APPENDIX TO CHAPTER II EXAMPLES

We give here some worked examples of the methods described in the preceding chapter, to illustrate and clarify the different situations which arise. The first example is N = 11, which is the first non-trivial level; here we give most detail. Then we consider N = 33, where we encounter oldforms and more complicated M-symbols, and N = 37, where there are two newforms, one of which has L(f, 1) = 0, necessitating a different method of computing Hecke eigenvalues. Finally we look at a square level, N = 49, to illustrate the direct method of computing periods.

Example 1: N = 11

For simplicity we will only work in H(11), rather than the smaller quotient space H^{+}(11). The M-symbols for N = 11 are (c : 1) for c modulo 11 and (1 : 0), which we abbreviate as (c) and (∞) respectively, with |c| ≤ 5. (Similarly with other prime levels). The 2-term and 3-term relations (2.2.6) and (2.2.7) are as follows.

\begin{aligned}(0) + (∞) & = 0 \\ (1) + (−1) & = 0 \\ (2) + (5) & = 0 \\ (−2) + (−5) & = 0 \\ (3) + (−4) & = 0 \\ (−3) + (4) & = 0 \end{aligned}  \begin{aligned}(0) + (∞) + (−1) & = 0 \\ (1) + (−2) + (5) & = 0 \\ (2) + (4) + (−4) & = 0 \\ (3) + (−5) + (−3) & = 0 \end{aligned}

Solving these equations we can express all 12 symbols in terms of A = (2), B = (3) and C = (0):

\begin{aligned}(0) & = C \\ (∞) & = −C \\ (1) = (−1) & = 0 \\ (2) = (−2) & = A \\ (5) = (−5) & = −A \end{aligned}  \begin{aligned}(3) & = B \\ (−3) & = A − B \\ (4) & = B − A \\ (−4) & = −B \end{aligned}

There are two classes of cusps, [0] and [∞], with [a/b] = [0] if 11 ∤ b and [a/b] = [∞] if 11 | b. Hence δ((c)) = δ({0, 1/c}) = [1/c] − [0] = 0 for c ≢ 0. It follows that

H(11) = ker(δ) = ⟨A, B⟩ ,

with 2g = dim H(11) = 2, so that the genus is 1. There is therefore one newform f. This makes the rest of the calculation simpler, as we do not have to find and split off eigenspaces.

The conjugation ∗ involution maps (c) ↦ (−c), so A^{∗} = A and B^{∗} = A − B. This has

EXAMPLE 1: N = 11 53

matrix ( \begin{matrix}1 & 1 \\ 0 & −1 \end{matrix} ) with respect to the basis A, B. The +1- and −1-eigenspaces are generated by A and A − 2B respectively, and we have left eigenvectors v^{+} = (2, 1) and v^{−} = (0, 1). Thus the period lattice is Type 1 (non-rectangular), and Ω(f) = Ω_{0}(f) = ⟨A, f⟩.

If we had worked in H^{+}(11), viewed as the quotient H(11)/H^{−}(11), by including relations (c) = (−c), the effect would be to identify (c) and (−c). This gives a 1-dimensional space generated by \overline{B} with \overline{A} = 2\overline{B}, where the bars denote the projections to the quotient. Notice that although \overline{B} is a generator here, the integral of f over B is not a real period; its real part is half the real period. However we do still have Ω(f) = ⟨B + B^{∗}, f⟩ = 2Re ⟨B, f⟩, so we could compute Ω(f) in this context without actually knowing whether it was 1 or 2 times the smallest real period.

To compute Hecke eigenvalues we may work in the subspace ⟨A⟩; since this subspace is conjugation invariant (being the +1-eigenspace) we will have T_{p}(A) = a_{p}A for all p ≠ 11. We first compute T_{2} explicitly. The first method, converting the M-symbol A = (2 : 1) to the modular symbol {0, 1/2}, gives:

T_{2}(A) = T_{2} (\{0, \frac{1}{2}\}) = \{0, 1\} + \{0, \frac{1}{4}\} + \{\frac{1}{2}, \frac{3}{4}\}

= \{0, 1\} + \{0, \frac{1}{4}\} + \{\frac{1}{2}, 1\} + \{1, \frac{3}{4}\}

= (1 : 1) + (4 : 1) + (1 : 2) + (−4 : 1)

= (1) + (4) + (−5) + (−4)

= 0 + (B − A) + (−A) + (−B) = −2A,

so that a_{2} = −2. Alternatively, using the Heilbronn matrices from Section 2.4, we compute:

T_{2}(A) = T_{2}((2 : 1)) = (2 : 1)R_{2}

= (2 : 1) ( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ) + (2 : 1) ( \begin{matrix}2 & 0 \\ 0 & 1 \end{matrix} ) + (2 : 1) ( \begin{matrix}2 & 1 \\ 0 & 1 \end{matrix} ) + (2 : 1) ( \begin{matrix}1 & 0 \\ 1 & 2 \end{matrix} )

= (2 : 2) + (4 : 1) + (4 : 3) + (3 : 2)

= (1) + (4) + (5) + (−4)

= 0 + (B − A) + (−A) + (−B)

= −2A.

Now (1 + 2 − a_{2})L(f, 1) = ⟨{0, 1/2} , f⟩ = ⟨A, f⟩ = Ω(f), giving

\frac{L(f, 1)}{Ω(f)} = \frac{1}{5}.

For all primes p ≠ 11 we will evaluate μ_{p} = ∑_{a=0}^{p−1} {0, a/p} = n_{p}A for a certain integer n_{p}, since then also 1/5 = n_{p}/(1 + p − a_{p}), giving

a_{p} = 1 + p − 5n_{p}.

At this stage we already know that the corresponding elliptic curve has rank 0, and that 1 + p − a_{p} ≡ 0 (mod 5) for all p ≠ 11, so that it will possess a rational 5-isogeny.

To save time, we can use the fact that {0, a/p}^{∗} = {0, −a/p}; thus for odd p we need only evaluate half the sum, say

μ′_{p} = ∑_{a=1}^{(p−1)/2} \{0, \frac{a}{p}\} ,

and then set μ_{p} = μ′_{p} + (μ′_{p})^{∗}.

For p = 3, we have μ′_{3} = {0, 1/3} = (3 : 1) = (3) = B, so μ_{3} = B + B^{∗} = A, giving n_{3} = 1 and a_{3} = 1 + 3 − 5n_{3} = −1.

For p = 5 we compute:

\{0, \frac{1}{5}\} = (5 : 1) = (5) = −A;
\{0, \frac{2}{5}\} = \{0, \frac{1}{2}\} + \{\frac{1}{2}, \frac{2}{5}\}
= (2 : 1) + (−5 : 2) = (2) + (3) = A + B;
μ′_{5} = (−A) + (A + B) = B;
μ_{5} = B + B^{∗} = A,  so that n_{5} = 1;
a_{5} = 1 + 5 − 5n_{5} = 1.

Similarly, with p = 7 we have n_{7} = 2, so that a_{7} = 1 + 7 − 5n_{7} = −2, and with p = 13 we have n_{13} = 2 so that a_{13} = 4.

These computations can also be carried out using Heilbronn matrices, by applying the Hecke operators directly to C = (0 : 1) = {0, ∞}, as follows. Once we know that a_{2} = −2, we have

−2C = T_{2}(C) = T_{2}((0 : 1)) = (0 : 1)R_{2} = (0 : 2) + (0 : 1) + (0 : 1) + (1 : 2) = 3C − A,

giving C = \frac{1}{5}A in agreement with the ratio L(f, 1)/Ω(f) found earlier. Similarly, using the Heilbronn matrices R_{3} listed in Section 2.4, we find

a_{3}C = T_{3}(C) = (0 : 1)R_{3}
= (0 : 3) + (0 : 1) + (1 : 3) + (0 : 1) + (0 : 1) + (1 : −3)
= C + C + (B − A) + C + C + (−B)
= 4C − A = −C,

giving a_{3} = −1 again.

For the prime q = 11 we compute the involution W_{11} induced by the action of the matrix

( \begin{matrix}0 & −1 \\ 11 & 0 \end{matrix} ):

W_{11}(A) = ( \begin{matrix}0 & −1 \\ 11 & 0 \end{matrix} ) \{0, \frac{1}{2}\} = \{∞, \frac{−2}{11}\}

= \{∞, 0\} + \{0, \frac{−1}{5}\} + \{\frac{−1}{5}, \frac{−2}{11}\}

= (1 : 0) + (−5 : 1) + (11 : 5)

= (∞) + (−5) + (0)

= −A,

so that the eigenvalue ε_{11} of W_{11} is −1. In fact, this was implicit earlier, since L(f, 1) ≠ 0 implies that the sign of the functional equation is +1, which is minus the eigenvalue of the Fricke involution W_{11}.

The Fourier coefficients a(n) = a(n, f) for 1 ≤ n ≤ 16 are now given by the following table.

\begin{matrix}n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ a(n) & 1 & −2 & −1 & 2 & 1 & 2 & −2 & 0 & −2 & −2 & 1 & −2 & 4 & 4 & −1 & −4 \end{matrix}

Here we have used multiplicativity, and:

a(11) = −ε_{11} = +1;
a(4) = a(2)^{2} − 2a(1) = 2;
a(8) = a(2)a(4) − 2a(2) = 0;
a(16) = a(2)a(8) − 2a(4) = −4;
a(9) = a(3)^{2} − 3a(1) = −2.

We know that the period lattice Λ_{f} has a ℤ-basis of the form [ω_{1}, ω_{2}] = [2x, x + iy], where ω_{1} = P_{f}(A) and ω_{2} = P_{f}(B). We can compute the real period ω_{1} = Ω(f) = 5L(f, 1) by computing L(f, 1):

L(f, 1) = 2 ∑_{n=1}^{∞} \frac{a(n)}{n}t^{n}

where t = exp(−2π/\sqrt{11}) = 0.15 . . .. Using the first 16 terms which we have, already gives this to 13 decimal places:

L(f, 1) = 0.2538418608559 …;

thus

ω_{1} = Ω(f) = 1.269209304279 ….

For the imaginary period y we twist with a prime l ≡ 3 (mod 4). Here l = 3 will do, since

γ_{3} = \{0, \frac{1}{3}\} − \{0, \frac{−1}{3}\} = (3) − (−3) = −A + 2B ≠ 0.

To project onto the minus eigenspace we take the dot product of this cycle (expressed as a row vector (−1, 2)) with v^{−} = (0, 1) to get m^{−}(3) = 2. Hence

y = \frac{1}{2i}P (3, f) = \frac{\sqrt{3}}{2}L(f ⊗ 3, 1).

Summing the series for L(f ⊗ 3, 1) to 16 terms gives only 4 decimals:

L(f ⊗ 3, 1) = 1.6845 ….

This is less accurate than L(f, 1) since this series is a power series in exp(−2π/3\sqrt{11}) = 0.53 …, compared with 0.15 . . .. Hence y = 1.4588 …, so that

ω_{2} = 0.634604652139 … + 1.4588 … i.

So far we have only used the Hecke eigenvalues a_{p} for p ≤ 13, and only 16 terms of each series. If we use these approximate values for the period lattice generators ω_{1} and ω_{2} we already find the approximate values c_{4} = 495.99 and c_{6} = 20008.09 which round to the integer values c_{4} = 496 and c_{6} = 20008. Taking the first 25 a_{p} and the first 100 terms of the series gives

c_{4} = 495.9999999999954 …  and  c_{6} = 20008.0000000085.

The exact values c_{4} = 496 and c_{6} = 20008 are the invariants of an elliptic curve of conductor 11, which is in fact the modular curve E_{f}:

y^{2} + y = x^{3} − x^{2} − 10x − 20.

This is the first curve in the tables, with code 11A1 (or Antwerp code 11B). The value L(f, 1)/Ω(f) = 1/5 agrees with the value predicted by the Birch–Swinnerton-Dyer conjecture for L(E_{f}, 1)/Ω(E_{f}) , provided that E_{f} has trivial Tate–Shafarevich group.

We now illustrate the shortcut method presented in Section 2.11, where we guess the imaginary period and lattice type without computing H(11). Having computed P (3, f) = 2.9176 … i which is certainly non-zero, we consider the lattices ⟨x, yi⟩ and ⟨2x, x + yi⟩, where 2x = 1.2692 … (from above) and yi = P (3, f)/m^{−}, for m^{−} = 1, 2, 3, . . .. With m^{−} = 1 we do not find integral invariants, but for m^{−} = 2 and lattice type 1 we find the curve E_{f} = [0, −1, 1, −10, −20] given above^{1}.

Using the first variant of the method, where we do not even know x, we can take l^{+} = 5 since P (5, f) = 6.346 … ≠ 0. The correct value of m^{+} here is 10; if we do not know this, but try m^{+} = 1, 2, 3 … in a double loop with m^{−}, the first valid lattice we come across is with (m^{+}, m^{−}) = (2, 2) and type 1, which leads to the curve E′ = [0, −1, 1, 0, 0], also of conductor 11; this is 5-isogenous to the “correct” curve E_{f}, which comes from (m^{+}, m^{−}) = (10, 2) and type 1.

We may also consider the ratios P (l, f)/P (3, f) for other primes l ≡ 3 (mod 4); we restrict to those l satisfying (\frac{−11}{l}) = (\frac{l}{11}) = +1, since otherwise P (l, f) is trivially 0 (since the sign of the functional equation for the corresponding L(f ⊗ χ, s) is then −1). We find the following table of values (rounded: they are only computed approximately):

\begin{matrix}l & 3 & 23 & 31 & 47 & 59 & 67 & 71 & 103 & 163 & 179 & 191 & 199 & 223 & 251 \\ \frac{P (l, f)}{P (3, f)} & 1 & 1 & 1 & 0 & 1 & 9 & 1 & 0 & 4 & 25 & 1 & 4 & 1 & 1 \end{matrix}

The zero values for l = 47 and l = 103 indicate that the corresponding twists of the newform f have positive even analytic rank (one can check that the corresponding twists of the curve E_{f} do indeed have rank 2). As all these values are integral here (a priori they are only known to be rational) we do not find any nontrivial divisor of m^{−} (which we know in fact equals 2). The fact that all the integers are perfect squares is an amusing observation, but has a simple explanation in terms of the numbers appearing in the Birch–Swinnerton-Dyer conjecture for the twists of E_{f}.

There is one other curve E′′ isogenous to E_{f} in addition to E′ (found above). If the period lattice of E_{f} = [0, −1, 1, −10, −20] is ⟨2x, yi⟩ with x = 0.6346... and y = 1.4588..., then E′ = [0, −1, 1, 0, 0] has period lattice ⟨10x, 5x + yi⟩, and E′′ = [0, −1, 1, −7820, 263580] has lattice ⟨x/5, 2x/5 + yi⟩. These curves are linked by 5-isogenies E_{f} ↔ E′ and E_{f} ↔ E′′.

Finally, we compute the degree of the modular parametrization φ: X_{0}(11) → E_{f}. Of course, this is obviously 1, since the modular curve X_{0}(11) has genus 1, so that φ is the identity map in this case; but this example will serve to illustrate the general method.

The twelve M-symbols form 4 triangles which we choose as follows:

(1, 0), (−1, 1), (0, 1);  (1, 1), (−2, 1), (−1, 2);
(1, 2), (−3, 1), (−2, 3);  (1, 3), (−4, 1), (−3, 4).

1Here, [a_{1}, a_{2}, a_{3}, a_{4}, a_{6}] denote the Weierstrass coefficients of the curve; see Chapter 3.

There are two τ-orbits, corresponding to the two cusps at ∞ (of width 1) and at 0 (of width 11). The first contributes nothing. The second is as follows:

(1, 0) ↦ (1, 1) ↦ (1, 2) ↦ (1, 3) ↦ (1, 4) ≡ (−2, 3) ↦ (−2, 1) ↦ (−2, −1) ≡ (−3, 4) ↦ (−3, 1) ↦ (−3, −2) ≡ (−4, 1) ↦ (−4, −3) ≡ (−1, 2) ↦ (−1, 1) ↦ (1, 0).

There are four jump matrices coming from the above sequence. From (1, 4) ≡ (−2, 3) we obtain

δ_{1} = ( \begin{matrix}0 & −1 \\ 1 & 4 \end{matrix} ) ( \begin{matrix}1 & −1 \\ −2 & 3 \end{matrix} )^{−1} = ( \begin{matrix}−2 & −1 \\ 11 & 5 \end{matrix} );

the others are δ_{2} = ( \begin{matrix}4 & 1 \\ 11 & 3 \end{matrix} ), δ_{3} = ( \begin{matrix}−5 & −1 \\ 11 & 2 \end{matrix} ) and δ_{4} = ( \begin{matrix}−3 & 1 \\ 11 & −4 \end{matrix} ). Using modular

symbols, we can compute the coefficients of P_{f}(δ_{i}) with respect to the period basis ω_{1}, ω_{2}, to obtain P_{f}(δ_{1}) = −ω_{1}, P_{f}(δ_{2}) = ω_{2}, P_{f}(δ_{3}) = ω_{1}, and P_{f}(δ_{4}) = −ω_{2}. Hence

deg(φ) = \frac{1}{2} (| \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} | + | \begin{matrix}0 & 0 \\ −1 & 1 \end{matrix} | + | \begin{matrix}0 & −1 \\ −1 & 0 \end{matrix} | + | \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} | + | \begin{matrix}1 & −1 \\ 0 & 0 \end{matrix} | + | \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} |)

= \frac{1}{2}(1 + 0 − 1 + 1 + 0 + 1) = 1,

as expected.

Example 2: N = 33

Since 33 = 3·11, the number of M-symbols is 48 = 4·12, consisting of 33 symbols (c) = (c : 1), 13 symbols (1 : d) with gcd(d, 33) > 1, and the symbols (3 : 11) and (11 : 3). (In fact, whenever N is a product pq of 2 distinct primes, the M-symbols have this form, with exactly two symbols, (p : q) and (q : p) not of the form (c : 1) or (1 : d)).

There are four cusp classes represented by 0, 1/3, 1/11 and ∞, with the class of a cusp a/b being determined by gcd(b, 33). (Similarly, whenever N is square-free, the cusp classes are in one-one correspondence with the divisors of N).

Using the two-term and three-term relations, and including the relations (c : d) = (−c : d), we can express all the M-symbols in terms of six of them, and ker(δ^{+}) = ⟨(7), (2), (15) − (9)⟩. Hence H^{+}(33) is three-dimensional. We know there will be a two-dimensional oldclass coming from the newform at level 11, so there will also be a single newform f at this level.

If we compute the images of the basis modular symbols {0, 1/7}, {0, 1/2} and {1/9, 1/15} under T_{2} and W_{33}, we find that they have matrices

T_{2} = ( \begin{matrix}−2 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & −2 \end{matrix} )  and  W_{33} = ( \begin{matrix}1 & 0 & 0 \\ 0 & −1 & 0 \\ 1 & 0 & −1 \end{matrix} ) .

T_{2} has a double eigenvalue of −2, coming from the oldforms, which we ignore, and also the new eigenvalue a_{2} = 1 with left eigenvector v = (0, 1, 0). The corresponding eigenvalue for W_{33} is ε_{33} = −1. Hence the sign of the functional equation is +, and the analytic rank is even. Moreover since the eigencycle for a_{2} is the second basis element, which is {0, 1/2} = μ_{2}, we have 2(1 + 2 − a_{2})L(f, 1) = Ω(f), so that

\frac{L(f, 1)}{Ω(f)} = \frac{1}{4}.

In particular, L(f, 1) ≠ 0, so that the analytic rank is 0. Note that because we have factored out the pure imaginary component, we do not usually know at this stage whether the least real period Ω_{0}(f) is equal to Ω(f) or half this; all we can say is that Ω(f)/2 = Re ⟨{0, 1/2}, f⟩ is the least real part of a period (up to sign). But in this case, {0, 1/2} is certainly an integral cycle, and since ⟨{0, 1/2}, f⟩ is real, we can in fact deduce already that the period lattice is of type 2 (rectangular) with Ω(f) = 2Ω_{0}(f).

To compute more a_{p} we express each cycle μ_{p} as a linear combination of the basis and project to the eigenspace by taking the dot product with the left eigenvector v, which just amounts in this case to taking the second component. In this way we find a_{5} = −2, a_{7} = 4, a_{13} = −2, and so on. For the involutions W_{3} and W_{11} we can either compute their 3 × 3 matrices or just apply them directly to the eigencycle {0, 1/2}, and we find that ε_{3} = +1 and ε_{11} = −1. In fact we already knew that the product of these was ε_{33} = −1, so we need not have computed ε_{11} directly, though doing so serves as a check.

Now we go back and compute the full space H(33), which is six-dimensional, with basis

\{0, \frac{1}{7}\} , \{0, \frac{1}{4}\} , \{0, \frac{−1}{4}\} , \{\frac{1}{12}, \frac{−1}{6}\} , \{\frac{1}{12}, \frac{−1}{3}\} , \{0, \frac{1}{10}\} .

By computing the 6 × 6 matrices of conjugation and T_{2}, we may pick out the left eigenvectors

v^{+} = (0, 1, −1, 1, 2, 0)  and  v^{−} = (−1, 0, 0, 2, 1, 1).

Since these vectors are independent modulo 2, it follows (as expected) that the period lattice is type 2, with a ℤ-basis of the form [ω_{1}, ω_{2}] = [x, yi].

Firstly, x = Ω_{0}(f) = Ω(f)/2 = 2L(f, 1). Summing the series for L(f, 1) we obtain L(f, 1) = 0.74734 …, so that ω_{1} = x = 1.49468 … and Ω(f) = 2x = 2.98936 . . .. Then we use the twisting prime l = 7: the twisting cycle

γ_{7} = ∑_{a=1}^{6} (\frac{a}{7}) \{0, \frac{a}{7}\}

is evaluated in terms of our basis to be (2, 2, 0, −2, 0, 0), whose dot product with v^{−} is −6. Hence y = \sqrt{7}L(f ⊗7, 1)/6. The value of L(f ⊗7, 1) is determined by summing the series to be 3.11212 …, so that y = 1.37232 … and ω_{2} = 1.37232 … i. If we evaluate these from the first 100 terms of the series, using a_{p} for p < 100, we find the approximate values c_{4} = 552.99999 … and c_{6} = −4084.99947 . . .. These round to c_{4} = 553 and c_{6} = −4085, which are the invariants of the curve 33A1: y^{2} + xy = x^{3} + x^{2} − 11x. Notice that this curve has four rational points, which we could have predicted since the ratio L(f, 1)/Ω(f) = 1/4 implies that 1 + p − a_{p} ≡ 0 (mod 4) for all p ≠ 2, 3, 11.

Example 3: N = 37

Since 37 is prime the M-symbols are simple here, as for N = 11. We find that H^{+}(37) is two-dimensional, generated by A = (8) and B = (13). With this basis the matrices of T_{2} and W_{37} are

T_{2} = ( \begin{matrix}−2 & 0 \\ 0 & 0 \end{matrix} )  and  W_{37} = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .

Thus we have two one-dimensional eigenspaces, generated by A and B respectively, with eigenvalues (a_{2} = −2, ε_{37} = +1) for A and (a_{2} = 0, ε_{37} = −1) for B. The left eigenvectors are simply v_{1} = (1, 0) and v_{2} = (0, 1). Let us denote the corresponding newforms by f and g respectively. Now {0, 1/2} = 2B, so

\frac{L(f, 1)}{Ω(f)} = 0  and  \frac{L(g, 1)}{Ω(g)} = \frac{1}{3}.

The fact that ε_{37}(f) = +1 implies that f has odd analytic rank, while the previous line shows that g has analytic rank 0..

To compute Hecke eigenvalues, the method we used previously would only work for g, so instead we use the variation discussed in Section 2.9. The cycle {1/5, ∞} projects non-trivially onto both eigenspaces. In fact (1 + 2 − T_{2}){1/5, ∞} = −5A − B, so the components in the two eigenspaces are (−5)/(1 + 2 − (−2)) = −1 and (−1)/(1 + 2 − 0) = −1/3. Hence by computing (1 + p − T_{p}){1/5, ∞} = n_{1}(p)A + n_{2}(p)B for other primes p ≠ 37, we may deduce that

a(p, f) = 1 + p + n_{1}(p)  and  a(p, g) = 1 + p + 3n_{2}(p).

In this way we find that the first few Hecke eigenvalues are as follows:

\begin{matrix}p & 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & … \\ a(p, f) & −2 & −3 & −2 & −1 & −5 & −2 & 0 & 0 & … \\ a(p, g) & 0 & 1 & 0 & −1 & 3 & −4 & 6 & 2 & … \end{matrix}

Two things can be noticed here: the preponderance of negative values amongst the first few a(p, f) means that the curve E_{f} has many points modulo p for small p, which we might expect heuristically since we know that its analytic rank is odd, and hence positive. Secondly, since 1 + p − a(p, g) ≡ 0 (mod 3) for all p ≠ 37, we know that E_{g} will have a rational 3-isogeny.

Turning to the full space H(37), we find that it has basis ⟨(8), (16), (20), (28)⟩. Conjugation and W_{37} have matrices

( \begin{matrix}0 & −1 & 0 & 0 \\ −1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{matrix} )  and  ( \begin{matrix}0 & −1 & 0 & 0 \\ −1 & 0 & 0 & 0 \\ 1 & 0 & 0 & −1 \\ 0 & −1 & −1 & 0 \end{matrix} )

respectively.

For the A eigenspace corresponding to f we find left eigenvectors v_{1}^{+} = (−1, 1, 0, 0) and v_{1}^{−} = (−1, 0, −1, 1). These are independent modulo 2, so the period lattice is rectangular, say [x, yi]. To find x we must twist by a real quadratic character, using a prime l ≡ 1 (mod 4). Here l = 5 will do: the twisting cycle is {0, 1/5}−{0, 2/5}−{0, 3/5}+{0, 4/5} = (2, −2, 0, 2), whose dot product with v_{1}^{+} is −4, so that x = \sqrt{5}L(f ⊗5, 1)/4. For the imaginary period we use l = 3 with twisting cycle (0, 0, −1, 1) and a dot product of 2 with v_{1}^{−}, so that y = \sqrt{3}L(f ⊗ 3, 1)/2. Evaluating numerically, using 100 terms of the series and a_{p} for p < 100, we find the values

L(f ⊗ 5, 1) = 5.35486 …,  so that  x = 2.99346 …;
L(f ⊗ 3, 1) = 2.83062 …,  so that  y = 2.45139 …;

and finally,

c_{4} = 47.9999999996 …,
c_{6} = −216.000000004 ….

The rounded values c_{4} = 48 and c_{6} = −216 are those of the curve 37A1, with equation y^{2} + y = x^{3} − x. This curve does have rank 1. We may also check that the analytic rank is 1 by computing L′(f, 1) by summing the series given in Section 2.13: we find that L′(f, 1) = 0.306 …, which is certainly non-zero.

The B eigenspace is handled similarly to the example at level 33. We find v_{2}^{+} = (0, 1, 1, 1) and v_{2}^{−} = (1, 1, 0, 0). The period lattice is [x, iy] with x = 3L(g, 1)/2 and y = \sqrt{19}L(g⊗19, 1)/4. The latter needs more terms to compute to sufficient accuracy, as 19 is larger than the twisting primes we have previously used. Using p < 100 as before we find c_{4} = 1119.878 …, which rounds to the correct (with hindsight) value 1120, but for c_{6} we get 36304.495, and neither 36304 nor 36305 is correct. Going back to compute a_{p} for 100 < p < 200 we reevaluate the series to 200 terms, and find

L(g, 1) = 0.72568 …,  so that  x = 1.08852 …;
L(g ⊗ 19, 1) = 1.62207 …,  so that  y = 1.76761 …;

and hence

c_{4} = 1120.000008 …,  and  c_{6} = 36295.99943 ….

Now the rounded values c_{4} = 1120 and c_{6} = 36296 are the invariants of the curve 37B1 with equation y^{2} + y = x^{3} + x^{2} − 23x − 50. As expected, this curve does admit a rational 3-isogeny.

Example 4: N = 49

H(49) is two-dimensional, with a basis consisting of the M-symbols (11), (2). Hence there is a unique newform f at this level, which must be its own −7-twist, or in other words have

complex multiplication by −7. The conjugation matrix with respect to this basis is ( \begin{matrix}−1 & 0 \\ −1 & 1 \end{matrix} ),

so we take v^{+} = (1, −2) and v^{−} = (1, 0). Hence the period lattice has the form [2x, x+yi] with 2x = Ω_{0}(f) = Ω(f). Also a_{2} = 1, so we have L(f, 1)/Ω(f) = 1/2. Hence we may compute the real period via L(f, 1) as before, and find L(f, 1) = 0.96666 …, so that Ω(f) = 1.9333 . . .. But the method we have used in the earlier examples to find the imaginary period will not work here, since for every prime l ≡ 3 (mod 4), l ≠ 7, we have L(f ⊗l, 1) = 0, since χ(−49) = χ(−1) = −1 where χ is the associated quadratic character modulo l.

Instead, we compute periods directly, as in Section 2.10. The cycle (5) = {0, 1/5} is equal to (11)+(2), from which it follows that ⟨(5), f⟩ = −x+yi; the coefficients are the dot products of

the vector (1, 1) with v^{±}. Now \{0, 1/5\} = \{0, M(0)\} with M = ( \begin{matrix}10 & 1 \\ 49 & 5 \end{matrix} ). Hence the simpler

formula (2.10.5) gives

P_{f}(M) = ⟨\{0, \frac{1}{5}\} , f⟩ = −x + yi = ∑_{n=1}^{∞} \frac{a(n)}{n}e^{−2πn/49} (e^{2πinx_{2}} − e^{2πinx_{1}})

where x_{1} = −5/49 and x_{2} = 10/49. Summing the first 100 terms as before, we find the values

x = 0.96666 …  and  y = 2.557536 ….

Of course, the value of x merely confirms the value we had previously obtained a different way. These values give, in turn,

c_{4} = 104.99992 …  and  c_{6} = 1322.9994 …,

which round to the exact invariants c_{4} = 105 and c_{6} = 1323 of the curve 49A1, which has equation y^{2} + xy = x^{3} − x^{2} − 2x − 1.

Using the improved formula (2.10.8) with better convergence, gives (also using 100 terms)

x = 0.96665585 …  and  y = 2.55753099 …,

which lead to the better values

c_{4} = 104.9999992 …  and  c_{6} = 1322.99998 ….

In this computation, we have not exploited the presence of complex multiplication. Notice that, in fact, y/x = \sqrt{7}. Obviously if we had known this it would have given us an easier way of computing y from x, and hence from L(f, 1). However not all newforms at square levels have complex multiplication. Some are twists of forms at lower levels (for example, 100A is the 5-twist of 20A, and 144B is the −3-twist of 48A), which means that we could find the associated curves more easily by twisting the earlier curve. Others first appear at the square level in pairs which are twists of each other (for example, 121A and 121C are −11-twists of each other, and 196B is the −7-twist of 196A). One could probably find both periods of all such forms by looking at suitable twists to moduli not coprime to the level, but we have not done this systematically, as the more direct method was adequate in all the cases we came across in compiling the tables.

In practice we always computed the Hecke eigenvalues for p < 1000 at least, with a larger bound for higher levels. In some cases, particularly when the target values of c_{4} or (more usually) c_{6} were large, and especially when a large twisting prime was needed, we needed to sum the series to several thousand terms before obtaining the vales of c_{4} and c_{6} to sufficient accuracy.

These four examples exhibit essentially all the variations which can occur. The only problem with the larger levels is one of scale, as the number of symbols and the dimensions of the spaces grow. A large proportion of the computation time, in practice, is taken up with Gaussian elimination. This is why we have tried wherever possible to reduce the size of the matrices which occur: first by carefully using the 2-term symbol relations to identify symbols in pairs as early as possible, and secondly by working in H^{+}(N) during the stage where we are searching for Hecke eigenvalues. The symbol relation matrices are very sparse (with at most three entries per row); sparse matrix techniques, which we use in our implementation, help greatly here. For finding eigenvectors of the Hecke algebra, however, we use a completely general purpose exact Gaussian elimination procedure.

The second time-consuming stage is when we are computing a large number of Hecke eigenvalues, where we call a very large number of times the procedures to convert rational numbers (cusps) to M-symbols and look these up in tables to find their coordinates with respect to the symbol basis. It is vital that these procedures are written efficiently; during the preparation of the tables, many great improvements in the efficiency of the program were achieved over a period of several months.