# Modular forms Jens Funke University of Durham Lecture 19 Jens Funke Modular forms SECTION 18 Hecke theory III In the previous lecture we have seen that the L function of Hecke eigenfo

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```					Modular forms

Jens Funke

University of Durham

Lecture 19

Jens Funke     Modular forms
SECTION 18: Hecke theory III
In the previous lecture, we have seen that the L-function of Hecke eigenforms has an
Euler product similar to the one for the Riemann zeta function. Thus such modular
forms are/should be of particular (arithmetic) interest.

Question
Can one always ﬁnd an "eigenbasis" for Mk (N, χ); i.e., a basis consisting of
eigenforms for all Hecke operators Tn ?
Can one always ﬁnd eigenforms for all Hecke operators Tn ?
Can one diagonalize an individual Hecke operators Tn operating on Mk (N, χ)?

Remark 18.1
As we have seen in an example the previous lecture, Eisenstein series are often
eigenforms. In fact, one can build an eigenbasis for the space of Eisenstein series.
Hence we will concentrate on cusp forms in what follows.

We will ﬁrst show that the Hecke operator Tn for n coprime to N is indeed
diagonalizable.
We slightly modify the deﬁnition of the Petersson scalar product and set
1
Z
f , g :=                   f (τ )g(τ )v k dudv
v2
[PSL2 (Z) : Γ] FΓ
for f , g ∈ Sk (Γ) with Γ ⊆ SL2 (Z) a congruence subgroup. Note that with this
normalization the deﬁnition of the Petersson scalar product is independent of whether
one views f , g in Sk (Γ ) with Γ ⊂ Γ.

Jens Funke   Modular forms
Proposition 18.2
Let f , g ∈ Sk (Γ) and let α ∈ GL+ (Q). Then
2
1

f |k α, g|k α = f , g .

2

f |k α, g = f , g|k α−1 .

Proof.
For (1), we view f , g, f |k α, g|k α as forms on Γ := Γ ∩ α−1 Γα. Then
Z                                   Z
[PSL2 (Z) : Γ ] f |k α, g|k α =         f |k α(τ )g|k α(τ )v k dudv =
v 2     f (ατ )g(ατ )Im(ατ )k dudv
v2
FΓ                                        FΓ

So                                                 Z
f |k α, g|k α =        1
f (τ )g(τ )v k dudv = f , g ,
v2
[PSL2 (Z):Γ ]       αFΓ

since αFΓ is a fundamental domain for the action of Γ as well. (2) follows by
considering (1) with g replaced by g|k α−1 and Γ replaced by Γ ∩ αΓα−1 .

Jens Funke       Modular forms
Proposition 18.3
Let f , g ∈ Sk (N, χ). Let n be coprime to N. Then
Tn f , g = χ(n) f , Tn g .
That is, the adjoint map    ∗
Tn   with respect to the Petersson scalar product is given by
∗
Tn = χ(n)Tn .
Finally, we also have (for all n)
∗    −1
T n = ωN T n ωN .
“          ”
0 −1
where ωN denotes (the weight k -action of) the Fricke element                  N 0
.

Proof.
If α ∈ ∆n (N), then n n σn α−1 ∈ ∆n (N) as well. Here as before σn ∈ Γ0 (N) such
` 0´
1 “       ” 0              1
that σn ≡ 1/n n
0
0
mod N. Note that at this point we need (n, N) = 1. From that it is
easy to see if α runs through a system of right coset representatives of Γ1 (N)\∆n (N),
1
then so does n n σn α−1 . Since n n acts trivially, we get
` 0´               ` 0´
0                   0
X                                X
Tn f , g = nk /2−1              f |k α, g = nk /2−1                f , g|k α−1
α∈Γ1 (N)\∆n (N)
1
α∈Γ1 (N)\∆n (N)
1
X
= nk /2−1                                            σn α−1 = f , χ(n)Tn g .
`n 0´
f , χ(n)g|k    0n
α∈Γ1 (N)\∆n (N)
1

For the last statement argue similarly. For α ∈ ∆n (N), we have
1
` n 0 ´ −1 −1
0 n ωN α      ωN ∈ ∆n (N) and so on...
1
Jens Funke      Modular forms
Corollary 18.4
For n coprime to N, the Hecke operator Tn is normal, that is, it commutes with its
adjoint. In fact, scaling Tn with cn , a square root of χ(n), we have that cn Tn is
Hermitian, that is, equal to its adjoint. In particular, Tn can be diagonalized.

Theorem 18.5
There exists a simultaneous eigenbasis on Sk (N, χ) for all Tn with (n, N) = 1.

Proof.
This is linear algebra. A (possibly inﬁnite) family of commuting diagonalizable linear
operators Φ1 , Φ2 , . . . on a vector space V can be simultaneously diagonalized.
Indeed, since Φ1 can be diagonalized we can decompose V into eigenspaces for the
ﬁrst operator Φ1 . Then Φ2 preserves each eigenspace for Φ1 . (Here we need that Φ1
and Φ2 commute). Hence we can decompose V into simultaneous eigenspaces for Φ1
and Φ2 . Continue in this fashion.

Remark 18.6
For (n, N) > 1, the Hecke operator Tn is in general not normal, that is, cannot be
diagonalized in general. (For more, see below).In that case one has Tn = Un and
Tn f , g = nk f , Vn g ,
which we leave as an exercise!

Jens Funke    Modular forms
Remark 18.7
So far, we have not achieved what we want, namely a basis consisting of eigenforms
for all Hecke operators. However, we already do have an Euler product for the partial
L − series
X            Y              1
LN (f , s) :=    an n−s =
1 − ap p−s + χ(p)pk −1−2s
(n,N)=1                  p N
P∞       n
if f =    n=1 an q is an eigenform for all Tn with (n, N) = 1.

Remark 18.8
Since all the remaining Hecke operators Tp with p | N commute with the others, they
preserve the simultaneous eigenspaces for Tn with (n, N) = 1. Hence one way of
concluding that we have an eigenbasis for all Hecke operators would if all the
simultaneous eigenspaces for Tn with (n, N) = 1 were one-dimensional.
This assertion is called Strong Multiplicity One. That is, a Hecke eigenform is uniquely
determined (up to scalars) by its eigenvalues for (n, N) = 1.

Remark 18.9 (Strong Multiplicity One does NOT hold on Sk (N, χ))
For example, let f = ∞ an q n ∈ Sk (SL2 (Z)) be a normalized eigenform for all
P
n=1
Hecke operators (such a form always exists by our work so far). Then consider
g(τ ) = Vp f (τ ) = f (pτ ) ∈ Sk (Γ0 (p)). Now take n coprime to p. Since the Hecke
(1)                    (p)
operators Tn        for level 1 and Tn     for level p coincide on Sk (SL2 (Z)) and since Vp
(p)
commutes with          we see that both f and g are eigenforms with eigenvalues an for
Tn ,
all Tn with (n, p) = 1. (homework!) Hence we cannot distinguish f and g by their Hecke
eigenvalues for (n, p) = 1.

Jens Funke         Modular forms
Remark 18.10
Of course, this construction works in more generality for any level N. We see that
forms coming from lower level are in the way of Strong Multiplicity One. We will see
now that this is the only obstruction.

Deﬁnition 18.11 (Old and new forms)
old
We let Sk (N, χ) be the space of old forms generated by
[[
{f ( τ ); f ∈ Sk (M, χ)},
M

where M runs over all divisors of N with M = N such that the conductor of χ divides M,
old
and runs over all divisors of N/M. In other words, Sk (N, χ) is the subspace of
Sk (N, χ) generated by forms of lower level.
The space of new forms is the orthogonal complement of the space of old forms:
Sk (N, χ) := Sk (N, χ)⊥ .
new          old

Lemma 18.12
1   The Fricke element ωN maps Sk (N, χ) isomorphically to Sk (N, χ) preserving old
and new spaces.
2          new           old
Both, Sk (N, χ) and Sk (N, χ) are stable under all Hecke operators.

Jens Funke   Modular forms
Proof.
For (1), let f (τ ) := −k /2 f ( τ ) ∈ Sk (N, χ) be an old form where f ∈ Sk (N/ , χ) and
old
old
> 1. Then one easily sees f |k ωN = f |k ωN/ ∈ Sk (N/ , χ) ⊂ Sk (N, χ). Similarly,
for f ∈ Sk (M, χ) with M | N, one has f |k ωN (τ ) = −k /2 f |k (ωN/M )( τ ). using the
(essential) self-adjointness of ωN wrt the Petersson product then gives the assertion for
new forms. This shows (1).
For (2), ﬁrst we consider the case (n, N) = 1. Then Tn commutes with all V operators
for dividing N. This shows preservation of the old space. The (essential)
self-adjointness of Tn then gives preservation of the new space. Now assume
(n, N) > 1 when Tn = Un . Using Um Vm = 1, one easily obtains the assertion for old
forms. For new forms, use the adjointness of Un with nk Vn = ωN Un ωN , and use the
stability of old forms for the Fricke element.

Theorem 18.13 (Atkin-Lehner)
1                                      new
Strong Multiplicity One holds for Sk (N, χ).
2                                                             new
In particular, there exists a unique orthogonal basis of Sk (N, χ) consisting
normalized eigenforms for all Hecke operators. Such forms are called primitive
forms (or sometimes also new forms, which does cause confusion).
For f = ∞ an q n primitive, there exists η(f ) of absolute value 1 such that
P
3
n=1
new
f |k ωN = η(f )f ∈ Sk (N, χ), where f is the primitive form given by
∞
X
f (τ ) =   an q n .
n=1
In particular Tn f = an f .

Jens Funke   Modular forms
Proof.
(1) is the fundamental result which we do not prove. (2) follows from that. (3) follows
−1           ∗
from that the adjoint of Tn is given by ωN Tn ωN and that Tn f = an f .

Example 18.14
Let f (τ ) ∈ S2 (Γ0 (11)) = η 2 (τ )η 2 (11τ ) be the normalized eigenform we have
encountered before. Note T2 f = −2f . Then consider the space generated by
f , V2 f , V4 f , V8 f ∈ S2 (Γ0 (88)). These are all eigenforms for Tn with n odd, but the
matrix for T2 = U2 acting on S2 Γ0 (88)) is given by (exercise!)
!
−2    1   0   0
−2    0   1   0   ,
0    0   0   1
0    0   0   0

which is not diagonalizable. Since V ⊥ inside S2 (Γ0 (11)) is also Hecke stable, we
conclude that T2 = U2 cannot be diagonalized on S2 (Γ0 (88)).

Example 18.15 (Hecke)
1                                     new
If N = 1 or χ is primitive, then Sk (N, χ) = Sk (N, χ).
2   For p a prime, we have
new
Sk (Γ0 (p)) = Sk (SL2 (Z)) ⊕ Vp Sk (SL2 (Z)) ⊕ Sk (Γ0 (p)).

Jens Funke           Modular forms

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