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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.

F. C.
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Seewoo Lee
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    H. Maass constructed his "special waveforms" in effect by a theta correspondence from real-indefinite, but rationally anisotropic, $O(2)$'s. – paul garrett May 18 '19 at 19:52
  • @paulgarrett Could you elaborate more, or give some references? – Seewoo Lee May 18 '19 at 19:55
  • I'd have to look up the precise citations for Maass' papers. But, at least in contemporary terms, use a binary quadratic form $Q$ made from the norm form from a real quadratic extension $k$ of $\mathbb Q$, and look at the theta pairing $O(Q)\times SL_2$. The automorphic forms on $SO(Q)$ are given by Hecke characters on the extension, and their images under the theta correspondence are Maass' special waveforms. – paul garrett May 18 '19 at 20:25
  • @paulgarrett Is it automorphic induction of Hecke character of a real quadratic field to $\mathrm{GL}_{2}/\mathbb{Q}$? Tbh, I don't know what rationally anisotropic means, and how that related to eigenvalues. Could you write as an answer to this question if you have time? – Seewoo Lee May 18 '19 at 22:20
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    "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. – GH from MO May 18 '19 at 22:26
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    @GHfromMO Is there any result about algebraic independence about eigen values? – Seewoo Lee May 18 '19 at 23:29
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    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. – GH from MO May 18 '19 at 23:34
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    @GHfromMO That makes sense to me...thanks. – Seewoo Lee May 19 '19 at 00:11
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    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. – Will Sawin May 19 '19 at 00:30
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    Maass constructed the "complex multiplication" case (Hecke characters of real quadratic fields) back in the 1949. Bump's 1.9 has a sketch of it. I must admit I find the original writing (and German language) hard to parse. As you note, the spectral parameters are essentially $\pi n$ divided by log of the square of the fundamental unit (see after 1.9.17 in Bump). https://link.springer.com/article/10.1007/BF01329622 – MyNinthAccount May 19 '19 at 01:47
  • I may write a note giving a modern treatment of the Hecke-Maass correspondence... but it won't fit as an answer here. But/and part of the point of that might be to see that already in that simplest case, the eigenvalues (for the real-quadratic, not complex-quadratic, over \Q) case are logs of quadratic irrationals (so-called fundamental units). By results of A. Baker and others on "transcendence theory", we know (and "expect", in further cases) that these are algebraically independent, I think...? – paul garrett May 19 '19 at 03:19
  • Oop, to let the other shoe drop: the proven algebraic independence of various exponentials/logs of algebraic things would seem (though I'd need to look at the specifics) to already contradict the conjecture of the question... although one could argue either for-or-against about the special-ness of ... special wave-forms. – paul garrett May 19 '19 at 03:38
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    Just to chime in quickly —- I believe linear independence over the algebraic numbers of certain logarithms of algebraic numbers is given by Baker’s method, rather than algebraic independence. But why would algebraic independence be relevant? The question notes that \log{\eps} (and 1/\log{\eps} etc.) is OK, no? Pardon my confusion! – alpoge May 19 '19 at 19:38
  • @alpoge, yes, you are certainly correct! I was just free-associating, and not thinking too carefully about what I was writing. – paul garrett May 19 '19 at 20:59
  • To be honest, I don't get it why algebraically independence of eigenvalues may disprove the conjecture. Why is it? – Seewoo Lee May 19 '19 at 21:33

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