most recent 30 from http://mathoverflow.net 2013-05-22T23:16:44Z http://mathoverflow.net/feeds/question/121399 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121399/refining-the-moderate-growth-condition-on-automorphic-forms Valerie 2013-02-10T15:49:02Z 2013-02-24T17:22:00Z

<p>Let $G$ be a simple algebraic group defined over $\mathbb Q$. In their Corvallis article (automorphic forms and automorphic representations), Borel and Jacquet define an automorphic form to be a smooth function on $G(\mathbb A)$ which is left-$G(\mathbb Q)$-invariant, satisfies some "finiteness" conditions, and also the following moderate growth condition: </p> <ul> <li>For every $y\in G(\mathbb A)$, the map $x\mapsto f(x\cdot y)$ on $G(\mathbb R)$ is slowly increasing.</li> </ul> <p>I have a couple of questions about this condition:</p> <ol> <li><p>Assume that $f$ appears in the discrete spectrum (but is not necessarily a cusp form). Is it true that the map $x\mapsto f(x\cdot y)$ is bounded on unipotent subgroups of $G(\mathbb R)$?</p></li> <li><p>What can be said about the "growth" of the map $x\mapsto f(x\cdot y)$ for a fixed $y\in G(\mathbb A)$ when $x$ varies in $G(\mathbb Q_p)$? For example, will this map be bounded on unipotent subgroups of $G(\mathbb Q_p)$?</p></li> </ol>

http://mathoverflow.net/questions/121399/refining-the-moderate-growth-condition-on-automorphic-forms/121404#121404 anton 2013-02-10T16:24:09Z 2013-02-10T16:24:09Z

<p>If the unipotent subgroup is defined over $\mathbb Q$, then its quotient by the congruence subgroup will be compact, hence the function in question be periodic, hence bounded.</p> <p>To the second question, lets say the function is fixed by a compact open subgroup of the finite adelic-valued points, which is to say that the real component is fixed by an arithmetic group. Further, this infinite component is an eigenfunction of the algebra of invariant differential operators. Say it is $K$-finite, maybe of a fixed $K$-type, then it satisfies a differential equation of the form $\Delta f=\lambda f$, where $\Delta$ is induced by the Casimir, normalize it to have positive definite principle symbol, then $\Delta+C$ is invertible for some $C>0$ and the operator $(\Delta+C)^{-N}$ has kernel of any order of differentiablility as $N$ increases. So you have $$ f=(\lambda+C)^N(\Delta+C)^{-N}f $$ from which moderate growth of $f$ is concluded as the resolvent kernel shows moderate growth.</p>