CONGRUENCES. 1. History

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1 CONGRUENCES HAO BILLY LEE Abstract. These are notes I created for a seminar tak, foowing the papers of On the -adic Representations and Congruences for Coefficients of Moduar Forms by Swinnerton-Dyer and Congruences et Formes Moduaires by Serre. I caim no credit to the originaity of the contents of these notes. Nor do I caim that they are without errors, nor readabe. n=1 1. History We have the Ramanujan -function, weight 1 cusp-form = q 1 q n 4 = τnq n. Like a moduar forms, τn s are known once we know τp. For a positive v, denote σ v n = d n dv, for exampe: σ v q = 1 + p v. Ramanujan computed a bunch of congruences between τn and σ v n mod some prime power. In particuar the primes are, 3, 5, 7, 3 and 691. Here 3 = k 1, and 691 b 1 Bernoui number Question: Can we find/prove these even tho we are not Ramanujan? Are these a the possibe primes? Let χ : Ga Q/Q Z be the cycotomic character, then we know that for a p, χ F rob p = p. We know that given a normaized eigenform of weight k, we have a continuous homomorphism everything is eve 1 n=1 ρ : Ga K /Q GL Z technicay, just need K maxima agebraic extension of Q ramified ony at, satisfying x a p x + p k 1 for a p. This says that det ρ = χ k 1. The question now becomes, can we find congruence reations between the trace and the determinant of every eement in the image of ρ. I don t know much about the group theory of GL Z p... One natura thing, is to consider the image of ρ : Ga K /Q GL F. Lemma 1.1. Suppose that > 3, G is a subgroup of GL Z which is cosed in the -adic topoogy which we have because of compactness. If the image of G under reduction mod contains SL F then G contains SL Z. If it does not, ca an exceptiona prime when it does not contains SL Z. For = or 3, need to repace F with Z/8 or Z/9. Proof. Idea: et G n be the image of G in GL Z/ n Z. Use induction to prove that G n SL Z/ n Z, which is aso sufficient to prove that G n contains the kerne H n = ker SL Z/ n Z SL Z/ n 1 Z. 0 1 Then you can write H to be generated by I + u for some u. Ie. for n =, u = 0 0 The image of I + u SL Z generaize to H n 1 and n u, so there s σ G such that 0 0, or 1 1. σ = I + u + v 1

2 CONGRUENCES where v has Z -entries. Then σ = I + u + v u + v I + u mod because u = 0 and a other terms have. Why happens to non-exceptiona primes ie. does contain SL Z. Lemma 1.. Suppose f is normaized eigenform, not exceptiona for. Let N, N be non-empty open sets in Z and Z respectivey. Then the set of primes p for which p N and a p N has positive density are infinite sets. Proof. Goa is to show that the image of the map ρ, χ : Ga K /Q GL Z Z contains SL Z 1. When we project to the first factor, we get a of SL Z. Therefore, the image of the commutator subgroup must contain Comm SL Z 1. When > 3, Comm SL Z = SL Z, because the reduction map SL Z SL F maps the commutator to the commutator. Fact: Comm SL F = SL F, and by the emma. When = or 3, pick σ G such that ρ σ SL Z. Then χ k 1 σ = 1. Since Z does not contain non-trivia roots of unity of odd order, χ σ = 1. Since χ is surjective, the image wi then be exacty a the α β with det α = β k 1. Then the map T r ρ, χ : Ga K /Q Z Z Tr 1 is surjective take. This is a map F robp a p p. det 0 Use Chebotarev density theorem, and uniform distributivity of the Frobenius eements. In fact, Serre defines exceptiona prime to be exacty those, whose image is not surjective. Now, the goa is to study the exceptiona primes. In particuar, what are the possibe subgroups of GL F that does not contain SL F. Swinnerton-Dyer ists a bunch of cases of such a subgroup G GL F want to see that they are sma enough for non-triva agebraic reation. Definition 1.3. Bore subgroup is one that s conjugate to invertibe upper trianguar matrix. This is in 1-1 correspondence with 1-dimsensiona subspace W V and group is those with W as an eigenspace. Cartan subgroup is a maxima semisimpe diagonaizabe over agebraic cosure commutative subgroup. Spit means diagonaizabe over F p, is in 1-1 corr with 1-dimensiona subspaces W 1, W V, and group is those with these two as eigenspace non-spit is not diagonaizabe over F p. We can base change to quadratic extension K of F, and et W be any one dimensiona eigenspaces of V K that is not induced by a subspace of V, and W its conjugate. Have 1 1 correspondence of the Cartan whch have W, W as eigenspaces. An eement in the group is then uniquey determined by its eigenvaue associated to W. In fact, an eement It s cycic group isomorphic to F p Cartan means that we have two eigenspaces after base extension. So an eement of the normaizer, wi either fix or interchange the two eigenspaces. If it fixes, it s aready in the Cartan subgroup by maximaity. Theorem 1.4. Suppose ρ : Ga K /Q GL Z is a continuous homomorphism such that det ρ = χ k 1 for some even integer k. Let G GL F be the image of ρ, and suppose that it does not contain SL F. These cases give you congruences mod. Then

3 CONGRUENCES 3 1 G is contained in a Bore subgroup of GL F. There is an integer m such that a n n m σ k 1 m n mod for a n prime to Ony happens if m < < k or m = 0 and divides the numerator of b k Bernouii number G is contained in the normaizer of a Cartan subgroup, but not in the Cartan subgroup itsef a n 0 mod whenever n is a quadratic non-residue mod Can ony happen if < k 3 The image of G in P GL F is isomorphic to S 4 specia to eve 1 p 1 k a p 0, 1, or 4 mod for a p Harder to describe, but a finite set Proof. We can write ρ σ = ασ βσ 0 δσ α is a continuous homomorphism Ga K /Q F and so it s equa to χm which is αδ, so δ = χ k 1 m. Let σ = F robp for p then. for some integer m. The determinant is χ k 1 The genera thing for n foows. a p = p m + p k 1 m mod. Let C be the Cartan subgroup, and N its normaizer. Consider the homomorphism This is surjective by hypothesis. continuous homomorphism to Z Ga K /Q N N/C = {±1}. Since the image is commutative, we factor through Ga K ab/q = Z. to {±1} that s surjective, is the one whose kernes are squares. Therefore, p ρ F robp C = 1. Suppose α N but not in C. Then α interchanges the two eigenspaces, so we can write α = Hence, a p 0 mod whenever p is a quadratic non-residue mod. Let. Moduar forms mod P = E = 1 4 σ 1 nq n Q = E 4 = σ 3 nq n R = E 6 = σ 5 nq n. 0 0 The ony, so trace is 0. We know that any moduar form of weight k can be expressed as an isobaric poynomia in Q and R. In particuar, et A be the isobaric poynomia such that A Q, R = E 1. Let M k be space of moduar forms of weight k with O coefficients, where O = {a Q : v p a 0}. Let M k F q be the F -vector space generated by the reduction map. The space M of mod moduar forms is the sume of Mk. Fact.1. The reduction map M k M k is injective, but eements of different weight can have the same q-expansion Define the operators θ = q d dq : n a n q n n na n q n On the Gaois representation, it corresponds to a twist by χ. In particuar, θ 1 : a n q n n,=1 a nq n. This means that θ 1 acts as the identity when restricted to the kerne of U : a n q n a n q n.

4 CONGRUENCES 4 Theorem [ ].. Suppose > 3 when p = or 3, Q = R = 1, from the fact that 178 = Q 3 R, we get F p then 1 à Q, R = 1 so it has q-expansion 1 M is naturay isomorphic to F [Q, R] / à 1, with a natura grading with vaues in Z/ 1 M is just Proof. 1 anayze some congruences of Bernouii numbers. If à Q, R = 1, then ceary, à 1 is in the kerne of F [Q, R] M = F q. Since M is an integra domain, the kerne is prime. The dimension is, so we just need to check that à 1 is not maxima, and is irreducibe. Maxima because ese M is a finite fied, which means that Q and R are then agebraic over F... but they have q-terms. Remark.3. The à is the Hasse invariant Theorem.4. Suppose f and f are moduar forms of weight k and k respectivey. If f f mod and f 0 mod, then we have k k mod 1. Definition.5. Define the fitration ωf = inf which is a sum of eements where a the reevant k are congruence make them a beong to the same M k. {j : f M j }. Define wf = if f = 0. If f is a a graded eement, Theorem.6. Let f M. Then ωθ f ω f + + 1, with equaity iff ω f 0 mod. mod 1, can mutipy by suitabe powers of à to Proof. First, we know that n = n 1. So then if n is a quadratic non-residue mod, then This is equivaent to a n 0 mod a n n n +1 mod. θ f = θ +1/ f. If we assume that > k, then ω f = k and ω θ f = k Simiary, we see that if k n 0 mod then k + n 0 mod. This means that n > k. Therefore, ω θ +1/ f = k We can do more. Since is odd and k is even, = k and = k are not possibe. Let s suppose k < < k. Simiar to the above, ω θ n f = k + n + 1 if n k. Therefore, ω θ k+1 f = k k + 1 n 1 = k + k k + 1 n 1 = k n n +1/ What we wi see, is that to get to ω θ f, we wi no onger drop again.

5 CONGRUENCES 5 Suppose f is in the kerne of the operator U, then θ 1 f = f. Ca f = f 1 and et f i = θ A f 1 for some A where ω θ A 1 f 1. Let ci 1 be the number of times we add + 1 to ωf i before reaching something divisibe by. Let b i 1 be the amount we fa. Since 1 forms a cyce, we get i c i = 1. In tota, I ve added 1-number of + 1 so = i c i + 1 = i b i 1 where the ast equaity is because I ended up where I started. This shows that b i = + 1. But since this has to be w θ f = k < + 1, we see that we can ony appy θ once to the above. Therefore, k + 1 = + 1 k + = + 1 = = k 3 = = k 1.