# Cusp form becomes a bounded function on $\operatorname{SL}_2(\mathbb R)$

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

$$f(x+iy)y^{k/2}$$

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?

• $\phi$ 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. – reuns Mar 9 '19 at 2:36
• I see, thanks $\space$ – 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 $$y \to \infty$$.

• What do you think of the case $f \in S_k(\Gamma_0(n))$ ? This time the fundamental domain looks like that – reuns Mar 9 '19 at 3:52
• 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 $\lim\limits_{y \to \infty} e^{-y}y = 0$. – D_S Mar 9 '19 at 4:21
• $SL_2(Z) = \bigcup \alpha_j \Gamma_0(n)$ 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 $z \to i \infty$ – reuns Mar 9 '19 at 4:27