What is the relationship between (g,K)-module and Maass forms? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:24:25Z http://mathoverflow.net/feeds/question/75854 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75854/what-is-the-relationship-between-g-k-module-and-maass-forms What is the relationship between (g,K)-module and Maass forms? 7-adic 2011-09-19T14:57:17Z 2012-04-06T13:26:54Z <p>What is the relationship between (g,K)-module and Maass forms for GL(2)?</p> <p>(g,K)-modules are defined in chapter 2 of Bump, Automorphic forms and representations.</p> <p>There is a classification of (g,K)-modules.</p> <p>What is the relationship between (g,K)-module and Maass forms for GL(2)? and what does the classification of (g,K)-module imply for GL(2) Maass forms?</p> http://mathoverflow.net/questions/75854/what-is-the-relationship-between-g-k-module-and-maass-forms/75868#75868 Answer by David Loeffler for What is the relationship between (g,K)-module and Maass forms? David Loeffler 2011-09-19T16:22:19Z 2011-09-19T16:22:19Z <p>So you've seen that there are essentially three types of (g, K)-modules: finite-dimensional ones; principal series; and discrete series. The finite-dimensional ones don't interest us, since they are never unitary. So that leaves two cases. And these correspond precisely to the two kinds of cuspidal automorphic representations for $GL(2, \mathbf{A}_\mathbf{Q})$: those whose factor at $\infty$ is discrete series (or a limit of discrete series) correspond to classical modular eigenforms; and those whose factor at $\infty$ is principal series correspond to Maass forms.</p> <p>You could say that the classification doesn't directly tell you anything new about Maass forms themselves, but rather it tells you about how Maass forms fit into the general picture of automorphic forms.</p> http://mathoverflow.net/questions/75854/what-is-the-relationship-between-g-k-module-and-maass-forms/93304#93304 Answer by Marc Palm for What is the relationship between (g,K)-module and Maass forms? Marc Palm 2012-04-06T11:05:25Z 2012-04-06T13:26:54Z <p>The classification of $(g,K)$-modules tells you that the Beltrami-Laplace eigenvalue </p> <p><code>$$\lambda = (1-s)s$$</code> </p> <p>of Maass form $f$ (of weight $0$) satsifies either $$\Re s = 1/2$$ or </p> <p><code>$$0 &lt; s &lt;1,$$</code> that for the adelic decomposition <code>$$f = f _\infty \otimes \bigotimes_p f_p,$$</code> the function $f_\infty$ is is a vector of the induced representation <code>$$Ind_{B}^G | \cdotp |^{s-1/2}$$</code> at the infinite place. Of course, this is slight overkill and follows also from the positivity of the Beltrami Laplace operator.</p> <p>However, it explains better the Selberg eigenvalue conjecture $0 \neq \lambda \geq 1/4$. The Selberg eigenvalue conjecture asserts that $0 &lt; s &lt;1$ or eq. $0 &lt; \lambda \leq 1/4$ should imply $s=1/2$ or eq. $\lambda =1/4$,. The eigenvalue $\lambda = 1/4$ does in fact occur, Bump gives an example in chapter 1.</p> <p>Selberg eigenvalue conjecture can be generalized also to the $p$-adic places, and is then called the Ramanujan-Petersson conjecture. </p> <p>The difference between $\Re s = 1/2$ and $\Re s \neq 1/2$ is that the former is tempered and the latter is not. </p> <p>There are some non trivial bounds towards the Selberg eigenvalues known. Functoriality of the $n$-th symmetric tensor would imply the result, and actually this is what Kim and Sarnak used to prove a non-trivial bound.</p> <p>Furthermore the representation $Ind_{B}^G | \cdotp |^{s-1/2}$ are not square integrable, but the discrete series are. That is why the name discrete series, they occur discretely in the right regular representation of $L^2(G)$.</p>