HIGHER HIDA THEORY NOTES - University of Chicagohaolee/Higher Hida Theory Notes.pdf · Q are modular (representations coming from class group). 2.1. Deformation. Let F be a number

HIGHER HIDA THEORY NOTES

HAO (BILLY) LEE

Abstract. These are notes I took in class, taught by Professor Frank Calegari and George Boxer. I claim no credit to the

originality of the contents of these notes. Nor do I claim that they are without errors, nor readable.

Part 1. Frank's Lectures

1. Class Field Theory

Let K/Q be a number �eld. Let I = IK be the group of fractional ideals. Let S be a �nite set of places of K, and

IS ⊆ I be the fractional ideals prime to S. Let U be the unit group of K. Let C = CK be the class group, then we have

1→ U → K× → I → C → 1.

Similarly, de�ne KS ⊆ K× be the elements so that its corresponding principal fractional ideal is prime to S. Then we

have

1 U K× I C 1

1 U KS IS C 1.lemma

Lemma 1.1. Any ideal class in C is represented by an ideal prime to S.

Proof. Let [a] ∈ C. We want G ∈ IS such that [a] = [G] in C. Same thing as �nding G−1 ∈ IS such that[aG−1

]is trivial

in C.

That is, need x ∈ K× such that

(x) =

∏p∈S

pvp(a)

× outside S.

Consider

OK∏

p∈S Ok,p

∏p∈S(OK/pk).

�

Example 1.2. Suppose Z[√−5], let p =

(2, 1 +

√−5). Then we want to �nd x ∈ Z

[√−5]with vp(x) = 1. This is just

1 +√−5.

De�nition 1.3. A modulus is m = m∞mf where mf ⊆ OK is an ideal, and mf is a subset of in�nite places.

Let

K× ⊇ Km,1 =

x ∈ K× :

v | ∞, v real, v ∈ m∞ xv > 0

v | mf x− 1 ∈ mfOKv

.

If S is the set of primes dividing mf , then1

HIGHER HIDA THEORY NOTES 2

1 Um,1 Km,1 Im Cm 1

1 U Km = KS(m) I C 1.

=

By the snake lemma, we derive

U →∏v|S

O×K,v(1 + mfOKv )

⊕∏

v|m∞, v real

R×

(R×)2 → Cm → C → 1.

This Cm is the ray class group of conductor m.

Example 1.4. Suppose K = Q. Suppose S = {p,∞}. Let mf = (p), m∞ = {∞}. Of course, C = {1}.Principal primes

(K×m,1

)are (x) where x ≡ 1 mod p and x > 0. There is a map I → (Z/pZ)× by (x) = (y) with y > 0,

this maps to y mod p. Therefore, Cm = (Z/pZ)×.

If we didn't add this conductor at ∞, this map is not well-de�ned. But we see that when m = (p), Cm = (Z/pZ)×

±1 . The

exact sequence we have is

{±1} →∏ Z×p

1 + pZp→ Cm → C = {1} → 1.

Let L/K be �nite Galois with Galois group G = Gal (L/K). Let p be a prime of K. Let P be a prime above p in L.

Then we have

L LP OLP `

K Kp OKpk.

We also have

G = Gal(L/K)←↩ Gal (LP /Kp) � Gal (`/k)

after making some choices. The second is called the decomposition group, Dp, and the last is Dp/Ip = 〈Frob〉.If L/K is abelian, p unrami�ed in L/K, then let (p, L/K) be the Frobenius endomorphism in G.

De�nition 1.5. Artin map. Let L/K be a �nite abelian extension unrami�ed outside S. The Artin map is

ISK → Gal (L/K)

given by p→ (p, L/K). Cebotarev density theorem says that this is surjective.

Theorem 1.6. Existence theorem. For all conductors m, there exists extension Lm/K abelian such that the Artin map

factors through Cm and

ISK Gal(Lm/K)

Cm.

∼=

Theorem 1.7. If L/K is abelian, there exists conductor m (dividing only primes rami�ed in L/K) such that L ⊆ Lm.

That is

ISK → Cm � Gal (L/K) .

How we should view these two theorems is that Galois representations are automorphic and vice versa.

1.1. Galois Representations. Let G = Gal (L/K) be �nite. We can just study the primes of good reduction (no

rami�cation) and these are dense, which is �ne. However, we may have another situation where we have a tower of

representations.


Gal (L1/K) → GLn(Fp)↑ ↑

Gal (Lm/K) → GLn (Z/pmZ)

↑ ↑... ...

Then we will �nd that the decomposition group at p of this tower captures a remarkable amount of information, even at

the bad primes. This somehow justi�es studying these things.

Example 1.8. Let K = Q and m =(pk)

(∞). Suppose Lk = Q(ζpk)then Gal (Lk/Q) ∼=

(Z/pkZ

)×. Then the

corresponding Artin map is ` : ζ → ζ` gets mapped to [`].

1.2. Local Class Field Theory. Suppose K/Qp is �nite. The Artin map is just K× → GabK . Additionally, we have the

commutative diagram

O×K I

K× GabK

Z Gabk Z.

∼=

val

1 7→Frob ∼=

Non-canonically, K× ∼= Z×O×K (by choosing a uniformizer πK). If L/K is a �nite abelian extension, then

K×/NL/K(L×)→ Gal (L/K) .

Recall that we have

Gal(Q(ζpk)/Q

)(Z/pkZ)×

Gal(Qp(ζpk)/Qp

)Q×p

∼=

De�nition 1.9. De�ne the adèles to be

AK =

{x ∈

∏v

Kv : x ∈ Ov for all but �nitely many v

}and

A×K =

{x ∈

∏v

K×v : x ∈ O×v for all but �nitely many v

}.

We have a natural map

A×K → IK

by (αv) 7→∏r p

valv(αv)v . Then

∏O×Kv is of course in the kernel. By the canonical embedding K× ↪→ A×K , we get

K×\A×K/∏

v �nite

O×Kv → C.


Example 1.10. Let K = Q. We can consider a general element of A×Q . Focus on the primes that are not units, then we

can divide by elements of K× to make every coordinate in O×Kv . This means that we can put everything in the double

quotient as (1, 1, ...., 1, ?) where there ? is the in�nite place. Similar to the general case, this is the kernel of the above

map.

Now, suppose m = (p). We want to similarly construct a natural map

Q×\A×Q/∏6=p

Z×` × (1 + pZp)→ Cm = C(p).

We can't do the same thing as above. If a = (av) has ap ∈ 1 + pZp then we can do (av) 7→∏` 6=p `

val`(α`) ∈ I(p). If not,

let y ∈ Q× be such that yαp ∈ 1 + pZp (this is because we want our map to be trivial on Q×). Now, map as we would

with (yαv).

2. Modularity of GL1

Goal: all 1-dimensional representations of GQ are modular (representations coming from class group).

2.1. Deformation. Let F be a number �eld, S a �nite set of places of F , and FS be the maximal extension of F

unrami�ed outside S. Let GF,S = Gal (FS/F ). Fix a prime p and k a �nite �eld of characteristic p.

We want to study families of representations of GF,S . Suppose G = 〈gi | ri〉 is a �nite presentable group mean. We can

just de�ne

ρ(gk) = [xi,j,k]i,j subjecting to the conditions ri(xk) = 0.

Now, suppose our group is Zp. Since the topology of this and C don't match, it's only interesting when the coe�cient

�eld is something like Qp. Consider the one dimensional representation

Zp → Q×p

mapping γ (a generator of Zp) to 1+x (it has to be something like this because higher powers of γ is smaller and smaller).

Since Zp is compact, this actually lies in Z×p , so we have a map Zp → F×p . However, the topology here is uninteresting, so

we do not expect some nice families. This is why we �x a residual representation, and try to consider the set of lifts.

Fix

ρ : GF,S → GLn(k).

Let C be the category of local Artinian rings (A,m) with A/m = k. A deformation ρ of ρ to GLn(A) is

GF,S GLn(A)

GLn(k).

ρ

ρ

We also say ρ ∼ ρ′ if there exists M ∈ ker (GLn(A)→ GLn(k)) such that ρ′ = MρM−1. Then de�ne the functor

D : C → Sets

by D(A) = {lifts ρ : GF,S → GLn(A)} / ∼. Notice that D(k) = {ρ}.

De�nition 2.1. The functor D is said to be representable if there exists R such that D(A) = hom (R,A). Really, we

should be talking about pro-representability with complete Noetherian local rings.

Return to the case n = 1, and ρ : G→ k× is trivial. This means that a lift ρ should land in 1 + m ⊆ A×. We can also

replace G by Gab = G/[G,G].

Let G be a �nite abelian p-group. Let

R = W (k)[G].


Example 2.2. If G = Z/pmZ then letting x = [1]− 1 then

R ∼= W (k)[X]/ (1 + x)pm − 1.

Consider the map ρuniv : G→ R× given by g 7→ [g]. Then given any map ρ : G→ A× we have the map R→ A given

by [g]→ ρ(g) making the following commute.

GLn(R)

GF,S GLn(A)

GLn(k).

ρ

triv

Example 2.3. Consider G = Zp. Then we can let

R = lim←W (k)[Z/pnZ] = W (k)JZpK ∼= W (k)JxK.

2.2. Galois deformation. Let R∅ be the universal lift of the trivial representation ρ : GF,∅ → k× and let RQ be the

universal lift of the trivial representation ρ : GF,Q → k× (for some �nite set of primes Q not containing v | p). Morally, we

would like R∅ ∼= W (k)[ClF ⊗ Zp] (the right hand side is something automorphic). Similarly, RQ = W (k) [RClF (Q)⊗ Zp](ray class group). The ⊗Zp is just so that we can focus on the p-part of these groups.

Assumptions (existence of Galois representations for automorphic forms of GL(1)):

(1) We have a surjective map RQ �W (k)[RCLF (Q)⊗ Zp](2) The maps F×v → RCl(Q)⊗ Zp coming from local and global class �eld theory coincide

We now explain the second assumption in greater details. Suppose v ∈ Q. Given ρ : GF,Q → RCl(Q)⊗Zp, which we can

restrict to the decomposition group Dv. This induces a map F×v → GFv → RCl(Q) ⊗ Zp. We want to worry about the

rami�cation at v (which is captured by O×v ), so we consider the map

O×v → F×v → RCl(Q)⊗ Zp.

Since O×v is a pro-` group, this factors through

O×v F×v RCl(Q)⊗ Zp.

k×v ⊗ ZpWe want to say that this is compatible with the map

F×v → F×\A×F → RClF (Q)⊗ Zp

coming from global class �eld theory.

Example 2.4. Let F = Q and Q = {q,∞}. Then RCL(q) ∼= (Z/qZ)×. This is the extension given by Q(ζq). Consider

Q×q → Q×\A×Q → (Z/qZ)×

which is the map x 7→ (1, ..., 1, x, 1, ...). The second map is just Z×q → (Z/qZ)× then extend to Q×q by q 7→ 1.

Now the class �eld theory map

Q×q /NL/Qq (L×)→ (Z/qZ)×

is given by (1 + x)q − 1. Since we are assuming rami�cation, the image of the norm map will not contain the whole O×.


Now, we have a map R∅ → T∅ = W (k) [ClF ⊗ Zp]. We know R∅ ∼= W (k)JΓK and is the quotient of someW (k)Jx1, ..., xqK.Here, q is such that Γ/[Γ,Γ]Γp ∼= (Z/pZ)q. Want to choose allow some rami�cations at some set of primes Q, with |Q| = q.

They will be chosen so that we still have control, in the sense that RQ (the universal deformation ring for lifts unrami�ed

outside Q) will still have the same rank. From class �eld theory, we have an exact sequence.

O×∏v|Q Zp ⊗ (O/vO)× Zp ⊗RCl(Q) Zp ⊗ Cl 1

We are assuming Cl has rank q.

First, assume F = Q or an imaginary quadratic �eld so that O× = 1, and the class group is cyclic and equal to [I],

with |Cl| = p. The goal is to choose Q so that Zp ⊗ RCl(Q) has the same rank. This is the same as saying this exact

sequence does not split. Additionally, we also want N(v) ≡ 1 mod p so that the product terms are non-zero.

Since Ip = (α), we want a non-trivial solution in (O/vO)× ⊗ Zp. What we will need, is α is not a p-th power mod v.

The �eld K = F (ζp, α1/p) works.

1→ Z/pZ→ Gal (K/F )→ (Z/pZ)× → 1.

Here, we want Frobv to have order p (so that the image of it in (Z/pZ)×

has order 1, which is to say it has norm 1

mod p). We are guaranteed to have these by Chebotarev.

Now, suppose Cl has two generators Ip = (α) and Jp = (β). Want {α, β} to generate∏v|Q(O/vO)× ⊗ Zp. Do

K = F(ζp, α

1/p, β1/p)then we have

1→ Z/p2Z→ Gal (K/F )→ (Z/pZ)× → 1,

and a similar argument su�ces.

Now, what if O× is not trivial. We will need to write Zr1+r2−1 on the left hand side, and we will then need to choose

q + (r1 + r2 − 1) number of primes.

To summarize, suppose F = Q or imaginary quadratic �eld. We know W (k)Jx1, ..., xqK � R where q is minimal, equal

to

dimmR/(m2R, p).

We showed that there exists in�nitely many sets QN of q-primes (v ≡ 1 mod pN ) with

RQN TQN

R∅ T∅,

where RQN and TQN have no additional generators as compared to R∅ and T∅. That is, W (k)Jx1, ..., xqK � RQN and

TQN , giving us the commutative diagram

W (k)Jx1, ..., xqK RQN TQN

⊗v|QNW (k) [k×v ⊗ Zp] R∅ T∅.

LCFT

These W (k)[k×v ⊗ Zp] is parametrizing the deformations of the inertia group inside the decomposition group, which is

what gives us the map W (k)[k×v ⊗ Zp]→ RQN .

Let S∞ = W (k)Jt1, ..., tqK. Let A = (t1, ..., tq) be the augmentation ideal, and let

AN =(

(1 + t1)pN

− 1, ..., (1 + tq)pN − 1

).

If every v ≡ 1 mod pN , then ⊗v|QW (k)[k×v ⊗Zp] ∼= S∞/AN . If we think about it, TQN � T∅ is just quotient by AN and

similarly for RQN � R∅.

When F is Q or imaginary quadratic, TQN is free over S∞/AN = W (k) [RCl(Q)]. This means that


W (k)/pN Jx1, ..., xqK RQN /pN TQN pN

S∞/(AN , pN ) R∅/p

N T∅/pN .

For the whole bottom row, and the top right, they all have �nite length. Call this part of the diagram DQ,N . For

N > M , we have DQN → DQM . Want to patch these diagrams together. There are two ways to do so. One can assume

that these primes are chosen to be close to 1 mod p∞ or use ultra-�lter argument by Scholze.

Our �nal picture, is then

W (k)Jx1, ..., xqK R∞ T∞

S∞ R∅ T∅.

/A

By de�nition, S∞ is free with q variables. This diagram implies that T∞, a �nite rank S∞-module, has a regular

sequence of q+1 in S∞, so also in W (k)Jx1, ..., xqK. Therefore, T∞ is free and equal to W (k)Jx1, ..., xqK. This sandwichingforces R∅ = T∅. That is, an abelian extension of degree p comes from the class group. Note, thus far, this only works for

F = Q and imaginary �eld. General case will be discussed next time.

3. Galois Module

3.1. Galois Module of local �elds. Let p and ` be primes (not necessarily ` 6= p). Suppose K/Qp is a �nite extension

with ring of integers OK , uniformizer ω and residue �eld k. Let M be a GK module, �nite with pM = 0. De�ne

M∨ = hom (M,Fp) and M? = M∨(1) = hom (M,µp)

which both have a canonical action of GK .

Tate local duality compares cohomology of M to that of its duals. More precisely, we have a perfect pairing given by

Hi (K,M)×H2−i (K,M?)→ H2 (K,µp) ∼= Fp.

This is expected, because M ×M? → µp.

Example 3.1. Let M = Fp with a trivial action of Galois. When i = 1, we have

H1 (K,Fp) = hom (GK ,Fp)LCFT∼= hom

(K×,Fp

)= hom

(K×/K×p,Fp

).

On the other hand, by Kummer theory,

H1 (K,µp) ∼= K×/K×p

by φα ← [ α where φα(σ) = σα1/p

α1/p ∈ µp.Tate local duality is then a map

hom(K×/K×,Fp

)×K×/K×p → Fp

which is just the evaluation map.

We have

H2 (K,µp) ∼= Br(K)[p]

where Br(K) is the Brauer group.

Example 3.2. Let p = 2, then ζp ∈ K. Pick a non-canonical isomorphism µp ∼= Fp. Then our pairing is

K×/K×2 ×K×/K×2 → Br[2].

Given an element (α, β) you map it to the quaternion algebra corresponding to x2 = α, y2 = β and xy = −yx.


Similarly, for general p, assume ζp ∈ K, then we get a quaternion algebra corresponding to xp = α, yp = β and

xy = ζpyx.

Now, for a general case, givenα ∈ hom(K×/K×2,Fp

), it will cut out some extension L/K of degree p. Fix a generator

σ ∈ Gal (L/K). The map

hom(K×/K×p,Fp

)×K×/K×p → Br[p].

is mapping (L, β) to the quaternion algebra L[z] where zp = β and xz = zσ(x).

When is this quaternion algebra is trivial. We have

K×/NK(√α)/K

(K(√α)×

)→ Gal

(K(√α/K

)and the quaternion algebra is trivial if β ∈ NK(

√α)/K (K(

√α)×) which is exactly when β = x2 − αy2.

H1unr (K,M)) ⊆ H1 (K,M) = H1 (GK ,M)

res→ H1 (I,M) and the unrami�ed ones are the just the kernel of this

restriction. In�ation-restriction says

0→ H1(k,M I

)→ H1 (GK ,M)→ H1 (I,M)

and so H1unr (K,M) ∼= H1

(k,M I

). One can check that

#H1(k,M I

)= #H0

(k,M I

)= #(M I)G/I = #MG = #H0 (K,M) .

Example 3.3. Consider

H1unr (K,Fp) = H1 (k,Fp)

which are just maps factoring through the maximal unrami�ed quotient (so just the residue �eld).

3.2. Selmer Groups. Let F be a number �eld, M �nite GF -module, pM = 0.

De�nition 3.4. A Selmer condition relative to (F,M) is

L = {Lv}places

where

Lv ⊆ H1 (Fv,Mv)

satisfying: for all but �nitely many v, Lv = H1unr (Fv,Mv). Then de�ne

H1L (F,M) =

{[c] ∈ H1 (F,M) : [cv] ∈ Lv

}.

Associated, we have L?the dual Selmer condition L? = {L?v} where L?v ⊆ H1 (Fv,M?). This is de�ne to be the perp of

Lv under the perfect pairing

H1 (Fv,Mv)×H1 (Fv,M?v )→ Fp.

When v - p,(H1ur

)⊥= H1

ur.

Theorem 3.5. Wiles-Greenberg (Tate).

#H1L (F,M)

#H1L? (F,M?)

=#H0 (F,M)

#H0 (F,M?)

∏v

#Lv#H0 (Fv,Mv)

.

Note, the quotient on the right, it is 1 if we have unrami�ed condition at v.

Example 3.6. Let M = Fp, L = unrami�ed, Lv = H1ur for all r. Then

H1L (F,M) = hom (Cl/pCl,Fp) .

Similarly,

H1L? (F,M?) = H1

L? (F, µp) ⊆ F×/F×p.


The unrami�ed class in hom (K×,Fp) are just the ones where O×K 's are trivial. Therefore,

Lv = hom(K×/O×K ,Fp

).

Then L⊥v = O×K/O×pK . Therefore, (assuming p > 2)

H1L? =

{α ∈ F×/F×p such that vv(α) = 0 mod p

}.

That is, (α) = Ip for some ideal. The element α doesn't quite determine the ideal I, but determines the ideal class, as

(αβ)p

= Ip(β)p. We have

1→ O×/O×p → H1L? → Cl[p]→ 1.

Therefore,

− logp#H1

L

#H1L?

= dimFp O×/O×p = r1 + r2 − 1 +

1 if p | wF

0 else.

The more direct approach, we getp

p

1 if p | wF

0 else

∏v|∞

1

H0 (Fv,Fp)

which we can �nd is equal.

Let Q be a �nite set of auxiliary primes q ≡ 1 mod p. Let Lv be unrami�eid for all v /∈ Q and Lv = H1 (Fv,Fp) forv ∈ Q, and so L?v = 0 for these q's.

|HQ|∣∣∣H1Q?

∣∣∣ = same as above, with more primes in Q.

For these, ∏v∈Q

#H1 (Fv,Fp)p = #H0 (Fv,Fp)

=∏ p2

p= pQ.

This above is p2 because p - v and so a map

O×v → Fp

factors through k×v . Now, assume ζp /∈ F . Then

#H1Q

#H1Q?

= p|Q|−(r1+r2−1).

because H1∅ ↪→ H1

Q and so H1Q? ↪→ H1

∅? which is the kernel of H1∅? → L?v which is given by H1

∅? ⊆ H1 (F,M?) →H1 (Fv,M

?).

Lemma 3.7. Let dimH1∅ = q, dimH1

φ? = q + r1 + r2 − 1. Then there exists in�nitely many sets Q ={q1 ≡ 1 mod pN

}with |Q| = q + r1 + r2 − 1 and such that #H1

Q = #H1∅ .

In the context of last time, we want the rank of the ray class groups to not get bigger. Let α ∈ H1∅? . We can think of

α ∈ F×/F×p. We want to �nd q ≡ 1 mod pN such that

[α] ∈ H1∅? (F, µp)→ H1 (Fq, µp)

not to land in H1Q? (F, µp). We know

H1 (Fq, µp) ∼= F×q /F×pq .

So we want to �nd [α] to not be a perfect p-th power. Last time, we used

1→ Z/pZ→ Gal(F (α1/p, ζpN )/F

)→ Gal

(F (ζpN )/F

)→ 1


We want α to be trivial in the projection. That is, we want N(α) = 1 mod pN . We have to worry about whether

α1/p ∈ F (ζpN ). By Kummer theory, we just need α 6= ζp for α1/p /∈ F (ζpN ). This leads to something called the

Taylor-Wiles condition.

Let's go back to ρ : GF → F×p the trivial representation, and consider the deformations of this. We have some universal

representation

ρQ : GF → R×Q(= ZpJΓQK×

)and RQ ∼= ZpJx1, ..., xrK/I where r = dimm/

(m2, p

). To �nd this r, �nd

hom(R,Fp[ε]/ε2

)= hom

(m/(m2, p),Fp

).

We know that Fp[ε]/ε2 = Fp ⊕ εFp, so it has to have the shape(1 ?

0 1

)and so the set of lifts to Fp[ε]/ε2 is just

H1 (F,Fp) = hom (GF ,Fp) .

After we restrict local conditions, we want H1Q (F,Fp) . If this was GLn, then we have(

ρ ?

0 ρ

)as blocks, so we get Z1 (GF , adρ), then up to conjugacy class, we get H1.

What we have shown is the following. Let q = dimH1∅ (F,Fp) and ζp /∈ F , then there exists

Q ={qi ≡ 1 mod pN

}, |Q| = q + r1 + r2 − 1

such that

ZpJx1, ..., xqK RQ TQ

R∅ T∅.

Suppose qi ≡ 1 mod p∞.

ZpJx1, ..., xqK R∞ T∞

ZpJt1, ..., t|Q|K R∅ T∅.

The ZpJt1, ..., t|Q|K is representing the deformation at inertia. If we take A to be the augmentation ideal of ZpJt1, ..., t|Q|K.To say that something is unrami�ed at q, it's to saying g acts like 1. That means RQ∞/A = R∅.

What we don't have compared to last time, is we don't have TQ is free. It doesn't even make sense to be free over

ZpJt1, ..., t|Q|K, because ZpJx1, ..., xqK surjects onto it. What we want, is the notion of it being of codimension r1 + r2 − 1.

Let F be a number �eld, p prime and ζp /∈ F . We showed that if dimH1∅ (F,Fp) = q then dimH1

∅? (F, µp) = q+`0, where

`0 = r1 + r2 − 1. Then we proved that there exists in�nitely many sets Q of primes N(q) = 1 mod pN and |Q| = q + `0

such that

dimH1Q (F,Fp) = q,

which is i� dimH1Q? (F, µp) = 0. Then if I = (ti) then


ZpJx1, ..., xqK RQ TQ

⊗Zp [k×v ⊗ Zp] R∅ T∅

ZpJt1, ..., tq+`0K/(

(1 + ti)pN − 1

)

/I

If `0 = 0 (no units) then TQN is free over Zp[ti]/ (1 + ti)pN − 1. This comes from

1→ Zp ⊗∏

(O/q)× ↪→ RCl ⊗ Zp → Cl ⊗ Zp.

This can't be the case for us this time because of dimensions. The goal is to account for these `0.

Let's think about GL2. We have

GL2(Z) ↪→ GL2(R).

In the modular form case, we study the lattices inside by studying H/Γ. One reason why we do this, is because H is

contractible and so

H? (Γ,Z) ∼= H? (Γ\H,Z) .

In fact, this was the classical de�nition of cohomology of groups. Note: since we may have torsion, we either need to look

at cohomology of orbifolds and stacks, or have high enough level (so that Γ is torsion free and injects into PGL2(R)). The

action on H by SL2(R) is SO2(R). The choice of H = SL2(R)/SO2(R) is like a quotient of a maximal compact.

For GL1/F , we should look at G = ResF/QGL1/F and look at the real points

G(R) = (F ⊗Q R)×

=∏r1

R× ×∏r1

C×.

Let K =∏

complex S1 which is a compact subgroup. Let UQ ⊆ A×F be an open subset of the idèles de�ned the following

way.

UQ,v =

S1 v complex

±1 v real, v /∈ Q

1 v real, v ∈ Q

O×v v �nite, v /∈ Q

1 + vOv v �nite, v ∈ Q

.

The analog of G(Z) is O×F and G(R) is (R⊗ F )×

=∏

R×∏

C×. By taking log |·| on G(R), this maps into Rr1+r2

(getting an extra copy of R because we don't care about product of components is 1. It doesn't a�ect cohomology, so we

don't care). The map

F×\A×F /UQ � RCl(Q)

is surjective. The �bers look like Rr1+r2/O×F which is basically R×(S1)r1+r2−1

.

This will be useful for us in the following way. As usual, for Q′ ⊆ Q, we have an exact sequence

?→ RCl(Q)→ RCl(Q′)→ 0

and the ? is some product of units. This is usually hard to compute. However, we have the commutative diagram


? RCl(Q) RCl(Q′) 0

F×\A×F /UQ F×\A×F /UQ′

∏RCl(Q)(S

1)r1+r2−1∏RCl(Q′)(S

1)r1+r2−1

In the bottom row, we have a cover of topological spaces. It can be that we have more factors, or just a double covering

(which we want).

Problem: control the size of TQ = Zp[RCl(Q)⊗ Z/pNZ

]over

S∞ = ZpJt1, ..., tq+`0K/IN

where IN =(

(1 + ti)pN − 1

). In the limit,

S∞ T∞

ZpJx1, ..., xqK

Imagine T∞ is free over S∞.

De�nition 3.8. A regular sequence (S →M) of length n is (s1, ..., sn) such that Ms1↪→ M , M/s1

s2↪→ M/s1,... are all

injective.

A regular sequence in S∞ is (t1, ..., tq+`0 , p). We see that since ZpJx1, ..., xqK surjects onto T∞, we can at most a regular

sequence of length q + 1. This suggests that T∞ has codimension `0.

We have G : RQ → TQ and TQ acts faithfully on MQ (think of space of modular forms).

• ZpJx1, ..., xqK � TQ and we have a regular sequence of length q+ 1. Since ZpJx1, ..., xqK is a domain, T can't have

regular sequence of this rank without being the whole thing already.

If S∞/IN acts on some module MQ of �codimension `0�. Suppose `0 = 1.

De�nition 3.9. Let R be a Noetherian local ring. A �nitely generated module M is balanced, if there exists partial

resolution Ra → Ra →M → 0. Equivalently, there exists a minimal resolution

Rm2 → Rm1 →M → 0

and m2 ≤ m1.

Example 3.10. R = ZpJt1, ..., trK/ ((1 + ti)p − 1) and M = Zp. Then

I = (t1, .., tr) = Rr → R→M → 0

so M is balanced i� r ≤ 1.

Suppose we have

S R

S/I.

If MS a balance S-module, then MR = M ⊗S S/I is a balance R-module.

Proof. We have

Sa → Sa →MS → 0


which gives

Ra → Ra →MR → 0.

�

First, we let `0 = 1. The goal is to prove balanced at every �nite stage. That is,

(S∞/A)r → (S∞/A)

r → MQ

↓ ↓ ↓Zrp → Zrp → M∅

and similarly mod pN . Taking limits, we get Sr∞ → Sr∞ →M∞ and so codimension of M∞ is at most 1.

Then either M∞ has positive rank (codimension 0) or there's a resolution of M∞ and implies existence of regular

sequence of length q + 1. The �rst case almost never happen, because............ Something here about depth and some

theorem will force it to be free over R∞ = ZpJx1, ..., xqK.Previously, we were consider ZpJRCl(Q)⊗ ZpK. The components of

XQ = F×\A×F /UQ

are RCl(Q) and the �bres are the tori. De�ne

TQ ⊆ EndZpH?(F×\A×F /UQ,Zp

)given by Hecke operators Tα. This just acts by multiplication by α in each of the ray class group component for H0.

Claim 3.11. H? (XQ,Zp) ∼= H0 (XQ,Zp)⊗ Λr1+r1−1Zr1+r2−1 by Kunneth (the Z's are coming for the S1's).

Proof. Recall that what we did was identify (F × R)×/K

log→ (S1)r1+r2−1. This means that the action of an adele that is

trivial on all the �nite places, act by multiplication on the left, and so translation on the right. Therefore, it is trivial on

cohomology. �

Last time, we were replacing ray class groups with these larger spaces XQ = F×\A×F /UQ, which are disjoint union of

tori (S1)`0 .

Fix p. De�ne

TQ ⊆ EndZpH? (XQ,Zp)

to be the subalgebra generated by Tα for α ∈ A×F . These Tα acts on cohomology via H0 = Zp[RCl(Q)]. In fact, it is

literally the multiplication on the left by α of XQ → XQ. This is not always the case.

Let f be a modular form, or π an automorphic representation. We may hope that πK or f may contribute to cohmology

H1 (X(N),Zp), if we are lucky. We know that T`f = α`f for ` - Np, and there is a Galois representation ρ : GQ → GL2(E)

such that

Tr (ρ(Frob`)) = a`.

For this eigenform, we have a Hecke action of T on H1 and (f). This gives a map T→ E given by T` → a`.

We have two �avors of Hecke algebra. Tan which are generated by T` with ` - Np and T generated by all. (The an

stand for anemic?). Usually we care about Tan here. For example, S12 (Γ0(q)) contains ∆(τ) and ∆(qτ) which have the

same eigenvalues. By considering the Tan, these cut out the same Galois representations and are counted only once.

Let fi for i = 1, ..., N be eigenforms are level N . Since GQ is compact, we get representations ρi : GQ → GL2(OEi)and so combining these, we get

ρ : GQ →∏

GL2(Oi) = GL2

(∏Oi)� GL2

(∏ki

).

Each (ρi) is given by uniquely by a map T→ ki. If T→ Fp ⊕ Fp → Fp to f ≡ b mod p ...


Fix residual ρ, which is given by some m : T ↪→∏O◦i → k so Tm →

∏ρi=ρOi,

ρ : GQ → GL2(∏

(ρi)ss 6=(ρ)ss

Oi)

and Tr (ρ (Frob`)) = T` ∈ Tm.

Theorem 3.12. (Carayol) If ρ is absolutely irreducible, then ρ : GQ → GL2(Tm).

Remark 3.13. This Tm is really Tanm , the source of Galois representations.

Tanm = Tvery anemicm if we take out a �nite set of Hecke operators.

Example 3.14. For S2 (Γ0(23)) which has genus 2, we have a pair of complex conjugate modular eigenforms

f = q +

(−1 +

√5

2

)q2 + ...

and the coe�cients of ap are in Z2

[√5](for 2 - p).

T = Z2

[1 +√

5

2

]= Z2 [x] /x2 − x− 1

and Tm = Z2[√

5] = Z2[x]/(x2 − 5).

Back to our discussion. H?(XQ,Zp) is T-module, with maps going through Zp [RCl(Q)]. The mod p maximal ideals of

this, are characters of RCl(Q). For us, when we impose ρi 6= trivial, our localization

Zp [RCl(Q)]m = Zp [RCl(Q)⊗ Zp]

because we are saying that when you are coprime to p, you must act trivially. We get a representation to Tanm , so get a

map RQ → TanQ,m. We want TanQ,m to have codimension `0 (we have power to make it bigger).

We want to exploit dimXQ = `0. We had a cover RCl(Q) � Cl(Q), but this cover is mysterious, and requires us to

know about the image of the global units in the exact sequence. However, XQ � X∅ is a Galois cover U/UQ.

For F real quadratic �eld. The cover is

q(Z/qZ)×/εR/ log∣∣εZk ∩ (1 + qO)

∣∣→ (Z/qZ)×/ε

both map down to R/ log |εK |S1 something with cover being q−1m . Somehow add up to (Z/qZ)

×???

Let C?∅ be a complex of �nite free Zp-module complex H? (X∅,Zp), we can pull back to get a complex C?q of �nite free

Zp [U/UQ]-module.

Recall q = dimH1∅ (F,Fp) and `0 = r1 + r2 − 1 = dimH1

∅? (F, µp)− q and |Q| = q + `0.

C?Q : P0 → P1 → ...→ P`0

complex for Zp [∆Q,N ] where ∆Q,N = (O/qO)× ⊗ Z/pNZ. These give a complex at the in�nite level

P∞0 → ...→ P∞`0

with action of

(1) S∞ = ZpJt1, ..., tq+`0K(2) T∞ acts on H? (P ?)

(3) H`0(P ?) = lim←H`0(P ?Q)

(4) ZpJy1, ..., yqK R∞ (WHAT?)

Since maps between free modules of S∞ must be injective or have positive rank kernel, so this implies that P0 ↪→ P1 must

be an injection. Something something codimension at most 1. By some dimension argument, the whole sequence must be

exact, giving some resolution to H`0(P ?) and codim ≤ `0, deptah ≥ q + 1 and something something so free over R∞.


(sorry, the above is just too confusing...)


Part 2. George's Lectures

The �rst part of this course, will be Hida theory for algebraic modular forms on de�nite quaternion algebras. The

second part will be about coherent Hida theory for classical and Hilbert modular forms.

4. Quaternion Algebras

De�nition 4.1. Let F be a �eld, of characteristic not 2. For a, b ∈ F×, de�ne(a,bF

)to be the F -algebra with basis

1, i, j, ij given by i2 = a, j2 = b and ji = −ji. We will usually write k = ij and so k2 = −ab.

Proposition 4.2. We have D/F is a quaternion algebra ⇐⇒ D is a central simple F -algebra of dimension 4 ⇐⇒either D = M2(F ) or D is a division algebra of dimension 4.

Example 4.3. It's easy to check that(

1,bF

)∼= M2(F ) by i 7→

(1 0

0 −1

)and j 7→

(0 b

1 0

). Additionally,

(ax2,by2

F

)∼=(

a,bF

).

Let D0 = span (i, j, k) ={α ∈ D : α2 ∈ F, α /∈ F

}. Then we can de�ne an involution α 7→ α by 1 = 1, i = −i, j = −j

and k = −k. This is a canonical involution of D, and αβ = βα.

If we have M2(F ), this is

g =

(a b

c d

)7→

(d −b−c a

)= det(g)g−1.

De�nition 4.4. Suppose α = x+ iy + jz + kw. The reduced norm is

n(α) = αα = αα = x2 − ay2 − bz2 + abw2

and the reduced trace

t(α) = α+ α = 2x.

It is easy to check that n(αβ) = n(α)n(β). Also, for all α ∈ D, α2 − t(α)α + n(α) = 0. If α /∈ F , then F [α] is a

quadratic algebra over F .

If D is a division algebra then F [α] is a �eld. The norm and trace here then corresponds to that on F [α].

Proposition 4.5. There is a bijection

{quaternion algebras over F}/isom ↔{ternary quadratic forms over F of discriminant 1 ∈ F×/F×2

}/equiv

.

This map is just D 7→ n |D◦ . That is, the quadratic form −ax2 − by2 + abz2.

Proof. Only need to check injectivity. This is because these quadratic forms can be diagonalized (Gram-Schmidt), and

giving us the a and b.

Suppose D is a quaternion algebra and n |D0is −ax2 − by2 + abz2. Then α ↔ (1, 0, 0) and β ↔ (0, 1, 0). Then

−α2 = −a− (α+ β)2 = −a− b and −β2 = −b. �

Corollary 4.6. TFAE:

•(a,bF

)splits

• The quadratic form ax2 + by2 − z2 = 0 has a non-trivial solution

• ax2 + by2 = 1 has a solution (Hilbert's criterion)

• x2 − ay2 = b has a solution (ie. b ∈ NF(√a)/F (F (

√a)×))

Corollary 4.7. The quaternion algebras(a,−aF

)and

(a,1−aF

)are split.


4.1. Local Fields.

Theorem 4.8. If F is a local �eld 6= C, then there is a unique quaternion division algebra over F . If F = R, then this

is(−1,−1

R). When F/Qp is �nite, this is

(a,πF

)where a ∈ O×F with F (

√a) is unrami�ed, and π is a uniformizer.

De�nition 4.9. Hilbert symbol. Supose a, b ∈ F×.

(a, b)F =

1 if(a,bF

)is split

−1 otherwise.

We can check that (·, ·)F : F×/(F×)2 × F×/ (F×)2 → {±1} is bilinear, because N (F (

√a)×) has index 2 in F×.

If p 6= 2, (a, b)F = 1 if a, b ∈ O×F . Meanwhile, (a, b)Q2= 1 if a, b ∈ Z×2 and a ≡ 1 mod 4.

De�nition 4.10. Suppose F is a number �eld, de�ne

Ram (D) = {v place of F : D ⊗ Fv is non-split} .

Proposition 4.11. #Ram(D) <∞.

Theorem 4.12. There is a bijection between

{quaternion algebras /F}/isom ↔ {even sets of non-complex places of F}

by D 7→ Ram(D).

There is three part of this.

• local-global i� Hasse-Minkowski theorem

• reciprocity (Hilbert), which says∏v (a, b)v = 1. Over Q, this is equivalent to quadratic reciprocity. We can

basically reduce to the case that a, b are distinct odd-primes. Also,

(a, b)v = 1 unless v = p, q, 2

(p, q)p =

(q

p

).

• Existence, use Dirichlet. Say Ram (D) = {∞, p}. Take

(−1,−1

Q

)(−1,−p

Q

)when p ≡ 3 mod 4(

−2,−pQ

)when p ≡ 5 mod 8(

−q,−pQ

)with some congruence condition on q.

De�nition 4.13. Let R be OF for F local or number �eld. A lattice L ⊆ V a �nite dimensional vector space is a �nitely

generated R-submodule spanning V .

An order O ⊆ D (where D/F is a quaternion algebra) is a lattice which is an R-submodule.

A right/left/two sided fractional ideal I ⊆ D for O is an R-lattice such that IO ⊆ I and similarly for the others.

Remark 4.14. Warning: if α, β ∈ D are integral (ie. R[α] is a �nitely generated R-module) then α + β and αβ need not

be (no commutivity). This is why we don't have a unique maximal order.

Theorem 4.15. (Local-Global for lattices) If R is a Dedekind domain, V is a �nite dimensional Frac(R) vector space,

L0 ⊆ V is any O-lattice,

{lattices L ⊆ V } ↔{R(p)-lattice L(p) ⊆ V such that L(p) = L0(p) for almost all p

}↔ {Rp-lattice Lp ⊆ V ⊗ Fp such that Lp = L0,p for almost all p}

by ∩L(p) ← L(p). This is the same for orders and ideals.


Locally, all maximal order are conjugate, and all right ideals over maximal orders are principal.

4.1.1. Split case. Let R be a DVR, F = Frac(R), V = F 2 and M2(F ) = End(V ). If L,M ⊆ V are R-lattices, then

homR (L,M) ⊆ End(V )

is an R-lattice. This means that every maximal order is of the form End(L) for some lattice L, so there is a bijection

{maximal orders in M2(F )} ↔ PGL2(F )/PGL2(R)

αM2(R)α−1 ← α.

Every right End(L) fractional ideal is of the form hom (L,M) for some lattice M . So

{right ideals of M2(R)} ↔ GL2(F )/GL2(R)

αM2(R) = hom(R2, αR2

)← α.

4.1.2. Non-split case. Suppose D/F is a non-split quaternion algebra. Consider v : D → Z∪{∞} given by α 7→ vF (n(α)).

This satis�es

• v(α) =∞ i� α = 0 (only works for non-split)

• v(αβ) = v(α)v(β)

• v(α+ β) ≥ min (v(α), v(β))

Let O = v−1 ([0,∞]), which is an order. Supose α ∈ D − O is not integral, so not in any order. Then O is the unique

maximal order, with O× = v−1(0). Left, right ideal or two sided ideals are of the form v−1 ([n,∞]).

What's the image of the valuation. We know 2Z ⊆ v (D − {0}) ⊆ Z. Let I = v−1 ((0,∞]) which is a maximal two-sided

ideal. Clearly, πF ∈ I. This means that O/I is a simple algebra over R/π. Every element satis�es quadratic polynomial,

so O/I has to be a quadratic extension of R/π (it's a �eld by Artin-Wedderburn, no division algebras over �nite �eld).

Therefore, I = (πD) with v(πD) = 1, this is why it's called rami�ed.

Globally, let O ⊆ D be a maximal order (exists because they exists locally). Consider

{right O-ideals I} ↔ {Ip ⊆ Dp right Op − ideals such that Ip = Op a.e. p}

↔ (D ⊗ A∞F )×/(O ⊗ OF

)×.

Since D× acts on the left of the last set, we have the same action on the left side.

Cl(O) =D×\ {right O-ideals} =D×\ (D ⊗ A∞F )×/(O⊗OF )

× .

4.2. Class Groups of maximal orders of quaternion algebra. Notation: let F be a number �eld, OF be the integral

adèles which is∏

pOF,p. Let S ⊆ {places of F}, ASF the adèles away from S, A∞F the �nite adeles, which is OF ⊗OF F .Let D/F be a quaternion algebra, O ⊆ D maximal order. Let D×(A) = (O ⊗A)

×for any OF -algebra A.

Example 4.16. D×(OF ) = O and D×(F ) = D×.

Let D×,1 ⊆ D be the kernel of the norm map n : D× → Gm. Let PD× = D×/Z (mod the center). We know that

{L ⊆ Fn : OF lattices} ↔{

(Lp) : Lp ⊆ Fnp of OF,p lattices such that Lp = OnF,p almost every p}

↔ GLn (A∞F ) /GLn(OF ).

Recall that

GLn(F )\GLn(A∞F )/GLn(OF ) ∼= Cl(OF )


for all n. We can see this, because this is lattice, up to change of variable. Therefore, this is

{projective rank n, OF -modules}

and it's an exercise that this is the class group for Dedekind domains. Basically, if P is a projective module of A, then

P ∼= An−1 ⊕ I for some fractional ideal I of A.

• all maximal orders are locally conjugate

• all right fractional ideals for maximal orders are locally principal

This means that

Cl(O) ∼= D×(F )\D× (A∞F ) /D×(OF ).

We also check that

PD×(F )\ {max orders} ∼= PD×(F )\PD×(A∞F )/PD×(OF ).

We have a surjective map

D×(F )\D× (A∞F ) /D×(OF ) � PD×(F )\PD×(A∞F )/PD×(OF ),

which we will try to describe.

By the norm map, we have

D×(A∞F )/D×(OF )n→ (A∞F )

×/O×F .

This is basically right O-fractional ideals to IF (the ideal of F ).

Here, we need to use a fact.

• If F is a local �eld n : D× → F×, n : O×p → O×F,p is surjective unless F = R and D = H. For the split case, thisjust says the determinant map is surjective

By quotienting by D×(F ) on the left, the previous map turns into a map

Cl(O)n� Cl(OF ).

Suppose O′ is another maximal order. There exists (αp) ∈ D×(A∞F ) such that αpOpα−1p = O′p. Multiplication by (αp)

on the right, gives a bijection

D×(A∞F )/D×(OF )·(αp)∼= D×(A∞F )/α−1D×(OF )α.

Since α−1D×(OF )α =(O ⊗ OF

)×. Now, quotient on the left by D×(F ) to get

Cl(O) ∼= Cl(O′).

4.3. Brandt Groupoid. Suppose I ⊆ D is a lattice, and let

O`(I) = {α ∈ D : αI ⊆ I}

Or(I) = {α ∈ D : Iα ⊆ I}

which are orders. One can check that Or(I) is maximal i� O`(I) is maximal, in which case we call I is normal.

If O is maximal, and I = αO for some α ∈ D×. Then

αOα−1 ⊆ O`(I).

Since αOα−1 is maximal, the above is an equality.

Suppose I, I ′ are normal, then we can de�ne

I · I ′ ={∑

xiyi : xi ∈ I, yi ∈ I ′}.


Technically, we can always de�ne this, but it usually behaves horribly. We can see this from the next example.

Example 4.17. We can check that

M2(Zp)

((p 0

0 1

)M2(Zp)

)= M2(Zp).

This is bad, because we would like M2(Zp) to be the identity element.

De�nition 4.18. Say I, I ′ are compatible if Or(I) = O`(I ′).

The set of normal ideals (for global, that is locally principal) has a groupoid structure, by the following. Given I, there

exists I−1 where

Or(I−1) = O`(I) and O`(I−1) = Or(I)

(so they are compatible). Then II−1 = O`(I) and I−1I = Or(I). For existence of this I−1, we check this locally. we write

(αO)−1

=(Oα−1

)=(α−1

(αOα−1

)).

Now we can de�ne

Cl(O)→ {maximal order} /conjugation

we can de�ne I 7→ O`(I). Through diagram chasing, this is the map we wanted from earlier. This does not descend to

D×\ {normal ideals}.

4.4. Strong Approximation. We want to show that F ↪→ ASF is dense if |S| > 0.

First, this is not true for Gm.

Example 4.19. Let F = Q, the ray class group is

Cln = (Z/nZ) /± 1 = A×,∞Q /Q×Kn

and the important thing is that Kn is open. If Q was dense in(A∞Q)×

, then we have killed everything, so it is important

that this do not work for Gm.

Example 4.20. The following also do not hold

GL2(F ) ↪→ GL2(ASF )→ AS,×F

and D×(F ) ↪→ D×(ASF )→ AS,×F .

Theorem 4.21. (Strong approximation for D) Let F be a number �eld, and D/F a quaternion algebra. Let S be a �nite

set of places such that there exists v ∈ S with Dv split, then

D×,1(F ) ↪→ D×,1(ASF )

is dense.

Remark 4.22. This is also true for semisimple algebraic groups with certain conditions.

Now, we want to explain why we need the condition that Dv is split. If Dv rami�es, D×,1v is compact. Since D× (AF )→D×(ASF ) has compact �bres we �nd that D×(F ) must be discrete. Then strong approximation can not hold.

De�nition 4.23. We say D/F is de�nite if Dv rami�es for all archimedian places (then F is not real). We say is is

inde�nite otherwise.

Recall we have the norm map

n : Cl(O) = D×(F )\D×(A∞F )/D×(OF )→ F×\A∞,×F /OF = Cl(F )


and this factors through

Cl+(F ) = n(D×(F )

)\A∞,×F /n

(D×(OF )

).

Note that n(D×(OF )

)= OF , but it is not true that n (D×(F )) = F×. Call this map f .

Theorem 4.24. If D is inde�nite then f is a bijection.

Proof. Let α, β ∈ D×(A∞F ) with n(α) = n(β). Now, we need to show α and β have the same class.

αβ−1 = rk

where r ∈ D×,1(F ) and k ∈ K an open compact subgroup. Then

α = rkβ = rβ(β−1kβ

)so take K = βD×(OF )β−1 then we are done.

�

4.5. Finiteness of class groups and class number formula. Suppose F = Q, D/Q de�nite. |O×| is �nite. So

n : O → Z, n−1(1) is a �nite set. De�ne d =∏p rami�es p.

Proposition 4.25. Let I be a right O-ideal. There exists 0 6= α ∈ I, such that n(α)2 ≤ 8π2 d · n(I)2.

Proof. Same as usual, using the Minkowski's theorem. This says the if Λ ⊆ Rn is a lattice, S ⊆ Rn is convex with −S = S

and vol (S) = 2nvol (Rn/Λ), then there exists α 6= 0 in S ∩ Λ.

In the de�nite quaternion algebra case (unlike real quadratic �eld) the n ≤ stu� set is convex. �

Corollary 4.26. Any ideal class contains an ideal I ⊆ O with n(I)2 ≤ 8π2 d.

Recall that for K/Q a quadratic imaginary �eld, the

ress=1ζK(s) =2πh

w√dK

where W =∣∣O×K∣∣ and h =class number. We did this by

ζK(s) =∑I⊆OK

1

N(I)s=

∑[b]∈Cl(OK)

∑I∈[b], I⊆OK

1

N(I)s

=∑

[b]∈Cl(OK)

1

w

∑α∈b−1

1

N (αb)s .

Then we compute the residue of∑α∈b−1

1N(αb)s .

Theorem 4.27. (Eichler Mass Formula) ∑[I]∈Cl(O)

1

|O`(I)×|=ϕ(d)

24.

The left hand side is called the mass. It is about h2 just because the units are usually just ±1. The ϕ is the Euler totient

function.

Suppose ram(D) = {p,∞}, ∑supersingular ellip curve E/Fp up to isom

1

|Aut(E)|=p− 1

24.

We will explain this later in the course on how it relates to our mass formula.

Example 4.28. Suppose p = 2, 3, 5, 6 and h = 1. The unique maximal order is

|O|× = 24, 12, 6, 4.


When p = 11, and h = 2,1

6+

1

4=

10

24.

Proof. Let

ζD(s) =∑

I⊆O right ideals

1

n(I)2s.

Here, n(I)2 = [O : I]. Then

ζD(s) =∑

[I]∈Cl(O)

1

|O`(I)×|∑α∈I−1

1

n(αI)2s.

The residue at s = 1 of∑α∈I−1

1n(αI)2s is 2π2

d . We can calculate this from the following lemma.

Lemma 4.29. For D(s) =∑ an

ns with an ≥ 0 and∑ni=1 ai = cn + O(n2) then D(s) is meromorphic at s = 1 and

ress=1

D(s) = c).

ζD(s) also has an Euler product

ζD(s) =∏p

ζD,p(s)

where ζD,p(s) =∑I⊆Op

1n(I)2s . We have this because we have I ↔ (Ip) and intersection of things behave naturally. This

is not because we have unique factorization of ideals (no di�erence for Dedekind domains).

For how many I ⊆ Ops with n(I) = pr, [Op : I] = p2r. If p rami�es, there's exactly one I ⊆ O with n(I) = pr for all r,

where I = (πrD). THerefore,

ζD,p(s) =(1− p−2s

)−1.

If p splits, then I ⊆M2(Zp)↔ L ⊆ Z2p where hom

(Z2p, L

)↔ L. If

#{I ⊆ O : with n(I)2 = pr

}= #

{L ⊆ Z2

p :[Z2p : L

]= pr

}= ar.

This ar satis�es a recurrence. First, a1 = p+ 1 (there p+ 1 lattices of index p)

ar+1 = a1ar − par−1

which is saying that index pr+1 lattices are index p lattices in index pr lattices, but we overcounted lattices in pZ2p.

Therefore,

ζD,p(s) =(1− p−2s

)−1 (1− p1−2s

)−1.

Therefore,

ζD(s) = ζ(2s)ζ(2s− 1)∏p|d

(1− p1−2s

).

Therefore, the residue

ress=1

ζD =π2

6

1

2d∏p|d

(1− p−1

)=π2

12ϕ(d).

�

Theorem 4.30. (Class number formula). Assume Ram(D) = {p,∞} then

|Cl(O)| = p− 1

12+ε24

+ε33

where ε2 = 1−(−4p

)and ε3 = 1−

(−3p

).

Recall thatx

2= genus (Γ\H)− 1 =

[PSL2(Z) : Γ]

12− ε2

4− ε3

3− c

2where ε2 and ε3 are the number of elliptic cusps, and c is the number of cusps. These two formula are extremely similar.

This is because when Ram(D) = {p,∞} then

|Cl(O)| = genus of X0(p) + 1.


Let F be a number �eld, and let

M = {f : Cl(F )→ C} .

We can de�ne Hecke operators (Spf) (I) = f ([pI]) . Then Sp's commute and the simultaneous eigenvectors are characters

up to scale. Additionally,

Spχ = χ(p)χ.

Now, for quaternion algebras. For F totally real, D/F de�nite quaternion algebras, O ⊆ D maximal order

M = {f : Cl(O)→ C} .

De�ne (Spf) (I) = f(pI) for p ⊆ OF . But there are lots of ideals I ′ with I ⊇ I ′ ⊇ pI. If p ∈ Ram(D), then there is one

πDI. Else, there are N(p) + 1 number of them. De�ne

(Tpf)(I) =∑

I⊇I′⊇pI

f ([I ′]) .

Remark 4.31. When p ∈ Ram(D), then T 2p = Sp.

The Tp and Sp's all commute and are semisimple. This means that M is spanned by Hecke eigenforms f . That is,

Spf = χf (p)f and Tpf = af,pf .

If f0 ([I]) = 1 is the constant function, then Spf = f and

Tpf =

(N(p) + 1) f p /∈ Ram(D)

f else.

Consider the norm map

Cl(O) Cl(F )

F×,>0\A∞,×F /O×F = Cl+(F )

Cl(O) → Cl(F )

%↗F×,>0\A∞,×F /O×F = Cl+(F )

where the bottom is called the narrow class group. As Frank said, we have

0→ (Z/2Z)r → Cl+(F )→ Cl(F )→ 0

where 0 ≤ r < [F : O], .......................

Fix χ : Cl+(F )→ C×, de�ne fχ ([I]) = χ (n+([I])). Then

Tpfχ = χ(p) (1 +N(p)) for p /∈ Ram(D) and Spfχ = χ(p)2.

Example 4.32. Let F = Q and D/Q de�nite. Then M has basis δ[I] for I ∈ Cl(O). Then

Tpδ[I] =∑

I⊆I′⊆p−1I

δ[I′] =∑

pI⊆I′⊆I

f ([I ′]) .

This equality is because p is principal when F = Q. This is also

=∑

cI′,Iδ[I′]


where

CI′,I = # {I ) J ) pI : J ∈ [I ′]} =1

|O`(I)×|#{I ) αI ′ ) pI : α ∈ D×

}=

1

|O`(I)×|#{α ∈ II ′−1 : n(I)p = n(I ′)α

}.

This II ′−1 is a Z-lattice. This n(α)n(I′)n(I) is an integral quadratic form.

∑I′

cI′I =

p+ 1 p split

1 p rami�es.

Example 4.33. Let D =(−1,−1

Q

). Then O = Z + Zi+ Zj + Z 1+i+j+k

2 is called the Hurwitz order. The norm n on O is

Q(x, y, z, w) = x2 + y2 + z2 + w2 + wx+ yw + zw

and rQ(p) = (p+ 1)24 when p 6= 2.

Example 4.34. Let D =(−1,−11

Q

)and O = Z + Zi+ Z 1+j

2 + Zi 1+j2 . Then |O×| = 4 and

Q(x, y, z, w) = x2 + xy + 3y2 + z2 + zw + 3w2.

Then

|Cl(O)| = 2

and rQ(p) = 14 times the number of solutions to Q(x, y, z, w) = p. We can check this order is maximal, because its trace

matrix is 2 0 1 0

0 −2 0 −1

1 0 −5 0

0 −1 0 −6

and so the order has discriminant −112.

For p 6= 11,

Tp =

(rQ(p) 3

2 (p+ 1− rQ(p))

p+ 1− rQ(p) − 12 (p+ 1) + 3

2rQ(p)

).

Also, rQ(2) = 1, and

T2 =

(1 3

2 0

), T2

(1

1

)= 3

(1

1

)

Tp

(1

−1

)= ap

(1

−1

), T2

(1

−1

)= 2

(1

−1

).

Then ap = 52rQ(p)− 3

2 (p+ 1) ≡ p+ 1 mod 5. Here, ap is the coe�cient of

f11(q) = q

∞∏n=1

(1− qn)2 (

1− q11)2,

and

θQ =∑

rQ(n)qn ∈M11 (Γ0(11)) .

This is spanned by E2(z)− 11E2(11z) and f11. Additionally,

f11 = (?)E(11)2 + (?)θQ.


The other maximal order is This has trace matrix2 1 0 −11

1 −1 −11 −11

0 −11 −88 −44

−11 −11 −44 33

which has discriminant −112. The norm form is

Q (x, y, z, w) = x2 + xy + y2 − xw + 5yw + 11yz + 14w2 + 44wz + 44z2.

Solving this, gives us that |O×| = 6.

Theorem 4.35. (Eichler) For F = Q,Ram(Q) = {p,∞}, there eixsts a Hecke equivariant isomorphism

M2 (Γ0(p)) = M.

Recall

M ={f : D×(F )\D×(A∞F )/D×(OF )→ C

}is the space of modular forms, and we de�ned Hecke operators Tp and Sp here. Here, we can replace C with Z or Fp,then Hecke operators can be de�ned in the exact same way. We want modular forms of more general weight and level.

By level, we mean replacing D×(OF ) with an open compact K ⊆ D× (A∞F ). Weight is a bit more troublesome, because

we somehow need the topology of C.

4.6. Algebraic Hecke characters.

De�nition 4.36. Ray class characters are {F×\A∞,×F /1 + mOF → C×

}.

Complex Hecke characters are {χ : F×\A×F → C× : continuous

}.

Meanwhile, p-adic Hecke characters

χ : F×\A∞,×F → Q×pcorrespond to 1-dimensional Galois representations.

Let χ = (χ∞, χ∞) be a character where χ∞ is a locally constant character, and χ∞ : (R⊗ F )× =

∏R×∏

C× → C×.Recall that characters R× → C× are exactly x 7→ |x|s sign(x)ε for s ∈ C and ε ∈ {0.1}. Meanwhile, characters C× → C×

are z 7→ zszs′

=∣∣z2∣∣s′ ( z

z

)s−s′where s, s′ ∈ C with s−s′ ∈ Z (we've made a choice of branch cut in the second expression).

De�nition 4.37. We say χ is algebraic if χ∞ |(R⊗F )×◦ is a polynomial (the identity component).

We can write a p-adic Hecke character as χ = (χp, χp). Then χp is locally constant (incompatibility of topologies) and

χp is not necessarily locally constant.

De�nition 4.38. We say χ a p-adic Hecke character is algebraic if there exists U ⊆ (Qp ⊗ F )×open compact such that

χp |U is a polynomial. That is,

χp |U (x) =∏

τ :F↪→Qp

τ(x)nτ

with nτ ∈ Z.

We have a correspondence,

{complex Hecke characters} {p-adic Hecke character} ↔ 1-dim Galois reps

⊆ ⊇ ⊇{algebraic complex Hecke characters} ↔ {p-adic algebraic Hecke characters } ↔ 1-dim Galois reps de Rham at p


However, the correspondence at the left bottom row, depends on an isomorphism

i : C ∼= Cp⊆ ⊆ .Q ∼= Q

Now, take χ : F×\A×F → C× algebraic. Let

χ(x) = i

(χ(x)

∏τ :F↪→C

τ(x∞)−nτ

) ∏τ :F↪→Qp

τ(xp)niτ .

Exercise 4.39. This element in the �rst bracket is in Q.

De�nition 4.40. An automorphic form on D× of level K ⊆ D× (A∞F ), is a function

f : D×(F )\D×(AF )/K → C.

such that for all g∞ ∈ D×(A∞F ), D× (F ⊗ R) → C is given by g∞ 7→ f(g∞, g∞), satisfying some regularity and growth

condition.

Fg∞(γg∞) = Fg∞(g∞)

for γ ∈ D×(F ) ∩ g∞K, dg. Therefore, D = M2(Q) and D× = GL/Q.

Miracle of compact connected Lie algebrais is that G a unrami�ed reductive algebraic group.

C[G] = ⊕V �nite cont. complex repsVp ⊗ V v = {f : K → C} .

Then

{atomorphic forms } ↔ {algebraic automorphism} .

Now let W be an algebraic representation of D×(F × R) and

S(K,V ) ={f : D×(F )\D×(AF )/K →W

}such that f(xg∞) = g−1

∞ f(x). That is, all algebraic automorphic form is equal to

⊕KS(K,W )⊗W∨.

Given f ∈ S(K,W ), let iW = W ⊗ Qp which is a representation of D× (Qp ⊗ F ).

f(g) = gp(g−1∞ f(g)

).

This is

• constant on

• D×(F ) invariant

• f(gk) = k−1p f(g) for all k ∈ K

4.7. p-adic modular form. From now on, assume

• Dp is split for all p | p• Fix an isomorphism

D× (F ⊗Qp) ∼= GL2(F ⊗Qp)

Let E/Qp a �nite dimensional representation. O ⊆ E ring of integers, λ its uniformizer, F residue �eld. Every τ : F ↪→ E

factors through E.

• (rational version) Let V be a �nite dimensional E vector space algebraic representation of GL2(F ⊗Qp)• (integral version) Let L be a �nitely generated O-module with continuous action of GL2 (OF ⊗ Zp)


For K ⊆ D× (A∞F ) open compact such that K ⊆ D× (A∞,pF )×GL2 (OF ⊗ Zp) in (integral) situation,

S (K,L) ={f : D×(F )\D×(A∞F )→ L

}such that f(gk) = k−1

p f(g) for all k ∈ K.

Example 4.41. Algebraic representations of GL2(C) are SymkC2 ⊗ detnC2 for k ≥ 0 and n ∈ Z. For a weight

(kτ , nτ )τ∈hom(F,E),

V(kτ ,nτ ) = ⊗τ∈hom(F,E)SymkτE2 ⊗ detnτE2,

which carries an action of GL2 (F ⊗Qp). Similarly,

L(kτ ,nτ ) = ⊗τ∈hom(F,E)SymkτO2 ⊗ detnτO2

which carries an action of GL2(OF ⊗Qp).We can also take things like L(kτ ,nτ )/λ

n or L be representation of GL2(Zp)/1 + pnM2(Zp).

D× (A∞F ) = qri=1D×(F )αiK. The algebraic modular form f ∈ S(K,L) is determined by f(αi) ∈ L. If αi = γαik for

some γ ∈ D×(F ) and k ∈ K then we must have f(αi) = k−1p f(αi). To have this double coset relation, k ∈ ∩α−1

i D×(F )αi.

We have just shown that f must be invariant by these guys, so:

Proposition 4.42. S (K,L) = ⊕ri=1LK∩α−1

i D×(F )αi

Since the center is F× ⊆ D×(F ), we have K ∩F× ⊆ K ∩α−1i D×(F )αi so these groups are always in�nite (F is totally

real).

Proposition 4.43. K ∩ F× ⊆ K ∩ α−1i D×(F )αi has a �nite index, and they are equal when K is su�ciently small.

Proof. Since PD×(F ) ↪→ PD×(AF ). We have an inclusion(K ∩ α−1

i D×(F )αi)Z/Z ↪→ KZ/Z · PD× (R⊗ F ) ⊆ PD× (AF ) .

This set is compact, and the embedding is discrete so it must be �nite.

If 1 6= k ∈(K ∩ α−1

i D×(F )αi)Z/Z, it has �nite order. Suppose for all k ∈ K, there exists prime q prime of F with Dq

is split, with kq ∈ 1 +M2(OFq ) then it can't have �nite order, even mod the center. Therefore, it must be trivial. �

In general, these centers cause a lot of trouble. We can avoid it by working with PD× or D×,1. However, we may not

have enough Hecke operators (?) and other bad things....

Let D/F be a de�nite quaternion algebra, p such that p | p and p /∈ Ram(D). For all v /∈ Ram(D), �x an isomorphism

GL2(Fv) ∼= D×v . Let E/Qp be a �nite extension, with ring of integer O, uniformizer λ, and residue �eld k. hom (F,E) =

[F : Q].

Let N = (kτ , ητ )τ∈hom(F,E), kτ ≥ 0 and ητ ∈ Z.

GL2 (OF ⊗ Zp) acts on LN = ⊗τ :F↪→ESymkτO2 ⊗ detητO2

and the action is view GL2(O). De�ne

M(K,L) ={f : D×(F )\D×(A∞F )→ L : f(gk) = k−1

p f(g) for all k ∈ K}

where L is any �nite O-module with continuous action of GL2(OF ⊗ Zp). Here,

K ⊆ D×(A∞,pF )×GL2(OF ⊗ Zp)

open compact (use GL2 for the second factor, because it's split at all the places above p).

S(K,L) = ⊕ri=1LK∩α−1

i D×(F )αi

where D×(A∞F ) = qD×(F )αiK.


K ∩ F× ⊆ O×F �nite index (F× is the center of D), K ∩ F× ⊆ α−1i D×αi ∩K.

De�nition 4.44. Let K be a �eld. We say g ∈ GLn(K) is neat, if 〈α1, ..., αn〉 ⊆ K× is torsion free (eigenvalues of g).

Example 4.45. Say g =

(a 0

0 −a

)with a not a root of 1 and the characteristic of K is not 2. Then

(a 0

0 −a

)and

(a 0

0 −a

)−1

has order 2, so g ∈ PGL2(k) has order 2.

If K is a p-adic �eld, 1 + πnO is torsion free for su�ciently large n. For example, if g ≡

1 ∗ ∗

... ∗1

mod πn then

g is neat.

We say level K is neat, if for all g ∈ K, there exists v /∈ Ram(D) such that gv is neat.

Proposition 4.46. K ∩ E× ⊆ α−1i D×(F )αi ∩K has �nite index and if K is neat, then they are equal. Also, −1 /∈ K if

K is neat.

If F = Q,

dimS(K,L(k,0)

)=∣∣D×(F )\D×(A∞F )/K

∣∣ (k + 1) =ϕ(d)

24[k0 : k]

for k neat, where k0 =(OD ⊗ Z

)(OD maximal order).

Let F be general, x ∈ K ∩ F× acts on LN via∏τ τ(x)kτ+2ητ . S(K,LN ) = 0 unless

∏τ τ(x)kτ+2ητ = 1 for all

x ∈ K ∩ F×.Dirichlet's theorem, if Γ ⊆ O×,>0

F �nite index, and if for all x ∈ Γ,∏τ :F↪→R τ(x)cτ = 1 for cτ ∈ R then cτ is independent

of τ .

De�nition 4.47. We say that µ = (kτ , ητ ) is paritious, if kτ + 2ητ is independent of τ .

Proposition 4.48. S(K,−) is exact in L's such that all the stabilizers act trivially. If K is neat and µ is paritious

S(K,Lµ/λN ) = S (K,Lµ) /λN .

But S(K,Lµ/λ

N)will be non-zero for k su�ciently small, even with K = Kp × GL2 (OF ⊗ Zp) for Kp su�ciently

small.

Example 4.49. D =(−1,−1

Q

), K = (OD ⊗OF )

×where OD is the maximal order. Then

S(K,L(k,0)) =(SymkC2

)⟨ 1 0

0 −1

⟩⊕(SymkC2

)⟨

ζ0

−ζ0

⟩.

Hecke operators,

lim→

K⊆...

S(K,L) = S(L)

which has an action of D× (A∞,pF )×GL2(OF ⊗ Zp), S(L)K = S(K,L). In the rational setting,

lim→

K⊆...

S(K,V ) = S(V )

with action of D×(A∞F ).

Given K,K ′ ⊆ D×(A∞,pF )×GL2(OF ⊗ Zp). Can de�ne [k′gk] : S(K,L)→ S(K ′, L) with k′gk = qgiK : f 7→∑gif .

There's also awful formula you can derive from composition.

Take S to be a �nite set of �nite places of F such that


• Ram(D) ⊆ S• If v | p then v ∈ S•∏v/∈S GL2(OFv ) ⊆ K (i� K = KS ×

∏v/∈S GL2(OFv ))

• For v /∈ S, Tv =

[K

(πv 0

0 1

)K

], Sv =

[K

(πv 0

0 πv

)K

], Tv, Sv acts on S(K,L) and commutes (spherical

Hecke algebra)

GL2(OFv )

(πv 0

0 1

)GL2(OFv )/GL2(OFv ) ⊆ GL2 (Fv) /GL2(OFv ).

This last thing is corresponding to lattices in F 2v , by g to gO2

Fv⊆ F 2

v .(πv 0

0 1

)O2Fv

= 〈πve1, e2〉, and so our set is the set of lattices L = 〈πve′1, e′2〉 for some basis e′1, e′2 of O2

Fv. This is

the same as lattices L ⊆ O2Fv

such that O2Fv/L = kv (residue �eld).

Elementary divisors, says that for any lattice L ⊆ F 2v , there exists a basis e

′1 and e

′2 for O2

Fvsuch that L =

⟨πae′1, π

be′2⟩,

a ≥ b.

GL2(Fv) = qa≥bGL2(OFv )

(πav 0

0 πbv

)GL2(OFv ).

Suppose K ⊆ GL2(OFv ). Then G/K is like lattices with extra structure. We have

G\ (G/K ′ ×G/K)↔ K ′\G/K

by (g1, g2) 7→ g−11 g2.

TS,univ = O [Tv, Sv]v/∈S and if M if TS,univ-module TS(M) = im(TS,univ → EndO(M)

).

F �eld , 〈·, ·〉 non-degenerate symmetric bilinear form, or E/F quadratic with Hermitian form. T is normal wrt 〈·, ·〉,if TT ? = T ?t. If 〈·, ·〉 is anisotropic (〈x, x〉 6= 0 unless x = 0) then T is semisimple.

〈[k, k′] v, v′〉K =⟨v,[k′g−1k

]v′⟩K′

with 〈·, ·〉 : S (K,W )2 → C (Petersson inner product).

Let M be a free O-module, with TS,univ action. TS(M) acting on M . Semisimplicity implies TS(M)[ 1p ] is a product of

�elds. These act on M [ 1p ], just from a system of Hecke eigenvalues with multiplicities.

TS(M) = im

(TS,univ → EndE(M [

1

p])

)+

only depends on M [ 1p ].

T(M) is generated by T , has eigenvalues 0 and p. This has to be equal to Zp[T ]/T (T−p), but there are di�erent modules

over this. We can have T =

(p 0

0 0

)and T =

(p 1

0 0

), which are non-isomorphic. This is because M1/pM1 has 2-

dimensional eigenspace, while M2/pM2 has 1-dimensional eigenspace. M [ 1p ] a free T[ 1

p ]-module of rank 1, dimM/pM = 1

i� M is free.

Hecke operators at p, K = Kp × GL2(OF ⊗ Zp). S (K,Lµ) ⊆ S (K,Vµ) acted on by Tv and Sv with v | p. Sv =(πv 0

0 πv

)

Sπv =1∏

τ :Fv↪→E τ(πv)kτ+2ητSv, Tπv =

1∏τ(πv)ηv

Tv

and Sπv , Tπv acts on S(K,Lv).


Part 3. Appendix

Theorem 4.50. Auslander Buchsbaum formula. If R is a commutative Noetherian local ring and M is a non-zero �nitely

generated R-module of projective dimension, then

pdR(M) + depth(M) = depth(R)

where pd stands for the projective dimension of a module (minimal length of a projective resolution), and depth for a depth

of a module.

De�nition 4.51. Let R be a commutative ring and I an ideal, M a �nite R-module with the property that IM ⊆ M .

Then the I-depth is

depthI(M) = min{i : Exti (R/I,M) 6= 0

}.

The depth of a local ring R is the m-depth as a module over itself.

Proposition 4.52. We have

depth(M) ≤ dim(M).

If R is a Cohen-Macaulay local ring, then the depth(R) = dimR.

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HIGHER HIDA THEORY NOTES - University of Chicagohaolee/Higher Hida Theory Notes.pdf · Q are modular (representations coming from class group). 2.1. Deformation. Let F be a number