# Computing Maass forms

## (with Hejhal's method)


In this presentation, I'll describe Maass forms, how to compute them, and why we care. This is a project I've begun since joining the Simons Collaboration, but it's already led to multiple collaborations with others. It's still a work in progress, but I intend to have some complete data very soon.

An overview of what's to come:

• What are Maass forms?
• Hejhal's Algorithm
1. Creating a linear system
2. Creating a nontrivial linear system
3. Locating eigenvalues

### What are Maass forms?

Maass forms are distinguished modular cuspforms. Let $\Gamma < \SL(2, \mathbb{Z})$ be a congruence subgroup and let $\chi$ be character on $\Gamma$. We'll call a function $f$ a Maass form for $(\Gamma, \chi)$ if

• $f$ is an eigenfunction: $\Delta f = \lambda f$
• $f$ is automorphic: $f(\gamma z) = \chi(\gamma)f(z)$ for all $\gamma \in \Gamma$
• $f$ is $L^2$: $\displaystyle \int_{\Gamma \backslash \mathcal{H}} \lvert f \rvert^2 d\mu < \infty$

We will write the eigenvalues as $\lambda = \frac{1}{4} + R^2$. Selberg conjectured that (on congruence subgroups), if $\lambda \neq 0$ then $R \in \mathbb{R}_{\geq 0}$, so that $\lambda \geq \frac{1}{4}$. Any eigenvalue $\lambda \in (0, \frac{1}{4})$ is called exceptional.

The Selberg Eigenvalue Conjecture (that $\lambda \geq \frac{1}{4})$ is the Ramanujan-Petersson Conjecture at $\infty$. The $p$-part of the $L$-function associated to a Maass form will (generically) be of the form \begin{align} L_p(s) &= (1 - \alpha_{p,1} p^{-s})^{-1} (1 - \alpha_{p,2} p^{-s})^{-1}, \\ L_\infty(s) &= \Gamma_\mathbb{R}(s - \mu_{\infty, 1}) \Gamma_\mathbb{R}(s - \mu_{\infty, 2}). \end{align} SEC is that $\Re \mu_{\infty, j} = 0$, and RPC is that $\lvert \alpha_{p,j} \rvert = 1$, or equivalently that $\log_p \lvert \alpha_{p,j} \rvert = 0$.

The best bound on this conjecture is the Kim-Sarnak bound, that $\lvert \Re \alpha_{\infty, j} \rvert$ and $\log_p \lvert \alpha_{p, j} \rvert$ are bounded above by $\frac{7}{64}$.

One reason I care is because of the spectral decomposition. Each $g \in L^2(\Gamma \backslash \mathcal{H})$ has a spectral expansion of shape \begin{align} g(z) = & \sum_{\mu_j \ \text{Maass}} \langle g, \mu_j \rangle \mu_j(z) \\ & + \sum_{\text{Eisenstein}} \int \langle g, E(\cdot, u) E(z, u) du. \end{align}

My most common hammer from my tool belt is to expand everything in terms of the spectral decomposition, average things out, and then roll up my sleeves and do complex analysis on what remains.

### Computing $\GL(2)$ Maass forms

I'll describe an approach for computing Maass forms first described by Hejhal, and since refined by Strömberg.

The basic idea is to construct a non-trivial linear system for the coefficients, and to solve it. The challenge is that the coefficients and the eigenvalue are deeply connected, and it's hard to find one without the other.

A Maass form with eigenvalue $\lambda = \frac{1}{4} + R^2$ has a Fourier expansion of the form $$f(z) = \sum_{n \neq 0} c(n) \sqrt{y} K_{iR}(2 \pi \lvert n \rvert y) e(nx).$$ Here, $e(x) = e^{2 \pi i x}$, $z = x + iy$, and $K_\alpha(u)$ is the modified $K$-Bessel function of the second kind.

Note that we can see how the unknown coefficients $c(n)$ and unknown eigenvalue $R$ interact from this expansion.

### Overview of strategy

We will be to guess an eigenvalue $R$, use linear algebra and modularity to compute a set of coefficients that would be approximately the "true" coefficients of a Maass form if $R$ were approximately an actual eigenvalue.

In practice, we won't guess the correct $R$. But if we are near an actual eigenvalue, we can try to adjust our guess for $R$ to be nearer an actual eigenvalue. Then we repeat to find a candidate $R$ at sufficient precision.

Once we've settled on an $R$, it is straightforward to compute very many coefficients. Heuristically we can examine the properties of the resulting form to see if it looks like a Maass form.

Let $\kappa_n(y) = \sqrt{y} K_{iR}(2 \pi \lvert n \rvert y)$, and write the Fourier of a Maass form $f$ as $$f(z) = \sum_{n \neq 0} c(n) \kappa_n(y) e(nx).$$ The coefficients are bounded by $O(\sqrt{n})$ (Hecke bound), while $\kappa_n(y)$ decreases exponentially in $\lvert n \rvert$. Thus for any $\epsilon \gt 0$, there is a constant $M(y, R)$ such that $$f(z) = \sum_{\lvert n \rvert \leq M(y)} c(n) \kappa_n(y) e(nx) + [[\epsilon]].$$ Further, the function $M(y, R)$ is a decreasing function in $y$.

The truncated Fourier series $$\widehat{f}(z) = \sum_{\lvert n \rvert \leq M(y)} c(n) \kappa_n(y) e(nx)$$ can be viewed as a discrete Fourier transform. If we choose points along a horocycle $\{ z_m = x_m + iY : x_m = \frac{1}{2Q}(m - \frac{1}{2}), 1 - Q \leq m \leq Q \}$ (with $Q \gt M(Y)$), then we can invert this transform to see $$c(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q \widehat{f}(z_m)e(-nx_m).$$

By un-truncating the series, we can "remove the hat", $$c(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q f(z_m)e(-nx_m) + [[\epsilon]].$$

For a fixed $R$, we can consider this equation for various $n$. The result is essentially a linear system in the coefficients $c(n)$. But the system is currently a tautology — we haven't don't anything to mix up the coefficients.

To make the system nontrivial, we will use the modularity of $f$.

### Making a nontrivial linear system

To include modularity, we make two choices.

First, we choose $Y$ small, so that part of our horocycle $z_m = x_m + iY$ will be outside of the fundamental domain. Then we can pullback each $z_m$ to a point $z_m^*$ in the fundamental domain. The result is that $$c(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q \chi_m f(z_m^*)e(-nx_m) + [[\epsilon]],$$ where $\chi_m$ is the character applied on the pullback matrix.

Second, we work not only with the typical Fourier expansion of $f$, but instead the Fourier expansions $f_\ell$ of $f$ at each cusp $\ell$. That is, in terms of $f_\ell(z) = f(\sigma_\ell z)$, where $\sigma_\ell \infty = \ell$.

For each point $z^*$ in the fundamental domain, we identify the nearest cusp $\ell = \ell(z^*)$ (i.e. the cusp with respect to which $z^*$ has the greatest height). We choose to represent the value of $f(z^*)$ in terms of the expansion of $f_\ell$.

As we are involving each cusp, we must enlarge our system to include each cusp, and solve for all expansions simultaneously. For each cusp $j$, we have an expansion $$f_j(z) = \sum_{n \neq 0} c_j(n) \kappa_n(y) e(nx)$$ and we can set up the system

$$c_j(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q f_j(z_m)e(-nx_m) + [[\epsilon]].$$

$$c_j(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q f_j(z_m)e(-nx_m) + [[\epsilon]].$$

Let $z_{mj} = \sigma_j z_m$, so that $f_j(z_m) = f(z_{mj})$, and let $z_{mj}^*$ be the pullback of $z_{mj}$ in coordinates of the nearest cusp $\ell$.

In total, $$c_j(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q \chi_{mj} f_\ell(z_{mj}^*)e(-nx_m) + [[\epsilon]].$$

Let $z_{mj} = \sigma_j z_m$, so that $f_j(z_m) = f(z_{mj})$, and let $z_{mj}^*$ be the pullback of $z_{mj}$ in coordinates of the nearest cusp $\ell$.

It is the nontrivial mixing coming from $f_j(z_m)$ and $f_\ell(z_{mj}^*)$ that gives a non-tautological system, allowing us to solve for the Fourier coefficients.

In total, $$c_j(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q \chi_{mj} f_\ell(z_{mj}^*)e(-nx_m) + [[\epsilon]].$$

### Lemma

Let $z_{mj}^*$ and $z_m$ be as above. It is possible to choose $Y$ such that $z_{mj}^* \neq z_m$ for all $j$ and $m$. Further, the imaginary parts of each $z_{mj}^*$ are bounded below by a number $Y_0$ (depending on the level).

### Locating eigenvalues

Given a guess for $R$, we thus have the system $$c_j(n) \kappa_n(Y) = \frac{1}{2Q} \sum_{1-Q = m}^Q \chi_{mj} f_\ell(z_{mj}^*)e(-nx_m) + [[\epsilon]].$$ As $\Im(z_{mj}^*) \gt Y_0$ for all $m$ and $j$, we can truncate each series $f_\ell$ on the right at the same point $M_0 = M(Y_0)$. Expanding this out and collecting coefficients of each coefficient $c_j(m)$, we get $$c_j(n) \kappa_n(Y) = \sum_{\text{cusps } \ell} \sum_{1 \leq \lvert k \leq M_0} c_j(k) V_{nkj\ell} + 2 [[\epsilon]].$$

Making the choice that $c_\infty(1) = 1$ (which can be true for newforms), we can solve for the list of coefficients (unless multiple forms share the same eigenvalue, but let's ignore that in this talk. This is resolvable).

Structurally, we have constructed a linear system $VC = 0$ for a gross-but-computable matrix $V = V(R, Y)$ consisting mostly of Bessel functions, and an unknown vector of coefficients $C$. We can solve for $C = C(R, Y)$.

If $R$ were a true eigenvalue, then $C$ would be independent of $Y$ (as long as $Y$ is chosen small enough for the lemma to apply). And if $R$ were a true eigenvalue, then the resulting list of coefficients would give a Hecke eigenform that satisfies Hecke relations.

From these, we can create a cost functional $\mathrm{cost}(R)$. For example, we can choose two values $Y_1$ and $Y_2$, solve, and take the sum of the differences of the computed coefficients.

In practice, we choose an interval $[R_1, R_2]$ and try to minize $\mathrm{cost}(R)$ within this interval by repeatedly creating and solving the linear system. Proceeding in a sufficiently fine mesh, we can find and compute many Maass forms.

### Thank you

I'm happy to answer any questions.

### References

• D. A. Hejhal. On the Calculation of Maass Cusp Forms , Proceedings of the "International School on Mathematical Aspects of Quantum Chaos II", Lecture Notes in Physics, Springer, 2004.
• D. A. Hejhal and S. Arno, On Fourier coefficients of Maass waveforms for $\mathrm{PSL}(2, \mathbb{Z})$ , Math. Comp. 61 (1993), no. 203, 245–267.
• H. Kim and P. Sarnak. Refined estimates towards the Ramanujan and Selberg Conjectures. J. Amer. Math. Soc, 16(1):175–181, 2003.
• R. P. Langlands. Problems in the theory of automorphic forms. In Lectures in modern analysis and applications, III, pages 18–61. Lecture Notes in Math., Vol. 170. 1970.
• F. Strömberg. Computational Aspects of Maass Waveforms. PhD Dissertation. 2005.