# Solutions to Selected Exercises¶

## Chapter 1¶

1. Exercise 1.1. Suppose is a matrix with positive determinant. Then is a linear fractional transformation that fixes the real line, so it must either fix or swap the upper and lower half planes. Now

so since , the imaginary part of is positive; hence sends the upper half plane to itself.

2. Exercise 1.2. Avoiding poles, the quotient rule for differentiation goes through exactly as in the real case, so any rational function () is holomorphic on . By the fundamental theorem of algebra, this set of poles is finite, and hence it is discrete. Write for each and let which is a polynomial nonzero at . Thus for each we have is holomorphic at and hence is meromorphic on .

1. The product of two meromorphic functions on the upper half plane is itself meromorphic. Also, for all we have

so is weakly modular.

2. If is meromorphic on the upper half plane, then so is . Now

so is a weakly modular form of weight .

3. Let and be modular functions. Then, as above, is a weakly modular function. Let and be their -expansions around any ; then their formal product is the -expansion of . But the formal product of two Laurant series about the same point is itself a Laurant series with convergence in the intersection of the convergent domains of the original series, so has a meromorphic -expansion at each and hence at each cusp.

4. We are in exactly the same case as in part (c), but because and are modular functions, and hence the function is holomorphic at each of its cusps.

3. Exercise 1.4. Let be a weakly modular function of odd weight . Since , we have so .

4. Exercise 1.5. Because is the trivial group, must be all of . As , we must have .

1. The group is the inverse image of the subgroup of generated by , and the inverse image of a group (under a group homomorphism) is a group.
2. The group contains the kernel of the homomorphism , and that kernel has finite index since the quotient is contained in , which is finite.
3. Same argument as previous part.
4. The level is at most since both groups contain . It can be no greater than since is in both groups.
5. Exercise 1.7. See [DS05, Lemma 1.2.2].

6. Exercise 1.8. Let , where and are relatively prime. By the Euclidian algorithm, we can find such that . Let . Note that and . Also let be the identity map on . Now sends to so we have which sends to .

## Chapter 2¶

1. Exercise 2.1. We have

Variation: Compute .

2. Exercise 2.2. Omitted.

3. Variation: Compute .

4. Exercise 2.4. Omitted.

5. Exercise 2.5. We have . A choice of with and is . A basis for is then

The Victor Miller basis is then

Variation: Compute the Victor Miller basis for .

6. Exercise 2.6. From the previous exercise we have . Then

7. Exercise 2.7. No, it is not always integral. For example, for , the coefficient of is . Variation: Find, with proof, the set of all such that the normalized series is integral (use that is eventually very large compared to ).

8. Exercise 2.8. We compute the Victor Miller basis to precision great enough to determine . This means we need up to .

Then the matrix of on this basis is

(The rows of this matrix are the linear combinations that give the images of the under .) This matrix has characteristic polynomial

## Chapter 3¶

1. Exercise 3.1. Write , so . Let be the isomorphism given by . We have

where the second equality can be verified easily by expanding out each side, and for the congruence we use that . Thus the subgroup of generated by is taken isomorphically to the subgroup of generated by .

2. Exercise 3.2. For any integer , we have , so . Thus

1. .
2. .
3. See [Cre97a, Prop. 2.2.1].
3. Exercise 3.4. We start with , . Then . Let . Since , we use the right coset representative and see that

Repeating the process, we have , which is in the same coset at . Thus

Putting it together gives

1. Coset representatives for in are

which we refer to below as and , respectively.

2. In terms of representatives we have

3. By the first three relations we have and .

4. Finally,

## Chapter 4¶

1. Exercise 4.1. Suppose is a Dirichlet character with modulus . Then , a contradiction.

1. Any finite subgroup of the multiplicative group of a field is cyclic (since the number of roots of a polynomial over a field is at most its degree), so is cyclic. Let be an integer that reduces to a generator of . Let ; by the binomial theorem

so has order . Since is odd, , so has order ; hence is cyclic.

2. By the binomial theorem , so has order in , and clearly has order . Since , is not a power of in . Thus the subgroups and have trivial intersection. The product of their orders is , so the claim follows.

2. Exercise 4.3. Write . The order of divides , so the condition implies that divides the order of for each . Thus the order of is divisible by the least common multiple of the , i.e., by .

1. The bijection given by is a homomorphism since

2. We have an exact sequence

so it suffices to solve the discrete log problem in the kernel and cokernel. We prove by induction on that we can solve the discrete log problem in the kernel easily (compared to known methods for solving the discrete log problem in ). We have an exact sequence

The first part of this problem shows that we can solve the discrete log problem in the kernel, and by induction we can solve it in the cokernel. This completes the proof.

3. Exercise 4.5. If , then since is nontrivial, Exercise 4.2 implies that factors through , hence has conductor , as claimed. If , then again from Exercise 4.2 we see that if has order , then factors through but nothing smaller.

1. Take .
2. The element has order .
3. A minimal generator for is , and the characters are , , , .
4. Each of the four Galois orbits has size .

## Chapter 5¶

1. Exercise 5.1. The eigenspace of with eigenvalue is preserved by , since if , then

Because is diagonalizable, its minimal polynomial equals its characteristic polynomial; hence the same is true for the restriction of to , i.e., the restriction of is diagonalizable. Choose basis for all so that the restrictions of to these eigenspaces is diagonal with respect to these bases. Then the concatentation of these bases is a basis that simultaneously diagonalizes and .

2. Exercise 5.2. When is the trivial character, the are defined by

Thus , and for , we have .

3. Exercise 5.3. Omitted.

4. Exercise 5.4. The Eisenstein series in our basis for are of the form or with . There are six characters with modulus such that , and we have the two series and associated to each of these. This gives a dimension of .

## Chapter 6¶

1. By Proposition 3.10, we have . By the Chinese Remainder Theorem,

So we are reduced to computing . We have with if and only if , so there are such pairs. The unit group has order . It follows that

2. Omitted.

1. Exercise 6.2. Omitted.

2. Exercise 6.3. Omitted.

3. Exercise 6.4. Omitted.

4. Exercise 6.5. See the source code to Sage.

## Chapter 7¶

1. Exercise 7.1. Take a basis of and let be the matrix whose rows are these basis elements. Let be the row echelon form of . After a permutation of columns, we may write , where is the identity matrix. The matrix , where is a different sized identity matrix, has the property that .

2. Exercise 7.2. The answer is no. For example if is times the identity matrix and if , then but .

3. Exercise 7.3. Let be an invertible matrix such that is in (reduced) echelon form and the are elementary matrices, i.e., the result of applying an elementary row operation to the identity matrix. If is a prime that does not divide any of the nonzero numerators or denominators of the entries of and any , then . This is because is in echelon form and can be transformed to via a series of elementary row operations modulo .

1. The echelon form (over ) is

2. The kernel is the -dimensional span of .

3. The characteristic polynomial is .

1. The answer is given in the problem.
2. See [Coh93, Section 2.4].

## Chapter 8¶

1. Exercise 1.11. Using the Chinese Remainder Theorem we immediately reduce to proving the statement when both and are powers of a prime . Then is represented by an integer with . That same integer defines an element of that reduces modulo to .

2. Exercise 1.12. See [Shi94, Lemma 1.38].

3. Exercise 1.13. Coset representitives for are in bijection with where and , so the following are representatives:

which we call , respectively. Now our Manin symbols are of the form and for modulo the relations

First, note that acts trivially on Manin symbols of odd weight because it sends to , to and to , so

Thus the last relation is trivially true.

Now and . Also and .

The first relation on the first symbol says that

and the second relation tells us that

4. Exercise 1.14. Let and . All that remains to be shown is that this pairing respects the relation for all modular symbols . By linearity it suffices to show the invariance of . We have

where the second to last simplification is due to invariance under , i.e.,

(The proof for works in

exactly the same way.)

1. Let . For any we have

First, if , then and

so . As , conjugation by is self-inverse, so it must be a bijection.

Now if , then (mod ), so (mod ), and so . Thus .

If , then (mod ) as before and also (mod ), so . Thus .

2. Omitted.

## Chapter 9¶

1. Exercise 9.1. Consider the surjective homomorphism

Notice that is the exact inverse image of the subgroup of matrices of the form and is the inverse image of the subroup of upper triangular matrices. It thus suffices to observe that is normal in , which is clear. Finally, the quotient is isomorphic to the group of diagonal matrices in , which is isomorphic to .

2. Exercise 9.2. It is enough to show for primes , since each can be written in terms of the . Since , we have that

so

By Dirichlet’s theorem on primes in arithmetic progression, there is a prime congruent to mod . Since and are relatively prime, there exist integers and such that . Then

3. Exercise 9.3. Take . The space is a direct sum of the two old subspaces coming from and the new subspace, which has dimension . If is a basis for and is a basis for , then is a basis for on which all Hecke operators , with , have diagonal matrix. However, the operator on does not act as a scalar on , so it cannot be in the ring generated by all operators with .

4. Exercise 9.4. Omitted.

## Chapter 10¶

1. Hint: Use either repeated integration by parts or a change of variables that relates the integral to the function.

2. Exercise 10.2. See [Cre97a, Section 2.8].

1. Let . Then , but the nontrivial conjugate of is , so has dimension .

2. Choose such that . Write

(1)

with . Let be the -span of the , and let . By considering the conjugates of (1), we see that the Galois conjugates of are in the -span of the , so

(2)

Likewise, taking the above modulo for any , we obtain a matrix equation

where the columns of are the -conjugates of , the matrix is the Vandermonde matrix corresponding to (and its conjugates), and has columns . Since is a Vandermonde matrix, it is invertible, so . Taking the limit as goes to infinity, we see that each is a linear combination of the , hence an element of . Thus , so (2) implies that . But so finally

3. Exercise 10.4. See the appendix to Chapter II in [Cre97a], where this example is worked out in complete detail.

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