The Generalisation of Theorem 4.0.1

We now no longer want to assume that one of our characters is trivial and the other quadratic. Suppose we have an Eisenstein series E₁ψ,ϕ∈ M1(Γ1(N ), ψϕ) . Here suppose

ψ is primitive with conductor u , ϕ is primitive with conductor v and ψϕ has conductor N = uv . We know by Proposition 4.1.4, that a dihedral representation comes from the induction of a character χ of GK = Gal( ¯Q/K) where K is a quadratic field. Associated to this dihedral representation is a cusp form f . We will first show that in order for a congruence to exist we must have ψ ≡ ηϕ (mod λ) where η is the character associated to the quadratic field K .

Proposition 4.2.6. Let ψ and ϕ be as above. Let K be an imaginary quadratic field of discriminant N with associated quadratic character η : (Z/N Z)× → C×. Let χ be a ray class character for K and fχ the associated cusp form. Then

aq(fχ) ≡ ψ(q) + ϕ(q) (mod λ),

where λ|` is a prime of Q({an(fχ)}) if and only if ψ ≡ ηϕ (mod λ) .

Proof. Recall that for an inert prime q in K , the Hecke eigenvalue of fχ is 0 . We

therefore would need ψ(q) + ϕ(q) ≡ 0 (mod λ) at such a prime. We immediately see that we must have ψ(q) ≡ −ϕ(q) (mod λ) . Hence we must have ψ = αϕ for some α such that α(q) ≡ −1 (mod λ) for all inert primes q . We may define α : (Z/N Z)× → C× by α = ψ_ϕ (where ψ and ϕ are viewed as functions on (Z/N Z)×). Hence α is a Dirichlet character modulo N .

Suppose we consider the reduction α viewed as a character α : G_Q → F×` . For any

τ ∈ G_Q\G_K we have α(τ ) = −1 . For σ ∈ GK we have σ = (στ−1)τ . Hence α(σ) =

α(στ−1)α(τ ) = (−1)(−1) = 1 . Hence we see that α is a character which takes the values ±1 . Also since the character factors through the quotient G_Q/GK = Gal(K/Q)

which is cyclic of order 2 , we see that α is in fact quadratic. In particular it is exactly the quadratic character η associated to K . Hence we deduce that ψ ≡ ηϕ (mod λ) .

Here we have proven that ψ ≡ ηϕ (mod λ) is a necessary condition. We will now try and construct a cusp form satisfying a congruence in a similar manner to the method used to prove Theorem 4.0.1 using this condition. That is, we will prove that the condition is sufficient. Rather than give separate results with proofs as we did when proving Theorem 4.0.1, we simply give a sketch proof detailing the differences. We first investigate the new Euler factors. Associated to the Eisenstein series E₁ψ,ϕ is the Euler factor ϕ(p)p − ψ(p) . We may consider the Euler factor modulo λ .

Suppose first that p is an inert prime in OK. Then η(p) = −1 and we see that

ψ(p) ≡ −ϕ(p) (mod λ) . Hence

ϕ(p)p − ψ(p) ≡ ϕ(p)p + ϕ(p) (mod λ) ≡ ϕ(p)(p + 1) (mod λ).

Suppose instead we consider a prime p that splits in OK. We now have η(p) = 1 hence

ψ(p) ≡ ϕ(p) (mod λ) . Hence

ϕ(p)p − ψ(p) ≡ ϕ(p)p − ϕ(p) (mod λ) ≡ ϕ(p)(p − 1).

Note that in both cases we see (p + 1) and (p − 1) appearing as before. We therefore see that ` divides (p + 1) in the inert case and (p − 1) in the split case since λ|` . We could also consider the Eisenstein series Eϕ,ψ₁ . However since ψ and ϕ are related via a congruence condition, we will end up with the same factors. We would therefore expect that dividing p − 1 in the split case or dividing p + 1 in the inert case should give a congruence just as in Theorem 4.0.1. Whether there is a congruence will depend on our choice of ray class character. It will turn out that we only need a slight modification. If we again consider a potential congruence, now using what we know about the char- acters ψ and ϕ , we will be able to determine which ray class character we will need to use. In the proof of Proposition 4.2.6, we considered a potential congruence at inert primes in order to determine conditions for the Dirichlet characters ψ and ϕ . We now want to consider the split primes. For a ray class character χ (i.e. a character of GK)

we want

ψ(q) + ϕ(q) ≡ χ(q1) + χ(q2) (mod λ) if q = q1q2 splits in OK.

We already know that ψ ≡ ηϕ (mod λ) . Hence at a split prime q we have ψ(q) ≡ η(q)ϕ(q) (mod λ) ≡ ϕ(q) (mod λ) . We therefore need

2ϕ(q) ≡ χ(q1) + χ(q2) (mod λ) if q = q1q2 splits in OK.

If we take χ = χ1χ2, where χ1 is the ray class character we use in Theorem 4.0.1, which

depends on which particular case we are in, we then need to determine χ2. Note that

in all cases χ1≡ 1 (mod λ). We note that before we simply had 2 on the LHS of the

to 1 (mod λ) in order for the congruence to be satisfied. However, since ϕ is arbitrary, we now have more possibilities for the LHS. If we take χ2 = ϕ ◦ NK/Q, where NK/Q is

the norm map, then we see that the congruence will be satisfied. This follows since χ(q1) + χ(q2) = χ1(q1)χ2(q1) + χ1(q2)χ2(q2)

≡ (ϕ ◦ N_K/Q)(q1) + (ϕ ◦ NK/Q)(q2) (mod λ)

≡ 2ϕ(q) (mod λ).

Here the second line follows since χ1(q1) ≡ χ1(q2) ≡ 1 (mod λ) and the third follows

since N_K/Q(q1) = NK/Q(q2) = q . The conductor of χ2 will depend on the conductor of

ϕ . Assume the conductor of ϕ is given by v = p1. . . pn. Since v|N , these are ramified

primes in K . In particular v factors as p2₁. . . p2_n in OK. The conductor of χ2 will

then be v = p1. . . pn. Taking the norm of any factor we see that NK/Q(pi) = pi. This

explains why we are adding a factor of v to the level of the cusp form in Theorem 4.0.2 since N_K/Q(v) = p1. . . pn.

We can also consider the congruence at the ramified primes. Since uv = N , the ramified primes are exactly those appearing in the prime factorisation of u and v . We again use the extended definition of a Dirichlet character. That is we have ψ : (Z/uZ) → C× and ϕ : (Z/vZ) → C×. For any non-invertible element q ∈ (Z/uZ) we have ψ(q) = 0 and for any non-invertible element q ∈ (Z/vZ) we have ϕ(q) = 0. We want to satisfy

ψ(q) + ϕ(q) ≡ χ(q) (mod λ) if q = q2 ramifies in OK.

Suppose first that q|u . We have ψ(q) = 0 but ϕ(q) depends on whether q|v We therefore need to satisfy

ϕ(q) ≡ χ(q) (mod λ).

Since χ1(q) ≡ 1 (mod λ) and NK/Q(q) = q , so χ2(q) = ϕ(q) , we see that the congru-

ence is satisfied for such a ramified prime regardless of whether q|v .

Now suppose q|v . We have ϕ(q) = 0 and ψ(q) depends on whether q|u . By the argument above the RHS of the congruence will be congruent to ϕ(q) (mod λ) . Hence the RHS is congruent to 0 (mod λ) . We therefore need

ψ(q) ≡ 0 (mod λ).

However we know that ψ(q) ≡ η(q)ϕ(q) (mod λ) ≡ 0 (mod λ) . It follows that the congruence holds for these primes as well.

In conclusion we see that the only modification we have made is to the ray class char- acter χ . We have simply multiplied by another ray class character χ2 = ϕ ◦ NK/Q with

conductor v = p1. . . pn. This has the effect of raising the level of the cusp form f in

each case by v = p1. . . pn since NK/Q(v) = v . Note that this choice is arbitrary, we

could just have easily had χ2 = ψ ◦ NK/Q due to the fact that ϕ ≡ ηψ (mod λ) as

well since η is quadratic. This would instead have the effect of raising the level by u . We note that in Theorem 4.0.1, χ2 was trivial since we had a trivial character (which

we could choose to be ϕ arbitrarily and so v = 1 in this case). This choice of χ2 for

Theorem 4.0.1 was enough because we had already chosen a ray class character that would guarantee the congruence held mod λ . We also note that in each case we still obtain the same conditions on when χ = χσ. This is because we can simply consider

what is happening to χ1. This follows since we have χ2 = χ2σ (This can be seen

by considering the norms of ideals in OK). If χ1 6= χ1σ, then we see that χ 6= χσ.

We also note that modulo λ the ray class character χ reduces to χ2 = ϕ ◦ NK/Q.

It follows that in this case χ_Q ≡ ϕ2 _{(mod λ) . Hence the character of the induced}

representation, and therefore the character of the cusp form, will be  = ηχ_Q ≡ ηϕ2

(mod λ) ≡ ψϕ (mod λ) . Other than these changes we could state results similar to Propositions 4.2.2, 4.2.3 and 4.2.4 along with proofs following largely the same steps. We have therefore given a (sketch) proof of Theorem 4.0.2.

Although we will not cover the details of the Bloch-Kato conjecture in the case of weight 1 , we note that the exact same argument as in Section 3.3 still applies in this case. Although the standard results of modular forms break down in the case of weight 1 , so in some sense the theory is harder, the Bloch-Kato conjecture is actually more straightforward. Here the conjecture is basically a consequence of Dirichlet’s analytic class number formula.

In order to prove Theorem 4.0.3, we will need to see some preliminaries on lifting projective representations.