2nd eigenvalues for cusp forms for $\Gamma_0(4)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:35:23Z http://mathoverflow.net/feeds/question/95914 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95914/2nd-eigenvalues-for-cusp-forms-for-gamma-04 2nd eigenvalues for cusp forms for $\Gamma_0(4)$ peter 2012-05-03T20:42:12Z 2012-05-06T17:12:15Z <p>Let f be a newform for $\Gamma_0(4)$ with a trivial character.</p> <p>I guess that the eigenvalue $\lambda(2)$ for $T_2$ is 0. I want to know that this is known result or not. If so, could you explain or give a reference?</p> <p>Edit: 1)As Ramsey pointed out, here $T_2$ means $U_2$ operator. 2)I am doing some research for automorphic L-functions for $\Gamma_0(4)$ and I could see that automorphic forms with trivial character has zero 2nd coefficient..so I asked the above question..</p> http://mathoverflow.net/questions/95914/2nd-eigenvalues-for-cusp-forms-for-gamma-04/95931#95931 Answer by paul garrett for 2nd eigenvalues for cusp forms for $\Gamma_0(4)$ paul garrett 2012-05-03T22:57:10Z 2012-05-06T17:12:15Z <p>This is certainly known.</p> <p>For any prime, any reasonable sense in which $T_p$ acts on cuspforms for $\Gamma_0(p^\ell N)$ would project to $\Gamma_0(N)$ (for $p$ prime to $N$), compatibly with everything else going on. One part of the characterization of newforms is being in the kernel of such a projection. </p> <p>The inevitability of this is better seen thinking of automorphic forms on adele groups, because then the corresponding $T_p$ is an integral operator that produces right $K_p=GL_2(\mathbb Z_p)$-invariant functions. The representation generated by a newform has no $K_p$-invariant vectors in it, so this projection is $0$.</p> <p>This specific statement may not be explicit, but the situation is described in a number of places: Gelbart's old Princeton book, Bump's book, for example. </p> <p>Edit: and there is some potential for misunderstanding about what is meant by $T_2$... Especially given this vanishing, and given that there is the $U_p$ operator ("Atkin-Lehner"), as mentioned in the other answer. Perhaps confirmation about the intention and context of the question would be good.</p> <p>Edit-edit: Yes, the above discussion applies equally to waveforms (whether or not it succeeds in addressing the intended question). This is already clear classically, and is even clearer looking at automorphic forms on adele groups, because the Hecke operators at finite places have no interaction with the archimedean phenomena (holomorphic discrete series, principal series, whatever type the repn is at archimedean places). </p> http://mathoverflow.net/questions/95914/2nd-eigenvalues-for-cusp-forms-for-gamma-04/95933#95933 Answer by Ramsey for 2nd eigenvalues for cusp forms for $\Gamma_0(4)$ Ramsey 2012-05-03T23:25:19Z 2012-05-03T23:25:19Z <p>I'm going to assume that the operator that you refer to as $T_2$ is the one often referred to as $U_2$ in this context: the one that has the effect $$\sum a_nq^n\mapsto \sum a_{2n}q^n$$ on $q$-expansions.</p> <p>For forms of level $\Gamma_0(4)$ and trivial character, this operator has the effect of decreasing the level, so the resulting form is a modular form for $\Gamma_0(2)$ with trivial character. In particular, if you had a newform on $\Gamma_0(4)$ that was an eigenform for $U_2$ with nonzero eigenvalue, then solving the equation $U_2f=\lambda f$ for $f$ would show that your form wasn't new at level $4$ at all.</p> <p>To see this "decreasing the level" statement, my tendency is to go to the geometric description of $U_2$, though there is a much more simple classical description using matrices (this must be in Miyaki's fine book). At the geometric level the $U_2$ operator involves replacing a cyclic subgroup of order $2^n$ of an elliptic curve by the collection of cyclic subgroups of order $2^{n+1}$ that contain it, so one can apply this to a level $2^{n+1}$ form and arrive at one of level $2^n$ in the construction.</p> http://mathoverflow.net/questions/95914/2nd-eigenvalues-for-cusp-forms-for-gamma-04/95963#95963 Answer by Marc Palm for 2nd eigenvalues for cusp forms for $\Gamma_0(4)$ Marc Palm 2012-05-04T07:42:40Z 2012-05-04T07:42:40Z <p>Let $\pi = \otimes_v \pi_v \otimes \pi_\infty$ be the associated automorphic representations to a newform of $\Gamma_0(p^2)$, then $\pi_p$ is super-cuspidal and all others $\pi_v$ for $v\neq p$ are unramified principal series representation. The Heckeeigenvalue at super cuspidal representation is zero, since the local Euler factor associated to it is a constant.</p>