4
$\begingroup$

Basically, as I know, we know almost nothing about Maass forms. For example, Cohen constructed first (maybe not) example of a Maass cusp form by using one of Ramanujan's $q$-series, as a non-definite theta series. (This is from his Inventiones paper, $q$-identities for Maass waveforms.) Also, this article from Buzzard gives some explicit examples, both "algebraic" (corresponds to even 2-dimensional Galois representations) and "non-algebraic" (something that mysterious?) Algebraic ones has an eigenvalue $1/4$, while non-algebraic ones have (conjecturally) transcendental eigenvalues (such as $$ \frac{1}{4} + \frac{\pi^{2}}{4\log^{2}(\sqrt{2} - 1)} $$ Also, this paper provides an algorithm to compute eigenvalues effectively, and the first eigenvalue for $\mathrm{SL}_{2}(\mathbb{Z})$ is $\lambda = \frac{1}{4} + r^{2}$, where $r = 9.5337...$.

I think the eigenvalues should be some special numbers, which can be transcendental, but maybe a little algebraic. More precisely, I hope that these numbers are related to periods. Periods are defined in Zagier-Kontsevich's article (this) as numbers that can be obtained by integral of rational functions over a domain defined by inequalities of rational coefficient polynomials. For example, any algebraic numbers, $\pi, \log 2, \zeta(3)$ are periods. Especially, $$ \log(\alpha) = \int_{1}^{\alpha} \frac{1}{x} dx $$ is a period for any $\alpha\in \overline{\mathbb{Q}}$. Periods form a proper subring of $\mathbb{C}$ that contains $\overline{\mathbb{Q}}$, denoted by $\mathcal{P}$, and $\mathcal{K} = \mathrm{Frac}(\mathcal{P})$ is a field between $\overline{\mathbb{Q}}$ and $\mathbb{C}$. So here is my conjecture:

If $\lambda$ is an eigenvalue of a Maass waveform (cusp form) on $\Gamma_{0}(N)$, then $\lambda \in \mathcal{K}$.

Obviously, there's no reason that this conjecture is true. The reason I belive this is because the eigenvalues shouldn't be just random transcendental numbers. To prove this conjecture, the first thing we have to figure out is the exact value of the smallest eigenvalue for $\mathrm{SL}_{2}(\mathbb{Z})$, which is approximately 91.14. Thanks in advance.

p.s. According to this question, if we assume some conjectures about motivic stuff, then $\mathcal{P}$ is not a field.

$\endgroup$
  • 4
    $\begingroup$ H. Maass constructed his "special waveforms" in effect by a theta correspondence from real-indefinite, but rationally anisotropic, $O(2)$'s. $\endgroup$ – paul garrett May 18 at 19:52
  • 1
    $\begingroup$ "We know almost nothing about Maass forms" is an understatement as there are several hundred papers about them (probably over a thousand). The eigenvalues are of course not random (they are eigenvalues), but surely they are (mostly) algebraically independent of each other and of all the periods. So I would bet that your conjecture is very far from the truth. $\endgroup$ – GH from MO May 18 at 22:26
  • 1
    $\begingroup$ @GHfromMO Is there any result about algebraic independence about eigen values? $\endgroup$ – Seewoo Lee May 18 at 23:29
  • 1
    $\begingroup$ There is no result about algebraic independence (of Maass newforms), but we have no reason to think otherwise. The Laplace eigenvalues are much like the zeta-zeros, which should be algebraically independent of each other as well. $\endgroup$ – GH from MO May 18 at 23:34
  • 4
    $\begingroup$ If non-algebraic Maass form’s eigenvalues were periods, then they would determine motives. But Langlands predicts no relationship between non-algebraic Maass forms and motives. So there are two correspondences, which have nothing to do with each other, between automorphic forms and motives? Doesn’t seem very likely. $\endgroup$ – Will Sawin May 19 at 0:30

Your Answer

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

Browse other questions tagged or ask your own question.