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49

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 $$ ...


27

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 ...


17

The theory of modular forms arose out of the study of elliptic integrals (as did the theory of elliptic curves, and much of modern algebraic geometry, and indeed much of modern mathematics). People understood that (complete) elliptic integrals (which we would think of as the number obtained by integrating a de Rham cohomology class, e.g. the one associated ...


15

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: ...


14

Let $\Delta$ denote the discriminant of the cubic curve given by the formula ($*$). (This is a somewhat complicated expression in the $a_i$s 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 $a_i$s, such that $1728 \Delta = c_4^3 - c_6^2$.) The formula ($*$) then defines ...


14

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 ...


14

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} ...


13

Let's start with $$ \sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1} $$ Differentiating $(1)$ and multiplying by $x$, we get $$ \sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2} $$ Taking the odd part of $(2)$ yields $$ \sum_{n=0}^\infty(2n+1)x^{2n+1}=\frac{x(1+x^2)}{(1-x^2)^2}\tag{3} $$ Using $(3)$, we get $$ \begin{align} ...


12

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 ...


12

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 ...


11

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 ...


11

When working with modular forms as analytic objects, the Galois structure is somewhat invisible and has to be rediscovered using the Hecke algebra. However, there is a purely algebraic notion of modular form, due to Katz, which makes the Galois structure (among other things) a lot more transparent. Katz's construction works well for level structures for ...


9

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 ...


9

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 ...


9

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 ...


8

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 ...


8

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 ...


8

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 $p$the 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 ...


8

To explain one aspect of the context of that interaction: Wigner was a very senior faculty member, had won a Nobel Prize, and was in his 60s. Shimura was a very young full professor. Another aspect: Wigner's 1939 paper on the representation theory of (in effect) $SO(2,1)$, written to address issues of quantum mechanics, was the first substantive result on ...


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

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

Some brief comments on your three questions. Why care about all modular forms and not just, say, cusp forms? Well, why in real analysis do we use all real numbers if, in practice, so few of them are really of direct interest (not so many classical transcendental constants come up, just $\pi$, $e$, Euler's constant,$\Gamma(1/4)$,...)? The answer is that we ...


7

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 ...


7

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 ...


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 ...


6

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 ...


6

Yes, it is possible. Here's the way to do it: choose your favorite elliptic curve $E/\mathbf C$ with $j(E) = i$. For example, the curve $$y^2 +xy = x^3 -36(j-1728)^{-1}x - (j-1728)^{-1}$$ has $j(E)=j$ for $j \neq 0, 1728$ (so, just plug $j=i$). Compute the periods $\omega_1$ and $\omega_2$. Then $\tau = \omega_2/\omega_1$ satisfies $j(\tau)=i$.


6

In each case (weight $1$ or weight $k \geq 2$), there is a number field $E$ (the number field generated by the prime-to-the-level Hecke eigenvalues of $f$) and a compatible family of $\lambda$-adic reps. $\rho_{\lambda}$ attached to $f$, where compatible means that if $p$ doesn't divide the level, and is prime-to-$\lambda$, then $\rho_{\lambda}$ is unram. at ...


6

Neither you nor Mathematica are to blame. If you change the coefficient $24$ in front of the sum to $12,$ the expression produces the desired series. Alternatively, following ccorn's comment, you can change the sum so that it runs from $1$ to $\infty$ rather than $-\infty$ to $\infty$. In the paper you cite, the authors write ...


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 ...



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