Modular Forms · Study tools

Step-throughDiamond–Shurman, Ch. 1: Modular Forms, Elliptic Curves, and Modular Curves

SL₂(ℤ) and its fundamental domain

Everything in this subject starts with one group acting on one space. $\mathrm{SL}_2(\Z)$ moves the points of the upper half-plane by fractional linear maps, and a modular form is a function that transforms in a fixed way under every one of those moves. Before studying such functions you need a map of the orbits, and that map is the fundamental domain.

The idea

The action.

\mathrm{SL}_2(\Z)

is the group of integer matrices

\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}

with

ad-bc=1

. It acts on the upper half-plane

\mathcal H=\{\tau:\operatorname{Im}\tau>0\}

by

\gamma\tau=\dfrac{a\tau+b}{c\tau+d}

. The one identity to remember is

\operatorname{Im}(\gamma\tau)=\frac{\operatorname{Im}\tau}{|c\tau+d|^2}.

The matrices $\gamma$ and $-\gamma$ act the same way, so the group that really acts is $\mathrm{PSL}_2(\Z)=\mathrm{SL}_2(\Z)/\{\pm I\}$ . It is generated by

S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}:\ \tau\mapsto-\frac1\tau, \qquad T=\begin{pmatrix}1&1\\0&1\end{pmatrix}:\ \tau\mapsto\tau+1.

Fundamental domain. Let

\mathcal D=\{\tau\in\mathcal H: |\operatorname{Re}\tau|\le\tfrac12,\ |\tau|\ge1\}

. Every point of

\mathcal H

is

\mathrm{SL}_2(\Z)

-equivalent to a point of

\mathcal D

. Two different points of

\mathcal D

can be equivalent only if both lie on its boundary: the two vertical edges are matched by

T

and the two halves of the arc by

S

. The only points of

\mathcal D

with a nontrivial stabiliser in

\mathrm{PSL}_2(\Z)

are

i

(stabiliser of order 2, generated by

S

) and

\rho=e^{2\pi i/3}

,

\rho+1=e^{\pi i/3}

(order 3).

The proof is an algorithm, and it is what the explorer runs. If $|\operatorname{Re}\tau|>\tfrac12$ , translate by a power of $T$ . If $|\tau|<1$ , apply $S$ : then $\operatorname{Im}(-1/\tau)=\operatorname{Im}\tau/|\tau|^2$ is strictly larger. Repeat. It stops because, for a fixed $\tau$ , only finitely many pairs $(c,d)$ have $|c\tau+d|<1$ , so the imaginary part can only grow finitely often.

Modular form of weight $k$ . A holomorphic

f:\mathcal H\to\C

with

f(\gamma\tau)=(c\tau+d)^k f(\tau)

for every

\gamma\in\mathrm{SL}_2(\Z)

, and holomorphic at

\infty

. Taking

\gamma=T

gives

f(\tau+1)=f(\tau)

, so

f(\tau)=\sum_n a_nq^n

with

q=e^{2\pi i\tau}

; holomorphic at

\infty

means

a_n=0

for

n<0

. If also

a_0=0

,

f

is a cusp form. The spaces are

\mathcal M_k(\mathrm{SL}_2(\Z))\supseteq\mathcal S_k(\mathrm{SL}_2(\Z))

.

Explore

Start at

Click or drag anywhere to choose a starting point

\tau_0

. the domain

\mathcal D

the path so far. Each tile is

\gamma\mathcal D

for one

\gamma\in\mathrm{PSL}_2(\Z)

.

Worked example

Reduce $\tau_0=\dfrac{7+i}{25}$ into $\mathcal D$ , and find a matrix $\gamma$ with $\gamma\tau_0\in\mathcal D$ . (It is the first preset above, so you can check each step.)

$\operatorname{Re}\tau_0=0.28$ is already in the strip. Should you apply $S$ ? What do you get, and how much does the imaginary part grow?

$|\tau_0|^2=50/625=2/25<1$ , so yes. $-1/\tau_0=-25/(7+i)=-25(7-i)/50=(-7+i)/2$ . The imaginary part goes from $1/25$ to $1/2$ , a factor of $1/|\tau_0|^2=12.5$ .
Continue from $(-7+i)/2=-3.5+0.5i$ .

Translate by $T^3$ : $-\tfrac12+\tfrac12 i$ . Its absolute value squared is $\tfrac12<1$ , so apply $S$ : $-1/\big(\tfrac{-1+i}{2}\big)=\dfrac{-2}{-1+i}=1+i$ . Then translate by $T^{-1}$ to reach $i$ , which is in $\mathcal D$ (on the arc).
Multiply the steps together to get $\gamma$ , and check it with the imaginary-part formula.

The last step goes on the left:

$\gamma=T^{-1}ST^{3}S=\begin{pmatrix}-4&1\\3&-1\end{pmatrix}.$

Here $3\tau_0-1=(-4+3i)/25$ , so $|3\tau_0-1|^2=1/25$ and $\operatorname{Im}\gamma\tau_0=(1/25)/(1/25)=1$ , which is $\operatorname{Im} i$ .
Is $\gamma$ the only matrix with $\gamma\tau_0=i$ ?

No. If $\gamma'\tau_0=i$ too, then $\gamma'\gamma^{-1}$ fixes $i$ , so it is $\pm I$ or $\pm S$ . Up to sign the answers are $\gamma$ and $S\gamma=\begin{pmatrix}-3&1\\-4&1\end{pmatrix}$ . The extra solution exists because $i$ is an elliptic point.

Check your understanding

The algorithm is at $\tau=0.3+0.5i$ and applies $S$ . By what factor does the imaginary part grow?

$0.34$ That would make it shrink. $S$ divides the imaginary part by $|\tau|^2$ , and $|\tau|^2=0.34<1$ .
$1/0.34\approx2.94$
$1/|\tau|\approx1.71$ The formula has $|c\tau+d|^2$ , squared. For $S$ , $c\tau+d=\tau$ .

\operatorname{Im}(S\tau)=\operatorname{Im}\tau/|\tau|^2=0.5/0.34\approx1.47

. The growth is exactly why the algorithm only applies

S

when

|\tau|<1

.

Why is every modular form of odd weight for $\mathrm{SL}_2(\Z)$ identically zero?

Because the q-expansion can only contain even powers of $q$ .Nothing forces that: periodicity $f(\tau+1)=f(\tau)$ allows every integer power.
Because $S$ has order 4.Applying the law four times along $S$ is consistent for every $k$ . The obstruction comes from a different element.
Because $-I\in\mathrm{SL}_2(\Z)$ acts trivially but contributes the factor $(-1)^k$ .

With

\gamma=-I

we get

\gamma\tau=\tau

and

c\tau+d=-1

, so

f(\tau)=(-1)^kf(\tau)

. For odd

k

this forces

f=0

. Smaller groups that do not contain

-I

, such as

\Gamma_1(N)

for

N\ge3

, do have forms of odd weight.

Let $f\in\mathcal M_6(\mathrm{SL}_2(\Z))$ . What can you say about $f(i)$ ?

$f(i)=0$ .
Nothing: $i$ is a point like any other. $i$ is fixed by $S$ . Put $\gamma=S$ and $\tau=i$ into the transformation law.
$f(i)=f(\rho)$ . $i$ and $\rho$ are not equivalent (both lie in $\mathcal D$ , not matched by the boundary identifications), so there is no reason for this.

f(i)=f(Si)=i^6f(i)=-f(i)

. The same argument gives

f(i)=0

for every weight

k\equiv2\pmod 4

. For example, the Eisenstein series

E_6

vanishes at

i

.

Practice

The algorithm terminatesFiniteness of small denominators

Fix $\tau=x+iy\in\mathcal H$ . Show that only finitely many coprime pairs $(c,d)$ satisfy $|c\tau+d|<1$ . Deduce that $\operatorname{Im}(\gamma\tau)$ takes a largest value as $\gamma$ runs over $\mathrm{SL}_2(\Z)$ , and that if $\gamma$ achieves it then a translate $T^n\gamma\tau$ lies in $\mathcal D$ .

Hint

Expand $|c\tau+d|^2=(cx+d)^2+c^2y^2$ .

Solution

If $|c\tau+d|<1$ then $c^2y^2<1$ , so $|c|<1/y$ , and for each such $c$ , $|cx+d|<1$ leaves at most two values of $d$ . So $|c\tau+d|$ takes only finitely many values below 1, and $\operatorname{Im}\gamma\tau=y/|c\tau+d|^2$ takes only finitely many values above $y$ : a maximum exists. Translate the maximising point into the strip; this keeps the imaginary part. If the result $\tau'$ had $|\tau'|<1$ , then $S\tau'$ would have a larger imaginary part, a contradiction. So $\tau'\in\mathcal D$ .

Forced zeros at ρUsing the stabiliser of order 3

Let $\rho=e^{2\pi i/3}$ and let $f\in\mathcal M_k(\mathrm{SL}_2(\Z))$ with $k$ even and $3\nmid k$ . Show that $f(\rho)=0$ . Which Eisenstein series must vanish at $\rho$ ?

Hint

Find a matrix with $c\tau+d=\tau+1$ that fixes $\rho$ .

Solution

$ST=\begin{pmatrix}0&-1\\1&1\end{pmatrix}$ sends $\tau$ to $-1/(\tau+1)$ , and $\rho+1=e^{\pi i/3}$ , so $-1/(\rho+1)=-e^{-\pi i/3}=e^{2\pi i/3}=\rho$ . The law gives $f(\rho)=(\rho+1)^kf(\rho)=e^{\pi ik/3}f(\rho)$ . For even $k$ with $3\nmid k$ , $e^{\pi ik/3}\ne1$ , so $f(\rho)=0$ . In particular $E_4(\rho)=0$ , $E_8(\rho)=0$ and $E_{10}(\rho)=0$ .

More practice: Diamond–Shurman, Exercises 1.1.1 and 1.1.2.

InteractiveDiamond–Shurman, Ch. 1: Modular Forms, Elliptic Curves, and Modular Curves

Lattices, tori and the j-invariant

A lattice in $\C$ gives a complex torus, and up to scaling every lattice is $\Z\tau\oplus\Z$ for some $\tau$ in the upper half-plane. Two values of $\tau$ give the same torus exactly when they are related by $\mathrm{SL}_2(\Z)$ . So the fundamental domain is also a catalogue of complex tori, and the j-invariant is a coordinate on that catalogue.

The idea

Lattices and tori. A lattice is

\Lambda=\omega_1\Z\oplus\omega_2\Z

with

\omega_1,\omega_2

linearly independent over

\R

. The quotient

\C/\Lambda

is a complex torus: a compact Riemann surface of genus 1 that is also an abelian group.

Classification.

\C/\Lambda\cong\C/\Lambda'

(holomorphically, as groups) exactly when

\Lambda'=m\Lambda

for some

m\in\C^\times

. Every lattice is homothetic to some

\Lambda_\tau=\tau\Z\oplus\Z

with

\tau\in\mathcal H

, and

\Lambda_\tau

,

\Lambda_{\tau'}

are homothetic exactly when

\tau'=\gamma\tau

for some

\gamma\in\mathrm{SL}_2(\Z)

.

The reason is a change of basis. If $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , then $a\tau+b$ and $c\tau+d$ are another basis of $\Lambda_\tau$ (the matrix has determinant 1), so

\Lambda_\tau=\Z(a\tau+b)\oplus\Z(c\tau+d)=(c\tau+d)\,\Lambda_{\gamma\tau}.

Invariants of a lattice. For even

k\ge4

,

G_k(\Lambda)=\sum_{\omega\in\Lambda\smallsetminus0}\omega^{-k}

, so

G_k(m\Lambda)=m^{-k}G_k(\Lambda)

. Put

g_2=60G_4

,

g_3=140G_6

. The Weierstrass function

\wp

of

\Lambda

satisfies

(\wp')^2=4\wp^3-g_2\wp-g_3

, and

z\mapsto(\wp(z),\wp'(z))

identifies

\C/\Lambda

with the cubic curve

y^2=4x^3-g_2x-g_3

(plus a point at infinity). Its discriminant

\Delta=g_2^3-27g_3^2

is never 0.

The j-invariant.

j(\Lambda)=1728\,g_2^3/\Delta

. Scaling

\Lambda

multiplies

g_2^3

and

\Delta

by the same factor

m^{-12}

, so

j

depends only on the torus. As a function of

\tau

,

j(\gamma\tau)=j(\tau)

, and

j

gives a bijection from

\mathrm{SL}_2(\Z)\backslash\mathcal H

onto

\C

. In terms of Eisenstein series (next modules),

g_2(\Lambda_\tau)=\tfrac{4\pi^4}{3}E_4(\tau)

,

g_3(\Lambda_\tau)=\tfrac{8\pi^6}{27}E_6(\tau)

, and

j(\tau)=\frac{1728\,E_4(\tau)^3}{E_4(\tau)^3-E_6(\tau)^2}=\frac1q+744+196884\,q+\cdots.

Explore

Jump to

Drag

\tau

.

\tau

its reduction

\tau'=\gamma\tau\in\mathcal D

The lattice

\Lambda_\tau

. basis

1,\tau

basis

c\tau+d,\ a\tau+b

. Gluing opposite edges of either cell gives the same torus.

Worked example

Compute $j$ for the square lattice $\Lambda_i=\Z i\oplus\Z$ and the hexagonal lattice $\Lambda_\rho$ , $\rho=e^{2\pi i/3}$ , without summing any series. Then find where $j$ is real.

Show that $i\Lambda_i=\Lambda_i$ and use it to compute $G_6(\Lambda_i)$ .

$i(mi+n)=-m+ni$ , so multiplication by $i$ permutes $\Lambda_i$ . Then $G_6(\Lambda_i)=G_6(i\Lambda_i)=i^{-6}G_6(\Lambda_i)=-G_6(\Lambda_i)$ , so $G_6=0$ and $g_3=0$ .
What is $j(i)$ ?

With $g_3=0$ , $\Delta=g_2^3$ (and $g_2\neq0$ because $\Delta\neq0$ ), so $j(i)=1728\,g_2^3/g_2^3=1728$ .
Do the same for $\Lambda_\rho$ . Which invariant vanishes?

$\rho^2=-1-\rho$ , so $\rho\Lambda_\rho=\Lambda_\rho$ . Then $G_4(\Lambda_\rho)=\rho^{-4}G_4(\Lambda_\rho)=\rho^{2}G_4(\Lambda_\rho)$ , and $\rho^2\neq1$ , so $g_2=0$ and $j(\rho)=0$ .
For which $\tau\in\mathcal D$ is $j(\tau)$ real? Try dragging $\tau$ along the edges of $\mathcal D$ above first.

The q-expansion of $j$ has real coefficients, and $-\bar\tau$ has $q$ -parameter $\bar q$ , so $j(-\bar\tau)=\overline{j(\tau)}$ . If $-\bar\tau$ is equivalent to $\tau$ , then $j(\tau)$ is real. That happens on the imaginary axis ( $-\bar\tau=\tau$ ), on the edges $\operatorname{Re}\tau=\pm\tfrac12$ ( $-\bar\tau=\tau\mp1$ ) and on the arc ( $-\bar\tau=-1/\tau$ ). Since $j$ is injective on the interior of $\mathcal D$ , these are the only real values: exactly the lattices with a reflection symmetry.

Check your understanding

Do $\tau$ and $\tau'=\dfrac{\tau}{\tau+1}$ give isomorphic complex tori?

Yes: $\tau'=\gamma\tau$ with $\gamma=\begin{pmatrix}1&0\\1&1\end{pmatrix}\in\mathrm{SL}_2(\Z)$ .
No: the map $\tau\mapsto\tau/(\tau+1)$ is neither a translation nor $S$ .Any element of $\mathrm{SL}_2(\Z)$ works, not only the generators. Check the determinant.
Only if $\tau$ lies in $\mathcal D$ .The classification holds for every $\tau\in\mathcal H$ ; $\mathcal D$ is just a convenient set of representatives.

\det\begin{pmatrix}1&0\\1&1\end{pmatrix}=1

, so

\Lambda_\tau=(\tau+1)\Lambda_{\tau'}

. In the explorer,

\tau

and

\tau'

reduce to the same point of

\mathcal D

and show the same

j

.

Which of these is unchanged when $\Lambda$ is replaced by $2\Lambda$ ?

$g_2(\Lambda)$ $g_2$ is a multiple of $G_4$ , and $G_4(2\Lambda)=2^{-4}G_4(\Lambda)$ .
$\Delta(\Lambda)$ $\Delta$ scales by $2^{-12}$ .
$j(\Lambda)$
The area of a fundamental parallelogramIt is multiplied by 4. The torus is isomorphic, but a lattice is more than its torus.

g_2^3

and

\Delta

both scale by

m^{-12}

, so their ratio is a function of the torus alone. That is the point of

j

.

$\C/\Lambda_\tau$ and $\C/\Lambda_{\gamma\tau}$ are the same torus. Why is $g_2(\Lambda_\tau)$ , as a function of $\tau$ , not $\mathrm{SL}_2(\Z)$ -invariant?

Because $\Lambda_{\gamma\tau}=(c\tau+d)^{-1}\Lambda_\tau$ is a rescaled lattice, and $g_2$ sees the scale: $g_2(\Lambda_{\gamma\tau})=(c\tau+d)^4g_2(\Lambda_\tau)$ .
Because $g_2$ is not holomorphic in $\tau$ .It is holomorphic: it is a multiple of $E_4$ . The issue is how it transforms.
Because the sum defining $G_4$ converges only on $\mathcal D$ .It converges absolutely for every lattice.

This rule

g_2(\gamma\tau)=(c\tau+d)^4g_2(\tau)

is exactly the statement that

g_2

is a modular form of weight 4. Modular forms are functions on lattices that are homogeneous under scaling;

j

has degree 0, so it is invariant.

Practice

The generators act on latticesTranslation and inversion

Show directly that $\Lambda_{\tau+1}=\Lambda_\tau$ and $\Lambda_{-1/\tau}=\tau^{-1}\Lambda_\tau$ . Conclude that $j(\tau+1)=j(\tau)$ and $j(-1/\tau)=j(\tau)$ , and why this proves $j(\gamma\tau)=j(\tau)$ for all $\gamma$ .

Solution

$\Z(\tau+1)\oplus\Z$ contains $\tau=(\tau+1)-1$ , so it equals $\Z\tau\oplus\Z$ . And $\tau\Lambda_{-1/\tau}=\Z(-1)\oplus\Z\tau=\Lambda_\tau$ . Since $j$ depends only on the lattice up to scaling, $j$ is invariant under $T$ and $S$ , and these generate $\mathrm{SL}_2(\Z)$ .

Every real number at least 1728An intermediate value argument

Show that $j(iy)$ is real for $y>0$ , and that every real number $\ge1728$ equals $j(iy)$ for some $y\ge1$ .

Hint

On the imaginary axis $q=e^{-2\pi y}$ is a positive real number. What does the q-expansion do as $y\to\infty$ ?

Solution

$j=1/q+744+\sum_{n\ge1}c_nq^n$ with real $c_n$ , so $j(iy)$ is real and continuous in $y$ . At $y=1$ it is $1728$ , and as $y\to\infty$ , $q\to0^+$ and $j(iy)\to+\infty$ . By the intermediate value theorem every value in $[1728,\infty)$ is taken on $[1,\infty)$ . (The explorer's preset $\tau=2i$ gives $j=287496=66^3$ .)

More practice: Diamond–Shurman, Exercises 1.4.3, 1.4.4 and 1.4.5.

InteractiveDiamond–Shurman, Ch. 2: Modular Curves as Riemann Surfaces

Congruence subgroups, cusps and elliptic points

The forms attached to elliptic curves are modular only for smaller groups, cut out of $\mathrm{SL}_2(\Z)$ by congruences mod $N$ . The quotient of $\mathcal H$ by such a group is a modular curve. To make it a compact Riemann surface you have to understand two kinds of special point: elliptic points, where the group has extra stabilisers, and cusps, the rational points added at the boundary.

The idea

Congruence subgroups. For

N\ge1

,

\Gamma(N)=\Big\{\gamma\equiv\begin{pmatrix}1&0\\0&1\end{pmatrix}\Big\}\subseteq\Gamma_1(N)=\Big\{\gamma\equiv\begin{pmatrix}1&*\\0&1\end{pmatrix}\Big\}\subseteq\Gamma_0(N)=\Big\{\gamma\equiv\begin{pmatrix}*&*\\0&*\end{pmatrix}\Big\},

congruences taken mod $N$ in $\mathrm{SL}_2(\Z)$ . A congruence subgroup is any subgroup containing some $\Gamma(N)$ ; it has finite index. For example $[\mathrm{SL}_2(\Z):\Gamma_0(N)]=N\prod_{p\mid N}(1+1/p)$ .

Modular curves.

Y(\Gamma)=\Gamma\backslash\mathcal H

, and

X(\Gamma)=\Gamma\backslash(\mathcal H\cup\mathbb P^1(\mathbb Q))

, where

\mathbb P^1(\mathbb Q)=\mathbb Q\cup\{\infty\}

. The

\Gamma

-orbits in

\mathbb P^1(\mathbb Q)

are the cusps. With suitable charts

X(\Gamma)

is a compact Riemann surface.

Elliptic points. A point

\tau\in\mathcal H

whose stabiliser in

\pm\Gamma/\{\pm I\}

is nontrivial. It is an

\mathrm{SL}_2(\Z)

-translate of

i

or of

\rho

, with stabiliser of order

h=2

or

3

. A neighbourhood is rotated by

2\pi/h

, so the chart there is "straighten, then raise to the power

h

".

Width of a cusp. If

s=\alpha(\infty)

with

\alpha\in\mathrm{SL}_2(\Z)

, the width

h_s

is the smallest

h>0

with

\pm\alpha T^h\alpha^{-1}\in\pm\Gamma

. Near

s

the chart is

\tau\mapsto e^{2\pi i\alpha^{-1}(\tau)/h_s}

.

Counting with cosets. Write

\mathrm{SL}_2(\Z)=\bigsqcup_{j=1}^{d}\pm\Gamma\gamma_j

. Then the tiles

\gamma_j\mathcal D

together form a fundamental domain for

\Gamma

, and right multiplication permutes the

d

cosets:

cusps correspond to the orbits of $T$ , and the size of each orbit is the width, so the widths add up to $d$ ;
elliptic points of period 2 correspond to cosets fixed by $S$ , and those of period 3 to cosets fixed by $ST$ . Their numbers are written $\varepsilon_2$ , $\varepsilon_3$ , and the number of cusps $\varepsilon_\infty$ .

The explorer builds the domain this way: it starts from $\mathcal D$ , keeps attaching neighbouring tiles $\gamma S\mathcal D$ , $\gamma T^{\pm1}\mathcal D$ , and adds a tile only if its coset is new.

Explore

Group

Level N 6

One tile per coset. Cusps on the real line share a colour when they are

\Gamma

-equivalent. elliptic point of period 2 period 3

Worked example

Work out the index, cusps and elliptic points of $\Gamma_0(2)$ . Then set the explorer to $\Gamma_0(2)$ and compare.

Two matrices lie in the same coset $\Gamma_0(N)\gamma$ exactly when their bottom rows agree in $\mathbb P^1(\Z/N\Z)$ , up to a unit. How many cosets are there for $N=2$ ?

$\mathbb P^1(\Z/2\Z)=\{(0:1),(1:0),(1:1)\}$ , so the index is 3, with representatives $I$ , $S$ , $ST$ . This agrees with $2(1+\tfrac12)=3$ .
Find the cusps and their widths.

A cusp $a/c$ in lowest terms is equivalent to $\infty$ if $c$ is even and to $0$ if $c$ is odd, so there are two. $T\in\Gamma_0(2)$ , so $\infty$ has width 1. The stabiliser of $0$ in $\mathrm{SL}_2(\Z)$ is $\pm\begin{pmatrix}1&0\\k&1\end{pmatrix}$ , which lies in $\Gamma_0(2)$ only for even $k$ : width 2. Check: $1+2=3$ .
Find an elliptic point of period 2.

An element of order 4 in $\mathrm{SL}_2(\Z)$ has trace 0. Try $\gamma=\begin{pmatrix}1&-1\\2&-1\end{pmatrix}\in\Gamma_0(2)$ . Its fixed point solves $2\tau^2-2\tau+1=0$ , so $\tau=(1+i)/2$ .
Why are there no elliptic points of period 3?

They would need an element of trace $\pm1$ . In $\Gamma_0(2)$ , $c$ is even, so $ad\equiv1\pmod 2$ : $a$ and $d$ are odd and $a+d$ is even. So $\varepsilon_3=0$ , $\varepsilon_2=1$ , $\varepsilon_\infty=2$ , and in the next module these give genus $0$ .

Check your understanding

What is the width of the cusp $0$ for $\Gamma_0(N)$ ?

1, like the cusp $\infty$ .Width is not the same at every cusp. Conjugate the stabiliser of $0$ and see which powers land in $\Gamma_0(N)$ .
$N$
$[\mathrm{SL}_2(\Z):\Gamma_0(N)]$ That is the sum of all the widths, over all cusps.

0=S(\infty)

, and

ST^hS^{-1}=\begin{pmatrix}1&0\\-h&1\end{pmatrix}

, which lies in

\Gamma_0(N)

exactly when

N\mid h

. The explorer's note lists the cusp

0

with width

N

for every level you try.

How many cusps does $\Gamma_0(p)$ have, for $p$ prime?

$p+1$ That is the index. Cusps are orbits of $T$ on the cosets, and one orbit can contain many cosets.
2
1 $0$ and $\infty$ are not equivalent: $\Gamma_0(p)$ preserves whether the denominator of a cusp is divisible by $p$ .

\infty

(width 1) and

0

(width

p

), and

1+p

is the index. Any

a/c

with

p\nmid c

is equivalent to

0

, and any with

p\mid c

to

\infty

.

How many elliptic points does $\Gamma_0(4)$ have?

One of period 2, as for $\Gamma_0(2)$ .It would need an element of trace 0 with $4\mid c$ . Then $-a^2\equiv ad-bc=1\pmod 4$ . Is $-1$ a square mod 4?
None.
One of each period.Period 3 needs trace $\pm1$ , but $a$ , $d$ are odd when $4\mid c$ , so the trace is even.

Period 2 would need

a^2\equiv-1\pmod 4

and period 3 an odd trace; both are impossible. A subgroup of

\Gamma_0(2)

can lose elliptic points: here the elliptic point of

\Gamma_0(2)

is not fixed by anything in the smaller group.

Practice

The index of Γ₀(p)Cosets and the projective line

Show that $\Gamma_0(p)\gamma\mapsto(c:d)\in\mathbb P^1(\Z/p\Z)$ , where $(c,d)$ is the bottom row of $\gamma$ , is a well-defined bijection. Deduce $[\mathrm{SL}_2(\Z):\Gamma_0(p)]=p+1$ .

Hint

Compute the bottom row of $h\gamma$ for $h\in\Gamma_0(p)$ . For surjectivity, lift a pair $(c,d)$ to coprime integers.

Solution

If $h=\begin{pmatrix}x&y\\pz&w\end{pmatrix}$ , the bottom row of $h\gamma$ is $pz(a,b)+w(c,d)\equiv w(c,d)$ , and $w$ is a unit mod $p$ . So the map is well defined. If $\gamma$ , $\gamma'$ have proportional bottom rows mod $p$ , then $\gamma'\gamma^{-1}$ has bottom-left entry $c'd-d'c\equiv0$ , so it lies in $\Gamma_0(p)$ : the map is injective. Every $(c:d)$ lifts to a coprime pair, which is the bottom row of some matrix in $\mathrm{SL}_2(\Z)$ . So the index is $|\mathbb P^1(\mathbb F_p)|=p+1$ .

The cusps of Γ₀(6)An invariant and some widths

Show that $\gcd(c,N)$ , for a cusp $a/c$ in lowest terms, is unchanged by $\Gamma_0(N)$ . Use it to show that $\infty,0,\tfrac12,\tfrac13$ are inequivalent for $\Gamma_0(6)$ , and find their widths. Why are there no other cusps?

Hint

For the widths, conjugate $T^h$ by $\alpha=\begin{pmatrix}1&0\\k&1\end{pmatrix}$ , which sends $\infty$ to $1/k$ .

Solution

A matrix $\begin{pmatrix}x&y\\Nz&w\end{pmatrix}$ sends $a/c$ to a fraction (still in lowest terms) with denominator $Nza+wc\equiv wc\pmod N$ , and $w$ is a unit mod $N$ , so $\gcd$ with $N$ is kept. The four cusps have $\gcd(c,6)=6,1,2,3$ . The computation $\alpha T^h\alpha^{-1}=\begin{pmatrix}1-kh&h\\-k^2h&1+kh\end{pmatrix}$ shows the width at $1/k$ is the least $h$ with $6\mid k^2h$ : 6 at $0$ ( $k=1$ ), 3 at $\tfrac12$ , 2 at $\tfrac13$ , and 1 at $\infty$ . The widths add to $1+6+3+2=12=6\cdot\tfrac32\cdot\tfrac43$ , the index, so there is no room for another cusp.

More practice: Diamond–Shurman, Exercises 1.2.3, 2.3.7 and 3.8.3.

CalculatorDiamond–Shurman, Ch. 3: Dimension Formulas

Genus and dimension formulas

Riemann–Hurwitz turns the counts from the last module into the genus of $X(\Gamma)$ , and Riemann–Roch turns the genus into the dimension of every space of modular forms. These formulas tell you, before computing anything, that weight 12 is the first weight with a cusp form for $\mathrm{SL}_2(\Z)$ , and that 11 is the first level with a weight-2 cusp form for $\Gamma_0(N)$ .

The idea

Genus. Let

d=[\mathrm{PSL}_2(\Z):\pm\Gamma/\{\pm I\}]

. The projection

X(\Gamma)\to X(1)

has degree

d

and ramifies only over the images of

i

,

\rho

and

\infty

. Riemann–Hurwitz gives

g\big(X(\Gamma)\big)=1+\frac{d}{12}-\frac{\varepsilon_2}{4}-\frac{\varepsilon_3}{3}-\frac{\varepsilon_\infty}{2}.

Where it comes from: $2g-2=-2d+\sum(e_x-1)$ . Over $i$ the $d-\varepsilon_2$ non-elliptic sheets pair up, contributing $(d-\varepsilon_2)/2$ ; over $\rho$ they come in threes, contributing $2(d-\varepsilon_3)/3$ ; over $\infty$ each cusp of width $h$ contributes $h-1$ , in total $d-\varepsilon_\infty$ .

Dimensions, $k\ge2$ even. With

g,\varepsilon_2,\varepsilon_3,\varepsilon_\infty

as above,

\dim\mathcal M_k(\Gamma)=(k-1)(g-1)+\Big\lfloor\frac k4\Big\rfloor\varepsilon_2+\Big\lfloor\frac k3\Big\rfloor\varepsilon_3+\frac k2\,\varepsilon_\infty,

\dim\mathcal S_k(\Gamma)=(k-1)(g-1)+\Big\lfloor\frac k4\Big\rfloor\varepsilon_2+\Big\lfloor\frac k3\Big\rfloor\varepsilon_3+\Big(\frac k2-1\Big)\varepsilon_\infty\quad(k\ge4),

while in weight 2, $\dim\mathcal S_2(\Gamma)=g$ and $\dim\mathcal M_2(\Gamma)=g+\varepsilon_\infty-1$ . The idea of the proof: $f(\tau)(d\tau)^{k/2}$ is a meromorphic differential on $X(\Gamma)$ , and its order at an elliptic point of period $h$ is fractional, which is where the floors come from. The gap $\dim\mathcal M_k-\dim\mathcal S_k$ is $\varepsilon_\infty$ for $k\ge4$ : one Eisenstein series for each cusp.

Level 1. For

\mathrm{SL}_2(\Z)

,

g=0

and

\varepsilon_2=\varepsilon_3=\varepsilon_\infty=1

, so

\dim\mathcal M_k=\lfloor k/12\rfloor

if

k\equiv2\pmod{12}

and

\lfloor k/12\rfloor+1

otherwise. In fact

\mathcal M(\mathrm{SL}_2(\Z))=\C[E_4,E_6]

.

Odd weights need more care: if $-I\in\Gamma$ there is nothing, and otherwise cusps split into regular and irregular ones. The calculator covers even $k$ .

Explore

Group

Level N 11 Weight k 2

\dim\mathcal S_k(\Gamma)

against the level

N

, at the chosen weight. the chosen level

Worked example

Compute the genus of $X_0(11)$ and the dimensions of $\mathcal S_2(\Gamma_0(11))$ , $\mathcal M_2(\Gamma_0(11))$ and $\mathcal S_4(\Gamma_0(11))$ . For a prime $p$ , $\varepsilon_2(\Gamma_0(p))$ is the number of solutions of $x^2\equiv-1\pmod p$ and $\varepsilon_3(\Gamma_0(p))$ the number of solutions of $x^2-x+1\equiv0\pmod p$ .

Find $d$ , $\varepsilon_2$ , $\varepsilon_3$ and $\varepsilon_\infty$ .

$d=12$ . Since $11\equiv3\pmod4$ , $-1$ is not a square mod 11: $\varepsilon_2=0$ . Since $11\equiv2\pmod3$ , $x^2-x+1$ has no roots mod 11: $\varepsilon_3=0$ . And $\varepsilon_\infty=2$ .
Compute the genus.

$g=1+\frac{12}{12}-0-0-\frac22=1.$
$X_0(11)$ is a torus. Module 7 shows which one.
Now the weight-2 dimensions.

$\dim\mathcal S_2=g=1$ and $\dim\mathcal M_2=g+\varepsilon_\infty-1=2$ : one cusp form and one Eisenstein series.
And $\dim\mathcal S_4(\Gamma_0(11))$ ?

$(4-1)(1-1)+\lfloor 4/4\rfloor\cdot0+\lfloor4/3\rfloor\cdot0+(2-1)\cdot2=2$ . Check it in the calculator with $N=11$ , $k=4$ .

Check your understanding

For which of these levels is $\mathcal S_2(\Gamma_0(N))=0$ ?

$N=11$ The worked example gives genus 1.
$N=13$
$N=14$ $d=24$ , $\varepsilon_2=\varepsilon_3=0$ , $\varepsilon_\infty=4$ : genus 1.
$N=15$ $d=24$ , $\varepsilon_2=\varepsilon_3=0$ , $\varepsilon_\infty=4$ : genus 1.

For

N=13

:

d=14

, and

13\equiv1

mod both 4 and 3, so

\varepsilon_2=\varepsilon_3=2

,

\varepsilon_\infty=2

. Then

g=1+\tfrac{14}{12}-\tfrac24-\tfrac23-1=0

. Primes with many elliptic points have smaller genus.

$\Gamma'\subseteq\Gamma$ are congruence subgroups. Which statement always holds?

$g(X(\Gamma'))=[\Gamma:\Gamma']\cdot g(X(\Gamma))$ That ignores ramification and the $-2$ in $2g-2$ . For $\Gamma_0(11)\subseteq\mathrm{SL}_2(\Z)$ it would give $12\cdot0=0$ , not 1.
$g(X(\Gamma'))\ge g(X(\Gamma))$
$g(X(\Gamma'))\le g(X(\Gamma))$ A smaller group gives a larger curve, which covers the other one.

X(\Gamma')\to X(\Gamma)

is a holomorphic map of degree

n\ge1

, so Riemann–Hurwitz gives

2g'-2\ge n(2g-2)

. If

g\ge1

this is at least

2g-2

; if

g=0

there is nothing to prove.

For $k\ge4$ even, $\dim\mathcal M_k(\mathrm{SL}_2(\Z))-\dim\mathcal S_k(\mathrm{SL}_2(\Z))=1$ . Why 1?

Because $\mathrm{SL}_2(\Z)$ has one cusp, and the complement of the cusp forms has one Eisenstein series per cusp.
Because $\dim\mathcal M_k(\mathrm{SL}_2(\Z))=1$ for all $k\ge4$ . $\dim\mathcal M_{12}=2$ and $\dim\mathcal M_{24}=3$ .
Because $\Delta$ is the only cusp form.There are more: $\Delta E_4$ has weight 16, and $\mathcal S_{24}$ is 2-dimensional. The difference counts the non-cusp part.

A form is a cusp form when it vanishes at every cusp, which is one linear condition per cusp; the conditions are independent once

k\ge4

. Try

\Gamma_0(N)

in the calculator:

\dim\mathcal M_k-\dim\mathcal S_k

equals the number of cusps.

Practice

Level one, every weightFrom the general formula

Use the formula for $\dim\mathcal M_k$ with $g=0$ , $\varepsilon_2=\varepsilon_3=\varepsilon_\infty=1$ to prove the level-one rule: $\lfloor k/12\rfloor$ if $k\equiv2\pmod{12}$ , and $\lfloor k/12\rfloor+1$ otherwise ( $k\ge2$ even).

Hint

Write $k=12m+r$ with $r\in\{0,2,4,6,8,10\}$ and evaluate the floors.

Solution

The formula reads $1-k+\lfloor k/4\rfloor+\lfloor k/3\rfloor+k/2$ . With $k=12m+r$ the floors are $3m+\lfloor r/4\rfloor$ and $4m+\lfloor r/3\rfloor$ , and $1-12m-r+3m+4m+6m=1+m-r$ . So the dimension is $m+1-r+\lfloor r/4\rfloor+\lfloor r/3\rfloor+r/2$ . For $r=0,2,4,6,8,10$ the correction $-r+\lfloor r/4\rfloor+\lfloor r/3\rfloor+r/2$ is $0,-1,0,0,0,0$ . So the answer is $m+1$ , except $m$ when $r=2$ .

The genus of X₀(37)A curve of genus 2

Compute $g(X_0(37))$ and $\dim\mathcal S_2(\Gamma_0(37))$ .

Solution

$d=38$ . Since $37\equiv1\pmod4$ and $37\equiv1\pmod3$ , $\varepsilon_2=\varepsilon_3=2$ ; $\varepsilon_\infty=2$ . So $g=1+\tfrac{38}{12}-\tfrac12-\tfrac23-1=\tfrac{38-6-8}{12}=2$ , and $\dim\mathcal S_2(\Gamma_0(37))=2$ . Module 7 returns to this example.

More practice: Diamond–Shurman, Exercises 3.1.4, 3.5.3 and 3.9.5.

Coefficient labDiamond–Shurman, Ch. 4: Eisenstein Series

Eisenstein series and q-expansions

The first modular forms anyone writes down are Eisenstein series: sum $(c\tau+d)^{-k}$ over all pairs of integers. Their q-expansions have divisor sums as coefficients. Because the spaces $\mathcal M_k$ are so small, matching two q-expansions proves identities between divisor sums that are hard to see directly.

The idea

Eisenstein series. For even

k\ge4

,

G_k(\tau)=\sum_{(c,d)\neq(0,0)}(c\tau+d)^{-k}

converges absolutely and lies in

\mathcal M_k(\mathrm{SL}_2(\Z))

. Its q-expansion is

G_k=2\zeta(k)E_k

, where

E_k(\tau)=1-\frac{2k}{B_k}\sum_{n\ge1}\sigma_{k-1}(n)\,q^n,\qquad \sigma_{k-1}(n)=\sum_{d\mid n}d^{k-1},

and the Bernoulli numbers come from $t/(e^t-1)=\sum B_kt^k/k!$ . So $E_4=1+240\sum\sigma_3(n)q^n$ , $E_6=1-504\sum\sigma_5(n)q^n$ , $E_8=1+480\sum\sigma_7(n)q^n$ , and $E_{12}=1+\tfrac{65520}{691}\sum\sigma_{11}(n)q^n$ .

The discriminant.

\Delta=(E_4^3-E_6^2)/1728=q\prod_{n\ge1}(1-q^n)^{24}=\sum\tau(n)q^n

is a cusp form of weight 12; it spans

\mathcal S_{12}(\mathrm{SL}_2(\Z))

.

Weight 2.

E_2=1-24\sum\sigma_1(n)q^n

is holomorphic and periodic but not modular:

E_2(-1/\tau)=\tau^2E_2(\tau)+\dfrac{12\tau}{2\pi i}

. The defining double sum does not converge absolutely, which is why the usual argument fails. Combinations such as

E_2(\tau)-NE_2(N\tau)

are modular of weight 2 for

\Gamma_0(N)

.

Four squares.

\theta(\tau)=\sum_{m\in\Z}q^{m^2}

satisfies

\theta^4\in\mathcal M_2(\Gamma_0(4))

, a space of dimension 2 spanned by Eisenstein series. Comparing two coefficients gives

\theta^4=\tfrac13\big(4E_2(4\tau)-E_2(\tau)\big)

, so the number of ways to write

n

as a sum of four squares is

r_4(n)=8\sum_{d\mid n,\ 4\nmid d}d.

The explorer computes these series exactly, with integer arithmetic, and checks each identity coefficient by coefficient. The log–log plot shows a second fact: Eisenstein coefficients grow like $n^{k-1}$ , while cusp form coefficients grow much more slowly, like $n^{(k-1)/2}$ up to small factors (Deligne's bound).

Explore

Series

Coefficients plotted 60

\log|a_n|

against

\log n

.

|a_n|

slope

k-1

slope

(k-1)/2

, both through

n=1

Worked example

Prove that for every $n\ge1$ ,

\sigma_7(n)=\sigma_3(n)+120\sum_{m=1}^{n-1}\sigma_3(m)\,\sigma_3(n-m).

Why must $E_4^2=E_8$ ?

Both lie in $\mathcal M_8(\mathrm{SL}_2(\Z))$ , which has dimension $\lfloor8/12\rfloor+1=1$ . So $E_4^2=cE_8$ , and comparing constant terms (both 1) gives $c=1$ .
Check the coefficient of $q$ on both sides.

$E_4^2=1+2\cdot240q+\cdots=1+480q+\cdots$ , and $E_8=1+480\sigma_7(1)q+\cdots$ . They agree.
Write down the coefficient of $q^n$ on each side and solve for $\sigma_7(n)$ .

On the left: $2\cdot240\,\sigma_3(n)+240^2\sum_{m=1}^{n-1}\sigma_3(m)\sigma_3(n-m)$ . On the right: $480\,\sigma_7(n)$ . Divide by 480 to get the identity.
Test it at $n=2$ .

$\sigma_7(2)=1+128=129$ , and $\sigma_3(2)+120\,\sigma_3(1)^2=9+120=129$ . The explorer's "E₄² = E₈?" check does this for every $n\le120$ .

Check your understanding

Why can $E_2=1-24\sum\sigma_1(n)q^n$ not be a modular form for $\mathrm{SL}_2(\Z)$ ?

$\mathcal M_2(\mathrm{SL}_2(\Z))=0$ , but $E_2$ has constant term 1.
Its coefficients are negative. $E_6$ has negative coefficients and is a modular form.
It is not holomorphic on $\mathcal H$ .The q-series converges for $|q|<1$ , so it is holomorphic. The problem is the transformation law.

The dimension formula gives

\dim\mathcal M_2=0

for level 1, so a nonzero weight-2 form is impossible. The correction term

12\tau/(2\pi i)

in its transformation law records exactly how it fails.

Write $f\in\mathcal M_{12}(\mathrm{SL}_2(\Z))$ as $f=cE_{12}+c'\Delta$ . You know that $a_n(f)=O(n^6)$ . What is $c$ ?

$c=0$
$c=1$ Then $a_n(f)$ would include $\tfrac{65520}{691}\sigma_{11}(n)\ge\tfrac{65520}{691}n^{11}$ , far bigger than $n^6$ .
It cannot be determined from growth alone.Compare the growth of the two pieces: $n^{11}$ against roughly $n^{11/2}$ .

\sigma_{11}(n)\ge n^{11}

, while

|\tau(n)|

grows like

n^{11/2}

up to a divisor-count factor. Any nonzero Eisenstein component would dominate. Compare E₄ and Δ in the plot: one hugs the slope

k-1

line, the other the slope

(k-1)/2

line.

Why is $r_4(n)$ given exactly by a divisor sum, with no "error term"?

Because $\theta^4$ is a modular form for $\mathrm{SL}_2(\Z)$ . $\mathcal M_2(\mathrm{SL}_2(\Z))=0$ . $\theta^4$ is modular only for the smaller group $\Gamma_0(4)$ .
Because $X_0(4)$ has genus 0, so $\mathcal S_2(\Gamma_0(4))=0$ and $\mathcal M_2(\Gamma_0(4))$ consists of Eisenstein series.
Because every modular form of weight 2 is an Eisenstein series. $\mathcal S_2(\Gamma_0(11))$ is 1-dimensional, so weight 2 has cusp forms in general.

\dim\mathcal M_2(\Gamma_0(4))=g+\varepsilon_\infty-1=0+3-1=2

, all Eisenstein. For eight squares a similar argument works; for many quadratic forms in more variables, a cusp form appears and the divisor-sum formula is only the main term.

Practice

τ(2) from E₄ and E₆Two coefficients by hand

Using $E_4=1+240q+2160q^2+\cdots$ and $E_6=1-504q-16632q^2+\cdots$ , compute the first two coefficients of $(E_4^3-E_6^2)/1728$ .

Solution

Coefficient of $q$ : $3\cdot240-2\cdot(-504)=720+1008=1728$ , giving $\tau(1)=1$ . Coefficient of $q^2$ : for $E_4^3$ , $3\cdot2160+3\cdot240^2=179280$ ; for $E_6^2$ , $2\cdot(-16632)+504^2=220752$ . The difference is $-41472=-24\cdot1728$ , so $\tau(2)=-24$ .

Four squares at a primeJacobi's formula

Show that $r_4(p)=8(p+1)$ for an odd prime $p$ , and check it by listing the representations of $3$ .

Solution

The divisors of $p$ are $1$ and $p$ , neither divisible by 4, so $r_4(p)=8(1+p)$ . For $p=3$ : $3=(\pm1)^2+(\pm1)^2+(\pm1)^2+0^2$ , with 4 places for the zero and $2^3$ choices of signs, giving $32=8\cdot4$ .

More practice: Diamond–Shurman, Exercises 1.1.7 and 1.2.8.

Coefficient labDiamond–Shurman, Ch. 5: Hecke Operators

Hecke operators and eigenforms

Hecke operators are linear maps on each space of modular forms. On q-expansions they act by a simple rule, they commute with one another, and their common eigenvectors have multiplicative coefficients. This is why $\tau(mn)=\tau(m)\tau(n)$ whenever $m$ and $n$ are coprime.

The idea

$T_p$ . For

f\in\mathcal M_k(\Gamma_0(N))

and a prime

p\nmid N

,

(T_pf)(\tau)=\frac1p\sum_{j=0}^{p-1}f\Big(\frac{\tau+j}{p}\Big)+p^{k-1}f(p\tau),\qquad a_n(T_pf)=a_{pn}(f)+p^{k-1}a_{n/p}(f),

where $a_{n/p}=0$ if $p\nmid n$ . The $p+1$ points $(\tau+j)/p$ and $p\tau$ correspond to the $p+1$ sublattices of index $p$ in $\Lambda_\tau$ , so $T_p$ is an average over sublattices. For $p\mid N$ only the first part survives: $a_n(U_pf)=a_{pn}(f)$ .

Eigenforms. The

T_p

commute, and for

p\nmid N

they are self-adjoint for the Petersson inner product

\langle f,g\rangle=\int_{\Gamma_0(N)\backslash\mathcal H}f\bar g\,y^k\,\tfrac{dx\,dy}{y^2}

. So

\mathcal S_k(\Gamma_0(N))

has a basis of simultaneous eigenforms. If

f

is an eigenform for all

T_n

with

a_1=1

, then

T_pf=a_pf

,

a_{mn}=a_ma_n

for

\gcd(m,n)=1

, and

a_{p^{r+1}}=a_pa_{p^r}-p^{k-1}a_{p^{r-1}}

for

p\nmid N

.

L-functions. For such

f

,

L(f,s)=\sum a_nn^{-s}=\prod_{p\mid N}(1-a_pp^{-s})^{-1}\prod_{p\nmid N}(1-a_pp^{-s}+p^{k-1-2s})^{-1}

. For

f\in\mathcal S_k(\mathrm{SL}_2(\Z))

,

\Lambda(f,s)=(2\pi)^{-s}\Gamma(s)L(f,s)

continues to all of

\C

and satisfies

\Lambda(f,k-s)=(-1)^{k/2}\Lambda(f,s)

.

Old and new. If

g\in\mathcal S_k(\Gamma_0(M))

and

dM\mid N

, then

g(d\tau)\in\mathcal S_k(\Gamma_0(N))

; these span the oldforms. Normalised eigenforms in their orthogonal complement are newforms, and a newform is determined by its

a_p

for all but finitely many

p

.

Explore

Form f

Prime p

The ratio

a_n(T_pf)/a_n(f)

for

1\le n\le30

(skipping

a_n(f)=0

). An eigenform gives a flat line at its eigenvalue.

Worked example

Given only $\tau(1)=1$ , $\tau(2)=-24$ , $\tau(3)=252$ and the fact that $\Delta$ spans $\mathcal S_{12}(\mathrm{SL}_2(\Z))$ , compute $\tau(4)$ and $\tau(6)$ .

Why is $\Delta$ an eigenform of $T_2$ , and what is the eigenvalue?

$T_2$ maps $\mathcal S_{12}$ to itself and that space is 1-dimensional, so $T_2\Delta=\lambda\Delta$ . The coefficient of $q$ gives $\lambda=a_1(T_2\Delta)=\tau(2)+0=-24$ .
Compare coefficients of $q^2$ in $T_2\Delta=-24\Delta$ .

$a_2(T_2\Delta)=\tau(4)+2^{11}\tau(1)$ , and this equals $-24\tau(2)=576$ . So $\tau(4)=576-2048=-1472$ .
Now compare coefficients of $q^3$ .

$a_3(T_2\Delta)=\tau(6)+0$ (since $2\nmid3$ ), and this equals $-24\tau(3)$ . So $\tau(6)=-24\cdot252=-6048=\tau(2)\tau(3)$ . Multiplicativity appears without being assumed.
Confirm in the explorer: choose Δ and $p=2$ . What does the table show?

Every row has ratio exactly $-24$ : $a_n(T_2\Delta)=-24\,\tau(n)$ for each $n$ listed, including $n=2$ and $n=3$ used above.

Check your understanding

$f=\sum a_nq^n$ is a cusp form with $a_1=1$ and $T_pf=\lambda f$ . What is $\lambda$ ?

$p^{k-1}$ That is the factor in front of $f(p\tau)$ , not the eigenvalue.
$a_p$
$a_p+p^{k-1}$ The second term contributes $p^{k-1}a_{1/p}$ , and $a_{1/p}=0$ .

Compare the coefficient of

q

:

a_1(T_pf)=a_p+p^{k-1}a_{1/p}=a_p

, while

a_1(\lambda f)=\lambda

. This is why the coefficients of a normalised eigenform are its eigenvalues.

$E_4$ is an eigenform for every $T_p$ . Is $E_4^3$ ?

Yes: a product of eigenforms is an eigenform. $T_p$ is linear, not multiplicative. Choose E₄³ in the explorer and look at the ratios.
No: $E_4^3=E_{12}+\tfrac{432000}{691}\Delta$ , a sum of two eigenforms with different eigenvalues.
Yes, with eigenvalue $(1+p^3)^3$ .For $p=2$ that is 729, but the ratio at $n=1$ is 249.

T_pE_{12}=(1+p^{11})E_{12}

and

T_p\Delta=\tau(p)\Delta

, and

1+p^{11}\ne\tau(p)

. The ratios in the explorer drift toward

1+p^{11}

as

n

grows, because the Eisenstein part dominates large coefficients.

$f$ is the newform in $\mathcal S_2(\Gamma_0(11))$ and $g(\tau)=f(2\tau)\in\mathcal S_2(\Gamma_0(22))$ . Is $g$ a newform of level 22?

Yes, because it has level 22.Being modular for $\Gamma_0(22)$ is not enough; newforms must be orthogonal to everything coming from lower levels.
No, it is an oldform: it comes from level 11. (Also $a_1(g)=0$ , so it cannot be normalised.)
It is not even in $\mathcal S_2(\Gamma_0(22))$ .If $\gamma\in\Gamma_0(22)$ then conjugating by $\mathrm{diag}(2,1)$ gives an element of $\Gamma_0(11)$ , so $f(2\tau)$ is modular for $\Gamma_0(22)$ .

f(\tau)

and

f(2\tau)

span a 2-dimensional space of oldforms in

\mathcal S_2(\Gamma_0(22))

, which has dimension

g(X_0(22))=2

. So level 22 has no newforms of weight 2 at all.

Practice

τ(8)The prime-power recursion

Use $a_{p^{r+1}}=a_pa_{p^r}-p^{k-1}a_{p^{r-1}}$ to compute $\tau(8)$ from $\tau(2)=-24$ and $\tau(4)=-1472$ .

Solution

$\tau(8)=\tau(2)\tau(4)-2^{11}\tau(2)=(-24)(-1472)+2048\cdot24=35328+49152=84480$ . Check it with the explorer's $T_2$ table: the row $n=4$ needs $\tau(8)$ .

Eisenstein series are eigenformsA divisor-sum identity

Show that $T_pE_k=\sigma_{k-1}(p)E_k=(1+p^{k-1})E_k$ for every prime $p$ .

Hint

You need $\sigma(pn)+p^{k-1}\sigma(n/p)=(1+p^{k-1})\sigma(n)$ with $\sigma=\sigma_{k-1}$ . Write $n=p^rm$ with $p\nmid m$ and use multiplicativity of $\sigma$ .

Solution

Constant term: $a_0(T_pE_k)=a_0+p^{k-1}a_0=(1+p^{k-1})a_0$ . For $n\ge1$ , put $x=p^{k-1}$ , so $\sigma(p^r)=1+x+\cdots+x^r$ . The identity becomes $\sigma(m)\big[(1+\cdots+x^{r+1})+x(1+\cdots+x^{r-1})\big]=\sigma(m)(1+x)(1+\cdots+x^r)$ , which is true term by term (for $r=0$ the second bracket is empty). Every coefficient of $E_k$ is a multiple of $\sigma_{k-1}(n)$ , so $T_pE_k=(1+p^{k-1})E_k$ .

More practice: Diamond–Shurman, Exercises 5.3.3, 5.8.3 and 5.9.3.

InteractiveDiamond–Shurman, Ch. 6: Jacobians and Abelian Varieties

Periods, Jacobians and an elliptic curve

A weight-2 cusp form is the same thing as a holomorphic differential on the modular curve. Integrating it along closed loops gives a lattice of periods, and $\C$ modulo that lattice is an elliptic curve. At level 11 you can watch it happen: the periods of a single cusp form land on a lattice whose j-invariant is that of $y^2+y=x^3-x^2-10x-20$ .

The idea

Forms are differentials. For

f\in\mathcal S_2(\Gamma)

and

\gamma\in\Gamma

,

d(\gamma\tau)=d\tau/(c\tau+d)^2

, so

f(\gamma\tau)\,d(\gamma\tau)=f(\tau)\,d\tau

. The differential

f(\tau)\,d\tau

descends to

X(\Gamma)

, and it is holomorphic there exactly because

f

is a cusp form. This gives

\mathcal S_2(\Gamma)\cong\Omega^1_{\mathrm{hol}}(X(\Gamma))

, of dimension

g

.

Jacobian. For a compact Riemann surface

X

of genus

g

, integration over loops embeds

H_1(X,\Z)

as a lattice in

\Omega^1_{\mathrm{hol}}(X)^\wedge\cong\C^g

, and

\mathrm{Jac}(X)=\C^g/H_1(X,\Z)

is a complex torus of dimension

g

. For

X_0(N)

this is

J_0(N)

, and Hecke operators act on it.

The period map. For

f=\sum a_nq^n\in\mathcal S_2(\Gamma_0(N))

put

\varphi(\tau)=2\pi i\int_{i\infty}^{\tau}f(z)\,dz=\sum_{n\ge1}\frac{a_n}{n}q^n

. Then

P(\gamma)=\varphi(\gamma\tau)-\varphi(\tau),\qquad\gamma\in\Gamma_0(N),

does not depend on $\tau$ , and $P:\Gamma_0(N)\to\C$ is a homomorphism that vanishes on parabolic elements and on elements of finite order. Its image $\Lambda_f$ is the set of periods of $2\pi i\,f(\tau)\,d\tau$ .

Elliptic curves from forms. If

f

is a newform with rational coefficients,

\Lambda_f

is a lattice and

\tau\mapsto\varphi(\tau)\bmod\Lambda_f

defines a surjective holomorphic map

X_0(N)\to\C/\Lambda_f

. More generally each newform

f

gives an abelian variety

A_f

, a quotient of

J_0(N)

of dimension equal to the degree of the field generated by its coefficients. Modularity, in the form stated in this chapter: for every complex elliptic curve

E

with

j(E)\in\mathbb Q

there are an

N

and a surjective holomorphic map

X_0(N)\to E

.

At level 11 the genus is 1, so $J_0(11)$ is $X_0(11)$ itself, and $\mathcal S_2(\Gamma_0(11))$ is spanned by $f=q\prod_{n\ge1}(1-q^n)^2(1-q^{11n})^2=q-2q^2-q^3+2q^4+q^5+\cdots$ .

Explore

Pick

\gamma\in\Gamma_0(11)

by its bottom row

(c,d)

; each pick adds its period

Near the real axis: the tiles of a domain for

\Gamma_0(11)

. Drag

\tau

;

P(\gamma)

should not change.

\tau

\gamma\tau

Periods found so far, in

\C

. periods a reduced basis of the lattice they generate

Worked example

Justify the claims about the period map for $f\in\mathcal S_2(\Gamma_0(11))$ , the ones the explorer lets you test.

Show that $P(\gamma)=\varphi(\gamma\tau)-\varphi(\tau)$ does not depend on $\tau$ .

Differentiate: $\frac{d}{d\tau}\varphi(\gamma\tau)=2\pi i\,f(\gamma\tau)\,(c\tau+d)^{-2}=2\pi i\,f(\tau)=\varphi'(\tau)$ , using the weight-2 law. So the difference has derivative 0 on the connected set $\mathcal H$ .
Show that $P$ is a homomorphism.

$P(\gamma\delta)=\big[\varphi(\gamma(\delta\tau))-\varphi(\delta\tau)\big]+\big[\varphi(\delta\tau)-\varphi(\tau)\big]=P(\gamma)+P(\delta)$ , evaluating the first bracket at the point $\delta\tau$ .
Why is $P(\gamma)=0$ for $\gamma=\begin{pmatrix}1&0\\11&1\end{pmatrix}$ ?

$\gamma$ is parabolic and fixes the cusp $0$ . Because $f$ is a cusp form, $\varphi$ extends continuously to the cusps, so letting $\tau\to0$ gives $P(\gamma)=\varphi(0)-\varphi(0)=0$ . The first button in the explorer confirms it numerically.
Choose $(c,d)=(11,2)$ and then $(11,4)$ in the explorer. What do the two periods generate, and what is $j$ ?

The periods are $\omega_1\approx-1.26921$ and $\omega_2\approx0.63460+1.45882i$ (up to sign). They are independent, the lattice they generate already contains every other period you add, and $j(\Z\omega_1\oplus\Z\omega_2)\approx-757.6726$ , which matches $j=-122023936/161051$ of $y^2+y=x^3-x^2-10x-20$ .

Check your understanding

Which $\gamma\in\Gamma_0(11)$ certainly has $P(\gamma)=0$ , without computing?

$\begin{pmatrix}1&1\\0&1\end{pmatrix}$
$\begin{pmatrix}6&1\\11&2\end{pmatrix}$ Its trace is 8, so it is hyperbolic, neither parabolic nor of finite order. The explorer shows its period is a basis vector of the lattice.
Every $\gamma$ , since $\varphi$ is $\Gamma_0(11)$ -invariant. $\varphi$ is only invariant up to periods; that is the whole point of $P$ .

\varphi(\tau+1)=\varphi(\tau)

because

\varphi

is a power series in

q

. In general

P

kills parabolic elements, elements of finite order and commutators, so it factors through

H_1(X_0(11),\Z)

.

What is the dimension of $J_0(23)$ ?

1That would need genus 1. Compute $g(X_0(23))$ with $d=24$ , $\varepsilon_2=\varepsilon_3=0$ , $\varepsilon_\infty=2$ .
2
23The dimension is the genus, not the level.

g=1+2-0-0-1=2

, so

J_0(23)

is a 2-dimensional complex torus, isomorphic to

\C^2

modulo the periods of the two-dimensional space

\mathcal S_2(\Gamma_0(23))

.

The periods you have found generate a lattice $\Lambda'\subsetneq\Lambda_f$ of index 4. How is $\C/\Lambda'$ related to $E=\C/\Lambda_f$ ?

They are isomorphic, since the lattices are commensurable.Isomorphic needs $\Lambda'=m\Lambda_f$ . Commensurable lattices usually have different $j$ ; the explorer shows this when you pick only $(33,-7)$ and $(33,-10)$ .
$z\mapsto z$ induces a surjective homomorphism $\C/\Lambda'\to E$ with kernel $\Lambda_f/\Lambda'$ of order 4: an isogeny of degree 4.
There is no holomorphic map between them.The identity of $\C$ descends, because $\Lambda'\subseteq\Lambda_f$ .

Isogenous curves need not be isomorphic. To identify

E

you need all of

\Lambda_f

, which is why the explorer keeps adding periods until the generated lattice stops growing.

Practice

What the period map forgetsTorsion and commutators

Show that $P(\gamma)=0$ if $\gamma$ has finite order, and that $P(\gamma\delta\gamma^{-1}\delta^{-1})=0$ for all $\gamma,\delta$ .

Solution

$P$ is a homomorphism to the torsion-free abelian group $\C$ . If $\gamma^n=\pm I$ then $nP(\gamma)=P(\pm I)=0$ ( $\pm I$ acts trivially), so $P(\gamma)=0$ . And $P(\gamma\delta\gamma^{-1}\delta^{-1})=P(\gamma)+P(\delta)-P(\gamma)-P(\delta)=0$ .

J₀(37) up to isogenyCounting newforms

$\mathcal S_2(\Gamma_0(37))$ is 2-dimensional (module 4) and is spanned by two newforms with rational coefficients, with $a_2=-2$ and $a_2=0$ . What does this say about $J_0(37)$ ?

Solution

Each newform $f_i$ gives a 1-dimensional quotient $A_{f_i}=E_i$ , an elliptic curve, and $J_0(37)$ is isogenous to $E_1\times E_2$ . The two curves are not isogenous to each other: isogenous curves have the same $a_p$ , and here $a_2$ differs.

More practice: Diamond–Shurman, Exercises 6.1.3 and 6.6.3.

Animated lessonDiamond–Shurman, Ch. 7: Modular Curves as Algebraic Curves

Elliptic curves as cubics: the group law

Chapter 7 moves from complex analysis to algebra. An elliptic curve becomes a cubic equation, its group law becomes the geometry of lines, and modular curves turn out to be algebraic curves with equations over $\mathbb Q$ . The animation shows the group law; you can drag both points and change the curve.

The idea

Weierstrass equation. Over a field of characteristic not 2 or 3,

E: y^2=x^3+ax+b

with

4a^3+27b^2\neq0

(so the curve has no singular point), together with one point

O

at infinity, where all vertical lines meet.

Group law. Three points on a line add to zero:

P+Q+R=O

. So

P+Q

is the reflection of the third intersection point of the line

PQ

, and

-(x,y)=(x,-y)

. In coordinates, with

\lambda

the slope of the chord (or tangent, if

P=Q

),

\lambda=\frac{y_Q-y_P}{x_Q-x_P}\ \Big(\text{or }\frac{3x_P^2+a}{2y_P}\Big),\qquad x_{P+Q}=\lambda^2-x_P-x_Q,\qquad y_{P+Q}=-\big(y_P+\lambda(x_{P+Q}-x_P)\big).

Over $\C$ nothing new happens: for $E=\C/\Lambda$ , the map $z\mapsto(\wp(z),\wp'(z))$ is a group isomorphism onto $y^2=4x^3-g_2x-g_3$ , so the chord rule is just addition in $\C/\Lambda$ . The point of the algebraic version is that it makes sense over $\mathbb Q$ and over $\mathbb F_p$ , which the next two modules need.

Modular curves over $\mathbb Q$ . The function field of

X_0(N)

is

\C(j,j_N)

with

j_N(\tau)=j(N\tau)

, and

j_N

satisfies a polynomial

\Phi_N(j,j_N)=0

with integer coefficients (the modular polynomial). So

X_0(N)

has a model over

\mathbb Q

, and so does the map

X_0(N)\to E

from the last module. Hecke operators and isogenies have algebraic descriptions too.

Watch

a −1.0 b 1.0

y^2=x^3+ax+b

. Drag P and Q along the curve. line third point

P+Q

2P

The curve and two points $P$ , $Q$ on it.
Draw the line through $P$ and $Q$ .
A cubic meets a line three times: the third point is $R$ , and $P+Q+R=O$ .
Reflect $R$ in the x-axis. That is $P+Q$ .
The same rule with the tangent at $P$ gives $2P$ . If the line is vertical, the third point is $O$ .

Worked example

On $E: y^2=x^3-x+1$ (the default curve above), take $P=(0,1)$ and $Q=(1,1)$ . Compute $P+Q$ and $2P$ .

Find the line through $P$ and $Q$ and all its intersections with $E$ .

The line is $y=1$ . Substituting, $x^3-x+1=1$ , so $x^3-x=0$ and $x\in\{0,1,-1\}$ . The third point is $R=(-1,1)$ .
What is $P+Q$ ? Check it against the formula.

$P+Q=-R=(-1,-1)$ . With $\lambda=0$ : $x=0-0-1=-1$ and $y=-(1+0)=-1$ .
Compute $2P$ with the tangent at $P$ .

$\lambda=(3\cdot0^2-1)/(2\cdot1)=-\tfrac12$ . Then $x_{2P}=\tfrac14-0=\tfrac14$ and $y_{2P}=-\big(1-\tfrac12\cdot\tfrac14\big)=-\tfrac78$ .
Verify that $2P$ lies on $E$ .

$\big(\tfrac14\big)^3-\tfrac14+1=\tfrac{1-16+64}{64}=\tfrac{49}{64}=\big(\tfrac78\big)^2$ . The coordinates stay rational: $E(\mathbb Q)$ is a group, and the chord rule is how you compute in it.

Check your understanding

Why is $P+Q$ defined as the reflection of the third point $R$ , rather than $R$ itself?

To keep the y-coordinate positive. $P+Q$ can have either sign of $y$ . The reason is structural.
The operation $(P,Q)\mapsto R$ has no identity element; with $O$ as identity and $P+Q+R=O$ , the reflection is what makes a group.
Because the line through $P$ and $Q$ meets $E$ twice more.A line meets a cubic in three points counted with multiplicity: $P$ , $Q$ and exactly one more.

If

P*Q

denotes the third point, then

P*O=-P

, not

P

. Defining

P+Q=O*(P*Q)

fixes this, and associativity is then a real theorem (one proof: over

\C

it is addition in

\C/\Lambda

).

Which points $P\ne O$ on $y^2=x^3+ax+b$ satisfy $2P=O$ ?

Exactly the points with $y=0$ .
Points where the tangent is horizontal.A horizontal tangent meets $E$ again at a finite point. You need the tangent to pass through $O$ .
None: only $O$ has order dividing 2.Over $\C$ , $E[2]\cong(\Z/2)^2$ has three nonzero points.

2P=O

means

P=-P

, so

(x,y)=(x,-y)

and

y=0

: the roots of the cubic. Over

\mathbb R

you can see one or three of them, depending on the sign of

4a^3+27b^2

.

Set $a=b=0$ in the explorer. Why is $y^2=x^3$ not an elliptic curve?

Because it has no rational points. $(1,1)$ and $(4,8)$ are on it.
$4a^3+27b^2=0$ : the curve is singular at $(0,0)$ , where both partial derivatives vanish and there is no tangent line.
Because it is symmetric about the x-axis.Every curve $y^2=x^3+ax+b$ is.

At a singular point the chord-tangent rule breaks down. The same happens on reductions mod

p

for the finitely many primes dividing the discriminant, the bad primes of the next module.

Practice

The 2-torsionThree roots, one group

Let $E:y^2=(x-e_1)(x-e_2)(x-e_3)$ with distinct $e_i$ . Show that $\{O,(e_1,0),(e_2,0),(e_3,0)\}$ is a subgroup isomorphic to $(\Z/2\Z)^2$ .

Solution

Each $(e_i,0)$ equals its own negative, so it has order 2. The three points lie on the line $y=0$ , so they add to $O$ , giving $(e_1,0)+(e_2,0)=-(e_3,0)=(e_3,0)$ . The set is closed under addition, every element has order at most 2, and it has 4 elements: it is $(\Z/2)^2$ .

A point of order 6Repeated doubling

On $y^2=x^3+1$ , show that $P=(2,3)$ has order 6.

Solution

Tangent at $P$ : $\lambda=3\cdot4/6=2$ , so $x_{2P}=4-4=0$ , $y_{2P}=-(3+2(0-2))=1$ : $2P=(0,1)$ . Tangent at $(0,1)$ has slope 0, so $4P=(0,-1)=-2P$ , hence $6P=O$ . Also $3P=P+2P$ : $\lambda=(1-3)/(0-2)=1$ , $x=1-2-0=-1$ , $y=-(3+1\cdot(-3))=0$ , so $3P=(-1,0)\neq O$ , and $2P\ne O$ . The order is 6.

More practice: Diamond–Shurman, Exercises 7.1.4 and 7.1.5.

InteractiveDiamond–Shurman, Ch. 8: The Eichler–Shimura Relation and L-functions

Counting points mod p

Reduce an elliptic curve over $\mathbb Q$ modulo a prime $p$ and count its points. The defect $a_p=p+1-\#E(\mathbb F_p)$ looks random, but for each curve below it equals the $p$ -th coefficient of one weight-2 newform, for every prime. The Eichler–Shimura relation explains why, and the Modularity Theorem says it happens for every elliptic curve over $\mathbb Q$ .

The idea

Reduction. Take a minimal integral Weierstrass equation for

E/\mathbb Q

,

y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6

, and reduce its coefficients mod

p

. Define

a_p(E)=p+1-\#\tilde E(\mathbb F_p)

, counting all solutions in

\mathbb F_p^2

plus the point at infinity. If the reduced curve is nonsingular,

p

is a prime of good reduction; this fails for the finitely many

p

dividing the discriminant.

Hasse's bound. At good primes,

|a_p(E)|\le2\sqrt p

. At bad primes,

a_p\in\{0,\pm1\}

:

1

or

-1

for a node (split or non-split multiplicative reduction),

0

for a cusp (additive).

Conductor and L-function. The conductor

N

is divisible exactly by the bad primes, to the first power for multiplicative reduction and to a higher power for additive.

L(E,s)=\prod_{p\mid N}(1-a_pp^{-s})^{-1}\prod_{p\nmid N}(1-a_pp^{-s}+p^{1-2s})^{-1}

.

Eichler–Shimura. On the reduction of

X_0(N)

mod

p\nmid N

, the Hecke correspondence

T_p

decomposes as Frobenius plus its transpose. As a consequence, if

f\in\mathcal S_2(\Gamma_0(N))

is a newform with rational coefficients and

E_f

is the curve from module 7 (which has a model over

\mathbb Q

), then

a_p(E_f)=a_p(f)

for all

p\nmid N

.

Modularity. For every elliptic curve

E/\mathbb Q

of conductor

N

there is a newform

f\in\mathcal S_2(\Gamma_0(N))

with

a_p(f)=a_p(E)

for all primes

p

. Equivalently

L(E,s)=L(f,s)

.

The explorer counts points by brute force and computes $f$ as an eta product, independently, and compares.

Explore

Curve and its newform

Prime p 13

Solutions

(x,y)\in\mathbb F_p^2

, with

0\le x,y<p

. the singular point, at a bad prime

a_p

for

p<200

. from counting (rings) from the newform (dots)

\pm2\sqrt p

Worked example

$E: y^2=x^3-x$ has conductor 32, and its newform is $f=q\prod_{n\ge1}(1-q^{4n})^2(1-q^{8n})^2=q-2q^5-3q^9+6q^{13}+\cdots$ . Check $a_5$ and $a_3$ by hand, then explain the zeros.

Count $E(\mathbb F_5)$ .

$x^3-x$ for $x=0,1,2,3,4$ is $0,0,1,4,0$ mod 5. The value 0 gives one $y$ ; the squares 1 and 4 give two each. So $1+1+2+2+1=7$ affine points, $8$ with $O$ , and $a_5=5+1-8=-2$ , the coefficient of $q^5$ .
Count $E(\mathbb F_3)$ .

$x^3\equiv x$ for every $x\in\mathbb F_3$ , so each $x$ gives only $y=0$ : 3 affine points, 4 in all, and $a_3=0$ . The $q^3$ coefficient of $f$ is also 0.
Show that $a_p=0$ for every prime $p\equiv3\pmod4$ .

$g(x)=x^3-x$ is odd, and $-1$ is not a square mod $p$ . For each pair $\{x,-x\}$ with $g(x)\ne0$ , exactly one of $g(x)$ , $g(-x)=-g(x)$ is a square, so the pair contributes 2 points. The three roots $0,\pm1$ contribute 1 each. Total $3+(p-3)=p$ affine points, so $a_p=0$ .
What does the formula for $f$ say about $a_p$ for $p\equiv3\pmod4$ ?

$f$ is $q$ times a series in $q^4$ , so only exponents $\equiv1\pmod4$ appear: $a_n=0$ for $n\equiv3\pmod 4$ . Point counting and the modular form agree for a structural reason (the curve has complex multiplication by $\Z[i]$ ); in the explorer they agree at every $p<200$ .

Check your understanding

An elliptic curve has good reduction at $p=101$ . What is the largest possible value of $\#E(\mathbb F_{101})$ ?

$101\cdot101+1$ That counts every point of the plane. Each $x$ gives at most two values of $y$ .
122
$2\cdot101+1=203$ That is a trivial bound (two $y$ per $x$ , plus $O$ ). Hasse is much sharper.

\#E=p+1-a_p\le p+1+2\sqrt p=102+20.1\ldots

, so at most 122. On average about half the values of

x^3+ax+b

are squares, which is why the count stays near

p+1

.

The curve 11a has $a_{11}=1$ . What does this say about its reduction at 11?

Good reduction.11 divides the discriminant: it is the conductor.
Additive reduction (a cusp).Then the count would be $p+1$ and $a_p=0$ .
Split multiplicative reduction: a node with tangent directions defined over $\mathbb F_{11}$ .

At a split node the nonsingular points form a group isomorphic to

\mathbb F_p^\times

, so there are

p-1

of them, plus the singular point:

p

in all, and

a_p=1

. Choose 11a and

p=11

in the explorer to see the singular point.

$y^2+y=x^3$ and $y^2+y=x^3-7$ have the same $a_p$ for every prime. Does that contradict the fact that a newform determines its elliptic curve?

Yes: one of the two curves cannot be modular.Every elliptic curve over $\mathbb Q$ is modular.
No: a newform determines its curve only up to isogeny, and these two curves are isogenous.
No: the equal $a_p$ are a coincidence of small primes.They agree at every prime, which is no coincidence.

Isogenous curves have the same L-function. By Faltings' isogeny theorem the converse holds: equal

a_p

for all

p

forces an isogeny. Both curves have conductor 27 and newform

q\prod(1-q^{3n})^2(1-q^{9n})^2

.

Practice

Count by hand at p = 7The curve y² = x³ + 1

Count $E(\mathbb F_7)$ for $E:y^2=x^3+1$ , and compare with the coefficient of $q^7$ in $\eta(6\tau)^4=q\prod(1-q^{6n})^4$ .

Solution

Cubes mod 7 are $0,1,6$ : $x=0$ gives $x^3+1=1$ (2 points); $x=1,2,4$ give $2$ , a square ( $3^2=9\equiv2$ ), so 2 points each; $x=3,5,6$ give $0$ , 1 point each. Total $2+6+3=11$ , so $\#E=12$ and $a_7=-4$ . The product is $q(1-4q^6+\cdots)$ , whose $q^7$ coefficient is $-4$ .

A rational point of order 5A congruence for every good prime

The point $(5,5)$ on $E:y^2+y=x^3-x^2-10x-20$ has order 5. Use the fact that reduction mod a good prime $p$ is injective on the torsion of order prime to $p$ to show that $a_p\equiv p+1\pmod5$ for $p\nmid55$ . Check $p=2,3,7$ .

Solution

The subgroup generated by $(5,5)$ maps injectively into $\tilde E(\mathbb F_p)$ , so 5 divides $\#\tilde E(\mathbb F_p)=p+1-a_p$ . Check: $a_2=-2$ and $3\equiv-2$ ; $a_3=-1$ and $4\equiv-1$ ; $a_7=-2$ and $8\equiv-2\pmod5$ . Module 10 reinterprets this as a statement about a Galois representation.

More practice: Diamond–Shurman, Exercises 8.3.6 and 8.3.7.

SimulationDiamond–Shurman, Ch. 9: Galois Representations

Galois representations and Frobenius

The absolute Galois group of $\mathbb Q$ acts on the torsion points of an elliptic curve, giving 2-by-2 matrix representations, and the strongest form of Modularity says these come from modular forms. The smallest case, 2-torsion, is already concrete: Frobenius at $p$ is the way $p$ permutes the roots of a cubic, and that permutation knows $a_p$ mod 2.

The idea

Frobenius. Let

F/\mathbb Q

be a Galois number field unramified at

p

. For a prime

\mathfrak p

of

F

over

p

there is a unique

\sigma\in\mathrm{Gal}(F/\mathbb Q)

with

\sigma(x)\equiv x^p\pmod{\mathfrak p}

for all algebraic integers

x

of

F

. Changing

\mathfrak p

conjugates

\sigma

, so

p

determines a conjugacy class

\mathrm{Frob}_p

.

Chebotarev density. For each conjugacy class

C\subseteq G=\mathrm{Gal}(F/\mathbb Q)

, the proportion of primes

p

with

\mathrm{Frob}_p=C

is

|C|/|G|

.

Representations from $E$ .

G_{\mathbb Q}=\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)

acts on

E[\ell]\cong(\Z/\ell)^2

, giving

\bar\rho_{E,\ell}:G_{\mathbb Q}\to\mathrm{GL}_2(\mathbb F_\ell)

, and on the Tate module

\varprojlim E[\ell^n]\cong\Z_\ell^2

, giving

\rho_{E,\ell}:G_{\mathbb Q}\to\mathrm{GL}_2(\Z_\ell)

.

Characteristic polynomial. For

p\nmid\ell N

,

\rho_{E,\ell}

is unramified at

p

and

\rho_{E,\ell}(\mathrm{Frob}_p)

has characteristic polynomial

x^2-a_p(E)x+p

. A newform

f

of weight 2 has representations

\rho_{f,\ell}

with the same property for

a_p(f)

. Modularity, Galois version:

\rho_{E,\ell}\cong\rho_{f,\ell}

for some newform

f

.

The case $\ell=2$ . For $E:y^2=g(x)$ with $g$ a cubic, $E[2]=\{O,(e_1,0),(e_2,0),(e_3,0)\}$ where the $e_i$ are the roots of $g$ , and $\mathrm{GL}_2(\mathbb F_2)\cong S_3$ is the group of permutations of the three nonzero points. For $p$ odd and good, the number of roots of $g$ mod $p$ tells you the cycle type of $\mathrm{Frob}_p$ :

3 roots: the identity, trace $0$ in $\mathbb F_2$ ;
1 root: a transposition, trace $0$ ;
0 roots: a 3-cycle, trace $1$ .

So $a_p$ is odd exactly when $g$ has no root mod $p$ . The simulation tests this, and Chebotarev, prime by prime.

A famous congruence.

\tau(n)\equiv\sigma_{11}(n)\pmod{691}

for all

n

. In the language of this chapter, the mod-691 representation attached to

\Delta

is reducible, with the same traces as

1\oplus\chi^{11}

where

\chi

is the mod-691 cyclotomic character.

Explore

Curve

y^2=g(x)

with

g

a cubic

Use primes up to 300

How often each Frobenius class occurs among good primes up to the bound. observed Chebotarev's

|C|/|G|

Worked example

For $E:y^2=x^3-2$ , find $\mathrm{Frob}_p$ in $\mathrm{GL}_2(\mathbb F_2)$ for $p=5,7,31$ , predict the parity of $a_p$ , and check it.

How many roots does $x^3-2$ have mod 5?

Since $\gcd(3,4)=1$ , cubing is a bijection of $\mathbb F_5$ , so there is exactly one cube root of 2 (namely 3, as $27\equiv2$ ). One root: $\mathrm{Frob}_5$ is a transposition, trace 0, so $a_5$ should be even. Counting gives $a_5=0$ .
Now $p=7$ .

The cubes mod 7 are $0,1,6$ , so 2 is not a cube: no roots, $\mathrm{Frob}_7$ is a 3-cycle with trace 1, and $a_7$ should be odd. Counting: $x=0$ gives $-2\equiv5$ , not a square; $x^3=1$ gives $6$ , not a square; $x^3=6$ (three values of $x$ ) gives $4$ , a square, 2 points each. So $\#E=6+1=7$ and $a_7=1$ .
Now $p=31$ .

$4^3=64\equiv2\pmod{31}$ , and since $31\equiv1\pmod3$ there are three cube roots of unity, so three roots ( $4,7,20$ ). $\mathrm{Frob}_{31}$ is the identity, and $a_{31}=4$ is even.
What proportions should the simulation show for this curve?

The splitting field $\mathbb Q(\sqrt[3]2,e^{2\pi i/3})$ has Galois group $S_3$ : identity $1/6$ , transpositions $3/6$ , 3-cycles $2/6$ . Choose $x^3-2$ and raise the bound to watch the bars approach these heights.

Check your understanding

For $p\nmid\ell N$ , what is $\det\bar\rho_{E,\ell}(\mathrm{Frob}_p)$ ?

$a_p\bmod\ell$ That is the trace.
$p\bmod\ell$
$1$ The image need not lie in $\mathrm{SL}_2$ . The Weil pairing identifies $\det$ with the action on $\ell$ -th roots of unity.

The characteristic polynomial is

x^2-a_px+p

. The determinant is the mod-

\ell

cyclotomic character, because the Weil pairing

E[\ell]\times E[\ell]\to\mu_\ell

is Galois-equivariant, and

\mathrm{Frob}_p

acts on

\mu_\ell

by

\zeta\mapsto\zeta^p

.

$E$ has a rational point of order $\ell$ . What follows for $\bar\rho_{E,\ell}$ ?

It is trivial.Only the line through the rational point is fixed pointwise; the rest of $E[\ell]$ can move.
It is reducible, of the form $\begin{pmatrix}1&*\\0&\chi\end{pmatrix}$ , so $a_p\equiv1+p\pmod\ell$ .
Nothing: rational points say nothing about Galois actions.A rational point is exactly a point fixed by every element of $G_{\mathbb Q}$ .

The rational point spans a Galois-stable line on which

G_{\mathbb Q}

acts trivially. The other diagonal entry is then forced by the determinant,

\chi(\mathrm{Frob}_p)=p

. Trace:

1+p

. This is the congruence from the last module's practice problem.

For $y^2=x^3+1$ , in what proportion of primes does $x^3+1$ split into three linear factors mod $p$ ?

$1/2$
$1/6$ That is the answer for Galois group $S_3$ . Here $-1$ is always a root, so the group is smaller.
$1/3$ The other roots come from $x^2-x+1$ , whose splitting field is $\mathbb Q(\sqrt{-3})$ , of degree 2.

G=\mathrm{Gal}(\mathbb Q(\sqrt{-3})/\mathbb Q)

has order 2, so Frobenius is trivial (3 roots) for half the primes, those

\equiv1\pmod 3

, and a transposition (1 root) for the other half. The simulation shows bars at

1/2

,

1/2

,

0

.

Practice

Ramanujan's congruence at small primesCheck before you believe

Check $\tau(p)\equiv1+p^{11}\pmod{691}$ for $p=2$ and $p=3$ , using $\tau(2)=-24$ and $\tau(3)=252$ .

Solution

$1+2^{11}=2049$ and $2049-(-24)=2073=3\cdot691$ . $1+3^{11}=177148$ and $177148-252=176896=256\cdot691$ .

Parity of a_p from 2-torsionWhy the simulation always agrees

Let $E:y^2=g(x)$ with $g$ a cubic, and $p$ an odd prime of good reduction. Show directly that $\#E(\mathbb F_p)$ is even if and only if $g$ has a root in $\mathbb F_p$ , and conclude that $a_p$ is odd exactly when $g$ has no root mod $p$ .

Hint

Pair each point with its negative.

Solution

$(x,y)\mapsto(x,-y)$ pairs off the points with $y\ne0$ . The points left over are $O$ and the points $(e,0)$ with $g(e)=0$ . So $\#E\equiv1+r\pmod2$ , where $r\in\{0,1,3\}$ is the number of roots. $\#E$ is even iff $r$ is odd iff $r\ge1$ . Since $a_p=p+1-\#E$ and $p+1$ is even, $a_p$ is odd iff $r=0$ , matching the trace of a 3-cycle in $\mathrm{GL}_2(\mathbb F_2)$ .

More practice: Diamond–Shurman, Exercises 9.1.1, 9.4.3 and 9.4.4.