 ⟹ x = 0.

Also

Gal(L_{S}/L) ↪ Hom_{𝔽}(S, A_{p^{k}}), 𝔽 = ℤ/p.

This is an isomorphism when f is defined over ℚ. No longer works otherwise. Step 6 - Proof (second step) Idea of the proof is: If x ∈ Sel(K, A_{p}) consider some P (ℓ) for a good ℓ. We have (global reciprocity)

∑_{v}⟨x_{v}, P (ℓ)_{v}⟩ = 0.

EULER SYSTEMS FOR HIGHER-WEIGHT MODULAR FORMS 5

If λ is the unique prime of K dividing ℓ, x_{v} ∈ H_{f}^{1},P (ℓ)_{v} ∈ H_{f^{∗}}^{1} for all v ≠ λ. So

⟨x_{λ}, P (ℓ)_{λ}⟩ = 0.

Want to deduce x_{λ} = 0, and from that for enough λ that x = 0. Lemma: 1. The pairing ⟨ , ⟩ induces nondegenerate pairings:

⟨ , ⟩^{±} : H_{f}^{1}(K_{λ}, A_{p})^{±} × H_{sin}^{1}(K_{λ}, A_{p})^{±} → ℤ/p.

2. Both H_{f}^{1}(K_{λ}, A_{p}) and H_{sin}^{1}(K_{λ}, A_{p}) are direct sums of their ±1 eigenspaces, all of which are 1 dimensional. Proof: the essential point is that

H_{f}^{1}(K_{λ}, A_{p}) ≅ H^{1}(K_{λ}^{ur}/K_{λ}, A_{p}) ≅ A_{p}

because the conditions imposed on ℓ are equivalent to Frob(ℓ) being the same as τ on K(A_{p^{k}}) and Frob(λ) = Frob(ℓ)^{2}. The rightmost isomorphism is just evaluation at Frob(λ), so for x ∈ Sel(K, A_{p}), [x, Frob(λ)] = 0 iff x_{λ} = 0. Corollary: If x ∈ Sel and P (n) are in the same eigenspace, then P (n)_{λ,sin} ≠ 0 and ⟨x_{λ}, P (n)_{λ}⟩ = 0 imply x_{λ} = 0. Proposition: Any σ ∈ Gal(L_{S}/L) is a Frob(λ) for some “good” ℓ. Step 7 - End of proof This is done separately for x ∈ Sel(K, A_{p})^{±}. First case: x in the opposite eigenspace of P (1): Let S = span(x, P (1)). Consider ℓ s.t. P (1)_{λ} ≠ 0. Their Frob’s will still span Gal(L_{S}/L). Notice P (ℓ) and x in the same eigenspace.

P (1)_{λ} ≠ 0 ⟹ P (ℓ)_{λ,sin} ≠ 0 ⟹ x_{λ} = 0

So x = 0 Second case: x in the same eigenspace as P (1). Choose some ℓ s.t. P (1)_{λ} ≠ 0, hence P (ℓ) ≠ 0. Let S = span(x, P (1), P (ℓ). Take α s.t. x′ = x − αP (1) has x′_{λ} = 0. Choose ℓ′ ≠ ℓ s.t. P (ℓ)_{λ′} ≠ 0 hence P (ℓℓ′)_{λ′,sin} ≠ 0.

x′_{λ} = 0 ⟹ ⟨x′_{λ′}, P (ℓℓ′)_{λ′}⟩ = 0 ⟹ x′_{λ′} = 0

So x′ = 0 hence x = αP (1). The f^{∗} condition in the case of bad reduction Recall definitions H_{f}^{1}(K_{v}, A_{p^{k}}) - preimage in H^{1}(K_{v}, A_{p^{k}}) of H_{f}^{1}(K_{v}, A).

6 AMNON BESSER

H_{f^{∗}}^{1}(K_{v}, A_{p^{k}}) - annihilator of H_{f}^{1}(K_{v}, A_{p^{k}}) in H^{1}(K_{v}, A_{p^{k}}) under ⟨, ⟩. Consider L_{v} an unramified extension of K_{v}. Fact: V^{I} has no weight 0 ⟹ V_{I}(−1) has no weight 0. Proposition (v ∤ p): H_{f}^{1}(L_{v}, V) = H^{1}(L_{v}, V). Proof: H^{1}(I, V) ≅ V_{I}(−1) Corollary: H_{f^{∗}}^{1}(L_{v}, A_{p^{k}}) is the image of the reduction from H^{1}(L_{v}, T). Let Δ = Gal(L_{v}, K_{v})

0 ⟶ H^{1}(K_{v}, T)/p^{k} ⟶^{red_{p^{k}}} H^{1}(K_{v}, A_{p^{k}}) ⟶ H^{2}(K_{v}, T)_{p^{k}} ⟶ 0
res↓ ^{res}↓ ^{res}↓
0 ⟶ (H^{1}(L_{v}, T)/p^{k})^{Δ} ⟶^{red_{p^{k}}} H^{1}(L_{v}, A_{p^{k}})^{Δ} ⟶ H^{2}(L_{v}, T)_{p^{k}}^{Δ}

In application L = K_{n}. We have

res_{L_{v}} D_{n}y_{n} ∈ Im (H^{1}(L_{v}, T)/p^{k})^{Δ} → H^{1}(L_{v}, A_{p^{k}})^{Δ}

and we can assume it lifts to z_{n}. Question: When does z_{n} come from H^{1}(K_{v}, T)/p^{k} Answer depends on

Ker H^{2}(K_{v}, T)_{p^{k}} → H^{2}(L_{v}, T)_{p^{k}}^{Δ}.

Proposition: The kernel is annihilated by a constant p^{M} independent of k and L_{v} (as long as it’s unramified). The proof relates the kernel to H^{1}(L_{v}^{ur}, T)^{Gal(L_{v}^{ur}/L_{v})}. This is killed by some p^{M}. What is p^{M} in the case of a Tate curve E = K_{v}/q^{ℤ}? Answer: M = ord_{p}(v(q)) = ord_{p} of the number of components of the Neron model of E/K. A basis for T_{p}(E) is given by (ξ_{p^{n}}) and (q^{1/p^{n}}). Can assume that q^{1/p^{M}} ∈ L The action of Gal(L̄_{v}/L_{v}^{ur}) factors through the p-part of the tame quotient ≅ ℤ_{p}. A generator fixes (ξ_{p^{n}}), sends (q^{1/p^{n}}) to p^{M}(ξ_{p^{n}}) + (q^{1/p^{n}}). H^{1}(L_{v}^{ur}, T) =Coinvariants(−1) - consists of ℤ_{p}(ξ_{p^{n}})/p^{M} and ℤ_{p}(q^{1/p^{n}}). Last part is not invariant after the twisting.