What's the best record toward Selberg's eigenvalue conjecture: a Maass form on $\Gamma_0(N)$ has eigenvalue greater than or equal to 1/4?
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5$\begingroup$ See terrytao.wordpress.com/2011/12/16/… for some comments by Lior Silberman on numerical verification of the conjecture by Booker and Strombergsson (and also mentioning the Kim-Sarnak result in quid's answer). $\endgroup$– Terry TaoCommented Oct 23, 2013 at 17:32
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4$\begingroup$ Greater than or equal to! $\endgroup$– David HansenCommented Oct 24, 2013 at 17:09
1 Answer
\[\frac{1}{4} - \left(\frac{7}{64}\right)^2 = \frac{975}{4096} \approx 0.238037\ldots\] as far as I know. Established in an appendix by Kim and Sarnak to Kim's "Functoriality for the exterior square of $GL_{4}$ and the symmetric fourth of $GL_{2}$", Journal of the American Math. Society, 16 (2003).
The above is "over the rationals" (which was/is my reading of the question), in comments work on the problem for general number fields got mentioned by Kim and Shaidi, Cuspidality of symmetric powers with applications. Duke Math. J. 112 (2002) and the more recent work of Blomer and Brumley, On the Ramanujan conjecture over number fields. Annals of Math. 174 (2011) which among others establishes the above for general number fields.
For generalizations in a different direction let me also mention Bourgain, Gamburd, Sarnak, Generalization of Selberg's 3/16 Theorem and Affine Sieve, Acta Math. 207 (2011) establishing 'a generalization of Selberg's theorem for infinite index "congruence" subgroups of SL(2,Z).'
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2$\begingroup$ And it might worth mentioning related work of Kim-Shaidi over arbitrary number field. $\endgroup$– AsafCommented Oct 23, 2013 at 18:32
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2$\begingroup$ It is perhaps useful to note that the Kim-Sarnak result is over $\mathbb{Q}$. The same result over an arbitrary number field was established by Blomer-Brumley (Annals of Math. 174 (2011), 581-605). $\endgroup$ Commented Oct 24, 2013 at 13:51
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$\begingroup$ Thanks for the edit and the additions! I edited in links to the two additional papers, and an additional one on a different way to generalize the problem. $\endgroup$– user9072Commented Oct 24, 2013 at 15:53