The paper says "Let $f(z)$ be a Maass form of $SL(2, \mathbb{Z}) \backslash \mathbb{H}$" -- what are some examples of this? Are there any explicit examples?

Here's an answer from What is the relationship between modular forms and Maass forms?

In the more common terminology modular forms on the upper half-plane fall into two categories: holomorphic forms and Maass forms. In fact there is a notion of Maass forms with weight and nebentypus, which includes holomorphic forms as follows: if $f(x+iy)$ is a weight $k$ holomorphic form, then $y^{k/2}f(x+iy)$ is a weight $k$ Maass form.

what about $\theta(z)^4$? That is a weight 2 form. Then is $y^{-1}\theta^4(z)$ weight 2 Maass form?

There seems to be a Siegel-Maass correspondence takes theta functions and Maass forms. Is that what's here?

Here is another point of confusion:

Are these two concepts related? Are Eisenstein series examples of Mass forms?

Other resources (e.g. [1]) are written so succinctly I have no idea what is going on.

Sources seem to indicate that explicity Maass forms are scarce. I am hoping to understand a bit why subjects might be important even if we can never actually write them down.

  • $\begingroup$ is it the same as a modular form? $\endgroup$ – cactus314 Oct 28 '16 at 21:30
  • 1
    $\begingroup$ Does this help? I am not an expert. $\endgroup$ – Will R Dec 25 '16 at 19:12
  • $\begingroup$ Maass forms are a generalization of Hecke eigenforms (those modular forms with multiplicative Fourier coefficients) discarding the holomorphic condition. Yes Eiseinstein series are Mass forms. They are important because as eigenforms (and when everything works well, Selberg 1/4 conjecture and Ramanjuan conjecture) they are associated to a Dirichlet series in the Selberg class : with a Riemann hypothesis. $\endgroup$ – reuns Dec 25 '16 at 23:16
  • $\begingroup$ @user1952009, Maass forms need not be Hecke eigenforms, just as modular forms need not be. And I think very few number theorists think they are interesting because they are in the Selberg class! Rather, the converse is closer to the truth: the Selberg class is interesting because an element in the Selberg class should correspond to the $L$-function of an automorphic form. $\endgroup$ – Peter Humphries Dec 26 '16 at 21:26
  • $\begingroup$ @PeterHumphries I don't see what you wanted to say. And "Classical Maass forms are a type of modular form" doesn't seem right to me. Did you mean (weakly) modular functions ? $\endgroup$ – reuns Dec 27 '16 at 18:51

The canonical reference for learning about Maass forms is probably Spectral Methods of Automorphic Forms by Henryk Iwaniec. Good references for the more general theory of Maass forms are the first half of the paper The Subconvexity Problem for Artin $L$-Functions by Duke, Friedlander, and Iwaniec, and Chapter 3 of the book Automorphic Representations and $L$-Functions for the General Linear Group by Goldfeld and Hundley. I don't know a great reference for Maass forms of half-integral weight or for harmonic (weak) Maass forms.

Classical Maass forms are a type of modular form. They are functions $f$ on the upper half-plane $\mathbb{H}$ that are automorphic, of moderate growth, and are eigenfunctions of the weight $0$ Laplacian with eigenvalue $\lambda_f$.

  • Automorphic means that $f(\gamma z) = f(z)$ for all $z \in \mathbb{H}$ and $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})$, where $\gamma z = \frac{az + b}{cz + d}$.
  • Moderate growth means that $f(z)$ grows at most polynomially at the cusp at infinity, so that $f(z) = O(y^N)$ as $y \to \infty$ for some positive integer $N$.
  • The weight $0$ Laplacian is \[\Delta = -y^2\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right).\]

An explicit example of a classical (weight zero, level one) Maass form is the (real analytic) Eisenstein series \[E(z,s) = \sum_{\gamma \in \Gamma_{\infty} \backslash \Gamma} \Im(\gamma z)^s = \frac{1}{2} \sum_{\substack{m,n = 1 \ (m,n) = 1}}^{\infty} \frac{\Im(z)^s}{|mz + n|^s}, \] where $s \in \mathbb{C}$. This has Laplacian eigenvalue $\lambda = s(1 - s)$.

If we demand that not only is $f$ of moderate growth, but that its constant term \[\rho_f(0) = \int_{0}^{1} f(x + iy) \, dx\] vanishes for all $y > 0$, then $f$ is said to be a Maass cusp form. Eisenstein series are not cusp forms. Maass cusp forms are very similar to holomorphic cusp forms of weight $k \geq 2$: they have a basis of Hecke eigenforms to which one can associate $L$-functions that are entire and satisfy a functional equation.

With all this being said, it is very hard to explicitly write out a Maass cusp form. The only cases we can really write out explicitly are due to Maass; they arise from Hecke characters of a real quadratic extension of $\mathbb{Q}$. Here by explicit, I mean that we can write down the Fourier coefficients (or equivalently the Heck eigenvalues) of the Maass cusp form.

In general, when dealing with Maass cusp forms, you really shouldn't be thinking about a specific example, and specific examples of modular forms are rarely why number theorists find them interesting. Even for holomorphic cusp forms that are Hecke eigenforms, we can rarely say anything particularly detailed about the Hecke eigenvalues.

If you want to see numerical examples of (the Hecke eigenvalues of) Maass cusp forms, then you should browse the $L$-Functions and Modular Forms Database.

More generally, one can talk of Maass forms $f$ of weight $k$, level $q$, and nebentypus $\chi$, where $k$ is an integer and $\chi$ is a Dirichlet character modulo $q$. This means we replace the automorphy condition with $f(\gamma z) = \chi(\gamma) j_{\gamma}(z)^k f(z)$ for all $\gamma \in \Gamma_0(q)$, the moderate growth condition is now required at every singular cusp of $\Gamma_0(q) \backslash \mathbb{H}$, and $f$ must be an eigenfunction of the weight $k$ Laplacian \[\Delta_k = -y^2\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) + iky \frac{\partial}{\partial x}.\] In this case, a holomorphic modular form $F(z)$ of weight $k \geq 2$ is a Maass form $f(z) = y^{k/2} F(z)$ of weight $k$, and a classical Maass form is a Maass form of weight $0$.

One can also define half-integral weight Maass forms, though the automorphy condition is a little more complicated.

The function $y^{-1} \theta(z)$ is not a Maass form of weight $2$, because it does not satisfy the right automorphy condition (try to work out exactly the automorphy condition it does satisfy). The Siegel-Maass correspondence is a relation between classical Maass forms (of weight $0$) and half-integral weight Maass forms (of weight $-1/2$ or $1/2$). The theta function is an example of a Maass form of weight $1/2$, but there are plenty of other such Maass forms (and they are very hard to explicitly describe).

Harmonic (weak) Maass forms are a generalisation of classical Maass forms, where we weaken the moderate growth condition.

  • $\begingroup$ can you explain why number theorists find them interesting despite being able to pin them down? i started learning number theory with very explicit examples and now I'm told you can never see these objects :-/ $\endgroup$ – cactus314 Dec 27 '16 at 0:24
  • 1
    $\begingroup$ @cactus314 You can "see them" when applying the spectral theorem to the hyperbolic (weight $k$) Laplacian $\Delta_k$. And since $\Delta_k$ commutes with the Hecke operators (those having a self-commuting property $gcd(n,m)=1 \implies T_n T_m = T_m T_n$), it means the "Fourier" coefficients of Maass forms are multiplicative, so they are closely related to the traditional Hecke eigenforms / generalized Riemann hypothesis theory. $\endgroup$ – reuns Dec 27 '16 at 18:57
  • $\begingroup$ What @user1952009 said. Maass cusp forms and Eisenstein series give a spectral decomposition of $L^2(\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H})$, so if you want to prove anything involving functions in this space, then the problem usually reduces to proving something about Maass cusp forms, and these are nice to work with. $\endgroup$ – Peter Humphries Dec 28 '16 at 16:13
  • $\begingroup$ More generally, have a look at this MO question and its answers: mathoverflow.net/questions/24604/… $\endgroup$ – Peter Humphries Dec 28 '16 at 16:13
  • $\begingroup$ @PeterHumphries Hi. With my tools, except Brauer's theorem I heard about, it is very hard to imagine how an Artin L-function $L(s,\rho)$ can have a functional equation and come from an automorphic form. Do you have a better picture ? $\endgroup$ – reuns Aug 23 '17 at 20:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.