Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

First, "bounded" means pointwise bounded by some constant multiple of $y^N$.

Your suspicions are correct, that boundedness of a function does not imply boundedness of its derivative, in a great variety of circumstances.

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.)

Alternatively, one can define a space of uniformly moderate growth automorphic forms, by requiring directly that all their derivatives (by the Lie algebra) be of moderate growth... perhaps with exponents predictable from the exponent of the original, etc. This approach doesn't avoid the issue, really, because one must still prove that there are many such things, etc., and it comes back to the argument that eigenfunctions of moderate growth are of uniform moderate growth.

share|improve this answer
    
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
    
@Niccolo' : Oh, yes, complex norm. Sorry to misunderstand. –  paul garrett Dec 29 '11 at 16:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.