Newforms with inner twists

K. Buzzard and W. Stein

December 10, 1998

Abstract

1 Introduction

Doi, Koike, Ribet, Shimura, and Yamauchi have studied newforms f which possess an inner twist. That is, for which there exists a nontrivial primitive character χ so that f ⊗ χ = σf for some σ ∈ Gal(\overline{Q}/Q). Let A_{f} be the abelian variety associated to f (considered only up to isogeny). Then A_{f} is simple over Q. Inner twists of f are of interest because they can give rise to nontrivial decompositions of A_{f} over a finite extension of Q. For example, Koike used the theory to see that the 2 dimensional abelian variety J_{0}(81)^{new} is isogeneous over Q(\sqrt{−3}) to E × E^{τ} where E is an explicitely determined elliptic curve over Q(\sqrt{−3}) and τ is the nontrivial automorphism.

In this paper we work out a single example. It is hoped that the ideas used in working out the example can be used to both find more examples and prove a converse to some of Ribet’s theorems.

2 Theorems of Shimura and Ribet

Theorem 2.1 (Shimura). Let N, r, s, k > 0 be integers such that s|N and let M be the least common multiple of N, r^{2}, and rs. Let χ (resp. Ψ) be a primitive character mod r (resp. s). If f = ∑ a_{n}q^{n} ∈ S_{k}(Γ_{0}(N), Ψ) then

f ⊗ χ := ∑ χ(n)a_{n}q^{n} ∈ S_{k}(Γ_{0}(M), Ψχ^{2}).

Proof. Proposition 3.64 of [S].

As a special case we have Corollary 2.2. Let 2^{n} ≥ 64 and let χ be a primitive character of conductor dividing 8. Then “twisting by χ” (i.e., the map f ↦ f ⊗ χ) preserves S_{k}(Γ_{0}(2^{n}), Ψ).

Proof. Put r = 8 in the theorem and note that χ^{2} = 1.

We introduce some notation in order to recall Ribet’s Theorem. Let E = Q(f) = Q(a_{1}, a_{2}, a_{3}, …). Let

Γ = {γ ∈ Aut(E) : γf = f ⊗ χ_{γ}, some χ_{γ}}.

It is known that Γ is an abelian subgroup of Aut(E). Let F be the subfield of E fixed by Γ. Remark 2.3. What is [F : Q] in terms of Γ and # Aut(E)? Embed E in its normal closure K. Let H ⊂ G = Gal(K/Q) be the subgroup of automorphisms which preserve E. Since every automorphism extends, Aut(E) is a quotient of H with kernel J, the subgroup of G corresponding to E via Galois theory. Note that J is normal in H but need not be normal in G. The inverse image Γ′ of Γ in H is a subgroup of the same index as the index of Γ in Aut(E). Unless I make a mistake computing indices, we get

[F : Q] = [G : Γ′ + J] = \frac{[G : J]}{[Γ′ + J : J]} = \frac{[E : Q]}{#Γ}

which is just what we would expect.

For a primitive character φ of conductor c define

g(φ) = ∑_{u=1}^{c} φ(u)e^{2πiu/c}.

For γ, δ ∈ Γ, let

c(γ, δ) = \frac{g(χ_{γ}^{−1})g(γ(χ_{δ}^{−1}))}{g(χ_{γδ}^{−1})}.

In [R] there is an explicit description of End(A_{f}) ⊗ Q in terms of generators and relations. Let X be the E-vector space

X = ∑_{γ∈Γ} E · X_{γ}

where the X_{γ} are formal symbols. By imposing on the X_{γ} the rules

X_{γ} · e = γ(e)X_{γ}, for e ∈ E and γ ∈ Γ
X_{γ}X_{δ} = c(γ, δ)X_{γδ}

we make X into an associative algebra.

Theorem 2.4 (Ribet[R]). There algebra X is a central simple algebra over F which is isomorphic to (End A) ⊗ Q. Furthermore, the 2-cocycle c corresponds to the class of (End A) ⊗ Q in Br(F) = H^{2}(F, \overline{F}^{×}).

3 Example in which the endomorphism ring has order 2 in the Brauer group

The characteristic polynomial of T_{3} on S_{2}^{new}(Γ_{0}(512)) is

(x^{2} − 6)^{2}(x^{2} − 2)^{4}(x^{2} + 4x + 2)(x^{2} − 4x + 2)

Let f = ∑ a_{n}q^{n} be one of the newforms in the four dimensional kernel V of (T_{3}^{2} − 6)^{2}. and let E = Q(f) = Q(a_{1}, a_{2}, a_{3}, a_{4}, …).

Proposition 3.1. We have E = Q(\sqrt{2}, \sqrt{3}) and the compa-nions of f coincide with the twists of f by the four Dirichlet characters of conductor dividing 8. In particular, Γ = Gal(E/Q).

Proof. The newform f and its companions lie inside of V. By “multiplicity one”, V has dimension 4 so [Q(f) : Q] ≤ 4. The characteristic polynomial of T_{7} on V is (x^{2} − 8)^{2} so a_{3} = ±\sqrt{6} and a_{7} = ±2\sqrt{2} so E = Q(\sqrt{2}, \sqrt{3}).

Let χ be a character of conductor dividing 8. By Theorem 2.1, f ⊗ χ lies in S_{2}(Γ_{0}(512)). In fact, it must lie in S_{2}^{new}(Γ_{0}(512)) because if f ⊗ χ lies in S_{2}(Γ_{0}(2^{n})) for n ≥ 6 then Theorem 2.1 implies that f = (f ⊗ χ) ⊗ χ lies in S_{2}(Γ_{0}(2^{n})), and certainly we can take n ≥ 6. So we’re done.

Alternatively, since the image of χ is contained in {±1}, the T_{3} eigenvalue of f equal ±\sqrt{6}. The factor (x^{2} − 6)^{2} exactly divides the characteristic polynomial of T_{3} on S_{2}(Γ_{0}(512)) so f ⊗ χ must be a companion of f.

The q-expansion of f can be computed to be

f = q −\sqrt{6}q^{3} +2\sqrt{3}q^{5} +2\sqrt{2}q^{7} +3q^{9} −\sqrt{6}q^{11} −2\sqrt{3}q^{13} −6\sqrt{2}q^{15} +4q^{17} +⋯

Remark 3.2. f is the first example of a weight 2 newform having trivial character for which Q(a_{n}) is a proper subfield of E for all n.

Theorem 3.3. The endomorphism ring of A_{f} is a 2 × 2 matrix ring over a quaternion division algebra with center Q.

Proof. Write G = Gal(E/Q) = \{γ_{1} = 1, γ_{2}, γ_{3}, γ_{6}\} where γ_{d} fixes \sqrt{d}, d = 2, 3, 6. Let χ_{d}, d = ±1, ±2 be the four Dirichlet characters of conductor dividing 8, where χ_{d} corresponds to the field Q(\sqrt{d}). We have

f = f ⊗ χ_{1}
γ_{2}f = q + \sqrt{6}q^{3} − 2\sqrt{3}q^{5} + 2\sqrt{2}q^{7} + ⋯ = f ⊗ χ_{2}
γ_{3}f = q + \sqrt{6}q^{3} + 2\sqrt{3}q^{5} − 2\sqrt{2}q^{7} + ⋯ = f ⊗ χ_{−1}
γ_{6}f = q − \sqrt{6}q^{3} − 2\sqrt{3}q^{5} − 2\sqrt{2}q^{7} + ⋯ = f ⊗ χ_{−2}

The sums are

g(χ_{1}) = 1
g(χ_{−1}) = e^{2πi/4} − e^{2πi3/4} = i − (−i) = 2i
g(χ_{2}) = e^{2πi/8} − e^{2πi3/8} − e^{2πi5/8} + e^{2πi7/8} = 2\sqrt{2}
g(χ_{−2}) = e^{2πi/8} + e^{2πi3/8} − e^{2πi5/8} − e^{2πi7/8} = 2i\sqrt{2}.

Thus we can compute

c(γ_{i}, γ_{j}) = \frac{g(χ_{γ_{i}})g(χ_{γ_{j}})}{g(χ_{γ_{i}γ_{j}})}.

For example,

c(γ_{2}, γ_{3}) = \frac{2\sqrt{2} · 2i}{2i\sqrt{2}} = 2

By Theorem 2.4 we obtain a presentation of (End A) ⊗ Q. Note that c(γ_{i}, γ_{j}) = c(γ_{j}, γ_{i}) so that the generators X_{γ} commute. The endomorphism ring is not commutative as E does not commute with X_{γ} for nontrivial γ.

The 2-cocycle c represents the element of the Brauer group Br(Q) corresponding to (End A) ⊗ Q. We ask, does this element have order 1 or order 2? Let K = Q_{2}(\sqrt{2}, \sqrt{3}). Then, by inf-res, c arises from an element of

H^{2}(Gal(K/Q_{2}), K^{×}) → Br(Q).

Note that inf is injective because of Hilbert’s Theorem 90; see Proposition 6 on page 156 of [SE].

I think that c must have order dividing 2 and

H^{2}(Gal(Q_{2}(\sqrt{2})/Q_{2}), Q_{2}(\sqrt{2})^{×}) ⊂ H^{2}(Gal(K/Q_{2}), K^{×})

is the unique subgroup of order 2 of Br(Q), hence c must lie in it. Using the Local symbol in Chapter XIV of we can write down the nonzero element of H^{2}(Gal(Q_{2}(\sqrt{2})/Q_{2}), Q_{2}(\sqrt{2})^{×}).

Let G′ = Gal(Q_{2}(\sqrt{2})/Q_{2}) = \{1, τ\}. What is c as a map G′ × G′ → Q_{2}(\sqrt{2})^{×}? We have:

c(1, 1) = 1, c(1, τ) = c(τ, 1) = 1, c(τ, τ) = −4.

This is precisely the 2-cocycle given by the local symbol (2, −4); see Washington’s article [W]. Thus this cocycle is trivial iff −4 is a norm from Q_{2}(\sqrt{2}). If −4 is a norm then so is −1 and hence −1 ∈ Q_{2}(\sqrt{2}). But this is impossible as the ramification degree of 2 in Q(\sqrt{2}, i) is e = 4, so that the local degree of 2 is 4. Thus we finally conclude that c is nontrivial and hence obtain the theorem.

References

[D] K. Doi, M. Yamauchi, On the Hecke operators for Γ_{0}(N) and class fields over quadratic number fields, J. Math. Soc. Japan, 25 (1973), 629–643.

[K] M. Koike, On certain abelian varieties obtained from new forms of weight 2 on Γ_{0}(3^{4}) and Γ_{0}(3^{5})., Nagoya Math. J., 62 (1976), 29–39.

[R] K. Ribet, Twists of Modular Forms and Endomorphisms of Abelian Varieties, Math. Ann. 253, (1980), 43–62.

[SE] J.P. Serre, Local Fields, Springer-Verlag, (1979).

[S] G. Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Princeton University Press, (1994).

[W] L.C. Washington, Galois Cohomology, In “Modular Forms and Fermat’s Last Theorem”, Ed.’s Cornell-Silverman-Stevens, (1997).