# 2nd eigenvalues for cusp forms for $\Gamma_0(4)$

Let f be a newform for $\Gamma_0(4)$ with a trivial character.

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?

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..

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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.

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.

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.

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Dear Nick, The elementary matrix argument can also be found in the paper of Atkin and Lehner. Best wishes, Matt –  Emerton May 4 '12 at 0:33
@Matt - Indeed! Although, (1) I've always found it easier to wrap my head around the geometric side of the coin than the double coset stuff, and (2) my recent experience has taught me that its sometimes easier to get springer texts at your local library than Mathematische Annalen. Plus, I've always dug Miyaki's book as a great reference. –  Ramsey May 4 '12 at 0:47
thank you for your detailed explanation. It was a great help to me. –  peter May 4 '12 at 0:48
...continuing on point (1) in my previous comment: Maybe this speaks to an oddly specific point of view on these things (which I often feel I have, and often to my detriment), but when I try to recall how $U_p$ and $V_p$ mess with the level, the geometric picture on the Tate curve is how I remember these things. –  Ramsey May 4 '12 at 0:52
Dear Emerton, thank you for the exact reference. Lemma 7 in Atkin-Lehner is what I wanted.. –  peter May 4 '12 at 2:41

This is certainly known.

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.

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 $0$.

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.

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.

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).

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Thank you for your answer and comments. when I read your writing, it implies that the analogous question for Maass forms is true.. Right? –  peter May 4 '12 at 1:01

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.

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Note I do not impose anything on $\pi_\infty$, so works Maass form or modular form equally well, and actually over any global field. The supercuspidal in question will be a depth zero supercuspidal representation, i.e it has a $\Gamma(\p)$-invariant vector. –  Marc Palm May 4 '12 at 7:47
In the situation you describe, $\pi_p$ could also be a ramified principal series. –  David Hansen May 7 '12 at 1:06
@David Hansen, what characters do you have in mind. –  Marc Palm May 7 '12 at 6:21