Solutions to Selected Exercises

Chapter 1

  1. Exercise 1.1. Suppose \gamma = \abcd{a}{nb}{c}{d} \in \GL_2(\R) is a matrix with positive determinant. Then \gamma is a linear fractional transformation that fixes the real line, so it must either fix or swap the upper and lower half planes. Now

    \gamma(i) = \frac{ai+b}{ci+d} = \frac{ac+bd + (ad-bc)i}{d^2+c^2},

    so since \det \gamma = ad-bc > 0, the imaginary part of \gamma(i) is positive; hence \gamma 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 f(z) = p(z)/q(z) (p,q \in \C[z]) is holomorphic on \C-\{\alpha : q(\alpha) = 0 \}. By the fundamental theorem of algebra, this set of poles is finite, and hence it is discrete. Write q (z) = a_n(z-\alpha_1)^{r_1}\cdots(z-\alpha_k)^{r_k} for each \alpha_i and let q_i(z)=q(z)/(z-\alpha_i)^{r_i} which is a polynomial nonzero at \alpha_i. Thus for each i we have (z-\alpha_i)^{r_i}f(z) = p(z)/q'(z) is holomorphic at \alpha_i and hence f(z) is meromorphic on \C.

  3. Exercise 1.3.

    1. The product fg of two meromorphic functions on the upper half plane is itself meromorphic. Also, for all \gamma \in \SL_2(\Z) we have

&= \frac{1}{(cz+d)^{k+j}}((fg) \circ \gamma)\\
&= \frac{1}{(cz+d)^k}(f \circ \gamma) \frac{1}{(cz+d)^j}(g \circ \gamma)
= fg,

      so fg is weakly modular.

    2. If f is meromorphic on the upper half plane, then so is 1/f. Now

      \qquad\qquad\frac{1}{f} = \frac{1}{(cz+d)^{-k} f \circ \gamma} = (cz+d)^k ((1/f) \circ \gamma) = \frac{1}{f}^{[\gamma]_{-k}},

      so 1/f is a weakly modular form of weight -k.

    3. Let f and g be modular functions. Then, as above, fg is a weakly modular function. Let \sum_{n=m}^\infty a_nq^n and \sum_{n=m'}^\infty b_nq^n be their q-expansions around any \alpha \in \P^1(\Q); then their formal product is the q-expansion of fg. 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 fg has a meromorphic q-expansion at each \alpha \in
\P^1(\Q) and hence at each cusp.

    4. We are in exactly the same case as in part (c), but because f and g are modular functions, m, m'\ge 0 and hence the function is holomorphic at each of its cusps.

  4. Exercise 1.4. Let f be a weakly modular function of odd weight k. Since \gamma = \abcd{-1}{0}{0}{-1} \in \SL_2(\Z), we have f(z) =
(-1)^{-k}f(\gamma(z)) = -f(z) so f = 0.

  5. Exercise 1.5. Because \SL_2(\Z/1\Z) is the trivial group, \Gamma(1) =
\ker(\SL_2(\Z) \rightarrow \SL_2(\Z/1\Z)) must be all of \SL_2(\Z). As \SL_2(\Z) = \Gamma(1) \subset \Gamma_1(1)
\subset \Gamma_0(1) \subset \SL_2(\Z), we must have \Gamma(1) =
\Gamma_1(1) = \Gamma_0(1) = \SL_2(\Z).

  6. Exercise 1.6.

    1. The group \Gamma_1(N) is the inverse image of the subgroup of \SL_2(\Z/N\Z) generated by \abcd{1}{1}{0}{1}, and the inverse image of a group (under a group homomorphism) is a group.
    2. The group contains the kernel of the homomorphism \SL_2(\Z) \to \SL_2(\Z/N\Z), and that kernel has finite index since the quotient is contained in \SL_2(\Z/N\Z), which is finite.
    3. Same argument as previous part.
    4. The level is at most N since both groups contain \Gamma(N). It can be no greater than N since \abcd{1}{N}{0}{1} is in both groups.
  7. Exercise 1.7. See [DS05, Lemma 1.2.2].

  8. Exercise 1.8. Let \alpha = p/q \in \Q, where p and q are relatively prime. By the Euclidian algorithm, we can find x, y \in \Z such that px + qy = 1. Let \gamma_\alpha = \abcd{p}{-y}{q}{x}. Note that \gamma_\alpha \in
\SL_2(\Z) and \gamma_{\alpha}(\infty) = \alpha. Also let \gamma_\infty be the identity map on \P^1(\Q). Now \gamma_\beta^{-1} sends \beta to \infty so we have \gamma_\alpha \circ \gamma_\beta^{-1} which sends \alpha to \beta.

Chapter 2

  1. Exercise 2.1. We have

    \zeta(26) = \frac{1315862 \cdot \pi^{26}}{11094481976030578125}.

    Variation: Compute \zeta(28).

  2. Exercise 2.2. Omitted.

  3. Exercise 2.3.

    E_8 &=  -\frac{B_8}{16} + q + \sum_{n=2}^{\infty} \sigma_7(n) q^n\\
&= \frac{1}{480} + q + 129q^{2} + 2188q^{3} + \cdots.

    Variation: Compute E_{10}.

  4. Exercise 2.4. Omitted.

  5. Exercise 2.5. We have d=\dim S_{28} = 2. A choice of a,b with 4a+6b\leq 14 and 4a+6b \con 4\pmod{12} is a=1, b=0. A basis for S_{28} is then

    \qquad \qquad g_1 &=  \Delta F_6^{2(2-1) + 0} F_4 =q - 792q^{2} - 324q^{3} + 67590208q^{4} + \cdots,\\
g_2 &=  \Delta^2 F_6^{2(2-2)+0} F_4 = q^{2} + 192q^{3} - 8280q^{4} + \cdots.

    The Victor Miller basis is then

    f_1 &= g_1 +729g_2 = q + 151740q^{3} + 61032448q^{4} + \cdots, \\
f_2 &= g_2 = q^{2} + 192q^{3} - 8280q^{4} + \cdots.

    Variation: Compute the Victor Miller basis for S_{30}.

  6. Exercise 2.6. From the previous exercise we have f = \Delta^2 F_4. Then

    \qquad \qquad f = \Delta^2 F_4 &= \left(\frac{F_4^3 - F_6^2}{-1728}\right)^2 \cdot F_4 \\
&= \left(\frac{\left(-\frac{8}{B_4} E_4\right)^3 -
\left(-\frac{12}{B_6} E_6\right)^2}{-1728}\right)^2
\cdot \left(-\frac{8}{B_4} E_4\right) \\
&= 5186160 E_4E_6^{4} - 564480000 E_4^{4}E_6^{2} + 15360000000 E_4^{7}.

  7. Exercise 2.7. No, it is not always integral. For example, for k=12, the coefficient of q is -2\cdot 12/B_{12} = 65520/691 \not\in\Z. Variation: Find, with proof, the set of all k such that the normalized series F_k is integral (use that B_k is eventually very large compared to 2k).

  8. Exercise 2.8. We compute the Victor Miller basis to precision great enough to determine T_2. This means we need up to O(q^5).

    f_0 &= 1 + 2611200q^{3} + 19524758400q^{4} + \cdots,\\
f_1 &= q + 50220q^{3} + 87866368q^{4} +\cdots,\\
f_2 &= q^{2} + 432q^{3} + 39960q^{4} + \cdots.

    Then the matrix of T_2 on this basis is


    (The rows of this matrix are the linear combinations that give the images of the f_i under T_2.) This matrix has characteristic polynomial

    (x - 2147483649) \cdot (x^{2} - 39960x - 2235350016).

Chapter 3

  1. Exercise 3.1. Write g=\abcd{a}{b}{c}{d}, so \lambda' = \frac{a\lambda + b}{c\lambda + d}. Let f be the isomorphism \C/\Lambda \to \C/\Lambda' given by f(z) = z/(c\lambda + d). We have

    \qquad\quad f\left(\frac{1}{N}\right) =
\frac{1}{N(c\lambda + d)} = \frac{a}{N} - \frac{c}{N} \cdot \frac{a \lambda + b}{c\lambda + d}
\cong \frac{a}{N} \pmod{\Z + \Z \lambda'},

    where the second equality can be verified easily by expanding out each side, and for the congruence we use that N\mid c. Thus the subgroup of \C/\Lambda generated by \frac{1}{N} is taken isomorphically to the subgroup of \C/\Lambda' generated by \frac{1}{N}.

  2. Exercise 3.2. For any integer r, we have \abcd{1}{r}{0}{1}\in \Gamma_0(N), so \{0,\infty\} = \{r, \infty\}. Thus

    \qquad  0 = \{0,\infty\} - \{0,\infty\} =
\{n,\infty\} - \{m,\infty\} =
\{n,\infty\} + \{\infty, m\} = \{n,m\}.

  3. Exercise 3.3.

    1. (0:1), (1:0), (1:1), \ldots, (1,p-1).
    2. p+1.
    3. See [Cre97a, Prop. 2.2.1].
  4. Exercise 3.4. We start with b=4, a=7. Then 4\cdot 2 \con 1\pmod{7}. Let \delta_1 = \abcd{4}{1}{7}{2}\in\SL_2(\Z). Since \delta_1\in \Gamma_0(7), we use the right coset representative \abcd{1}{0}{0}{1} and see that

    \{0,4/7\} = \{0,1/2\} + \abcd{1}{0}{0}{1}\{0,\infty\}.

    Repeating the process, we have \delta_2 = \abcd{1}{1}{2}{0}, which is in the same coset at \abcd{0}{-1}{1}{\hfill0}. Thus

    \{0,1/2\} = \abcd{0}{6}{1}{0} \{0,\infty\} + \{0,0\}.

    Putting it together gives

    \{0,4/7\} = \abcd{1}{0}{0}{1} \{0,\infty\} + \abcd{0}{6}{1}{0}\{0,\infty\}
= [(0,1)] + [(1,0)].

  5. Exercise 3.5.

    1. Coset representatives for \Gamma_0(3) in \SL_2(\Z) are

      \mtwo{ 1 }{ 0  }{ 0  }{ 1 },\quad
\mtwo{ 1 }{ 0 }{ 1  }{ 1 },\quad
\mtwo{ 1 }{ 0  }{ 2  }{ 1 },\quad
\mtwo{ 0 }{ -1  }{ 1  }{ 0 },

      which we refer to below as [r_0], [r_1], [r_2],f and [r_3], respectively.

    2. In terms of representatives we have

& [r_0]+[r_3]=0, & [r_0]+[r_3]+[r_2]=0, \\
& [r_1]+[r_2]=0, & [r_1]+[r_1]+[r_1]=0, \\
& [r_2]+[r_1]=0, & [r_2]+[r_0]+[r_3]=0, \\
& [r_3]+[r_0]=0, & [r_3]+[r_2]+[r_0]=0.

    3. By the first three relations we have [r_2] = [r_1] = 0 = 0[r_0] and [r_3] = -1[r_0].

    4. Finally,

      \qquad\quad T_2([r_0]) & =
[r_0] \abcd{1}{0}{0}{2}
+ [r_0]\abcd{2 }{ 0 }{ 0 }{ 1}
+ [r_0]\abcd{2 }{ 1 }{ 0 }{ 1}
+ [r_0]\abcd{1 }{ 0 }{ 1 }{ 2} \\
& =
\left[ \abcd{1 }{ 0 }{ 0 }{ 2} \right]
+ \left[ \abcd{2 }{ 0 }{ 0 }{ 1} \right]
+ \left[ \abcd{2 }{ 1 }{ 0 }{ 1} \right]
+ \left[ \abcd{1 }{ 0 }{ 1 }{ 2} \right] \\
& =  [r_0] + [r_0] + [r_0] + [r_2] \\
& =  3[r_0].

Chapter 4

  1. Exercise 4.1. Suppose f is a Dirichlet character with modulus N. Then -1=f(-1) = f(-1+N)=1, a contradiction.

  2. Exercise 4.2.

    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 (\Z/p\Z)^* is cyclic. Let g be an integer that reduces to a generator of (\Z/p\Z)^*. Let x=1+p \in (\Z/p^n\Z)^*; by the binomial theorem

      \qquad\qquad x^{p^{n-2}} = 1 + p^{n-2} \cdot p + \cdots
\con 1 + p^{n-1} \not\con 0 \pmod{p^{n}},

      so x has order p^{n-1}. Since p is odd, \gcd(p^{n-1},p-1) = 1, so xg has order p^{n-1}\cdot (p-1) = \vphi(p^n); hence (\Z/p^n\Z)^* is cyclic.

    2. By the binomial theorem (1+2^2)^{2^{n-3}} \not\con 1 \pmod{2^n}, so 5 has order 2^{n-2} in (\Z/2^n\Z)^*, and clearly -1 has order 2. Since 5\con 1\pmod{4}, -1 is not a power of 5 in (\Z/2^n\Z)^*. Thus the subgroups \langle -1 \rangle and \langle 5 \rangle have trivial intersection. The product of their orders is 2^{n-1} = \vphi(2^n) = \#(\Z/2^n\Z)^*, so the claim follows.

  3. Exercise 4.3. Write n=\prod p_i^{e_i}. The order of g divides n, so the condition implies that p_i^{e_i} divides the order of g for each i. Thus the order of g is divisible by the least common multiple of the p_i^{e_i}, i.e., by n.

  4. Exercise 4.4.

    1. The bijection given by 1+p^{n-1}a\pmod{p^n} \mapsto a\pmod{p} is a homomorphism since

      \hspace{6em}(1+p^{n-1}a)(1+p^{n-1}b) \con 1 + p^{n-1}(a+b) \pmod{p^n}.

    2. We have an exact sequence

      \qquad\qquad  1 \to 1 + p \Z/p^{n}\Z \to (\Z/p^n\Z)^* \to (\Z/p\Z)^*\to 1,

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

      \qquad\qquad  1 \to 1 + p^{n-1} \Z/p^{n}\Z \to (\Z/p^n\Z)^*  \to
(\Z/p^{n-1}\Z)^* \to 1.

      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.

  5. Exercise 4.5. If \eps(5)=1, then since \eps is nontrivial, Exercise 4.2 implies that \eps factors through (\Z/4\Z)^*, hence has conductor 4 = 2^{1 + 1}, as claimed. If \eps(5)\neq 1, then again from Exercise 4.2 we see that if \eps has order r, then \eps factors through (\Z/2^{r+2}\Z)^* but nothing smaller.

  6. Exercise 4.6.

    1. Take f=x^2+2.
    2. The element 2 has order 4.
    3. A minimal generator for (\Z/25\Z)^* is 2, and the characters are [1], [2], [3], [4].
    4. Each of the four Galois orbits has size 1.

Chapter 5

  1. Exercise 5.1. The eigenspace E_{\lambda} of A with eigenvalue \lambda is preserved by B, since if v\in E_{\lambda}, then

    ABv = BAv = B(\lambda v) = \lambda B v.

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

  2. Exercise 5.2. When \eps is the trivial character, the B_{k,\eps} are defined by

    \qquad \sum_{a=1}^{1} \frac{\eps(a) x e^{ax}}{e^{x}-1}
= \frac{x e^x}{e^x - 1} = x + \frac{x}{e^x - 1}
= \sum_{k=0}^{\infty} B_{k,\eps} \frac{x^k}{k!}.

    Thus B_{1,\eps} = 1 + B_1 = \frac{1}{2}, and for k>1, we have B_{k,\eps} = B_k.

  3. Exercise 5.3. Omitted.

  4. Exercise 5.4. The Eisenstein series in our basis for E_3(\Gamma_1(13)) are of the form E_{3,1,\eps} or E_{3,\eps,1} with \eps(-1)=(-1)^3=-1. There are six characters \eps with modulus 13 such that \eps(-1)=-1, and we have the two series E_{3,1,\eps} and E_{3,\eps,1} associated to each of these. This gives a dimension of 12.

Chapter 6

  1. Exercise 6.1.

    1. By Proposition 3.10, we have [\SL_2(\Z):\Gamma_0(N)] = \#\P^1(\Z/N\Z). By the Chinese Remainder Theorem,

      \#\P^1(\Z/N\Z) = \prod_{p\mid N} \#\P^1(\Z/p^{\ord_p(N)}\Z).

      So we are reduced to computing \#\P^1(\Z/p^{\ord_p(N)}\Z). We have (a,b)\in (\Z/p^n\Z)^2 with \gcd(a,b,p)=1 if and only if (a,b)\not\in (p\Z/p^n\Z)^2, so there are p^{2n} - p^{2(n-1)} such pairs. The unit group (\Z/p^n\Z)^* has order \vphi(p^n) = p^n - p^{n-1}. It follows that

      \#\P^1(\Z/N\Z) = \frac{p^{2n} - p^{2(n-1)}}{p^n - p^{n-1}}
= p^{n} + p^{n-1}.

    2. Omitted.

  2. Exercise 6.2. Omitted.

  3. Exercise 6.3. Omitted.

  4. Exercise 6.4. Omitted.

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

Chapter 7

  1. Exercise 7.1. Take a basis of W and let G be the matrix whose rows are these basis elements. Let B be the row echelon form of G. After a permutation p of columns, we may write B = p_i(I | C), where I is the identity matrix. The matrix A = p^{-1}(-C^t | I), where I is a different sized identity matrix, has the property that W = \Ker(A).

  2. Exercise 7.2. The answer is no. For example if A=nI is n times the identity matrix and if p \mid n, then \rref(A\pmod{p})=0 but \rref(A)\pmod{p}=I.

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

  4. Exercise 7.4.

    1. The echelon form (over \Q) is


    2. The kernel is the 1-dimensional span of \left(1,-2,1\right).

    3. The characteristic polynomial is x \cdot (x^{2} - 15x - 18).

  5. Exercise 7.5.

    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 M=p^r and N=p^s are powers of a prime p. Then [a]\in(\Z/p^s\Z)^* is represented by an integer a with \gcd(a,p)=1. That same integer a defines an element of (\Z/p^r\Z)^* that reduces modulo p^s to [a].

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

  3. Exercise 1.13. Coset representitives for \Gamma_1(3) are in bijection with (c,d) where c,d \in \Z/3\Z and \mathrm{gcd}(c,d,N)=1, so the following are representatives:

    \abcd{1 }{0}{0}{1},\,
\abcd{2 }{0}{0}{2}, \,
\abcd{0 }{2}{1}{0},\,
\abcd{1 }{0}{1}{1},\,
\abcd{2 }{0}{1}{2},\,
\abcd{1 }{1}{2}{0},\,
\abcd{1 }{0}{2}{1},\,
\abcd{2 }{0}{2}{2},

    which we call r_1, \ldots, r_8, respectively. Now our Manin symbols are of the form [X,r_i] and [Y, r_i] for 1 \le i \le 8 modulo the relations

    x+x\sigma = 0, \quad x+x\tau+x\tau^2 = 0, \quad\text{ and } x-xJ = 0.

    First, note that J acts trivially on Manin symbols of odd weight because it sends X to -X, Y to -Y and r_i to -r_i, so

    [z,g]J = [-z,-g] = [z,g].

    Thus the last relation is trivially true.

    Now \sigma^{-1}X = -Y and \sigma^{-1}Y = X. Also \tau^{-1}X = -Y, \tau^{-1}Y = X-Y, \tau^{-2}X = -X+Y and \tau^{-2}Y = -X.

    The first relation on the first symbol says that

    [X, r_1] = -[-Y, r_3] = [Y, r_3]

    and the second relation tells us that

    [X, r_1] + [-Y, r_5] + [-X+Y, r_6] = 0.

  4. Exercise 1.14. Let f \in S_k(\Gamma) and g \in \Gamma. All that remains to be shown is that this pairing respects the relation x=xg for all modular symbols x. By linearity it suffices to show the invariance of \langle f, X^{k-i-2}Y^i\{\alpha, \beta\} \rangle. We have

    \qquad&\left\langle f, (X^{k-2-i}Y^i \{\alpha, \beta\})g^{-1} \right\rangle\\
& =  \left \langle f, (aX+bY)^{k-i-2}(cX+dY)^i \{g^{-1}(\alpha), g^{-1}(\beta)\} \right \rangle \\
& =  \int_{g^{-1}(\alpha)}^{g^{-1}(\beta)} f(z) (az+b)^{k-i-2}(cz+d)^i \; dz \\
& =  \int_{g^{-1}(\alpha)}^{g^{-1}(\beta)} f(z) \frac{(az+b)^{k-i-2}}{(cz+d)^{k-i-2}}(cz+d)^{k-2} \; dz \\
& =  \int_{g^{-1}(\alpha)}^{g^{-1}(\beta)} f(z) \, g(z)^{k-i-2} (cz+d)^{k-2} \; dz \\
& =  \int_{\alpha}^{\beta} f(g^{-1}(z)) \, g(g^{-1}(z))^{k-i-2} (cg^{-1}(z)+d)^{k-2} \; d(g^{-1}(z)) \\
& =  \int_{\alpha}^{\beta} f(g^{-1}(z)) \, z^{k-i-2} (cg^{-1}(z)+d)^{k-2} \; (cg^{-1}(z)+d)^2 \; dz \\
& =  \int_{\alpha}^{\beta} f(z) \, z^{k-i-2} \; dz \\
& =  \left \langle f, X^{k-i-2}Y^i \{ \alpha , \beta \} \right \rangle,

    where the second to last simplification is due to invariance under [g]_k, i.e.,

    \qquad f(g^{-1}(z)) = f^{[g]_k}(g^{-1}(z)) = (cg^{-1}(z)+d)^{-k} f(g(g^{-1}(z))).

    (The proof for f \in \overline{S}_k(\Gamma) works in

    exactly the same way.)

  5. Exercise 1.15.

    1. Let \eta = {\scriptsize \left(\begin{array}{rr}-1 & 0 \\ 0 & 1 \end{array}\right)}. For any \gamma = {\scriptsize \left(\begin{array}{cc}a & b \\c & d \end{array}\right)} we have

      \qquad\qquad \gamma \eta =  {\scriptsize  \left(\begin{array}{rr}-a & b \\ -c & d \end{array}\right)},~~~~
\eta \gamma = {\scriptsize \left(\begin{array}{rr}-a & -b \\ c & d
\end{array}\right)},~~~~ \mbox{and}~~~~~ \eta \gamma \eta =
{\scriptsize \left(\begin{array}{rr} a & -b \\ -c & d

      First, if \gamma \in SL_2(\Z), then \eta \gamma \eta \in GL_2(\Z) and

      \det (\eta \gamma \eta) = \det \eta \det \gamma \det \eta =
(-1)(1)(-1) = 1

      so \eta \gamma \eta \in SL_2(\Z). As \eta^2 = 1, conjugation by \eta is self-inverse, so it must be a bijection.

      Now if \gamma \in \Gamma_0(N), then c \equiv 0 (mod N), so -c
\equiv 0 (mod N), and so \eta \gamma \eta \in \Gamma_0(N). Thus \eta\Gamma_0(N)\eta = \Gamma_0(N).

      If \gamma \in \Gamma_1(N), then -c \equiv 0 (mod N) as before and also a \equiv d \equiv 1 (mod N), so \eta \gamma \eta \in
\Gamma_1(N). Thus \eta\Gamma_1(N)\eta = \Gamma_1(N).

    2. Omitted.

Chapter 9

  1. Exercise 9.1. Consider the surjective homomorphism

    r:\SL_2(\Z)\to \SL_2(\Z/N\Z).

    Notice that \Gamma_1(N) is the exact inverse image of the subgroup H of matrices of the form \abcd{1}{*}{0}{1} and \Gamma_0(N) is the inverse image of the subroup T of upper triangular matrices. It thus suffices to observe that H is normal in T, which is clear. Finally, the quotient T/H is isomorphic to the group of diagonal matrices in \SL_2(\Z/N\Z)^*, which is isomorphic to (\Z/N\Z)^*.

  2. Exercise 9.2. It is enough to show \dbd{p}\in\Z[\ldots,T_n,\ldots] for primes p, since each \dbd{d} can be written in terms of the \dbd{p}. Since p\nmid N, we have that

    T_{p^2} = T_p^2 - \dbd{p}p^{k-1},


    \dbd{p}p^{k-1} = T_p^2 - T_{p^2}.

    By Dirichlet’s theorem on primes in arithmetic progression, there is a prime q\neq p congruent to p mod N. Since p^{k-1} and q^{k-1} are relatively prime, there exist integers a and b such that a p^{k-1} + b
q^{k-1} = 1. Then

    \qquad\dbd{p}=\dbd{p}(a p^{k-1} + b q^{k-1})
= a({T_p}^2-T_{p^2}) + b({T_q}^2-T_{q^2})
\in \Z[\ldots, T_n,\ldots].

  3. Exercise 9.3. Take N=33. The space S_2(\Gamma_0(33)) is a direct sum of the two old subspaces coming from S_2(\Gamma_0(11)) and the new subspace, which has dimension 1. If f is a basis for S_2(\Gamma_0(11)) and g is a basis for S_2(\Gamma_0(33))_{\new}, then \alpha_1(f), \alpha_3(f), g is a basis for S_2(\Gamma_0(33)) on which all Hecke operators T_n, with \gcd(n,33)=1, have diagonal matrix. However, the operator T_3 on S_2(\Gamma_0(33)) does not act as a scalar on \alpha_1(f), so it cannot be in the ring generated by all operators T_n with \gcd(n,33)=1.

  4. Exercise 9.4. Omitted.

Chapter 10

  1. Exercise 10.1.

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

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

  3. Exercise 10.3.

    1. Let f = \sqrt{-1} \sum q^n. Then d=2, but the nontrivial conjugate of f is -f, so V_f has dimension 1.

    2. Choose \alpha \in K such that K=\Q(\alpha). Write

      (1)\qquad\qquad f=\sum_{i=0}^{d-1} \alpha^i g_i

      with g_i \in \Q[[q]]. Let W_g be the \Q-span of the g_i, and let W_f = V_f \cap \Q[[q]]. By considering the \Gal(\Qbar/\Q) conjugates of (1), we see that the Galois conjugates of f are in the \C-span of the g_i, so

      (2)\qquad\qquad d = \dim_\C V_f \leq \dim_\Q W_g.

      Likewise, taking the above modulo O(q^n) for any n, we obtain a matrix equation

      \qquad\qquad  F = A G,

      where the columns of F are the \Gal(\Qbar/\Q)-conjugates of f, the matrix A is the Vandermonde matrix corresponding to \alpha (and its \Gal(\Qbar/\Q) conjugates), and G has columns g_i. Since A is a Vandermonde matrix, it is invertible, so A^{-1}F = G. Taking the limit as n goes to infinity, we see that each g_i is a linear combination of the f_i, hence an element of V_f. Thus W_g \subset W_f, so (2) implies that \dim_\Q W_f \geq d. But W_f \tensor_\Q \C \subset V_f so finally

      \qquad\qquad  d\leq \dim_\Q W_f = \dim_\C (W_f \tensor_\Q \C) \leq \dim_\C V_f = d.

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