New examples of modular icosahedral Artin representations

William A. Stein

(joint work with Kevin Buzzard)

September 16, 1999

§1. Artin’s conjecture

Consider irreducible

ρ : G_{Q} = Gal(\overline{Q}/Q) → GL_{n}(C), n ≥ 2.

Theorem 1 (Artin, 1924). L(ρ, s) is meromorphic.

Conjecture 2 (Artin, 1924). L(ρ, s) is entire.

Conjecture 3. Assume n = 2 and ρ is odd. Then ρ arises from a classical weight one modular form.

By Hecke, this conj. implies L(ρ ⊗ ε, s) entire for all ε.

What we know about Conjecture 3: Image of proj ρ : G_{Q} → PGL_{2}(C) solvable or ≈ A_{5}. Solvable case: Langlands, Tunnel, Hecke prove ρ = ρ_{f} for f classical weight one eigenform. Icosahedral case:

— Buhler (1970): the first example; conductor 800.

— Frey, Kiming, Wang (1994): 7 more, by systematizing Buhler’s work. Use Deligne-Serre + Langlands-Weil. Compute the number of weight 1 newforms and the number of rep’s ρ and get the same number.

— Buzzard, Dickinson, Shepherd-Barron, Taylor (1999): infinitely many examples. Use a frightening amount of deep mathematics, including Shimura-Taniyama.

§2. Our strategy

Theorem 4 (Buzzard-Taylor). Suppose p ≥ 5, and ρ : G_{Q} → GL_{2}(O_{L}). If ρ satisfies the following, then ρ is modular:

— finitely many ramified primes

— \overline{ρ} is modular and abs. irred.

— ρ is unramified at p and ρ(Frob_{p}) has distinct eigenvalues mod λ.

Strategy:

1. Choose an A_{5}-extension K/Q such that — 5 is unramified, and

— the order of Frob_{5} ∈ Gal(K/Q) is not 5.

2. Lift to ρ : G_{Q} → GL_{2}(C) of conductor N.

3. Pretend ρ comes from a weight one form f_{?} ∈ S_{1}(N, ε; C).

4. Reduce to get \overline{f}_{?} ∈ S_{1}(N, ε; \overline{F}_{5}).

5. Multiply \overline{f}_{?} by E_{4} ≡ 1 ∈ S_{4}(N; \overline{F}_{5}).

6. Search for f = \overline{f}_{?} · E_{4} ∈ S_{5}(N, ε; \overline{F}_{5}).

7. Twist to get a form g = f ⊗ε′ ∈ S_{5}(N ·M, εε′^{2}; F_{5}).

8. Then proj ρ_{g} : G_{Q} → PSL(2, F_{5}) ≈ A_{5}.

9. Prove that ρ_{g} is unramified at 5.

10. Prove that image of proj ρ_{g} is A_{5}.

11. Prove that the fixed field L of proj ρ_{g} has root field of small discriminant ≤ 2083^{2}; hence L is in Kiming’s table, so has no choice but to be K.

12. Apply Buzzard-Taylor to a lift of \overline{ρ} = ρ_{g}; conclude that ρ is modular.

Motivation: Maybe find a theorem: “the existence of a mod 5 representation that plausibly looks like it is \overline{ρ} must be \overline{ρ}.”

W. A. Stein III

§3. A new example of conductor 1376

The splitting field K of

h = x^{5} + 2x^{4} + 6x^{3} + 8x^{2} + 10x + 8 has group A_{5} and discriminant 2^{6} · 43^{2}. Since

h ≡ (x − 1)(x^{2} − x + 1)(x^{2} − x + 2) (mod 5), the order of Frob_{5} is 2. A minimal lift ρ : G_{Q} → GL_{2}(C) has conductor

N = 1376 = 2^{5} · 43.

Theorem 5 (-, Buzzard). ρ is modular.

Let

ε : (Z/2^{5}Z)^{∗} × (Z/43Z)^{∗} → C^{∗}

be one of the two characters of order 3 at 43 and conductor 4 at 2. Then ρ conjecturally comes from some f_{?} ∈ S_{1}(1376, ε; C). Search for f = f_{?} · E_{4}: If Frob_{p} has order 2 then a_{p} = 0. The first few such p are: 19, 31, 97, 107, 127. Search for f by finding ker(T_{19}) ∩ ker(T_{31}) ∩ ker(T_{97}), etc.

§4. Step 1: Find f using modular symbols

Using modular symbols, we compute a space S_{5}(1376, ε; \overline{F}_{5}) isomorphic to a lattice in S_{5}(1376, ε; Q[ε]) tensor \overline{F}_{5}; plus possibly some spurious torsion. Use the Hijikata trace formula to rule out the torsion.

Using modular symbols, we compute, in under 4 minutes on a 450Mhz Intel processor, S_{5}(1376, ε; \overline{F}_{25}) and

V := ker(T_{19}) ∩ ker(T_{31}) ∩ ker(T_{97}).

Find dim V = 8. T_{2}|_{V} = 0. Let α^{2} + 4α + 2 = 0 ∈ F_{25}. T_{3}|_{V} satisfies (T_{3}|_{V} + α^{4})^{4}(T_{3}|_{V} + α^{16})^{4}. V_{1} := ker(T_{3} + α^{4}) has dim 2. T_{5}|_{V_{1}} satisfies (T_{5}|_{V_{1}} + α^{10})(T_{5}|_{V_{1}} + α^{22}).

The newform in V_{1} with a_{5} + α^{10} = 0 is

f = q + α^{16}q^{3} + α^{22}q^{5} + α^{14}q^{7} + α^{14}q^{9} + 4q^{11} + · · · .

We check: the p ≤ 997 such that a_{p} = 0 are the same as those for which Frob_{p} has order 2. Morally we have found f_{?} · E_{4}.

§5. Step 2: Twist down to GL_{2}(F_{5})

No way to show

ρ_{f} : G_{Q} → GL_{2}(F_{25})

has projective image A_{5}!

Write ε = ε_{2} · ε_{43}.

g := f ⊗ ε_{43}

Proposition 6. For p ∤ 5N, a_{p}(g) ∈ F_{5}.

Proof sketch. Lift f to f̃ ∈ S_{5}(1376, ε; \overline{Q}). The corresponding automorphic representation π_{f} is ramified principal series because 43||1376. Thus

ρ_{f̃,λ}|D_{43} ∼ (\begin{matrix}Ψ_{1} & 0 \\ 0 & Ψ_{2} \end{matrix}) ,

with Ψ_{2} unramified. Since det(ρ_{f̃,λ}) = χ^{k−1}ε and χ only ramified at 5, get Ψ_{1}|_{I_{43}} = ε_{43}|_{I_{43}}:

ρ_{f̃,λ}|I_{43} ∼ (\begin{matrix}ε_{43} & 0 \\ 0 & 1 \end{matrix})

Thus

ρ_{f̃⊗ε_{43}^{−1},λ} ∼ (\begin{matrix}1 & 0 \\ 0 & ε_{43}^{−1} \end{matrix})

has the same conductor at 43 as f̃. By level lowering,

f ⊗ ε_{43}^{−1} ∈ S_{5}(1376, ε_{2}ε_{43}^{2}; F_{25}).

Claim: f ⊗ ε_{43}^{−1} = σ(f), where σ is the 5th power map F_{25} → F_{25}.

Proof. We have a_{p}(f ⊗ ε_{43}^{−1}) = 0 for p = 19, 31, 97. Also, ε_{43}(3) = α^{8}, ε_{43}(5) = α^{8}, so

a_{3}(f ⊗ ε_{43}^{−1}) = α^{16}/α^{8} = α^{8} = (α^{16})^{5}
a_{5}(f ⊗ ε_{43}^{−1}) = α^{22}/α^{8} = α^{14} = (α^{22})^{5}

f ⊗ε_{43}^{−1} and σ(f) both lie in S_{5}(1376, ε_{2}ε_{43}^{2}; F_{25}) and have same a_{p} for p = 3, 5, 19, 31, 97, so they are equal by how we found f.

So g = f ⊗ε_{43} = σ(f)⊗ε_{43}^{2}. Thus a_{p}(g) = a_{p}(f)^{5}ε_{43}^{2} has fifth power a_{p}(g)^{5} = a_{p}(f)^{25}ε_{43}^{10} = a_{p}(f)ε_{43} = a_{p}(g).

We have

proj ρ_{g} : G_{Q} → PSL(2, F_{5}) ≈ A_{5}.

§6. Step 3: Prove ρ_{g} unramified at 5

Proposition 7. Prove that ρ_{g} is unramified at 5.

Proof. V_{1} is spanned by the eigenforms

f = q + α^{16}q^{3} + α^{22}q^{5} + α^{14}q^{7} + α^{14}q^{9} + 4q^{11} + ⋯

and

f_{1} = q + α^{16}q^{3} + α^{10}q^{5} + α^{14}q^{7} + α^{14}q^{9} + 4q^{11} + ⋯

It looks like a_{p}(f_{1}) = a_{p}(f) for all p ≠ 5; finding a proof of this stumped us for a while. Suppose this is so for now.

Since f is ordinary, over \overline{F}_{5} Deligne’s theorem gives

ρ_{f}|_{D_{5}} ∼ (\begin{matrix}α & ∗ \\ 0 & β \end{matrix})

Since χ_{5}^{5−1} = 1, α(Frob_{5}) = ε(5)/a_{5} = α^{8}/α^{22} = α^{10} and β(Frob_{5}) = α^{22}. (The values of α and β on Frob_{5} are part of the theorem, as proved in Gross’s paper.) Assuming a_{p}(f_{1}) = a_{p}(f), for p ≠ 5, we have

ρ_{f}|_{D_{5}} ∼ ρ_{f_{1}}|_{D_{5}} ∼ (\begin{matrix}α′ & ∗ \\ 0 & β′ \end{matrix})

with α′(Frob_{5}) = α^{8}/α^{10} = α^{22} and β′(Frob_{5}) = α^{10}; in particular, α′ = β. Thus ρ_{f}|_{D_{5}} contains α ⊕ β, so ρ_{f}|_{D_{5}} = α ⊕ β and ∗ = 0.

Remains to check: a_{p}(f) = a_{p}(f_{1}) for p ≠ 5. Is the difference f −f_{1} a q-expansion involving only powers of q^{5}? Yes; if f−f_{1} is in the kernel of θ = q\frac{d}{dq}. Using present theorems, to compute θ : S_{5}(1376, ε; F_{25}) → S_{11}(1376, ε; F_{25}) using q-expansions requires knowing thousands and thousands of terms of the q-expansion – not feasible. Instead use one of the following:

— Define θ on modular symbols and compute it.

— Compute the intersection

⋂_{all p} ker(T_{p} − p · a_{p}(f)) ⊂ S_{11}(1376, ε; F_{25}).

The moment the dimension is 1, it follows that θ(f − f_{1}) = 0.

We carried out both computations and they worked. For the first, θ is multiplication by X^{ℓ}Y − Y^{ℓ}X. For the second, we find that

dim ⋂_{p≤11} ker(T_{p} − p · a_{p}(f)) = 1.

§7. Step 4: The image of proj ρ_{g} is A_{5}

List the first few hundred a_{p}(g). Most are nonzero, so the image can’t be cyclic or dihedral. It can’t be S_{4} since #S_{4} does not divide 60. Another trick (suggested by Lenstra) shows that it can’t be A_{4}.

W. A. Stein X

§8. Step 5: Bound the ramification

This involves considering the order of the image of inertia, and then bounding the discriminant.

§9. Other examples

I have repeated the computation of V and V_{1} for A_{5} representations of the following conductors

2416 = 2^{4} · 151
3184 = 2^{4} · 199
3556 = 2^{2} · 7 · 127
3756 = 2^{2} · 3 · 313.

In each case, the first three primes such that a_{p} = 0 were enough to obtain an 8-dimensional space. After that, using a few Hecke operators, we find a form f and a form that is (almost certainly) a companion of f.

In the 3556 and 3756 examples Frob_{5} has order 3; this presents interesting new twists. The dimensions are also large, e.g.,

dim S_{5}(3756, ε; F_{25}) = 2506.