# Moderate growth and Maass raising operators

Consider the Maass raising operator of weight $k$:

\begin{equation} R_k=iy {\partial \over \partial x}+ y {\partial \over \partial y} + \frac{k}{2}. \end{equation}

Fix now integers $k,N$ where $N \ge 1$, a Dirichlet character $\chi (\bmod N)$ and define the space $A_{k,\chi}(N)$ as the space of smooth functions $f:H \longrightarrow \mathbb{C}$ ($H$ the upper open half-plane) which respects the following conditions:

1. \begin{equation} \quad f ( \alpha z)=\chi(d) \Big( \frac{cz+d}{|cz+d|} \Big)^k f(z) \quad \forall \alpha \in \Gamma_0(N) \end{equation} where $\Gamma_0(N)$ is the congruence subgroup having $c \equiv 0 \bmod N$,

2. Let $a \in \mathbb{Q} \cup \infty$ be a cusp and $\sigma_a \in SL_2(\mathbb{Z})$ map $\infty$ to $a$: then we impose that $f(\sigma_a(x+iy))$ is bounded by a power of $y$, when $y \to \infty$. This is the condition of moderate growth, as defined by Goldfeld and Hundley.

G&H want to prove that $R_k$ maps $A_{k,\chi}(N)$ to $A_{k+2,\chi}(N)$.

It is a routine computation to check that if $f$ respects condition 1 for the weight $k$, then $R_kf$ respects condition 1 for the weight $k+2$, everything else being equal.

However, how do you show that if $f$ has moderate growth, the same can be said of $R_kf$? I do not know if the esteem on the growth of $f$ gives esteems on the growth of its partial derivatives, moreover we do not have olomorphy for $f$, just smoothness.

By the way, the whole definition of moderate growth (which I copied as it is) appears a little confusing to me: I am supposing that with "bounded" we mean bounded in norm.

First, "bounded" means pointwise bounded by some constant multiple of $y^N$.
In the present situation, we should add the hypothesis that $f$ is an eigenfunction (or generalized eigenfunction) for the Casimir operator (sometimes called "weight $k$ Laplacian). (In general, the condition would be $\mathfrak z$-finiteness, where $\mathfrak z$ is the center of the universal enveloping algebra.) Then moderate growth does imply moderate growth of all derivatives (given by the Lie algebra, which is what the Maass operator $R_k$ is, after all.)
• Thanks a lot. I still have a doubt about the "bounded" thing: $f(\sigma_a(x+iy))$ takes values in $\mathbb{C}$ so which partial order are you using to give a pointwise bound with a power of $y$? This is why I originally thought it meant $|f(\sigma_a(x+iy))|<My^N$ as $y$ goes to $infty$. Maybe the misunderstanding was about the type of norm, I meant complex norm, not $L^2$-norm. – Niccolo' Dec 29 '11 at 16:26