By a Maass form I just mean--maybe a bit loosely--any real analytic $\Bbb C$-valued function $f$ on the upper halfplane $\cal{H}$ which is automorphic of weight $k\in\Bbb Z$ with respect to a discrete subgroup $\Gamma<{\rm SL}_2(\Bbb R)$ such that $\Gamma\backslash\cal H$ has finite volume, and an eigenfunction for the Laplacian operator corresponding to the Casimir element in the universal enveloping algebra of the complexified $\rm{sl}_2$.

Is it true that the zeroes of these forms are isolated?

The answer is obviously affirmative in the case of holomorphic modular forms.

$\textbf{Edit}$: Scott's comment and Matt's answer below show that the answer is generally negative when the weight is $0$. Then, one can construct real valued Maass forms which have nodal curves.

Thus, let's make the assumption that $k\neq0$ and in particular that the Maass form $f$ belongs to a discrete series representation space (generated by a holomorphic modular form).

To make things as explicit as possible assume also that $\Gamma$ is either a congruence subgroup of ${\rm SL}_2(\Bbb Z)$ or the group of norm 1 elements in an Eichler order of an indefinite quaternion algebra over $\Bbb Q$

  • $\begingroup$ I may be missing something, but it looks like the real part of a Maass form is still a Maass form (maybe I need weight zero?). Then you get domains of positivity and negativity with walls formed by zeroes. $\endgroup$
    – S. Carnahan
    Aug 4, 2010 at 15:52
  • $\begingroup$ Scott, good observation. Yet, if the weight is not zero the automorphy relation rules out the possibility of real valued Maass forms. $\endgroup$ Aug 4, 2010 at 16:18

1 Answer 1


Maass forms will generally have nodal lines rather than zeros. Let's take the case where $\Gamma = SL_2(\mathbb{Z})$ and the weight $k=0$.

First recall that a Maass form has a Fourier expansion of the form $$f(x+iy) = \sum_{n \neq 0} a_n \sqrt{y} K_{\nu}(2 \pi |n| y) e^{2 \pi i n x},$$ where $K_{\nu}$ is the usual $K$-Bessel function. Here $\nu$ is purely imaginary. It is not difficult to show that $K_{it}(v)$ is real for $t, v$ real. If $f$ is an eigenfunction of all the Hecke operators then the $a_n$'s are real also, which is a consequence of the fact that the Hecke operators are self-adjoint with respect to the Petersson inner-product. Furthermore, by considering the reflection operator $T_{-1}f(x+iy) = f(-x+iy)$ which commutes with all the Hecke operators as well as the Laplacian, we should suppose $f$ is an eigenfunction of $T_{-1}$ also. The only possible eigenvalues of $T_{-1}$ are $\pm 1$ so in the above Fourier expansion we can replace $e^{2 \pi i n x}$ by $\cos(2 \pi n x)$ or $\sin(2 \pi n x)$. Then in either case $f$ is patently real-valued.

One can find some nice pictures of Maass forms at Fredrik Strömberg's homepage

Edit: I should also point out the simple fact that if $f$ is odd (i.e. $f(-x+iy) = -f(x+iy)$) then $f(iy) = 0$ for all $y$. This can be clearly seen in Strömberg's pictures.

  • $\begingroup$ Matt, thanks for your reply. Your arguments seem to depend strongly on the $k=0$ assumption: if $k\neq0$ there are no real valued Maass forms simply because the automorphy relation doesn't allow them. \par I think I posed the question in a form too general. In fact, I'm interested in Maass forms that belong to the space of a discrete series representation (i.e. those generated by holomorphic modular forms). Any hint about the behaviour of these? $\endgroup$ Aug 4, 2010 at 16:36
  • $\begingroup$ It's hard to imagine there being nodal lines when the Maass forms are genuinely complex, but I'm not familiar with any results in this direction so I hesitate to say anything definite. $\endgroup$
    – Matt Young
    Aug 4, 2010 at 17:27
  • $\begingroup$ Ok, thanks. I will edit the question so to make clear that I'm really looking at the $k\neq0$ situation. $\endgroup$ Aug 4, 2010 at 17:41

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