Timeline for Product of two cuspforms is not a cuspform

5 events

when toggle format	what		by	license	comment
May 4, 2012 at 1:31	vote	accept	Eren Mehmet Kiral
May 3, 2012 at 0:41	comment	added	Matt Young		At the last stage of my proof one has to interchange a limit and an infinite sum, so there needs to be some kind of continuity/smoothness assumption. As a counterexample, you can change $f$ on a set of measure zero, say making it be $1$ on some vertical line.
May 2, 2012 at 21:49	comment	added	Eren Mehmet Kiral		I think that the proof you have given in fact shows that no such $f$ could exist in $L^2(\Gamma \backslash \mathbb{H})$, not only smooth ones. I also realize that I can find an automorphic function $f \in L^2(\Gamma \backslash \mathbb{H})$ such that $\int_0^1 f(x+iy) dx = 0$ for $y >2$ say, but having $\lim_{y\to \infty}f(x+iy) \neq 0$. That would be obtained by averaging the function $\phi(z) = e^{2\pi i x} \mathbf{1}_{[-1/2,1/2]\times [2,\infty]}(z)$ with respect to the group. $$f(z) = \sum_{\gamma \in \Gamma} \phi(\gamma z).$$ With some more work I think $f$ can be made smooth.
May 2, 2012 at 19:59	history	edited	Matt Young	CC BY-SA 3.0	edited body
May 2, 2012 at 19:11	history	answered	Matt Young	CC BY-SA 3.0