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Some Maass form can be written ($K_{iR}$ is the K-Bessel function):

$$f(x+iy)=\sum_{n \ne 0}^{\infty} a_n \sqrt{y} \;K_{iR}(2\pi |n| y) \; e^{2 i\pi nx}$$

with the $a_n$ multiplicative, but inversly if I fix multiplicative coefficients, for example $a_n=\frac{\chi(n)}{\sqrt{|n|}}$ can I expect the function defined by following sum to have some properties of Maass forms?

$$g(x+iy)=\sum_{n\ne 0}^{\infty} \frac{\chi(n)}{\sqrt{|n|}} \sqrt{y} \;K_{iR}(2\pi |n|y) \; e^{2 i\pi nx}$$

For example to have $g(\frac{az+b}{cz+d})$ linked with $g(z)$ for certain $a,b,c,d$ like for a Maass form ? Or there is nothing to expect ?

Any reference on the demonstration of multiplicative property of Maass form coefficients ? (I found plenty references for classic modular forms but less for Maass forms and generaly there is no detail)

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    $\begingroup$ If you just fix a random multiplicative function (of which there are uncountably many) your chances of it being the Fourier coefficients of a Maass eigenform (of which there are only countably many) is zero. As for the proof of multiplicativity, it's the same for Maass forms as for the holomorphic case; it's a formal group-theoretic argument to check that the Hecke eigenvalues are multiplicative and then a basic trick involving an explicit calculation of what a Hecke operator does to Fourier expansions to relate the Hecke evals to the Fourier coefficients. $\endgroup$
    – znt
    Commented Apr 8, 2016 at 17:48
  • $\begingroup$ One perhaps-important quibble is that the Fourier coefficients are not "strictly" multiplicative, but only "weakly": $a_{mn}=a_m\cdot a_n$ generally only when $m,n$ are coprime. This is considerably different from the strict multiplicativity of Dirichlet characters. The Hecke operators give the second-order recursion for prime-power coefficients $a_{p^n}$, in contrast to the in-effect first-order recursion for Dirichlet characters. $\endgroup$
    – paul garrett
    Commented Apr 9, 2016 at 0:00
  • $\begingroup$ I understand that with "random" multiplicative coefficients the function is not anymore a Maass form, the question was more do we loose ALL the properties ? There is no more matrix (not one) to link $g(\frac{az+b}{cz+d})$ and $g(z)$ ? Or a discrete set of matrix not beeing a group ? (my question was maybe not very well formulated) $\endgroup$
    – Bertrand
    Commented Apr 9, 2016 at 5:04

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A few things.

  • All Maass forms can be written in the Bessel form that you mention. It is deduced from the growth condition of the Fourier coefficients of an arbitrary Maass forms.

  • If you want to twist a Maass form $f$ by a primitive Dirichlet character $\chi$, the natural object to consider is the twisted L-function associated to the Maass form:

$$L(s,f,\chi)=\sum \frac{\chi(n)a_n}{n^s}$$

  • If you directly modify the coefficients of the Maass form, I think you shouldn't except it to be automorphic anymore (that is, automorphic with probability 0). See znt's comment above. After all, all the data of the Maass form is carried in its coefficients, and you have deleted it (the $a_n$) and you have replaced it with unreleated data from a Dirichlet coefficient mod n.

  • I can't think of any reference. One usually sees the proof for modular forms, and then for all automorphic forms. But the idea is still the same (see again znt's comment).

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    $\begingroup$ Perhaps it is under-appreciated that the fact that those particular Bessel functions are the only ones that appear (as opposed to another solution of the same second-order ODE) is due to the asymptotics at infinity, which is a singular point, etc. Understanding those asymptotics in that case is nearly 100 years old by now, but in higher rank (e.g., $SL_3(\mathbb Z)$...), this is Casselman's subrepresentation theory, etc., which is pretty serious business. $\endgroup$
    – paul garrett
    Commented Apr 9, 2016 at 0:02

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