Zeros of the Modular Form Ek El − Ek+l
Sarah Reitzes and Polina Vulakh
Tufts University, Bard College
pvulakh@gmail.com
July 23, 2015
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Zeros of Modular Forms
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Zeros of Modular Forms
A complex-valued function f is a modular form if:
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Zeros of Modular Forms
A complex-valued function f is a modular form if:
1
a b
for all z ∈ H ∪ {∞}, γ =
∈ SL2 (Z)
c d
az+b
we have f (γ(z)) = f ( cz+d ) = (cz + d)k f (z). Then k is called the
weight of f . (Transformation Law)
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Zeros of Modular Forms
A complex-valued function f is a modular form if:
1
2
a b
for all z ∈ H ∪ {∞}, γ =
∈ SL2 (Z)
c d
az+b
we have f (γ(z)) = f ( cz+d ) = (cz + d)k f (z). Then k is called the
weight of f . (Transformation Law)
f is holomorphic (complex differentiable) for every point z ∈ H ∪ {∞},
so f can be expanded in a power series in z around any point z0 ∈ H
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
2/1
Zeros of Modular Forms
A complex-valued function f is a modular form if:
1
2
a b
for all z ∈ H ∪ {∞}, γ =
∈ SL2 (Z)
c d
az+b
we have f (γ(z)) = f ( cz+d ) = (cz + d)k f (z). Then k is called the
weight of f . (Transformation Law)
f is holomorphic (complex differentiable) for every point z ∈ H ∪ {∞},
so f can be expanded in a power series in z around any point z0 ∈ H
We look at modular forms of positive, even weight.
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
2/1
Zeros of Modular Forms
A complex-valued function f is a modular form if:
1
2
a b
for all z ∈ H ∪ {∞}, γ =
∈ SL2 (Z)
c d
az+b
we have f (γ(z)) = f ( cz+d ) = (cz + d)k f (z). Then k is called the
weight of f . (Transformation Law)
f is holomorphic (complex differentiable) for every point z ∈ H ∪ {∞},
so f can be expanded in a power series in z around any point z0 ∈ H
We look at modular forms of positive, even weight.
The valence formula tells us how many zeros f has.
X
k
1
1
= vi (f ) + vρ (f ) +
vz (f )
12
2
3
z6=i,ρ
z∈H
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Zeros of the Eisenstein Series in F
Eisenstein series of weight k:
∞
1 X
1
2k X
Ek (z) =
=
1
−
σk−1 (n)e 2πinz
2
Bk
(cz + d)k
(c,d)=1
n=1
c,d∈Z
It has been proven that the zeros of the Eisenstein series lie on the arc of
the fundamental domain F = {z = x + iy ∈ H : x ∈ (− 12 , 21 ), |z| ≥ 1}
(RSD, 1970).
Figure: F
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
Figure: Zeros of E70
July 23, 2015
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Zeros of Ek El − Ek+l
Conjecture:
The zeros of Ek El − Ek+l , a modular form of weight k + l, lie on the
boundary of F.
2
Figure: Zeros of E50
− E100
Figure: Zeros of E60 E8 − E68
Conjecture:
The zeros of Ek2 − E2k , a modular form of weight 2k, lie on the lines
x = ± 12 in F.
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Proving the zeros of Ek2 − E2k
Since Ek ( 12 + iy ) is real-valued, we prove the desired number of zeros
(b k6 c − (1 + n)) via IVT using points of the form 12 + iym where
m)
k
k
for θm = mπ
ym = tan(θ
2
k where m ∈ Z such that d 3 e ≤ m < 2 − n..
Why −n?
√
We run into problems for y ≥ √c0logkk , so n is the number of zeros with y
past this range.
However, there exists a method involving the√Fourier expansion that
proves the location of zeros for which y > c1 k log k, so we lose very few
zeros altogether.
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
2
July 23, 2015
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Approximating Ek2 − E2k
Write Ek ( 21 + iy ) = Mk ( 12 + iy ) + Rk ( 12 + iy ) where Mk corresponds to
c 2 + d 2 ≤ 1 - except for (c, d) = (1, 1) - and Rk corresponds to all other
(c, d).
Then
Ek2 ( 12 + iy ) − E2k ( 12 + iy ) = (Mk ( 12 + iy ) + Rk ( 12 + iy ))2
− (M2k ( 12 + iy ) + R2k ( 12 + iy ))
= Mk ( 12 + iy )2 + 2Mk ( 12 + iy )Rk ( 21 + iy )
+ Rk ( 12 + iy )2 − M2k ( 21 + iy ) − R2k ( 12 + iy )
9+12y
k , which is decreasing in k, and since
9
( 4 +y 2 ) 2
1
+ 1 1 k , we know |Mk ( 12 + iy )| ≤ 3. Then
1
( 2 +iy )k
(− 2 +iy )
We know |Rk ( 12 + iy )| <
Mk ( 12 + iy ) = 1 +
we want to show
|Mk ( 21 + iym )2 − M2k ( 12 + iym )| > 8
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
9+12ym
k
9
2)2
( 4 +ym
July 23, 2015
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Approximating Ek2 − E2k (cont.)
For our points
1
2
+ iym , we have a lower bound
k
|Mk ( 21
2
+ iym ) −
M2k ( 12
4( 1 + y 2 ) 2 − 2
+ iym )| ≥ 4 1 m 2 k
( 4 + ym )
so we want to show
2 ) k2 − 2
4( 14 + ym
9+12ym
>
8
k
9
2 )k
2)2
( 4 +ym
( 41 + ym
√
For large y , this is not true: specifically for y ≥ c0 √logk k where c0 ≤
so we work with ym <
√
c0 √logk k .
By simplifying further, we have
9 +y 2 k
4 m
1
2
4 +ym
2
> c2 ym where c2 =
38
√
3
√1 ,
8
+ 24.
This is true for k ≥ c2 , so we have proved
m
.
|Mk ( 12 + iym )2 − M2k ( 12 + iym )| > 8 9+12y
k
9
2)2
( 4 +ym
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Sign changes from Mk ( 21 + iym )2 − M2k ( 12 + iym )
If we rewrite
1
2
+ iym = re iθm , we have
Mk (re iθm )2 − M2k (re iθm ) =
4r k (−1)m + 2
r 2k
k
k
For θm = mπ
k where m ∈ Z such that d 3 e ≤ m < 2 − n, this yields
b k6 c − n sign changes corresponding to b k6 c − n − 1 zeros by IVT.
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Extending this to general Ek El − Ek+l
Recall that Bk,l =number of zeros of Ek El − Ek+l for which x = 12 .
Conjecture:
(k ≥ l) The number of zeros Ek El − Ek+l for which x =
of El2 − E2l . In other words, Bk,l ≥ Bl,l .
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
1
2
is at least that
July 23, 2015
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Example: Ek E34 − Ek+34
Figure: k=34
Figure: k=40
Figure: k=44
Figure: k=50
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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Extending this to general Ek El − Ek+l (cont.)
Our main term becomes
Mk (re iθ )Ml (re iθ ) − Mk+l (re iθ ) =
r 2l+k 2 cos(θk)+r 2k+l 2 cos(θl)+r k+l 2 cos(θ(k−l))
r 2(k+l)
If we rewrite k = l + d and let θm =
mπ
l
for d 3l e ≤ m < 2l ,
3l+2d 2(−1)m + r 2l+d 2 cos( mπ d)
r 3l+d 2(−1)m cos( mπ
l d) + r
l
r 4l+2d
as our main term instead.
By splitting this up into three cases for d ≡ 0, 2, 4 (mod 6), we follow a
similar method to show that Ek El − Ek+l has at least b 6l c − n − 1 zeros or
which x = 21 .
Sarah Reitzes and Polina Vulakh (Tufts University, Bard College)
Eisenstein Series
July 23, 2015
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