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32

We will use the Mellin transform technique. Recalling the Mellin transform and its inverse $$F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds.$$ Now, let's consider the function $$f(x)= \frac{x}{e^{\pi x}+1}.$$ Taking the Mellin transform of $f(x)$, we get $$F(s)={\pi ... 25 Here are some answers to your various questions: For the group GL_2, every cuspidal automorphic representation of GL_2(\mathbb A) (here \mathbb A is the adele ring of \mathbb Q) is generated by a uniquely determined newform, which (when normalized in a suitable fashion) is either a classical newform in the sense of Atkin and Lehner, i.e. a ... 13 The one-dimensional generalization of quadratic reciprocity is class field theory (over \mathbb Q, if you want to restrict to that case, where it is known as the Kronecker--Weber theorem). Here is a formulation which is useful for comparing with the two-dimensional version; it is helpful to split it into two parts: Given a Dirichlet character \chi: ... 10 Let \Delta denote the discriminant of the cubic curve given by the formula (*). (This is a somewhat complicated expression in the a_is which I won't write down here; but let me note that there are standard expressions c_4 and c_6 which are certain polynomials in the a_is, such that 1728 \Delta = c_4^3 - c_6^2.) The formula (*) then defines ... 10 Consider \Gamma = SL_2(\mathbb{Z}). Remember that \gamma = \begin{pmatrix} a & b \cr c& d \end{pmatrix} \in \Gamma operates on the upper half plane by T_\gamma : z \mapsto \dfrac{az+b}{cz+d}. Write \pi: H \to H/\Gamma for the quotient map. What is a (meromorphic) differential form \omega on H/\Gamma? It should be nothing but a ... 10 Here is how I tackled this when I was given it as a homework problem: We prove that \def\SL{\text{SL}} \def\Z{\mathbb{Z}} \def\GL{\text{GL}}$$|\SL_2(\Z/p^e\Z)|=p^{3e}\left(1-\frac{1}{p^2}\right)$$by induction. For the base case, e=1, note that we have the exact sequence$$1\to\SL_2(\Z/p\Z)\to\GL_2(\Z/p\Z)\xrightarrow{\det}(\Z/p\Z)^\times\to ...

10

The modular discriminant $\Delta (z) \in M_{12,0},$ the linear space of cusp forms of weight $12.$ Now dimension $M_{12,0} = 1$ (see, for example, T.M. Apostol: Moldular Functions and Dirichlet Series in Number Theory) Every modular form for the full modular group is a polynomial in $E_4$ and $E_6,$ where $E_{2k}$ are the Eisenstein series defined by ...

8

The proof is not at all obvious if you begin simply with the formula $$\Delta(q) = q \prod_{n=1}^{\infty} (1-q^n)^{24}.$$ However, as Derek Jennings explains in his answer, if you use the (absolutely crucial!) fact that $\Delta$ is a cusp form of weight twelve and level one, the proof is actually not very difficult. As Derek explains, the ring of modular ...

8

It is possible to express the finite-dimensionality in representation-theoretic terms: it is amounts to the fact that automorphic forms on $\mathrm{GL}_2(\mathbb A)$ form an admissible representation of $\mathrm{GL}_2(\mathbb A)$. (Actually, this statement is a little stronger, because it includes the Maass form case.) But I'm not sure that one can prove ...

8

You are asking if any conjugate of a modular curve is again a modular curve, and the answer is yes. This is a very special case of the general theory of conjugation of Shimura varieties, which says that any algebraic conjugate of a Shimura variety is again a Shimura variety, but which in this case can be verified directly. Firstly, just to explain why I ...

8

I don't have very specific answers to your questions (some of these might be better answered by someone with more background in complex geometry), but I think that I can address some aspects of the importance of modular forms for number theory. To understand, I think that a little historical perspective is always good to have. This is not exactly a direct ...

8

Note to begin with that the symbol $\circ$ does not mean composition; rather, it means the operation given by the formula further on in your question, namely: $$f\circ[\alpha]_k(\tau) = (c\tau + d)^k f(\alpha \tau).$$ Now to say that $f$ is a modular form of weight $k$ is to say that $$f\circ[\alpha]_k(\tau) = f(\tau),$$ for all $\alpha \in SL_2(\mathbb ... 7 Congruences between Hecke eigenforms are outward, "physical" manifestations of corresponding relationships between the associated two-dimensional Galois representations. In this particular case, Ramanujan's congruence is related to the fact that$691$is an irregular prime (in the sense of Kummer). Roughly the idea is that Eisenstein series relate to ... 7 The calculation of the Mellin transform of$f(x)$is not present in the above answer, so I will show it here. $$\mathfrak{M}\left(\frac{1}{e^{\pi x}+1};s \right) = \int_0^\infty \frac{1}{e^{\pi x}+1} x^{s-1} dx = \int_0^\infty \frac{1}{e^{\pi x}} \frac{1}{1+e^{-\pi x}} x^{s-1} dx \\= \int_0^\infty \frac{1}{e^{\pi x}} \sum_{q\ge 0} (-1)^q e^{-\pi q x} ... 7 For X(N), one can give a \mathbb{Q}-model, but there's a sense in which doing so is "cheating". The moduli interpretation of X(N) only makes sense over \mathbb{Q}(\zeta_N); if R is an algebra over this field, then X(N) classifies triples (E, P_1, P_2) where E is an elliptic curve over R and P_1, P_2 are R-points of E order N which ... 6 The question has some implicit hypotheses, possibly not clear to the questioner, and this implicitness and ambiguities about it complicate matters. First, the more natural descriptions of Eisenstein series for GL(2) of weights k>2 are not of the form in the question, but are \sum_{c,d} 1/(cz+d)^k. This does not converge for k=2, so k=2 has to be ... 6 No, it does not, because FLT does not hold for general cyclotomic fields. Here's a totally stupid counterexample: take a Pythagorean triple, e.g. 3^2 + 4^2 = 5^2. Then the number field generated by \sqrt{3}, \sqrt{4} = 2, and \sqrt{5} is an abelian extension of \mathbb{Q} and hence contained in a cyclotomic field \mathbb{Q}(\zeta_m) for some m ... 6 The construction of Eichler-Shimura associates to a weight 2 newform f\in S_2(\Gamma_0(N)) (trivial character) an abelian variety quotient A_f of the modular Jacobian J_0(N) (Jacobian of X_0(N)). Explicitly one takes the kernel I_f of the homomorphism from the Hecke algebra to \mathbf{C} determined by f and then A_f is J_0(N)/I_fJ_0(N) ... 6 The elliptic curve is not a modular form. The idea that there is a modular form f associated to E. It satisfies$$f(z)=\sum_{n=1}^\infty c_n q^n$$where q=\exp(2\pi i z), c_1=1 and (with finitely many exceptions) for prime p, the equation y^2=x^3+ax=b has p-c_p solutions (x,y) considered modulo p. It also satisfies various other conditions ... 6 As Qiaochu has pointed out, the answer to this question depends on what exactly you mean by "modular function". If you mean a function that satisfies the weight k functional equation for the action of \Gamma and is holomorphic everywhere on the upper half-plane and the cusps, then this is what is generally called a modular form; and the only weight 0 ... 6 You should identify the upper half plane with a subspace of \mathbb{C} with a subspace of the Riemann sphere. In this identification \mathbb{P}^1(\mathbb{R}) is a great circle separating the upper half plane and lower half plane, and \mathbb{P}^1(\mathbb{Q}) is the orbit of \infty under \text{PSL}_2(\mathbb{Z}). This orbit breaks up into a union ... 6 This is a layman answer. Hecke operators commute and are self-adjoint, hence the modular forms which are eigenvectors wrt. all Hecke operators form a basis of the space of all modular forms (and the same for cusp forms). If f is such an eigenvector then the L-series corresponding to f has multiplicative coefficients, i.e. it can be represented by an ... 5 It might help to go back to the definition of Hecke operators in level 1 in Serre's Course in arithmetic. For a prime p and a lattice \Lambda, the pthe Hecke corresondence (I forget if Serre uses exactly this terminology) takes \Lambda to \sum \Lambda', where \Lambda' runs over all index p sublattices of \Lambda. This is a multi-valued ... 5 The key to this is a little theorem. Let \mathcal{O}(\mathbb{D}-\{0\}) denote holomorphic functions on the punctured disc, and let X=\left\{f\in\mathcal{O}(\mathfrak{h}):f(z+1)=f(z)\text{ for all }z\right\} be the set of all 1-periodic holomorphic functions on the upper half-plane. Then, the function e(z)=\exp(2\pi i z) is a map ... 5 You fixed the main error I pointed out in the comments and used Jacobi's formula. To summarize:$$\sum_{n=1}^\infty\tau(n)q^n=q\prod_{n=1}^\infty(1-q^n)^{24}=q \left(\sum_{k=0}^\infty(-1)^k(2k+1)q^{k(k+1)/2}\right)^8\equiv q\sum_{k=0}^\infty \left((-1)^k(2k+1)q^{k(k+1)/2}\right)^8\mod2 \equiv\sum_{m=0}^\infty q^{(2m+1)^2}\mod2,$$QED. 5 Ford circles are the orbit of a horocycle under the action of the modular group$\Gamma$on the upper half plane$\frak h$(see e.g. this entry of SBS). Modular forms of weight zero (of which the$j$-invariant is an instance) are fully$\mathrm{SL}_2(\Bbb Z)$-invariant. So the periods of$j$(representing the quotient$H/\Gamma$) tile the hyperbolic plane ... 4 Just compute the$q$-expansion! We have$\Delta(z) / \Delta(pz) = (q + \dots) / (q^p + \dots) = q^{1-p} + \dots$, so there is a pole of order$(p-1)$at the cusp$\infty$. Since the function is obviously non-vanishing away from the cusps, and there are only two cusps, then (since the divisor of a function has degree 0) we see that the divisor is$(p-1)\{ (0) ...

4

There are basically 3 senses of "Hecke algebra", and they are related to each other. The modular-form sense is a special case of all three. The oldest version is that motivated by modular forms, if we think of modular forms as functions on (homothety classes of) lattices: the operator $T_p$ takes the average of a $\mathbb C$-valued function over lattices ...

4

Yes, your comparison with "homogenous" functions on the Riemann sphere (that is, complex projective one-space), is good, insofar as the objects are not really functions on the thing, but "global sections of a vector bundle". Automorphic forms on the upper half-plane $H$ are (mostly) not $\Gamma$-invariant, so do not "live" on the quotient space ...

4

This answer collects my comments above, and relates them to David Loeffler's answer: In the examples given, you are writing down a uniformizer for the genus $0$ modular curve $X_0(N)$ (with $N = 2, 3, 5$, and $7$), as well as a uniformizer for some genus $0$ cover of $X_0(N)$, and the algebraic relationship between the two uniformizers on the two different ...

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