Recovering modular forms from squares

Dinakar Ramakrishnan

The purpose of this appendix^{1} is to provide a proof of the fact that a holomorphic newform f of weight 2k, level N and trivial character, with Hecke eigenvalues {a_{p} | (p, N) = 1}, is determined up to a quadratic twist, in fact on the nose if N is square-free, by the knowledge of a_{p}^{2} for all primes p in a set of sufficiently large density. We will in fact prove a more general statement below, including the case of odd weight and non-trivial character, and also establish a mod ℓ analog. We found this result in the summer of 94, and we have since learned that it has also been known to others, including Don Blasius and J.-P. Serre. Also, Siman Wong has recently come up with a different proof in the weight 2 case (with trivial character). So we do not intend any display of great achievement by this write-up, and we give all the details for ease of use by those working in classical modular forms and number theory. We have also found a non-trivial extension of this result (in characteristic zero) to Maass forms using an array of results on automorphic L−functions, and this is the subject matter of a paper under preparation. This work was partially supported by an NSF grant. We thank Serre for his helpful comments on an earlier version which led to a finer result.

For every pair of integers N, k ⩾ 1, and character ω : (Z/N)^{∗} → C^{∗}, denote by S_{k}^{new}(N, ω) the set of normalized newforms f of weight k, level N and character ω, with Hecke eigenvalues a_{p}(f), for all p not dividing N, and corresponding p−Euler factors

L_{p}(s, f) = (1 − α_{p}p^{−s})^{−1}(1 − β_{p}p^{−s})^{−1},

where α_{p} = α_{p}(f) and β_{p} = β_{p}(f) are non-zero algebraic integers satisfying

a_{p}(f) = α_{p} + β_{p},  and  ω(p)p^{k−1} = α_{p}β_{p}.

Let us set

L_{p}(s, Ad(f)) = (1 − \frac{α_{p}}{β_{p}} p^{−s})^{−1}(1 − p^{−s})^{−2}(1 − \frac{β_{p}}{α_{p}} p^{−s})^{−1}.

Theorem A Let f ∈ S_{k}^{new}(N, ω) and g ∈ S_{k′}^{new}(N′, ω′), k ⩾ k′, be such that, for all primes p outside a set S of Dirichlet density δ(S) < \frac{1}{18}, we have

(∗)  L_{p}(s, Ad(f)) = L_{p}(s, Ad(g)).

1To appear in Inv. Math. as an appendix to A problem of Linnik for eliptic curves and mean-value estimates for automorphic representations by W. Duke and E. Kowalski

Then k = k′, and there exists a Dirichlet character χ of conductor M dividing NN′ such that

a_{p}(f) = a_{p}(g)χ(p),

all p prime to NN′. In particular, ω = ω′χ^{2}.

If f, g are not of CM type and have weights k, k′ ⩾ 2, then the same conclusion results if (*) is assumed to hold only for a set of primes of positive density.

When f and g have the same character, we can deduce the stronger result below:

Corollary Let f ∈ S_{k}^{new}(N, ω) and g ∈ S_{k}^{new}(N′, ω) be such that, for all primes p outside a set S of density δ(S) < \frac{1}{18}, we have

a_{p}(f)^{2} = a_{p}(g)^{2},

Then there exists a quadratic character χ of conductor M dividing NN′ such that

a_{p}(f) = a_{p}(g)χ(p),

for all p not dividing NN′. Moreover, if ω = 1 and N, N′ square-free, then f = g.

When f, g are not of CM type and of weight ⩾ 2, we get the same conclusion assuming only that δ(S) is < 1.

Theorem A ⟹ Corollary. The hypotheses imply that (α_{p}(f)/β_{p}(f)) + (β_{p}(f)/α_{p}(f)) + 1 equals (α_{p}(g)/β_{p}(g)) + (β_{p}(g)/α_{p}(g)) + 1), for all p outside S. It is then easy to see that L_{p}(s, Ad(f)) equals L_{p}(s, Ad(g)), for all such p. So we may apply the Theorem and deduce the existence of a χ such that a_{p}(f) = a_{p}(g)χ(p), for all p prime to NN′. Comparing squares, we see that χ must be quadratic.

Next let N, N′ be square-free, and ω trivial. Suppose χ is non-trivial. Denote by π, π′ the cuspidal automorphic representations of GL(2, A_{Q}) of trivial central character associated to f, g respectively. Then, up to exchanging f and g if necessary, N = N(π) must be N(π′ ⊗ χ), the conductor of π′ ⊗ (χ ∘ det). (Here we are identifying χ with the idèle class character of Q it defines.) Since N′ = N(π′) is square-free, and since π′ has trivial central character, one sees easily from the description of local representations and their conductors in [Ge], p.73, that the p−component π′_{p} must be the unramified special (Steinberg) representation at every prime p dividing N′. One sees then, by using the same theorem (loc. cit.) that ord_{p}(N(π′ ⊗ χ)) ⩾ 2, for any p dividing the conductor M of χ. Since Q has class number 1, there are no unramified characters χ. In other words, N = N(π′ ⊗ χ) is not square-free, giving the desired contradiction. QED.

Proof of Theorem A. We will in fact give two proofs. We fix a prime ℓ not dividing NN′, and begin with the theorems of Deligne ([De], for k ⩾ 3), Eichler-Shimura ([Sh], for k = 2), and Deligne-Serre ([DS] for k = 1), giving the existence, for h = f or g, of an irreducible, continuous representation

σ_{ℓ}(h) : Gal(\overline{Q}/Q) ⟶ GL_{2}(\overline{Q}_{ℓ}),

such that, for any prime p not dividing Nℓ,

tr(σ_{ℓ}(h)(F r_{p})) = a_{p}(h) = α_{p}(h) + β_{p}(h), |α_{p}(h)| = |β_{p}(h)| = p^{(k(h)−1)/2},

and

det(σ_{ℓ}(h)) = ω(h)χ_{cyc}^{k(h)−1}.

Here F r_{p} denotes the Frobenius conjugacy class at p, \overline{Q}_{ℓ} a fixed algebraic closure of Q_{ℓ}, and χ_{cyc} the cyclotomic character given by the Galois action on the inverse system of ℓ^{m}−th roots of unity. (ω(h) is ω or ω′ depending on whether h is f or g; similarly for k(h).) If we consider the field E generated by the coefficients of f, and a place λ of E above ℓ, then one has in fact a representation of Gal(\overline{Q}/Q) into GL_{2}(E_{λ}), and our σ_{ℓ} is its extension to \overline{Q}_{ℓ}. We work over \overline{Q}_{ℓ} because we will need to appeal to Schur’s lemma.

For any two dimensional \overline{Q}_{ℓ}−representation σ_{ℓ} of Gal(\overline{Q}/Q), set

Ad(σ_{ℓ}) = sym^{2}(σ_{ℓ}) ⊗ det(σ_{ℓ})^{−1}.

Theorem B Let K be a number field, and let σ_{ℓ} and σ′_{ℓ} be irreducible two dimensional \overline{Q}_{ℓ}-representations of Gal(\overline{Q}/K) with Frobenius traces a_{P}, a′_{P} (for almost all primes P) and conductors N, N′ respectively. Suppose Ad(σ_{ℓ}) ≃ Ad(σ′_{ℓ}). Then there exists ψ_{ℓ} ∈ Hom_{cont}(Gal(\overline{Q}/K), \overline{Q}_{ℓ}^{∗}) such that

σ_{ℓ} ≃ σ′_{ℓ} ⊗ ψ_{ℓ}.

Next let K = Q. Suppose we know either that σ_{ℓ} and σ′_{ℓ} are Hodge-Tate (see [Se1]) or that the ratio of their determinants is a finite order character times an even power of χ_{cyc}. Then

(∗∗)  ψ_{ℓ} = χ_{cyc}^{r}ν_{ℓ},

where r is an integer, and ν_{ℓ} the ℓ−adic character defined by a Dirichlet character ν.

Theorem B ⟹ Theorem A. Let f, g be as in Theorem A. Since σ_{ℓ}(f) and σ_{ℓ}(g) are simple, Ad(σ_{ℓ}(f)) and Ad(σ_{ℓ}(g)) are semisimple, and we claim that they are isomorphic.

Modulo this claim, we proceed as follows. Applying the first part of Theorem B, we get a character ψ_{ℓ} such that σ_{ℓ}(f) ≃ σ_{ℓ}(g)⊗ψ_{ℓ}. Comparing determinants, we get for almost all p,

(I)  ψ_{ℓ}(F r_{p})^{2} = χ_{cyc}(F r_{p})^{k−k′}ω(p)ω′(p)^{−1}.

At this point, one can use (at least) three different methods to finish the argument. The first uses a theorem of Faltings [Fa], which says that σ_{ℓ}(h) is Hodge-Tate for any newform h of conductor prime to ℓ. So, by the second part of Theorem B, k and k′ are of the same parity, and we get (**) with r = (k−k′)/2. Let H(f) (resp. H(g)) be the Q-Hodge structure of weight k − 1 (resp. k′ − 1) associated to (the motive of) f (resp. g). Then we must have H(f) ≃ H(g)(r), where H(g)(r) denotes the Tate twist H(g) ⊗ Q(r). Then r must be zero, since the Hodge type of H(f) (resp. H(g)) is {(k − 1, 0), (0, k − 1)} (resp. {(k′−1, 0), (0, k′−1)}), while that of H(g)(r) is {(k′−1−r, −r), (−r, k′−1−r)}. Done.

The second method uses L-functions. Let ν be the finite order character defined as ψ_{ℓ}χ_{cyc}^{(k′−k)/2}. Then by (I) we have, for every Dirichlet character μ, an identity

L_{p}(s, f ⊗ μ) = L_{p}(s − (k − k′)/2, g ⊗ μν),

for all p in the set T of all primes not dividing ℓNN′ and the conductor of μ. We may fix a μ, sufficiently ramified at the primes in T, such that the local factors of f ⊗ μ and g ⊗ νμ at any prime in T are 1. Interchanging f and g if necessary, we may assume that k ⩽ k′. Since the archimedean factor attached to f ⊗ μ is (2π)^{−s}Γ(s), and since its product with (the global Euler product) L(s, f ⊗ μ) is entire, any pole of the Gamma factor results in a zero of L(s, f ⊗ μ), which is ∏_{p∉T} L_{p}(s, f ⊗ μ) by the choice of μ. This happens for example at s = 0, and consequently, by the identity above, L(s + (k′ − k)/2) has a zero at s = 0, even though its the archimedean factor does not have a pole there (as k′ > k). Then, by applying the functional equation for g ⊗ μν (which relates s to k′ − s), we see that L(s, \overline{g} ⊗ \overline{μν}) has a zero at s = (k′ + k)/2. This is absurd (see [JS]) as this point is in the region (resp. on the boundary) of absolute convergence if k > 1 (resp. k = 1). So we must have k = k′.

The third method is to appeal, for ℓ large enough, to the mod ℓ result proved later in this appendix.

Now we prove the claim. The identity (*) says that the characteristic polynomials of the Frobenius classes F r_{p} agree on Ad(σ_{ℓ}(f)) and Ad(σ_{ℓ}(g)), for all p outside a set S of density δ < \frac{1}{18}. If δ(S) = 0, then by the Tchebotarev density theorem, Ad(σ_{ℓ}(f)) and Ad(σ_{ℓ}(g)) would be equivalent, and our object is to get the same conclusion under the weaker hypothesis on δ. By [GJ], we know that, for h = f or g, there is an (isobaric) automorphic representation Ad(h) of GL(3, A_{Q}), whose standard L-function identifies, after removing the archimedean factors, with ∏_{p} L_{p}(s − 1, Ad(h)). It suffices to show that Ad(f) and Ad(g) are isomorphic. Suppose not. Then we can find (isobaric) automorphic representations π, π′ of GL(k, A_{Q}), k ⩽ 3, such that Ad(f) ≃ π ⊞ η and Ad(g) ≃ π′ ⊞ η, where η is an automorphic representation of GL(3 − k, A_{Q}), taken to be 0 if k = 3. Let Z_{S}(s) be as in equation (3) of [Ra]. In the present case, if m (resp. r) denotes the number of cuspidals occurring in the isobaric decomposition [La] of π (resp. π′), necessarily with multiplicity 1, we have −ord_{s=1}Z_{S}(s) = m^{2} + r^{2} (compare with (4) of [Ra]). Since one knows the Ramanujan conjecture for holomorphic forms by Deligne, it is easy to verify that Lemma 2 of [Ra] holds for π (resp. π′) with β less than k^{2}m^{2}δ (resp. k^{2}r^{2}δ). Then the argument of section 2 of [Ra] shows that we must have 1 ⩽ 2k^{2}δ. Since δ < 1/18 and k ⩽ 3, we get the desired contradictiction.

It remains to treat the case when f, g are not of CM type and have weights ⩾ 2, with δ assumed to be just < 1. One knows by the works of Serre and Ribet [Ri] that σ_{ℓ}(f) is absolutely irreducible under restriction to any open subgroup. We note then that the same must be true for Ad(σ_{ℓ}(f)), as otherwise the restriction σ_{ℓ}(f)_{K} will, for some number field K, be induced by a character of Gal(\overline{Q}/F), for a quadratic extension F/K (see below), making σ_{ℓ}(f)_{F} reducible. Now, applying Theorem 2 of [Raj] for example, we may conclude that, as δ < 1, Ad(σ_{ℓ}(f)) must be isomorphic to Ad(σ_{ℓ}(g)) ⊗ ν_{ℓ}, for some one-dimensional ν_{ℓ} of Gal(\overline{Q}/Q) defined by a Dirichlet character. Let K be the cyclic extension of Q corresponding to ν_{ℓ}, and let τ be a generator of Gal(K/Q). Then, since Ad(σ_{ℓ}(f)_{K}) and Ad(σ_{ℓ}(g)_{K}) are isomorphic, we may apply Theorem B and conclude that σ_{ℓ}(f)_{K} ≃ σ_{ℓ}(g)_{K} ⊗ λ_{ℓ}, for a character λ_{ℓ} of Gal(\overline{Q}/K). Since σ_{ℓ}(f)_{K} and σ_{ℓ}(g)_{K} are invariant under τ, we get

σ_{ℓ}(g)_{K} ⊗ (λ/λ^{[τ]}) ≃ σ_{ℓ}(g)_{K}.

Since σ_{ℓ}(g) is irreducible under restriction to any open subgroup, σ_{ℓ}(g)_{K} cannot admit any non-trivial self-twist, and λ must be invariant under τ and hence must extend to a character of Gal(\overline{Q}/Q). The rest of the argument goes through as above, and Theorem A follows.

Proof of Theorem B. First we need a simple

Lemma. Let ρ_{ℓ} be an irreducible, n−dimensional, self-dual \overline{Q}_{ℓ}-representation of Gal(\overline{Q}/K). Then there exists an invariant non-degenerate bilinear form B on (the space of) ρ_{ℓ}, which is symmetric or alternating, such that

• (i) B is unique up to a non-zero scalar; and

• (ii) If ρ′_{ℓ} is another irreducible, n−dimensional, self-dual \overline{Q}_{ℓ}-representa-tion of Gal(\overline{Q}/K) with invariant non-degenerate bilinear form B′, such that ρ_{ℓ} and ρ′_{ℓ} are isomorphic, then they are isometric relative to B and B′.

Indeed, (i) and the statement above it are immediate consequences of Schur’s lemma. Also, since \overline{Q}_{ℓ} is algebraically closed, cB is isometric to B for any c ∈ \overline{Q}_{ℓ}^{∗}; hence we get (ii) as well.

Now let σ_{ℓ} and σ′_{ℓ} be as in Theorem B. Suppose (the semisimple representation) Ad(σ_{ℓ}) is reducible. Then it must contain a one dimensional summand η_{ℓ}, say. Then η_{ℓ} occurs in the (self-dual) End(σ_{ℓ}) = σ_{ℓ} ⊗σ_{ℓ}^{∨} = Ad(σ_{ℓ})⊕1. Schur’s lemma above forces η_{ℓ} to be non-trivial. Either η_{ℓ} is quadratic, or otherwise η_{ℓ}^{∨} will also occur in End(σ_{ℓ}). In either case, we see that End(σ_{ℓ}) must contain a quadratic character δ_{ℓ}, say; let F be the corresponding quadratic extension of K with non-trivial automorphism θ. Denote by σ_{F,ℓ} the restriction of σ_{ℓ} to Gal(\overline{Q}/F). We claim (as is well known) that if τ_{ℓ} is another semisimple representation of Gal(\overline{Q}/K) whose restriction to Gal(\overline{Q}/F) is isomorphic to σ_{F,ℓ}, then τ_{ℓ} ≃ σ_{ℓ} ⊗ δ_{ℓ}^{j}, for j ∈ {0, 1}. Indeed, by the hypothesis, the restriction of η_{ℓ} := τ_{ℓ} ⊗ σ_{ℓ}^{∨} to Gal(\overline{Q}/F) contains the trivial representation; so by Frobenius reciprocity, there is a non-trivial homomorphism between η_{ℓ} and the representation of Gal(\overline{Q}/K) induced by the trivial representation of Gal(\overline{Q}/F), which decomposes as 1⊕δ_{ℓ}. So δ_{ℓ}^{j} occurs in η_{ℓ}, for j = 0 or 1. Equivalently, there is an intertwining operator between τ_{ℓ} and σ_{ℓ} ⊗ δ_{ℓ}^{j}, which implies the claim by virtue of the irreducibility of σ_{ℓ}. Next observe that σ_{F,ℓ} must be reducible as End(σ_{F,ℓ}) contains 1 with multiplicity 2 (as the restriction of δ_{ℓ} to Gal(\overline{Q}/F) is trivial). Write σ_{F,ℓ} = ν_{ℓ} ⊕ μ_{ℓ}, with ν_{ℓ}, μ_{ℓ} being one-dimensionals of Gal(\overline{Q}/F). We claim that ν_{ℓ} is not θ-invariant. Indeed, otherwise μ_{ℓ} would also be θ-invariant as σ_{F,ℓ} is, and both ν_{ℓ} and μ_{ℓ} would admit extensions to Gal(\overline{Q}/K) and result in a reducible extension of σ_{ℓ,F}, which is impossible by the claim above. Thus ν_{ℓ} is not fixed by θ, and so we must have σ_{F,ℓ} ≃ ν_{ℓ} ⊕ ν_{ℓ}^{[θ]}. This forces σ_{ℓ} to be the induced representation Ind_{F}^{K}(ν_{ℓ}), as this induced representation has the same restriction to Gal(\overline{Q}/F) as σ_{ℓ} and is moreover isomorphic to its twist by any character of Gal(\overline{Q}/K) trivial on Gal(\overline{Q}/F). Since End(σ_{ℓ}) = End(σ′_{ℓ}), σ′_{ℓ} must also be of the form Ind_{F}^{K}(ν′_{ℓ}), for some one-dimensional ν′_{ℓ} of Gal(\overline{Q}/F). Since the determinant of Ind_{F}^{K}(ν_{ℓ}) is the transfer of ν_{ℓ} to Gal(\overline{Q}/K) times δ_{ℓ}, we see that

Ad(σ_{ℓ}) ≃ Ind_{F}^{K}(ν_{ℓ}/ν_{ℓ}^{[θ]}) ⊕ δ_{ℓ},

and similarly for Ad(σ′_{ℓ}). This implies that, up to replacing ν_{ℓ} by ν_{ℓ}^{[θ]}, we have

ν_{ℓ}/ν_{ℓ}^{[θ]} = ν′_{ℓ}/(ν′_{ℓ})^{[θ]}.

Then ν_{ℓ}/ν′_{ℓ} is θ−invariant, and hence extends to a character ψ_{ℓ} of Gal(\overline{Q}/K). In other words, σ_{ℓ} ≃ σ′_{ℓ} ⊗ ψ_{ℓ}, as claimed.

We next consider the case when Ad(σ_{ℓ}) and Ad(σ′_{ℓ}) are irreducible. Let λ_{ℓ} denote the product of the determinants ω_{ℓ}, ω′_{ℓ} of σ_{ℓ}, σ′_{ℓ} respectively. Set

η_{ℓ} := σ_{ℓ} ⊗ σ′_{ℓ}.

Then

sym^{2}(η_{ℓ}) ⊗ λ_{ℓ}^{−1} ≃ Ad(σ_{ℓ}) ⊗ Ad(σ′_{ℓ}) ⊕ 1.

Since Ad(σ_{ℓ}) and Ad(σ′_{ℓ}) are irreducible, self-dual and isomorphic, 1 occurs in their tensor product. Hence the multiplicity of λ_{ℓ} is greater than 1 in sym^{2}(η_{ℓ}), showing that η_{ℓ} is reducible. Now suppose η_{ℓ} contains a two dimensional summand τ_{ℓ}, say. Then the one dimensional det(τ_{ℓ}) occurs in the exterior square of η_{ℓ}. But on the other hand, we have

Λ^{2}(η_{ℓ}) ≃ sym^{2}(σ_{ℓ}) ⊗ ω′_{ℓ} ⊕ ω′_{ℓ} ⊗ sym^{2}(σ′_{ℓ}),

showing that, as the symmetric squares of σ_{ℓ} and σ′_{ℓ} are irreducible, there can be no one dimensional summand of Λ^{2}(η_{ℓ}). This shows that η_{ℓ} has no two dimensional summand. Since it is reducible, it must then have a one dimensional summand ν_{ℓ}, say. Then

σ_{ℓ} ≃ σ′_{ℓ}^{∨} ⊗ ν_{ℓ} ≃ σ′_{ℓ} ⊗ ω′_{ℓ}^{−1}ν_{ℓ}.

So we get the desired ψ_{ℓ} by taking it to be ω′_{ℓ}^{−1}ν_{ℓ}.

Now let K = Q. Comparing determinants, we see that ψ_{ℓ}^{2} = det(σ_{ℓ})det(σ′_{ℓ})^{−1}. So we get (**) immediately if the ratio of the determinants is a finite order character times an even power of χ_{cyc}. Finally, suppose σ_{ℓ} and σ′_{ℓ} are Hodge-Tate. Then ψ_{ℓ} will also be Hodge-Tate as it occurs in σ_{ℓ} ⊗ (σ′_{ℓ})^{∨}. Consequently, it corresponds to an algebraic Hecke character ψ. Since we are working over Q, it must be a finite order character times a power of χ_{cyc}. Done.

For the second proof, we begin by recalling the fact that the adjoint representation Ad: PGL(2, \overline{Q}_{ℓ}) ⟶ GL(3, \overline{Q}_{ℓ}) is isomorphic onto the special orthogonal group SO(3, \overline{Q}_{ℓ}). Denote by \overline{σ}_{ℓ} (resp. \overline{σ}′_{ℓ}) the composite of σ_{ℓ} (resp. σ′_{ℓ}) with the natural homomorphism of GL(2, \overline{Q}_{ℓ}) onto PGL(2, \overline{Q}_{ℓ}). Then it is easy to see that Ad(\overline{σ}_{ℓ}) identifies with the Ad(σ_{ℓ}) defined earlier (above Theorem B). So, by our hypothesis, we get two representations, namely Ad(\overline{σ}_{ℓ}) and Ad(\overline{σ}′_{ℓ}), into SO(3, \overline{Q}_{ℓ}), which are equivalent in GL(3, \overline{Q}_{ℓ}). Suppose they are irreducible. Then we may apply part (ii) of the Lemma and deduce that they are in fact isometric. By changing the isometry by −I if necessary, we may assume that they are equivalent in SO(3, \overline{Q}_{ℓ}). Since Ad is an isomorphism, σ_{ℓ} and σ′_{ℓ} define equivalent homomorphisms into PGL(2, \overline{Q}_{ℓ}). Hence σ_{ℓ} must be equivalent to σ′_{ℓ} ⊗ ψ_{ℓ}, for some ψ_{ℓ} ∈ Hom(Gal(\overline{Q}/K), \overline{Q}_{ℓ}^{∗}). When Ad(σ_{ℓ}) is reducible, one uses explicit arguments as in the reducible case of the first proof to conclude that Ad(σ_{ℓ}) and Ad(σ′_{ℓ}) are isometric. The rest follows. QED.

The mod ℓ version. For each newform f, let K_{f} denote the number field generated by the coefficients of f. If g is another newform, let 𝔒_{f,g} denote the ring of integers of the compositum K_{f}K_{g}. For h = f or g, write for p not dividing the level,

Q_{h}(T) = (1 − \frac{α_{p}(h)}{β_{p}(h)}T)(1 − T)(1 − \frac{β_{p}(h)}{α_{p}(h)}T),

so that L_{p}(s, Ad(h)) = Q_{h}(p^{−s})^{−1}. Note that, since α_{p}(h)β_{p}(h) = ω(h)p^{k(h)−1}, α_{p}(h) and β_{p}(h) are invertible modulo any prime ℓ not dividing pN(h).

Theorem C Let ℓ be an odd prime number and N, N′ positive integers prime to ℓ. Let f (resp. g) be a newform of level N (resp. N′), weight k (resp. k′), and character ω (resp. ω′). Let λ be a prime ideal above ℓ in 𝔒_{f,g}. Suppose we have

(C)  Q_{f}(T) ≡ Q_{g}(T) (mod λ),

for all p outside a set S (containing the primes divisors of ℓNN′) of density 0. Then k ≡ k′ (mod ℓ − 1), and there exists a character β, unramified at ℓ, such that

a_{p} ≡ b_{p}β(p) (mod λ),

for all p not dividing ℓNN′.

Remark: Note that if ω and ω′ are the same mod λ, and if k − k′ ≡ 0(mod ℓ − 1), the hypothesis (C) is equivalent to the congruence

a_{p}^{2} ≡ b_{p}^{2} (mod λ).

In this case β is necessarily quadratic. Moreover, if N and N′ are in addition square-free, one can conclude (as in the characteristic zero case) that β is trivial.

Proof. Let F_{λ} denote the residue field 𝔒_{f,g}/λ. Reducing the (integrally defined) ℓ-adic representations associated to f, g modulo λ and extending scalars to \overline{F}_{λ}, we get representations

\overline{σ}_{λ} : Gal(\overline{Q}/Q) ⟶ GL_{2}(\overline{F}_{λ})

and

\overline{σ}′_{λ} : Gal(\overline{Q}/Q) ⟶ GL_{2}(\overline{F}_{λ})

such that, for all p not dividing NN′ℓ, tr(\overline{σ}_{λ}(Fr_{p})) (resp. tr(\overline{σ}′_{λ}(Fr_{p}))) is the image of a_{p} (resp. b_{p}) in \overline{F}_{λ}. Moreover, by hypothesis, det(\overline{σ}_{λ}) and det(\overline{σ}′_{λ}) both equal χ^{k−1}\overline{ω} (resp. χ^{k′−1}\overline{ω}′), where χ : Gal(\overline{Q}/Q) → F_{ℓ}^{∗} is the mod ℓ cyclotomic character and \overline{ω} (resp. \overline{ω}′) the reduction (mod λ) of ω (resp. ω′). Clearly, the images of Gal(\overline{Q}/Q) under these two representations are finite.

For any \overline{F}_{λ}-representation τ_{λ} of a finite group G of dimension d, let τ_{λ}^{ss} denote its semisimplification. Note that in characteristic ℓ, the semisimplification is determined by the characteristic polynomials of τ_{λ}(g) for all g in G when d > ℓ, and also when d = ℓ = 3 if τ_{λ} is orthogonal of determinant 1.

By the hypothesis (C), the characteristic polynomials of Fr_{p} in the adjoint representations of \overline{σ}_{λ} and \overline{σ}′_{λ} are the same for all p in a set of density 1. Thus, by the Tchebotarev density theorem and the remark above, we see that

Ad(\overline{σ}_{λ}^{ss}) ≃ Ad(\overline{σ}′_{λ}^{ss}).

Since End(\overline{σ}_{λ}^{ss}) (resp. End(\overline{σ}′_{λ}^{ss})) is Ad(\overline{σ}_{λ}^{ss}) ⊕ 1 (resp. Ad(\overline{σ}′_{λ}^{ss}) ⊕ 1), it follows that \overline{σ}_{λ} is irreducible iff \overline{σ}′_{λ} is.

First suppose that \overline{σ}_{λ} and \overline{σ}′_{λ} are irreducible. In this case the detailed ℓ-adic argument given in the proof of (the first part of) Theorem B goes through, with \overline{Q}_{ℓ} replaced everywhere by \overline{F}_{λ}, once one notes the availability of the relevant form of the Frobenius reciprocity in characteristic ℓ (cf. [A], chap. III, Lemma 6) and the fact that the tensor square of a simple Galois module is semisimple [Se3]. One deduces an isomorphism of \overline{σ}_{λ} with \overline{σ}′_{λ} ⊗ ν_{λ}, for some character ν_{λ} of Gal(\overline{Q}/Q) into \overline{F}_{λ}. Since ω_{λ} and ω′_{λ} are the same modulo λ, we see by comparing determinants that ν_{λ}^{2} is χ^{k−k′} \overline{ω}/\overline{ω}′. We may write ν_{λ} as χ^{j}β_{λ}, for some j ∈ {0, …, ℓ − 2}, and a character β_{λ} unramified at ℓ. Consequently, k − k′ ≡ 2j( mod ℓ − 1), β_{λ}^{2} = \overline{ω}/\overline{ω}′, and

(∗ ∗ ∗)  \overline{σ}_{λ} ≃ \overline{σ}′_{λ} ⊗ χ^{j}β_{λ}.

Let G_{ℓ} denote the decomposition group at ℓ of Gal(\overline{Q}/Q), and let I denote the inertia subgroup. When a_{ℓ} is not zero modulo λ, one knows by Deligne (cf. [E], Theorem 2.5, for example), that \overline{ρ}_{λ}|_{G_{ℓ}} is reducible, and its semisimplification is of the form χ^{k−1}μ_{1,λ} ⊕ μ_{2,λ}, where each μ_{j,λ} is unramified. When a_{ℓ} is divisible by λ, a result of Fontaine (see [E], Theorem 2.6) asserts that the restriction to G_{ℓ} is irreducible, while the restriction to I demposes as ψ^{k−1} ⊕ ψ^{′k−1}, where ψ, ψ′ are the two fundamental characters of level 2 [Se2]. Similarly for the restriction of \overline{σ}′_{λ} at ℓ. In either case, we see that the only way (***) can hold is for j to be 0 modulo ℓ − 1.

It remains to consider when \overline{σ}_{λ} (and hence \overline{σ}′_{λ}) is reducible. Here we may write

\overline{σ}_{λ}^{ss} ≃ η_{λ} ⊕ χ^{k−1}\overline{ω}/η_{λ},

and

\overline{σ}′_{λ}^{ss} ≃ η′_{λ} ⊕ χ^{k′−1}\overline{ω}/η′_{λ},

for some \overline{F}_{λ}^{∗}-valued characters η_{λ}, η′_{λ} of Gal(\overline{Q}/Q). Then we have

Ad(\overline{σ}_{λ}^{ss}) ≃ η_{λ}^{2}/\overline{ω}χ^{k−1} ⊕ 1 ⊕ \overline{ω}χ^{k−1}/η_{λ}^{2},

and

Ad(\overline{σ}′_{λ}^{ss}) ≃ η′_{λ}^{2}/\overline{ω}χ^{k′−1} ⊕ 1 ⊕ \overline{ω}χ^{k′−1}/η′_{λ}^{2}.

Since Ad commutes with semisimplification, it follows, after possibly replacing η_{λ} with χ^{k−1}\overline{ω}/η_{λ}, that η_{λ}^{2}/χ^{k} = η′_{λ}^{2}/χ^{k′}. Arguing as above, we see that η_{λ} is of the form η′_{λ}χ^{j}β_{λ}, for some j ∈ {0, …, ℓ − 2} with k − k′ ≡ 2j( mod ℓ − 1), and a character β_{λ} :Gal(\overline{Q}/Q) → \overline{F}_{λ}^{∗}, unramified at ℓ, such that β_{λ}^{2} = \overline{ω}/\overline{ω}′. We obtain

\overline{σ}_{λ}^{ss} ≃ η′_{λ}β_{λ}χ^{(k−k′)/2} ⊕ β_{λ}χ^{(k+k′)/2−1}\overline{ω}′/η′_{λ}.

The reducibility of \overline{σ}_{λ} (resp. \overline{σ}′_{λ}) forces a_{ℓ} (resp. b_{ℓ}) to be non-zero modulo λ, as the restriction of \overline{σ}_{λ}^{ss} (resp. \overline{σ}′_{λ}^{ss}) to I must then be given by a direct sum of characters of level 1 [Se2]. Applying Deligne’s result on the shape of the restriction to G_{ℓ} (see above), we see that the only possibility is for k and k′ to be congruent modulo ℓ − 1. Then \overline{σ}_{λ}^{ss} is isomorphic to \overline{σ}′_{λ}^{ss} ⊗ β_{λ}. Done.

References

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Dinakar Ramakrishnan (dinakar@cco.caltech.edu) 253-37 Caltech, Pasadena, CA 91125.