Modular Forms of Level 1¶

In this chapter we study in detail the structure of level $1$ modular forms, i.e., modular forms on $\SL_2(\Z)=\Gamma_0(1)=\Gamma_1(1).$ We assume some complex analysis (e.g., the residue theorem), linear algebra, and that the reader has read Chapter Modular Forms.

Examples of Modular Forms of Level 1¶

In this section we will finally see some examples of modular forms of level $1$ ! We first introduce the Eisenstein series and then define $\Delta$ , which is a cusp form of weight $12$ . In Section Structure Theorem for Level 1 Modular Forms we prove the structure theorem, which says that all modular forms of level $1$ are polynomials in Eisenstein series.

For an even integer $k\geq 4$ , the nonnormalized weight $k$ Eisenstein series is the function on the extended upper half plane $\h^*=\h\cup\P^1(\Q)$ given by

(1) $G_k(z) = \sum_{m,n\in\Z}^{*} \frac{1}{(mz+n)^k}.$

The star on top of the sum symbol means that for each $z$ the sum is over all $m,n\in\Z$ such that $mz+n\neq 0$ .

Proposition 2.1

The function $G_k(z)$ is a modular form of weight $k$ , i.e., $G_k \in M_k(\SL_2(\Z))$ .

Proof

See [Ser73, Section VII.2.3] for a proof that $G_k(z)$ defines a holomorphic function on $\h^*$ . To see that $G_k$ is modular, observe that

$G_k(z+1) = \sum^* \frac{1}{(m(z+1)+n)^k} = \sum^* \frac{1}{(mz+(n+m))^k} = \sum^* \frac{1}{(mz+n)^k},$

where for the last equality we use that the map $(m,n+m) \mapsto (m, n)$ on $\Z\times \Z$ is invertible. Also,

$G_k(-1/z) &= \sum^* \frac{1}{(-m/z+n)^k} \\ &= \sum^* \frac{z^k}{(-m+nz)^k}\\ &= z^k \sum^* \frac{1}{(mz+n)^k} = z^k G_k(z),$

where we use that $(n,-m) \mapsto (m,n)$ is invertible.

Proposition 2.2

$G_k(\infty) = 2\zeta(k)$ , where $\zeta$ is the Riemann zeta function.

Proof

As $z\to \infty$ (along the imaginary axis) in (1), the terms that involve $z$ with $m\neq 0$ go to $0$ . Thus

$G_k(\infty) = \sum^*_{n\in\Z} \frac{1}{n^k}.$

This sum is twice $\zeta(k) = \sum_{n\geq 1} \frac{1}{n^k}$ , as claimed.

The Cusp Form $\Delta$ ¶

Suppose $E=\C/\Lambda$ is an elliptic curve over $\C$ , viewed as a quotient of $\C$ by a lattice $\Lambda=\Z\omega_1+\Z\omega_2$ , with $\omega_1/\omega_2\in\h$ (see [DS05, Section 1.4]). The Weierstrass $\wp$ -function of the lattice $\Lambda$ is

$\wp = \wp_{\Lambda}(u) = \frac{1}{u^2} + \sum_{k=4,6,8,\ldots} (k-1) G_k(\omega_1/\omega_2) u^{k-2},$

where the sum is over even integers $k\geq 4$ . It satisfies the differential equation

$(\wp')^2 = 4\wp^3 - 60 G_4(\omega_1/\omega_2) \wp - 140 G_6(\omega_1/\omega_2).$

If we set $x=\wp$ and $y=\wp'$ , the above is an (affine) equation of the form $y^2 = ax^3 + bx + c$ for an elliptic curve that is complex analytically isomorphic to $\C/\Lambda$ (see [Ahl78, pg. 277] for why the cubic has distinct roots).

The discriminant of the cubic

$4x^3 - 60G_4(\omega_1/\omega_2) x - 140 G_6(\omega_1/\omega_2)$

is $16D(\omega_1/\omega_2)$ , where

$D(z) = (60 G_4(z))^3 - 27(140 G_6(z))^2.$

Since $D(z)$ is the difference of two modular forms of weight $12$ it has weight $12$ . Morever,

$D(\infty) &= \left(60 G_4(\infty)\right)^3 - 27\left(140 G_6(\infty)\right)^2\\ &= \left(\frac{60}{3^2\cdot 5} \pi^4\right)^3 - 27\left(\frac{140\cdot 2}{3^3\cdot 5\cdot 7} \pi^6\right)^2\\ &= 0,$

so $D$ is a cusp form of weight $12$ .

Let

$\Delta = \frac{D}{(2\pi)^{12}}.$

Lemma 2.3

If $z\in\h$ , then $\Delta(z)\neq 0$ .

Proof

Let $\omega_1=z$ and $\omega_2=1$ . Since $E=\C/(\Z\omega_1 + \Z\omega_2)$ is an elliptic curve, it has nonzero discriminant $\Delta(z) = \Delta(\omega_1/\omega_2)\neq 0$ .

Proposition 2.4

We have $\Delta = q\cdot \prod_{n=1}^{\infty}(1-q^n)^{24}$ .

Proof

See [Ser73, Thm. 6, pg. 95].

Remark 2.5

Sage computes the $q$ -expansion of $\Delta$ efficiently to high precision using the command delta_qexp:

sage: delta_qexp(6)
q - 24*q^2 + 252*q^3 - 1472*q^4 + 4830*q^5 + O(q^6)

Fourier Expansions of Eisenstein Series¶

Recall from (?) that elements $f$ of $M_k(\SL_2(\Z))$ can be expressed as formal power series in terms of $q(z)=e^{2\pi i z}$ and that this expansion is called the Fourier expansion of $f$ . The following proposition gives the Fourier expansion of the Eisenstein series $G_k(z)$ .

Definition 2.6

For any integer $t\geq 0$ and any positive integer $n$ , the sigma function

$G_k(z) = 2\zeta(k) + 2\cdot \frac{(2\pi i)^{k}}{(k-1)!}\cdot \sum_{n=1}^{\infty} \sigma_{k-1}(n) q^n.$

is the sum of the $t^{th}$ powers of the positive divisors of $n$ . Also, let $d(n) = \sigma_0(n)$ , which is the number of divisors of $n$ , and let $\sigma(n) = \sigma_1(n)$ . For example, if $p$ is prime, then $\sigma_t(p) = 1 + p^t$ .

Proposition 2.7

For every even integer $k\geq 4$ , we have

$G_k(z) = 2\zeta(k) + 2\cdot \frac{(2\pi i)^{k}}{(k-1)!}\cdot \sum_{n=1}^{\infty} \sigma_{k-1}(n) q^n.$

Proof

See [Ser73, Section VII.4], which uses clever manipulations of series, starting with the identity

$\pi \cot(\pi z) = \frac{1}{z} + \sum_{m=1}^{\infty} \left( \frac{1}{z+m} + \frac{1}{z-m}\right).$

From a computational point of view, the $q$ -expansion of Proposition 2.7 is unsatisfactory because it involves transcendental numbers. To understand these numbers, we introduce the Bernoulli numbers $B_n$ for $n\geq 0$ defined by the following equality of formal power series:

(2) $\frac{x}{e^x - 1} = \sum_{n=0}^{\infty} B_n \frac{x^n}{n!}.$

Expanding the power series, we have

$\frac{x}{e^x - 1} = 1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \frac{x^6}{30240} - \frac{x^8}{1209600} + \cdots.$

As this expansion suggests, the Bernoulli numbers $B_n$ with $n>1$ odd are $0$ (see Exercise 2.2). Expanding the series further, we obtain the following table:

$\ds B_{0}=1,\quad B_{1}=-\frac{1}{2},\quad B_{2}=\frac{1}{6},\quad B_{4}=-\frac{1}{30},\quad B_{6}=\frac{1}{42},\quad B_{8}=-\frac{1}{30},\quad$

$\ds B_{10}=\frac{5}{66},\quad B_{12}=-\frac{691}{2730},\quad B_{14}=\frac{7}{6},\quad B_{16}=-\frac{3617}{510},\quad B_{18}=\frac{43867}{798},\quad$

$\ds{}\!\!B_{20}=-\frac{174611}{330},\quad B_{22}=\frac{854513}{138},\quad B_{24}=-\frac{236364091}{2730},\quad B_{26}=\frac{8553103}{6}.$

See Section Fast Computation of Bernoulli Numbers for a discussion of fast (analytic) methods for computing Bernoulli numbers.

We compute some Bernoulli numbers in Sage:

sage: bernoulli(12)
-691/2730
sage: bernoulli(50)
495057205241079648212477525/66
sage: len(str(bernoulli(10000)))
27706

A key fact is that Bernoulli numbers are rational numbers and they are connected to values of $\zeta$ at positive even integers.

Proposition 2.8

If $k\geq 2$ is an even integer, then

$\zeta(k) = -\frac{(2\pi i)^{k}}{2\cdot k!}\cdot B_k.$

Proof

This is proved by manipulating a series expansion of $z \cot(z)$ (see [Ser73, Section VII.4]).

Definition 2.9

The normalized Eisenstein series of even weight $k\geq 4$ is

$E_k = \frac{(k-1)!}{2\cdot (2\pi i)^k}\cdot G_k.$

Combining Proposition 2.7 and Proposition 2.8, we see that

(3) $E_k = -\frac{B_k}{2k} + q + \sum_{n=2}^{\infty} \sigma_{k-1}(n) q^n.$

Warning

Our series $E_k$ is normalized so that the coefficient of $q$ is $1$ , but often in the literature $E_k$ is normalized so that the constant coefficient is $1$ . We use the normalization with the coefficient of $q$ equal to $1$ , because then the eigenvalue of the $n^{th}$ Hecke operator (see Section Hecke Operators) is the coefficient of $q^n$ . Our normalization is also convenient when considering congruences between cusp forms and Eisenstein series.

Structure Theorem for Level 1 Modular Forms¶

In this section we describe a structure theorem for modular forms of level $1$ . If $f$ is a nonzero meromorphic function on $\h$ and $w\in\h$ , let $\ord_w(f)$ be the largest integer $n$ such that $f(z)/(w-z)^n$ is holomorphic at $w$ . If $f = \sum_{n=m}^{\infty} a_n q^n$ with $a_m\neq 0$ , we set $\ord_{\infty}(f)=m$ . We will use the following theorem to give a presentation for the vector space of modular forms of weight $k$ ; this presentation yields an algorithm to compute this space.

Let $M_k=M_k(\SL_2(\Z))$ denote the complex vector space of modular forms of weight $k$ for $\SL_2(\Z)$ . The standard fundamental domain $\cF$ for $\SL_2(\Z)$ is the set of $z\in \h$ with $|z|\geq 1$ and $|\Re(z)|\leq 1/2$ . Let $\rho = e^{2\pi i /3}$ .

Theorem 2.11

Let $k$ be any integer and suppose $f \in M_k(\SL_2(\Z))$ is nonzero. Then

$\ord_{\infty}(f) + \frac{1}{2}\ord_i(f) + \frac{1}{3} \ord_{\rho}(f) + \sum_{w\in \cF}^* \ord_w(f) = \frac{k}{12},$

where $\ds \sum^*_{w \in \cF}$ is the sum over elements of $\cF$ other than $i$ and $\rho$ !.

Proof

The proof in [Ser73, Section VII.3] uses the residue theorem.

Let $S_k=S_k(\SL_2(\Z))$ index{ $S_k$ } denote the subspace of weight $k$ cusp forms for $\SL_2(\Z)$ . We have an exact sequence

$0 \to S_k \to M_k \xrightarrow{\iota_\infty} \C$

that sends $f\in M_k$ to $f(\infty)$ . When $k\geq 4$ is even, the space $M_k$ contains the Eisenstein series $G_k$ , and $G_k(\infty)=2\zeta(k)\neq 0$ , so the map $M_k\to \C$ is surjective. This proves the following lemma.

Lemma 2.12

If $k\geq 4$ is even, then $M_k = S_k \oplus \C G_k$ and the following sequence is exact:

$0 \to S_k \to M_k \xrightarrow{\iota_\infty} \C \to 0.$

Proposition 2.13

For $k<0$ and $k=2$ , we have $M_k=0$ .

Proof

Suppose $f\in M_k$ is nonzero yet $k=2$ or $k<0$ . By Theorem 2.11,

$\ord_{\infty}(f) + \frac{1}{2}\ord_i(f) + \frac{1}{3} \ord_{\rho}(f) + \sum_{w\in D}^* \ord_w(f) = \frac{k}{12}\leq \frac{1}{6}.$

This is not possible because each quantity on the left is nonnegative so whatever the sum is, it is too big (or $0$ , in which case $k=0$ ).

Theorem 2.14

Multiplication by $\Delta$ defines an isomorphism $M_{k-12}\to S_k$ .

Proof

By Lemma 2.3, $\Delta$ is not identically $0$ , so because $\Delta$ is holomorphic, multiplication by $\Delta$ defines an injective map $M_{k-12}\hra S_k$ . To see that this map is surjective, we show that if $f\in S_k$ , then $f/\Delta \in M_{k-12}$ . Since $\Delta$ has weight $12$ and $\ord_\infty(\Delta)\geq 1$ , Theorem 2.11 implies that $\Delta$ has a simple zero at $\infty$ and does not vanish on $\h$ . Thus if $f\in S_k$ and if we let $g=f/\Delta$ , then $g$ is holomorphic and satisfies the appropriate transformation formula, so $g \in M_{k-12}$ .

Corollary 2.15

For $k=0,4,6,8,10,14$ , the space $M_k$ has dimension $1$ , with basis $1$ , $G_4$ , $G_6$ , $G_8$ , $G_{10}$ , and $G_{14}$ , respectively, and $S_k=0$ .

Proof

Combining Proposition 2.13 with Theorem 2.14, we see that the spaces $M_k$ for $k\leq 10$ cannot have dimension greater than $1$ , since otherwise $M_{k'}\neq 0$ for some $k'<0$ . Also $M_{14}$ has dimension at most $1$ , since $M _2$ has dimension $0$ . Each of the indicated spaces of weight $\geq 4$ contains the indicated Eisenstein series and so has dimension $1$ , as claimed.

Corollary 2.16

$\dim M_k = \begin{cases} 0 & \text{if }k\text{ is odd or negative}, \\ \lfloor{} k/12 \rfloor{} & \text{if } k\con 2\pmod{12},\\ \lfloor{} k/12 \rfloor{}+1 & \text{if } k\not\con 2\pmod{12}. \end{cases}$

Here $\lfloor{}x \rfloor{}$ is the biggest integer $\leq x$ .

Proof

As we have already seen above, the formula is true when $k\leq 12$ . By Theorem 2.14, the dimension increases by $1$ when $k$ is replaced by $k+12$ .

Theorem 2.17

The space $M_k$ has as basis the modular forms $G_4^a G_6^b$ , where $a,b$ run over all pairs of nonnegative integers such that $4a+6b=k$ .

Proof

Fix an even integer $k$ . We first prove by induction that the modular forms $G_4^a G_6^b$ generate $M_k$ ; the cases $k\leq 10$ and $k=14$ follow from the above arguments (e.g., when $k=0$ , we have $a=b=0$ and basis $1$ ). Choose some pair of nonnegative integers $a,b$ such that $4a+6b=k$ . The form $g=G_4^a G_6^b$ is not a cusp form, since it is nonzero at $\infty$ . Now suppose $f\in M_k$ is arbitrary. Since $g(\infty)\neq 0$ , there exists $\alpha\in\C$ such that $f - \alpha g \in S_k$ . Then by Theorem 2.14, there is $h \in M_{k-12}$ such that $f - \alpha g = \Delta \cdot h$ . By induction, $h$ is a polynomial in $G_4$ and $G_6$ of the required type, and so is $\Delta$ , so $f$ is as well. Thus

$\{G_4^aG_6^b\ |\ a\geq 0,\ b\geq 0,\ 4a+6b=k\}$

spans $M_k$ .

Suppose there is a nontrivial linear relation between the $G_4^a G_6^b$ for a given $k$ . By multiplying the linear relation by a suitable power of $G_4$ and $G_6$ , we may assume that we have such a nontrivial relation with $k\con 0\pmod{12}$ . Now divide the linear relation by the weight $k$ form $G_6^{k/6}$ to see that $G_4^3/G_6^2$ satisfies a polynomial with coefficients in $\C$ (see Exercise 2.4). Hence $G_4^3/G_6^2$ is a root of a polynomial, hence a constant, which is a contradiction since the $q$ -expansion of $G_4^3/G_6^2$ is not constant.

Algorithm 2.18

Given integers $n$ and $k$ , this algorithm computes a basis of $q$ -expansions for the complex vector space $M_k$ mod $q^n$ . The $q$ -expansions output by this algorithm have coefficients in $\Q$ .

[Simple Case] If $k=0$ , output the basis with just $1$ in it and terminate; otherwise if $k<4$ or $k$ is odd, output the empty basis and terminate.
[Power Series] Compute $E_4$ and $E_6$ mod $q^n$ using the formula from (3) and Section Fast Computation of Bernoulli Numbers.
[Initialize] Set $b\set 0$ .
[Enumerate Basis] For each integer $b$ between $0$ and $\lfloor k/6\rfloor$ , compute $a=(k-6b)/4$ . If $a$ is an integer, compute and output the basis element $E_4^a E_6^b \mod{q^n}$ . When computing $E_4^a$ , find $E_4^m\pmod{q^n}$ for each $m\leq a$ , and save these intermediate powers, so they can be reused later, and likewise for powers of $E_6$ .

Proof

This is simply a translation of Theorem 2.17 into an algorithm, since $E_k$ is a nonzero scalar multiple of $G_k$ . That the $q$ -expansions have coefficients in $\Q$ follows from (3).

Example 2.19

We compute a basis for $M_{24}$ , which is the space with smallest weight whose dimension is greater than $1$ . It has as basis $E_4^6$ , $E_4^3 E_6^2$ , and $E_6^4$ , whose explicit expansions are

$E_4^6 &= \frac{1}{191102976000000} + \frac{1}{132710400000}q + \frac{203}{44236800000}q^2 + \cdots,\\ E_4^3 E_6^2 &= \frac{1}{3511517184000} - \frac{1}{12192768000}q - \frac{377}{4064256000}q^2 + \cdots, \\ E_6^4 &= \frac{1}{64524128256} - \frac{1}{32006016}q + \frac{241}{10668672}q^2 +\cdots.$

We compute this basis in Sage as follows:

sage: E4 = eisenstein_series_qexp(4, 3)
sage: E6 = eisenstein_series_qexp(6, 3)
sage: E4^6
1/191102976000000 + 1/132710400000*q
                  + 203/44236800000*q^2 + O(q^3)
sage: E4^3*E6^2
1/3511517184000 - 1/12192768000*q
                - 377/4064256000*q^2 + O(q^3)
sage: E6^4
1/64524128256 - 1/32006016*q + 241/10668672*q^2 + O(q^3)

In Section The Miller Basis, we will discuss the reduced echelon form basis for $M_k$ .

The Miller Basis¶

Lemma 2.20

The space $S_k$ has a basis $f_1,\ldots, f_{d}$ such that if $a_i(f_j)$ is the $i^{th}$ coefficient of $f_j$ , then $a_i(f_j) = \delta_{i,j}$ for $i=1,\ldots, {d}$ . Moreover the $f_j$ all lie in $\Z[[q]]$ . We call this basis the Miller basis for $S_k$ .

This is a straightforward construction involving $E_4$ , $E_6$ and $\Delta$ . The following proof very closely follows [Lan95, Ch. X,Thm. 4.4], which in turn follows the first lemma of V. Miller’s thesis.

Proof

Let $d=\dim S_k$ . Since $B_4 = -1/30$ and $B_6=1/42$ , we note that

$F_4 = -\frac{8}{B_4} \cdot E_4 = 1 + 240q + 2160q^2 + 6720q^3 + 17520q^4 + \cdots$

and

$F_6 = -\frac{12}{B_6}\cdot E_6 = 1 - 504q - 16632q^2 - 122976q^3 - 532728q^4 +\cdots$

have $q$ -expansions in $\Z[[q]]$ with leading coefficient $1$ . Choose integers $a,b\geq 0$ such that

$4a+6b\leq 14 \qquad\text{and}\qquad 4a + 6b \con k \pmod{12},$

with $a=b=0$ when $k\con 0\pmod{12}$ , and let

$g_j = \Delta^j F_6^{2(d-j)+b} F_4^a = \left(\frac{\Delta}{F_6^2}\right)^j F_6^{2d + b} F_4^a ,\qquad\qquad \text{for } j=1,\ldots,d.$

Then it is elementary to check that $g_j$ has weight $k$

$a_j(g_j) = 1 \qquad\text{and}\qquad a_i(g_j)=0\qquad\text{when}\qquad i<j.$

Hence the $g_j$ are linearly independent over $\C$ , so form a basis for $S_k$ . Since $F_4, F_6$ , and $\Delta$ are all in $\Z[[q]]$ , so are the $g_j$ . The $f_i$ may then be constructed from the $g_j$ by Gauss elimination. The coefficients of the resulting power series lie in $\Z$ because each time we clear a column we use the power series $g_j$ whose leading coefficient is $1$ (so no denominators are introduced).

Remark 2.21

The basis coming from Miller’s lemma is “canonical”“, since it is just the reduced row echelon form of any basis. Also the set of all integral linear combinations of the elements of the Miller basis are precisely the modular forms of level $1$ with integral $q$ -expansion.

We extend the Miller basis to all $M_k$ by taking a multiple of $G_k$ with constant term $1$ and subtracting off the $f_i$ from the Miller basis so that the coefficients of $q, q^2, \ldots q^d$ of the resulting expansion are $0$ . We call the extra basis element $f_0$ .

Example 2.22

If $k=24$ , then $d=2$ . Choose $a=b=0$ , since $k\con 0\pmod{12}$ . Then

$g_1 = \Delta F_6^2 = q - 1032q^2 + 245196q^3 + 10965568q^4 + 60177390q^5 - \cdots$

and

$g_2 = \Delta^2 = q^2 - 48q^3 + 1080q^4 - 15040q^5 + \cdots.$

We let $f_2=g_2$ and

$f_1 = g_1 + 1032 g_2 = q + 195660q^3 + 12080128q^4 + 44656110q^5 - \cdots.$

Example 2.23

When $k=36$ , the Miller basis including $f_0$ is

$f_0 &= 1 + \quad\,\,\, 6218175600q^4 + 15281788354560q^5 + \cdots,\\ f_1 &= \,\,\,\,\,\, q + \quad\,\, 57093088q^4 + \,\,\,\,\,\,\,\,\,\,37927345230q^5 +\cdots,\\ f_2 &= \qquad q^2 + \,\,\,\,\,\,194184q^4 + \quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 7442432q^5 + \cdots,\\ f_3 &= \qquad\quad\,\, q^3 -\,\,\,\,\,\,\,\,\,\, 72q^4 +\qquad\quad\,\,\,\,\,\,\,\,\,\,\,\, 2484q^5 + \cdots.$

Example 2.24

The Sage command victor_miller_basis computes the Miller basis to any desired precision given $k$ .

sage: victor_miller_basis(28,5)
[
1 + 15590400*q^3 + 36957286800*q^4 + O(q^5),
q + 151740*q^3 + 61032448*q^4 + O(q^5),
q^2 + 192*q^3 - 8280*q^4 + O(q^5)
]

Remark 2.25

To write $f \in M_k$ as a polynomial in $E_4$ and $E_6$ , it is wasteful to compute the Miller basis. Instead, use the upper triangular (but not echelon!) basis $\Delta^j F_6^{2(d-j)+a} F_4^b$ , and match coefficients from $q^0$ to $q^d$ .

Hecke Operators¶

In this section we define Hecke operators on level $1$ modular forms and derive their basic properties. We will not give proofs of the analogous properties for Hecke operators on higher level modular forms, since the proofs are clearest in the level $1$ case, and the general case is similar (see, e.g., [Lan95]).

For any positive integer $n$ , let

$X_n = \left\{\mtwo{a}{b}{0}{d} \in \Mat_2(\Z) \,\,:\,\, a\geq 1,\,\, ad=n, \text{ and } 0\leq b <d\right\}.$

Note that the set $X_n$ is in bijection with the set of subgroups of $\Z^2$ of index $n$ , where $\abcd{a}{b}{c}{d}$ corresponds to $L=\Z\cdot (a,b) + \Z\cdot (0,d)$ , as one can see using Hermite normal form, which is the analogue over $\Z$ of echelon form (see Exercise 7.5).

Recall from (?) that if $\gamma=\abcd{a}{b}{c}{d}\in\GL_2(\Q)$ , then

$f^{[\gamma]_k} = \det(\gamma)^{k-1} (cz+d)^{-k} f(\gamma(z)).$

Definition 2.26

The $n^{th}$ Hecke operator $T_{n,k}$ of weight $k$ is the operator on the set of functions on $\h$ defined by

$T_{n,k}(f) = \sum_{\gamma \in X_n} f^{[\gamma]_k}.$

Remark 2.27

It would make more sense to write $T_{n,k}$ on the right, e.g., $f|T_{n,k}$ , since $T_{n,k}$ is defined using a right group action. However, if $n,m$ are integers, then the action of $T_{n,k}$ and $T_{m,k}$ on weakly modular functions commutes (by Proposition 2.29 below), so it makes no difference whether we view the Hecke operators of given weight $k$ as acting on the right or left.

Proposition 2.28

If $f$ is a weakly modular function of weight $k$ , then so is $T_{n,k}(f)$ ; if $f$ is a modular function, then so is $T_{n,k}(f)$ .

Proof

Suppose $\gamma\in\SL_2(\Z)$ . Since $\gamma$ induces an automorphism of $\Z^2$ , X_n \cdot \gamma = \{ \delta \gamma : \delta\in X_n\} is also in bijection with the subgroups of $\Z^2$ of index $n$ . For each element $\delta\gamma \in X_n\cdot \gamma$ , there is $\sigma\in\SL_2(\Z)$ such that $\sigma\delta\gamma\in X_n$ (the element $\sigma$ transforms $\delta\gamma$ to Hermite normal form), and the set of elements $\sigma\delta\gamma$ is thus equal to $X_n$ . Thus

$T_{n,k}(f) = \sum_{\sigma\delta\gamma\in X_n} f^{[\sigma\delta\gamma]_k} = \sum_{\delta\in X_n} f^{[\delta\gamma]_k} = T_{n,k}(f)^{[\gamma]_k}.$

A finite sum of meromorphic function is meromorphic, so $T_{n,k}(f)$ is weakly modular. If $f$ is holomorphic on $\h$ , then each $f^{[\delta]_k}$ is holomorphic on $\h$ for $\delta \in X_n$ . A finite sum of holomorphic functions is holomorphic, so $T_{n,k}(f)$ is holomorphic.

We will frequently drop $k$ from the notation in $T_{n,k}$ , since the weight $k$ is implicit in the modular function to which we apply the Hecke operator. Henceforth we make the convention that if we write $T_{n}(f)$ and if $f$ is modular, then we mean $T_{n,k}(f)$ , where $k$ is the weight of $f$ .

Proposition 2.29

On weight $k$ modular functions we have

(4) $T_{mn} = T_{m} T_{n} \qquad\qquad\qquad\qquad\quad \text{if }(m,n)=1,$

and

(5) $T_{p^n} = T_{p^{n-1}} T_p\, -\, p^{k-1} T_{p^{n-2}} \qquad\!\! \text{if $p$ is prime}.$

Proof

Let $L$ be a subgroup of index $mn$ . The quotient $\Z^2/L$ is an abelian group of order $mn$ , and $(m,n)=1$ , so $\Z^2/L$ decomposes uniquely as a direct sum of a subgroup of order $m$ with a subgroup of order $n$ . Thus there exists a unique subgroup $L'$ such that $L\subset L'\subset \Z^2$ , and $L'$ has index $m$ in $\Z^2$ . The subgroup $L'$ corresponds to an element of $X_m$ , and the index $n$ subgroup $L\subset L'$ corresponds to multiplying that element on the right by some uniquely determined element of $X_n$ . We thus have

$\SL_2(\Z) \cdot X_{m} \cdot X_n = \SL_2(\Z) \cdot X_{mn}$

i.e., the set products of elements in $X_m$ with elements of $X_n$ equal the elements of $X_{mn}$ , up to $\SL_2(\Z)$ -equivalence. Thus for any $f$ , we have $T_{mn}(f) = T_{n}(T_m(f))$ . Applying this formula with $m$ and $n$ swapped yields the equality $T_{mn} = T_{m} T_{n}$ .

We will show that $T_{p^n} + p^{k-1} T_{p^{n-2}} = T_p T_{p^{n-1}}$ . Suppose $f$ is a weight $k$ weakly modular function. Using that $f^{[\abcd{p}{0}{0}{p}]_k} = (p^2)^{k-1}p^{-k} f = p^{k-2} f$ , we have

$\sum_{x\in X_{p^n}} f^{[x]_k}\,\, +\,\, p^{k-1}\!\!\! \sum_{x\in X_{p^{n-2}}} f^{[x]_k} = \sum_{x\in X_{p^n}} f^{[x]_k} \,\,\,+\, p\!\!\!\! \sum_{x\in pX_{p^{n-2}}} f^{[x]_k}.$

Also

$T_p T_{p^{n-1}}(f) = \sum_{y\in X_p} \sum_{x\in X_{p^{n-1}}} (f^{[x]_k})^{[y]_k} = \sum_{x \in X_{p^{n-1}}\cdot X_p} f^{[x]_k}.$

Thus it suffices to show that $X_{p^n}$ disjoint union $p$ copies of $p X_{p^{n-2}}$ is equal to $X_{p^{n-1}}\cdot X_p$ , where we consider elements with multiplicities and up to left $\SL_2(\Z)$ -equivalence (i.e., the left action of $\SL_2(\Z)$ ).

Suppose $L$ is a subgroup of $\Z^2$ of index $p^n$ , so $L$ corresponds to an element of $X_{p^n}$ . First suppose $L$ is not contained in $p\Z^2$ . Then the image of $L$ in $\Z^2/p\Z^2 = (\Z/p\Z)^2$ is of order $p$ , so if $L'=p\Z^2 + L$ , then $[\Z^2:L']=p$ and $[L:L']=p^{n-1}$ , and $L'$ is the only subgroup with this property. Second, suppose that $L\subset p\Z^2$ if of index $p^n$ and that $x\in X_{p^n}$ corresponds to $L$ . Then every one of the $p+1$ subgroups $L'\subset \Z^2$ of index $p$ contains $L$ . Thus there are $p+1$ chains $L\subset L'\subset \Z^2$ with $[\Z^2:L']=p$ .

The chains $L\subset L'\subset\Z^2$ with $[\Z^2:L']=p$ and $[\Z^2:L]=p^{n-1}$ are in bijection with the elements of $X_{p^{n-1}}\cdot X_p$ . On the other hand the union of $X_{p^n}$ with $p$ copies of $p X_{p^{n-2}}$ corresponds to the subgroups $L$ of index $p^{n}$ , but with those that contain $p\Z^2$ counted $p+1$ times. The structure of the set of chains $L\subset L'\subset\Z^2$ that we derived in the previous paragraph gives the result.

Corollary 2.30

The Hecke operator $T_{p^n}$ , for prime $p$ , is a polynomial in $T_p$ with integer coefficients, i.e., $T_{p^n}\in\Z[T_p]$ . If $n,m$ are any integers, then $T_n T_m = T_m T_n$

Proof

The first statement follows from (5) of Proposition 2.29. It then follows that $T_n T_m = T_m T_n$ when $m$ and $n$ are both powers of a single prime $p$ . Combining this with (4) gives the second statement in general.

Proposition 2.31

Let $f = \sum_{n\in\Z} a_n q^n$ be a modular function of weight $k$ . Then

$T_{n}(f) = \sum_{m\in\Z} \left(\sum_{1\leq d\,\mid\, \gcd(n,m)} d^{k-1} a_{mn/d^2}\right) q^m.$

In particular, if $n=p$ is prime, then

$T_p(f) = \sum_{m\in \Z} \left(a_{mp} + p^{k-1} a_{m/p}\right) q^m,$

where $a_{m/p}=0$ if $m/p\not\in\Z$ .

Proof

This is proved in [Ser73, Section VII.5.3] by writing out $T_n(f)$ explicitly and using that $\sum_{0\leq b<d} e^{2\pi i bm/d}$ is $d$ if $d\mid m$ and $0$ otherwise.

Corollary 2.32

The Hecke operators preserve $M_k$ and $S_k$ .

Remark 2.33

Alternatively, for $M_k$ the above corollary is Proposition 2.28, and for $S_k$ we see from the definitions that if $f(\infty)=0$ , then $T_n f$ also vanishes at $\infty$ .

Example 2.34

Recall from (3) that

$E_4 = \frac{1}{240} + q + 9q^2 + 28q^3 + 73q^4 + 126q^5 + 252q^6 + 344q^7 +\cdots.$

Using the formula of Proposition 2.31, we see that

$T_2(E_4) = (1/240 + 2^3\cdot (1/240)) + 9q + (73 + 2^3\cdot 1) q^2 + \cdots.$

Since $M_4$ has dimension $1$ and since we have proved that $T_2$ preserves $M_4$ , we know that $T_2$ acts as a scalar. Thus we know just from the constant coefficient of $T_2(E_4)$ that

$T_2(E_4) = 9 E_4.$

More generally, for $p$ prime we see by inspection of the constant coefficient of $T_p(E_4)$ that

$T_p(E_4) = (1+p^3)E_4.$

In fact

$T_n(E_k) = \sigma_{k-1}(n) E_k,$

for any integer $n\geq 1$ and even weight $k\geq 4$ .

Example 2.35

By Corollary Corollary 2.32, the Hecke operators $T_n$ also preserve the subspace $S_k$ of $M_k$ . Since $S_{12}$ has dimension $1$ (spanned by $\Delta$ ), we see that $\Delta$ is an eigenvector for every $T_n$ . Since the coefficient of $q$ in the $q$ -expansion of $\Delta$ is $1$ , the eigenvalue of $T_n$ on $\Delta$ is the $n^{th}$ coefficient of $\Delta$ . Since $T_{nm}=T_n T_m$ for $\gcd(n,m)=1$ , we have proved the nonobvious fact that the Ramanujan function $\tau(n)$ that gives the $n^{th}$ coefficient of $\Delta$ is a multiplicative function, i.e., if $\gcd(n,m)=1$ , then $\tau(nm)=\tau(n)\tau(m)$ .

Remark 2.36

The Hecke operators respect the decomposition $M_k = S_k \oplus \C E_k$ , i.e., for all $k$ the series $E_k$ are eigenvectors for all $T_n$ .

Computing Hecke Operators¶

This section is about how to compute matrices of Hecke operators on $M_k$ .

Algorithm 2.37

This algorithm computes the matrix of the Hecke operator $T_n$ on the Miller basis for $M_k$ .

[Dimension] Compute $d= \dim(M_k)-1$ using Corollary 2.16.
[Basis] Using Lemma 2.20, compute the echelon basis $f_0,\ldots, f_d$ for $M_k$ $\pmod{q^{dn+1}}$ .
[Hecke operator] Using Proposition 2.31, compute for each $i$ the image $T_n(f_i) \pmod{q^{d+1}}$ .
[Write in terms of basis] The elements $T_n(f_i) \pmod{q^{d+1}}$ determine linear combinations of

$f_0,f_1,\ldots, f_d \pmod{q^d}.$
These linear combinations are easy to find once we compute $T_n(f_i) \pmod{q^{d+1}}$ , since our basis of $f_i$ is in echelon form. The linear combinations are just the coefficients of the power series $T_n(f_i)$ up to and including $q^d$ .
[Write down matrix] The matrix of $T_n$ acting from the right relative to the basis $f_0, \ldots, f_d$ is the matrix whose rows are the linear combinations found in the previous step, i.e., whose rows are the coefficients of $T_n(f_i)$ .

Proof

By Proposition 2.31, the $d^{th}$ coefficient of $T_n(f)$ involves only $a_{dn}$ and smaller-indexed coefficients of $f$ . We need only compute a modular form $f$ modulo $q^{dn+1}$ in order to compute $T_n(f)$ modulo $q^{d+1}$ . Uniqueness in step (4) follows from Lemma 2.20 above.

Example 2.38

We compute the Hecke operator $T_2$ on $M_{12}$ using the above algorithm.

[Compute dimension] We have $d=2 - 1 = 1$ .

[Compute basis] Compute up to (but not including) the coefficient of $q^{dn+1} = q^{1\cdot 2 + 1} = q^3$ . As given in the proof of Lemma 2.20, we have

$F_4 = 1 + 240q + 2160q^2 + \cdots \quad\text{and}\quad F_6 = 1 - 504q - 16632q^2 + \cdots.$

Thus $M_{12}$ has basis

$f_0 = 1 + 196560q^{2} + \cdots \quad\text{and}\quad f_1 = q - 24q^{2} + \cdots.$

Sage does the arithmetic in the above calculation as follows:

sage: R.<q> = QQ[['q']]
sage: F4 =  240 * eisenstein_series_qexp(4,3)
sage: F6 = -504 * eisenstein_series_qexp(6,3)
sage: F4^3
1 + 720*q + 179280*q^2 + O(q^3)
sage: Delta = (F4^3 - F6^2)/1728; Delta
q - 24*q^2 + O(q^3)
sage: F4^3 - 720*Delta
1 + 196560*q^2 + O(q^3)

[Compute Hecke operator] In each case letting $a_n$ denote the $n^{th}$ coefficient of $f_0$ or $f_1$ , respectively, we have

$T_2(f_0) &= T_2 (1 + 196560q^{2} + \cdots) \\ &= (a_0 + 2^{11} a_0) q^0 + (a_2 + 2^{11} a_{1/2})q^1 + \cdots\\ &= 2049 + 196560q + \cdots,$
and

$T_2(f_1) &= T_2 (q - 24q^{2} + \cdots) \\ &= (a_0 + 2^{11} a_0) q^0 + (a_2 + 2^{11} a_{1/2})q^1 + \cdots\\ &= 0 -24 q + \cdots.$
(Note that $a_{1/2} = 0$ .)
[Write in terms of basis] We read off at once that

$T_2(f_0) = 2049 f_0 + 196560 f_1 \quad\text{and}\quad T_2(f_1) = 0 f_0 + (-24)f_1.$
[Write down matrix] Thus the matrix of $T_2$ , acting from the right on the basis $f_0$ , $f_1$ , is

$T_2 = \mtwo{2049}{196560}{0}{-24}.$

As a check note that the characteristic polynomial of $T_2$ is $(x-2049)(x+24)$ and that $2049 = 1 + 2^{11}$ is the sum of the $11^{th}$ powers of the divisors of $2$ .

Example 2.39

The Hecke operator $T_2$ on $M_{36}$ with respect to the echelon basis is

$\left( \begin{matrix} \hfill 34359738369&\hfill0&\hfill6218175600&\hfill9026867482214400\\ 0&\hfill0&\hfill34416831456&\hfill5681332472832\\ 0&\hfill1&\hfill194184&\hfill-197264484\\ 0&\hfill0&\hfill-72&\hfill-54528\\ \end{matrix}\right).$

It has characteristic polynomial

$(x - 34359738369) \cdot (x^3 - 139656x^2 - 59208339456x - 1467625047588864),$

where the cubic factor is irreducible.

The echelon_form command creates the space of modular forms but with basis in echelon form (which is not the default).

sage: M = ModularForms(1,36, prec=6).echelon_form()
sage: M.basis()
[
1 + 6218175600*q^4 + 15281788354560*q^5 + O(q^6),
q + 57093088*q^4 + 37927345230*q^5 + O(q^6),
q^2 + 194184*q^4 + 7442432*q^5 + O(q^6),
q^3 - 72*q^4 + 2484*q^5 + O(q^6)
]

Next we compute the matrix of the Hecke operator $T_2$ .

sage: T2 = M.hecke_matrix(2); T2
[34359738369   0       6218175600 9026867482214400]
[          0   0      34416831456    5681332472832]
[          0   1           194184       -197264484]
[          0   0              -72           -54528]

Finally we compute and factor its characteristic polynomial.

sage: T2.charpoly().factor()
(x - 34359738369) * (x^3 - 139656*x^2 - 59208339456*x - 1467625047588864)

The following is a famous open problem about Hecke operators on modular forms of level $1$ . It generalizes our above observation that the characteristic polynomial of $T_2$ on $M_k$ , for $k=12, 36$ , factors as a product of a linear factor and an irreducible factor.

Conjuecture 2.40

The characteristic polynomial of $T_2$ on $S_k$ is irreducible for any $k$ .

Kevin Buzzard observed that in several specific cases the Galois group of the characteristic polynomial of $T_2$ is the full symmetric group (see [Buz96]). See also [FJ02] for more evidence for the following conjecture:

Conjuecture 2.41

For all primes $p$ and all even $k\geq 2$ the characteristic polynomial of $T_{p,k}$ acting on $S_k$ is irreducible.

Fast Computation of Fourier Coefficients¶

How difficult is it to compute prime-indexed coefficients of

$\Delta = \sum_{n=1}^{\infty} \tau(n) q^n = q - 24q^2 + 252q^3 - 1472q^4 + 4830q^5 - 6048q^6 + \cdots?$

Theorem 2.42

Let $p$ be a prime. There is a probabilistic algorithm to compute $\tau(p)$ , for prime $p$ , that has expected running time polynomial in $\log(p)$

Proof

See [ECdJ+06].

More generally, if $f=\sum a_n q^n$ is an eigenform in some space $M_k(\Gamma_1(N))$ , where $k\geq 2$ , then one expects that there is an algorithm to compute $a_p$ in time polynomial in $\log(p)$ . Bas Edixhoven, Jean-Marc Couveignes and Robin de Jong have proved that $\tau(p)$ can be computed in polynomial time; their approach involves sophisticated techniques from arithmetic geometry (e.g., ‘etale cohomology, motives, Arakelov theory). The ideas they use are inspired by the ones introduced by Schoof, Elkies and Atkin for quickly counting points on elliptic curves over finite fields (see [Sch95]).

Edixhoven describes (in an email to the author) the strategy as follows:

We compute the mod $\ell$ Galois representation $\rho$ associated to $\Delta$ . In particular, we produce a polynomial $f$ such that $\Q[x]/(f)$ is the fixed field of $\ker(\rho)$ . This is then used to obtain $\tau(p) \pmod{\ell}$ and to do a Schoof-like algorithm for computing $\tau(p)$ .
We compute the field of definition of suitable points of order $\ell$ on the modular Jacobian $J_1(\ell)$ to do part (1) (see [DS05, Ch. 6] for the definition of $J_1(\ell)$ ).
The method is to approximate the polynomial $f$ in some sense (e.g., over the complex numbers or modulo many small primes $r$ ) and to use an estimate from Arakelov theory to determine a precision that will suffice.

Fast Computation of Bernoulli Numbers¶

This section, which was written jointly with Kevin McGown, is about computing the Bernoulli numbers $B_n$ , for $n\geq 0$ , defined in Section Fourier Expansions of Eisenstein Series by

(6) $\frac{x}{e^x - 1} = \sum_{n=0}^{\infty} B_n \frac{x^n}{n!}.$

One way to compute $B_n$ is to multiply both sides of (?) by $e^x-1$ and equate coefficients of $x^{n+1}$ to obtain the recurrence

$B_0=1\text{,} \qquad B_n=-\frac{1}{n+1}\cdot \sum_{k=0}^{n-1}\binom{n+1}{k}B_{k}.$

This recurrence provides a straightforward and easy-to-implement method for calculating $B_n$ if one is interested in computing $B_n$ for all $n$ up to some bound. For example,

$B_1 = - \frac{1}{2} \cdot \left( \binom{2}{0} B_0\right) = -\frac{1}{2}$

and

$B_2 = -\frac{1}{3} \cdot \left( \binom{3}{0} B_0 + \binom{3}{1} B_1 \right) = -\frac{1}{3} \cdot \left( 1 - \frac{3}{2}\right) = \frac{1}{6}.$

However, computing $B_n$ via the recurrence is slow; it requires summing over many large terms, it requires storing the numbers $B_0,\dots,B_{n-1}$ , and it takes only limited advantage of asymptotically fast arithmetic algorithms. There is also an inductive procedure to compute Bernoulli numbers that resembles Pascal’s triangle called the Akiyama-Tanigawa algorithm (see [Kan00]).

Another approach to computing $B_n$ is to use Newton iteration and asymptotically fast polynomial arithmetic to approximate $1/(e^x -1)$ . This method yields a very fast algorithm to compute $B_0, B_2,\ldots, B_{p-3}$ modulo $p$ . See [BCS92] for an application of this method modulo a prime $p$ to the verification of Fermat’s last theorem for irregular primes up to one million.

Example 2.43

David Harvey implemented the algorithm of [BCS92] in Sage as the command bernoulli_mod_p:

sage: bernoulli_mod_p(23)
[1, 4, 13, 17, 13, 6, 10, 5, 10, 9, 15]

A third way to compute $B_n$ uses an algorithm based on Proposition 2.8, which we explain below (Algorithm 2.45). This algorithm appears to have been independently invented by several people: by Bernd C. Kellner (see [Kel06]); by Bill Dayl; and by H. Cohen and K. Belabas.

We compute $B_n$ as an exact rational number by approximating $\zeta(n)$ to very high precision using Proposition 2.8, the Euler product

$\zeta(s)=\sum_{m=1}^\infty m^{-s} =\prod_{p\text{ prime}} (1-p^{-s})^{-1},$

and the following theorem:

Theorem 2.44

For even $n\geq 2$ ,

$\denom(B_n)=\prod_{p-1 \,\mid\, n} p.$

Proof

See [Lan95, Ch. X, Thm. 2.1].

The Number of Digits of $B_n$ ¶

The following is a new quick way to compute the number of digits of the numerator of $B_n$ . For example, using it we can compute the number of digits of $B_{10^{50}}$ in less than a second.

By Theorem 2.44 we have $d_n = \denom(B_n) = \prod_{p-1\mid n} p$ . The number of digits of the numerator is thus

$\lceil \log_{10}(d_n \cdot |B_n|) \rceil.$

But

$\log(|B_n|) &= \log\left(\frac{2\cdot n!}{(2\pi)^n}\,\zeta(n)\right)\\ &= \log(2) + \log(n!) - n\log(2) - n\log(\pi) + \log(\zeta(n)),$

and $\zeta(n)\sim 1$ so $\log(\zeta(n))\sim 0$ . Finally, Stirling’s formula (see [Ahl78, pg. 198–206]) gives a fast way to compute $\log(n!) = \log(\Gamma(n+1))$ :

(7) $\log(\Gamma(z)) ``=\text{''} \frac{\log(2\pi)}{2} + \left(z - \frac{1}{2}\right)\log(z) - z + \sum_{m=1}^{\infty} \frac{B_{2m}}{2m(2m-1)z^{2m-1}}.$

We put quotes around the equality sign because $\log(\Gamma(z))$ does not converge to its Laurent series. Indeed, note that for any fixed value of $z$ the summands on the right side go to $\infty$ as $m\to\infty$ ! Nonetheless, we can use this formula to very efficiently compute $\log(\Gamma(z))$ , since if we truncate the sum, then the error is smaller than the next term in the infinite sum.

Computing $B_n$ Exactly¶

We return to the problem of computing $B_n$ . Let

$K=\frac{2\cdot n!}{(2\pi)^n}$

so $|B_n|=K\zeta(n)$ . Write

$B_n=\frac{a}{d},$

with $a,d\in\Z$ , $d\geq 1$ , and $\gcd(a,d)=1$ . It is elementary to show that $a=(-1)^{n/2+1}\;|a|$ for even $n\geq 2$ . Suppose that using the Euler product we approximate $\zeta(n)$ from below by a number $z$ such that

$0\leq \zeta(m)-z<\frac{1}{Kd}.$

Then $0\leq |B_n| - zK < d^{-1};$ hence $0\leq |a| - zKd < 1.$ It follows that $|a|=\lceil zKd\rceil$ and hence $a=(-1)^{n/2+1}\;\lceil zKd\rceil.$

It remains to compute $z$ . Consider the following problem: given $s>1$ and $\varepsilon>0$ , find $M\in\Z_+$ so that

$z=\prod_{p\leq M} (1-p^{-s})^{-1}$

satisfies $0\leq\zeta(s)-z<\varepsilon$ . We always have $0\leq\zeta(s)-z$ . Also,

$\sum_{n\leq M} n^{-s}\leq\prod_{p\leq M} (1-p^{-s})^{-1},$

so

$\zeta(s)-z \leq \sum_{n=M+1}^\infty n^{-s} \leq \int_M^\infty x^{-s}\;dx = \frac{1}{(s-1)M^{s-1}}.$

Thus if $M>\varepsilon^{-1/(s-1)}$ , then

$\frac{1}{(s-1)M^{s-1}} \leq \frac{1}{M^{s-1}} < \varepsilon \,,$

so $\zeta(s)-z<\varepsilon$ , as required. For our purposes, we have $s=n$ and $\varepsilon=(Kd)^{-1}$ , so it suffices to take $M>(Kd)^{1/(n-1)}$ .

Algorithm 2.45

Given an integer $n\geq 0$ , this algorithm computes the Bernoulli number $B_n$ as an exact rational number.

[Special cases] If $n=0$ , return $1$ ; if $n=1$ , return $-1/2$ ; if $n\geq 3$ is odd, return $0$ .
[Factorial factor] Compute $\ds K=\frac{2 \cdot n!}{(2\pi)^n}$ to sufficiently many digits of precision so the ceiling in step (6) is uniquely determined (this precision can be determined using Section The Number of Digits of ).
[Denominator] Compute $\ds d=\prod_{p-1 | n} p$ .
[Bound] Compute $\ds M = \left\lceil (K d)^{1/(n-1)} \right\rceil.$
[Approximate $\zeta(n)$ ] Compute $\ds z = \prod_{p\leq M} (1-p^{-n})^{-1}$ .
[Numerator] Compute $\ds a = (-1)^{n/2+1}\,\lceil d K z \rceil$ .
[Output $B_n$ ] Return $\ds \frac{a}{d}$ .

In step (5) use a sieve to compute all primes $p\leq M$ efficiently (which is fast, since $M$ is so small). In step (4) we may replace $M$ by any integer greater than the one specified by the formula, so we do not have to compute $(Kd)^{1/(n-1)}$ to very high precision.

In Section Computing Generalized Bernoulli Numbers Analytically below we will generalize the above algorithm.

Example 2.46

We illustrate Algorithm 2.45 by computing $B_{50}$ . Using 135 binary digits of precision, we compute

$K=7500866746076957704747736.71552473164563479.$

The divisors of $n$ are $1, 2, 5, 10, 25, 50$ , so

$d = 2\cdot 3\cdot 11=66.$

We find $M=4$ and compute

$z = 1.00000000000000088817842109308159029835012.$

Finally we compute

$dKz = 495057205241079648212477524.999999994425778,$

so

$B_{50}=\frac{495057205241079648212477525}{66}.$

Exercises¶

Exercise 2.1

Using Proposition 2.8 and the table found here, compute $\sum_{n=1}^{\infty}\frac{1}{n^{26}}$ explicitly.

Exercise 2.2

Prove that if $n>1$ is odd, then the Bernoulli number $B_n$ is $0$ .

Exercise 2.3

Use (3) to write down the coefficients of $1$ , $q$ , $q^2$ , and $q^3$ of the Eisenstein series $E_8$ .

Exercise 2.4

Suppose $k$ is a positive integer with $k\equiv 0 \pmod{12}$ . Suppose $a,b\geq 0$ are integers with $4a + 6b = k$ .

Prove $3 \mid a$ .
Show that $\ds G_4^a \cdot G_6^b \,/\, G_6^{\frac{k}{6}} = \left(G_4^3 / G_6^2 \right)^{\frac{a}{3}}.$

Exercise 2.5

Compute the Miller basis for $M_{28}(\SL_2(\Z))$ with precision $O(q^8)$ . Your answer will look like Example 2.23.

Exercise 2.6

Consider the cusp form $f = q^2 + 192 q^3 - 8280 q^4 + \cdots$ in $S_{28}(\SL_2(\Z))$ . Write $f$ as a polynomial in $E_4$ and $E_6$ (see Remark 2.25).

Exercise 2.7

Let $G_k$ be the weight $k$ Eisenstein series from equation (1). Let $c$ be the complex number so that the constant coefficient of the $q$ -expansion of $g = c \cdot G_k$ is $1$ . Is it always the case that the $q$ -expansion of $g$ lies in $\Z[[q]]$ ?

Exercise 2.8

Compute the matrix of the Hecke operator $T_2$ on the Miller basis for $M_{32}(\SL_2(\Z))$ . Then compute its characteristic polynomial and verify it factors as a product of two irreducible polynomials.

What Next? Much of the rest of this book is about methods for computing subspaces of $M_k(\Gamma_1(N))$ for general $N$ and $k$ . These general methods are more complicated than the methods presented in this chapter, since there are many more modular forms of small weight and it can be difficult to obtain them. Forms of level $N>1$ have subtle connections with elliptic curves, abelian varieties, and motives. Read on for more!

Modular Forms of Level 1¶

Examples of Modular Forms of Level 1¶

The Cusp Form $\Delta$ ¶

Fourier Expansions of Eisenstein Series¶

Structure Theorem for Level 1 Modular Forms¶

The Miller Basis¶

Hecke Operators¶

Computing Hecke Operators¶

Fast Computation of Fourier Coefficients¶

Fast Computation of Bernoulli Numbers¶

The Number of Digits of $B_n$ ¶

Computing $B_n$ Exactly¶

Exercises¶

Table Of Contents

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Modular Forms of Level 1¶

Examples of Modular Forms of Level 1¶

The Cusp Form ¶

Fourier Expansions of Eisenstein Series¶

Structure Theorem for Level 1 Modular Forms¶

The Miller Basis¶

Hecke Operators¶

Computing Hecke Operators¶

Fast Computation of Fourier Coefficients¶

Fast Computation of Bernoulli Numbers¶

The Number of Digits of ¶

Computing Exactly¶

Exercises¶

Table Of Contents

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The Cusp Form $\Delta$ ¶

The Number of Digits of $B_n$ ¶

Computing $B_n$ Exactly¶