Component groups of optimal quotients

William A. Stein

September 24, 1999

Contents

1. Ribet’s letter

2. Computing Φ_{E} without finding Tate’s q

3. Component groups of abelian varieties

4. Lemmas

5. Proof of theorem

6. Analysis of the modular polarization

7. Surjectivity of π_{∗}

§1. Ribet’s letter

November, 1987: Ribet wrote a letter to Mestre describing component groups of elliptic curves at primes of multiplicative reduction.

• N = pM with (M, p) = 1.

• E := an elliptic curve of conductor N.

• π : J_{0}(N) → E with ker(π) connected.

• m_{E} := modular degree of E.

• T_{J} := torus of J_{0}(N)/F_{p}.

• X_{J} := Hom(T_{J}, G_{m}) ≅ Div^{0}(X_{0}(M)(\overline{F}_{p})^{ss})

• X_{E} := Hom(T_{E}, G_{m}) =: X_{E^{∨}}.

Monodromy: ⟨ , ⟩_{J} : X_{J} × X_{J} → Z,

⟨C, C′⟩ := \begin{cases}0 & if C ≠ C′, \\ w_{C} = \frac{# Aut(C)}{2} & otherwise. \end{cases}

⟨ , ⟩ : X_{E} × X_{E^{∨}} → Z

Component groups:

\xymatrix{0 \ar[r] & X_{J} \ar[r] & Hom(X_{J}, Z) \ar[r] & Φ_{J} \ar[r] & 0 \\ 0 \ar[r] & X_{E^{∨}} \ar[r] & Hom(X_{E}, Z) \ar[r] & Φ_{E} \ar[r] & 0}

Let π^{∗} : X_{E} → X_{J} be the natural map and write π^{∗}X_{E} = Z · ∑ n_{C}[C].

Theorem 1 (Ribet).

# coker(Φ_{J} → Φ_{E}) = gcd(… n_{C} …),
#Φ_{E} = gcd_{C,C′}(w_{C}n_{C} − w_{C′}n_{C′}),
m_{E} · #Φ_{E} = ∑ w_{C}n_{C}^{2} = ⟨∑ n_{C}[C], ∑ n_{C}[C]⟩.

§2. Computing Φ_{E} without finding Tate’s q

Let I ⊂ T be the kernel of T_{n} ↦ a_{n} ∈ Z.

\xymatrix{X_{E} \ar@{^{(}->}[r]^{π^{∗}} & X_{J}[I] ⊂ X_{J}}

Can compute X_{J}[I], but the index is often not 1:

[X_{J}[I] : π^{∗}X_{E}] = 1 ⟺ # coker(X_{J} → X_{E}) = 1.

Idea:

m_{E} = \frac{m_{E} · #Φ_{E}}{#Φ_{E}} = \frac{∑ w_{C}n_{C}^{2}}{gcd(w_{C}n_{C} − w_{C′}n_{C′})}.

We can compute m_{E} using modular symbols. We can compute λ ∑ n_{C}[C] for some unknown λ (secretly, λ = 1/# coker(X_{J} → X_{E})). Then

\frac{∑ w_{C}λ^{2}n_{C}^{2}}{gcd(w_{C}λn_{C} − w_{C′}λn_{C′})} = λ · m_{E}.

Thus we can compute λ.

Example:

• p = 3, M = 11

• f = q + q^{2} − q^{3} − q^{4} − 2q^{5} + ⋯ ∈ S_{2}(33)^{new}

• X_{0}(11)(\overline{F}_{3})^{ss} = ZC_{1} ⊕ ZC_{2}, w_{1} = w_{2} = 1

• X = X[I] = Z(C_{1} − C_{2})

• m_{E} = 3

Thus

λ · 3 = \frac{1^{2} + (−1)^{2}}{2} = 1;

so λ = \frac{1}{3}, and

π^{∗}X_{E} = Z(3C_{1} − 3C_{2}).

Conclusion: #Φ_{E} = 6 and # coker(Φ_{J} → Φ_{E}) = 3. Remark: f is congruent mod 3 to a form of level 11.

§3. Component groups of abelian varieties

Let f = ∑ a_{n}q^{n} ∈ S_{2}(Γ_{0}(N)) be a newform. Set I := ker(T → Q(… a_{n} …)).

\xymatrix{ & IJ_{0}(N) \ar[d] & X_{A} \ar[r]^{π^{∗}} \ar[ddr]^{θ^{∗}} & X_{J} \ar[dd]^{π_{∗}} \\ A^{∨} \ar@{^{(}->}[r]^{π^{∨}} \ar[dr]_{θ} & J \ar@{->>}[d]^{π} \\  & A &  & X_{A^{∨}} \ar[uul]^{θ_{∗}}}

Modular degree: m_{A} := \sqrt{deg(θ)}

Consider L ⊂ X_{J}[I] of finite index.

m_{L} := #(X_{A^{∨}}/π_{∗}L)
Φ_{L} := coker(X_{J} → Hom(L, Z))

Theorem 2 (—). For any L as above,

\frac{#Φ_{A}}{m_{A}} = \frac{#Φ_{L}}{m_{L}},
image(Φ_{J} → Φ_{A}) = Φ_{X[I]}, # coker(Φ_{J} → Φ_{A}) = \frac{m_{A}}{m_{X[I]}},
m_{A} = \sqrt{#(H_{1}(A)/π_{∗}H_{1}(J)[I])}.

Example: p = 3, N = 3 · 17:

• f = q + αq^{2} + ⋯ ∈ S_{2}(Γ_{0}(3 · 17)), α^{2} + α − 4 = 0.

• dim J_{0}(51) = 13 and dim A = 2.

• m_{A} = 8.

• Compute L = X[I]; find #Φ_{L} = 4 and m_{L} = 2.

\frac{#Φ_{A}}{8} = \frac{4}{2} ⟹ #Φ_{A} = 16 and # coker(Φ_{J} → Φ_{A}) = \frac{8}{2} = 4.

Remark: f is congruent to a newform of level 17 mod 3.

§4. Lemmas

The remainder of the lecture is devoted to proving that \frac{#Φ_{A}}{m_{A}} = \frac{#Φ_{L}}{m_{L}}.

\xymatrix{X_{A} \ar@{^{(}->}[r]^{π^{∗}} \ar[dr]^{θ^{∗}} & X_{J} \ar@{->>}[d]^{π_{∗}} \\  & X_{A^{∨}} \ar[ul]^{θ_{∗}}}

injectivity: θ_{∗}π_{∗}π^{∗} = θ_{∗}θ^{∗} = [deg(θ)] ≠ 0 and X_{A} is free. surjectivity: proved later.

Restrict ⟨ , ⟩_{J} to obtain

α : X_{J} → Hom(π^{∗}X_{A}, Z).

Lemma 3. ker(π_{∗}) = ker(α)

Proof. x ∈ ker(π_{∗}), z ∈ X_{A} α(x)(π^{∗}z) = ⟨x, π^{∗}z⟩ = ⟨π_{∗}x, z⟩ = ⟨0, z⟩ = 0 ⟹ x ∈ ker(α) Thus ker(π_{∗}) ⊂ ker(α). dim coker(π_{∗}) = dim A = dim coker(α) ⟹ ker(π_{∗}) = ker(α).

Lemma 4. Φ_{π^{∗}X_{A}} ≅ Φ_{A}.

Proof.

\xymatrix{0 \ar[r] & X_{J}/ ker(α) \ar[d]^{≅} \ar[r] & Hom(π^{∗}X_{A}, Z) \ar[r] \ar[d]^{≅} & coker(α) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & X_{A′} \ar[r] & Hom(X_{A}, Z) \ar[r] & Φ_{A} \ar[r] & 0}

The middle isomorphism is induced by the isomorphism X_{A} → π^{∗}X_{A}. Commutes: x ∈ X_{J}, z ∈ X_{A}, then ⟨x, π^{∗}z⟩_{J} = ⟨π_{∗}x, z⟩_{A}. Thus coker(α) ≅ Φ_{A}, as claimed.

W. A. Stein, Component groups of optimal quotients V

§5. Proof of theorem

Lemma 5. \frac{#Φ_{L}}{m_{L}} does not depend on L. Proof. Consider L′ ⊂ X_{J}[I] finite index; set a = [L : L′] ∈ Q. Then

m_{L′} = [X_{A^{∨}} : π_{∗}L′] = [X_{A^{∨}} : π_{∗}L] · [π_{∗}L : π_{∗}L′] = m_{L} · a since π_{∗}|_{X_{J}[I]} is injective. Also,

#Φ_{L′} = # coker(X_{J} → Hom(L′, Z))
= # coker(X_{J} → Hom(L, Z)) · [L : L′] = #Φ_{L} · a.

For this, think of lattices inside Hom(X_{J}[I], Z). Proof of theorem. By Lemma 5, we may assume that L = π^{∗}X_{A}. By Lemma 4, Φ_{L} ≅ Φ_{A}.

m_{L} = [X_{A^{∨}} : π_{∗}L]
= [X_{A^{∨}} : π_{∗}π^{∗}X_{A}]
= [X_{A^{∨}} : θ^{∗}X_{A}]
= # coker(θ^{∗})
= m_{A}.

The final equality is proved below.

W. A. Stein, Component groups of optimal quotients VI

§6. Analysis of the modular polarization

Uniformization: A is purely toric, so over Q_{p}^{ur}:

\xymatrix{0 \ar[r] \ar[d] & X_{A} \ar[r] \ar[d] & X_{A^{∨}} \ar[r] \ar[d] & coker(θ_{a}) \ar[d] \\ ker(θ_{t}) \ar[r] \ar[d] & T_{A^{∨}} \ar[d] \ar[r] & T_{A} \ar[r] \ar[d] & 0 \\ ker(θ) \ar[r] & A^{∨} \ar[r]^{θ} & A}

The snake lemma gives

0 → ker(θ_{t}) → ker(θ) → coker(θ^{∗}) → 0.

The dual of

\xymatrix{ & X_{A} \ar[r] \ar[d] & X_{A^{∨}} \ar[d] \ar[r] & coker(θ_{a}) \\ ker(θ_{t}) \ar[r] & T_{A^{∨}} \ar[r]^{θ} & T_{A}}

is

\xymatrix{ & T_{A} & T_{A^{∨}} \ar[l]_{θ^{∨}} & coker(θ_{a})^{∨} \ar[l] \\ ker(θ_{t})^{∨} & X_{A^{∨}} \ar[l] \ar[u] & X_{A} \ar[l] \ar[u]}

But θ^{∨} = θ, since θ arises from the θ-divisor, so

coker(θ^{∗})^{∨} ≅ ker(θ_{t}).

In particular,

# ker(θ) = (# coker(θ^{∗}))^{2},

so

m_{A} = # coker(θ^{∗}).

§7. Surjectivity of π_{∗}

There are easier ways, but in the interest of suggesting a more general theorem we use the Raynaud uniformization.

Lemma 6. The map π_{∗} : X_{J} → X_{A^{∨}} is surjective.

Proof. Because G_{J} is simply connected, π induces a map G_{J} → T_{A}. Because π is surjective and T_{A} is a torus, the map G_{J} → T_{A} is surjective. The snake lemma applied to the following diagram gives a surjective map from B = ker(π) to M = coker(X_{J} → X_{A^{∨}}).

\xymatrix{ & X_{J} \ar[r] \ar[d] & X_{A^{∨}} \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\  & G_{J} \ar[r] \ar[d] & T_{A} \ar[d] \ar[r] & 0 \\ B \ar[r] & J \ar[r]^{π} & A}

Because π is optimal, B is connected so M must also be connected. Since M is discrete it follows that M = 0.

Remark: Abbes pointed out that the condition that A be purely toric can be relaxed.

Remarks:

• In the above theorem, J_{0}(N) can be replaced by any semistable Jacobian and A by any purely toric optimal quotient.

• Mazur, Raynaud, and Edixhoven have given an explicit formula for Φ_{J_{0}(Mp)} for p ≥ 5.