I'm reading Gelbart's notes on the decomposition of $L^2(\operatorname{SL}_2(\mathbb Z) \backslash \operatorname{SL}_2(\mathbb R))$, and am stuck on a small detail about reinterpreting modular forms as automorphic forms. Let $f$ be a cusp form of weight $k$ for $\operatorname{SL}_2(\mathbb Z)$ on the upper half plane $\mathbb H$. Let $\phi: \operatorname{SL}_2(\mathbb R) \rightarrow \mathbb C$ be the function

$$\phi(g) = f(g.i)(ci+d)^{-k} \tag{$g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$}$$

I want to say that $\phi$ is a bounded function. By the Iwasawa decomposition, it suffices to show that the restriction of $\phi$ to the upper triangular matrices is bounded. That is, I just need to show that $\phi(g)$ is bounded for $g = \begin{pmatrix} y^{1/2} & xy^{-1/2} \\ & y^{-1/2} \end{pmatrix}$. So it suffices to show that


is bounded for all $x \in \mathbb R$ and $y \in (0,\infty)$. The boundedness for $y$ large is not a big deal: we may restrict $x$ to $[0,1]$ and consider the Fourier expansion

$$f(x+iy) = \sum\limits_{n=1}^{\infty} a_ne^{2\pi in (x+iy)}$$

We know that as $y$ goes to infinity, $f(x+iy)$ tends to zero uniformly in $x$. Factoring out a term $e^{-2\pi y}$, we see that $f(x+iy)y^{k/2}$ still tends to zero as $y \to \infty$, since $\lim\limits_{y \to \infty} e^{-2\pi y}y^{k/2} = 0$.

My issue is the boundedness for $y$ small. How do we know that $f(x+iy)y^{k/2}$ does not blow up as $y$ approaches zero? I would imagine if $x+iy$ is tending towards a rational number (cusp), we can translate that cusp to $\infty$ using some $\gamma \in \operatorname{SL}_2(\mathbb Z)$ and repeat the same argument. What if $x$ is irrational though?

  • $\begingroup$ $\phi$ is left $SL_2(Z)$ invariant and it is rotated by the orthogonal group on the right so it suffices to look at the fundamental domain $\{\pmatrix{a & b \\ 0 & 1/a}, |a^2i+ab| \ge 1, ab \in [-1,1]\}$ and since $f(a^2i+ab)$ has exponential decay as $a \to \infty$ the $(ci+d)^{-k} = a^k$ term isn't a problem. $\endgroup$ – reuns Mar 9 '19 at 2:36
  • $\begingroup$ I see, thanks $\space$ $\endgroup$ – D_S Mar 9 '19 at 3:27

Thanks to reuns for answering. This is a writeup of his suggestion. My question is how we know $f(x+iy)y^{k/2}$ is bounded even when $y$ is small. Letting $$g = \begin{pmatrix} y^{1/2} & xy^{-1/2} \\ & y^{-1/2} \end{pmatrix}$$ we have $\phi(g) = f(x+iy)y^{k/2}$. There exists $\gamma \in \operatorname{SL}_2(\mathbb Z)$ such that $\gamma.(x+iy) =: x'+iy'$ lies in the usual fundamental domain. Letting

$$g' = \begin{pmatrix} y'^{1/2} & x'y'^{-1/2} \\ & y'^{-1/2} \end{pmatrix}$$

we have $\gamma g.i = g'.i$, so $\gamma g = g'\kappa$ for some $\kappa \in \operatorname{SO}_2(\mathbb R)$. We have $\phi(g'\kappa) = \xi \phi(g')$ for some $\xi$ on the unit circle, so

$$f(x+iy)y^{k/2} = \phi(g) = \phi(\gamma g) = \phi(g' \kappa) = \xi\phi(g') = \xi f(x'+iy')y'^{k/2}$$

and so $|f(x+iy)y^{k/2}| = |f(x'+iy')y'^{k/2}|$. So the question of boundedness of $f(x+iy)y^{k/2}$ is determined by $x+iy$ in the fundamental domain, which is in particular away from the real axis. Then boundedness follows from what I wrote in my question about what happens when $y \to \infty$.

  • $\begingroup$ What do you think of the case $f \in S_k(\Gamma_0(n))$ ? This time the fundamental domain looks like that $\endgroup$ – reuns Mar 9 '19 at 3:52
  • $\begingroup$ The same idea should work, using the fundamental domain we can assume the only case where $x+iy$ is near the real axis is when it's heading towards a cusp, at which point we can translate to cusp $\infty$ and use $\lim\limits_{y \to \infty} e^{-y}y = 0$. $\endgroup$ – D_S Mar 9 '19 at 4:21
  • $\begingroup$ $SL_2(Z) = \bigcup \alpha_j \Gamma_0(n)$ gives the fundamental domain $F(\Gamma_0(n)) = \bigcup \alpha_j F(SL_2(Z))$ and cusp form means $f(\alpha_j z)$ has exponential decay when $z \to i \infty$ $\endgroup$ – reuns Mar 9 '19 at 4:27

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