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.
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
.
The product of two meromorphic functions on the upper half
plane is itself meromorphic. Also, for all
we have
so is weakly modular.
If is meromorphic on the upper half plane, then so is
.
Now
so is a weakly modular form of weight
.
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.
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.
Exercise 1.4.
Let be a weakly modular function of odd weight
. Since
, we have
so
.
Exercise 1.5.
Because is the trivial group,
must be all of
. As
, we must have
.
Exercise 1.7. See [DS05, Lemma 1.2.2].
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
.
Exercise 2.1. We have
Variation: Compute .
Exercise 2.2. Omitted.
Variation: Compute .
Exercise 2.4. Omitted.
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 .
Exercise 2.6.
From the previous exercise we have .
Then
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
).
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
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
.
Exercise 3.2.
For any integer ,
we have
,
so
.
Thus
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
Coset representatives for in
are
which we refer to below as and
, respectively.
In terms of representatives we have
By the first three relations we have and
.
Finally,
Exercise 4.1.
Suppose is a Dirichlet character with modulus
.
Then
, a contradiction.
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.
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.
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
.
The bijection given by
is a homomorphism since
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.
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.
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
.
Exercise 5.2.
When is the trivial character, the
are defined by
Thus ,
and for
, we have
.
Exercise 5.3. Omitted.
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
.
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
Omitted.
Exercise 6.2. Omitted.
Exercise 6.3. Omitted.
Exercise 6.4. Omitted.
Exercise 6.5. See the source code to Sage.
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
.
Exercise 7.2.
The answer is no. For example if is
times the
identity matrix and if
, then
but
.
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
.
The echelon form (over ) is
The kernel is the -dimensional span of
.
The characteristic polynomial is .
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
.
Exercise 1.12. See [Shi94, Lemma 1.38].
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
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.,
exactly the same way.)
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
.
Omitted.
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
.
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
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
.
Exercise 9.4. Omitted.
Hint: Use either repeated
integration by parts or a change of variables that relates the integral
to the function.
Exercise 10.2. See [Cre97a, Section 2.8].
Let . Then
, but the nontrivial
conjugate of
is
, so
has dimension
.
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
Exercise 10.4. See the appendix to Chapter II in [Cre97a], where this example is worked out in complete detail.