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Let $N\geq 1$ be an integer and let $S_2(\Gamma_0(N))$ be the cusp forms of weight 2 for the usual congruence subgroup $\Gamma_0(N)\subset SL_2(\mathbb Z)$. Let $a_n(f)$ denote the n-th Fourier coefficient of $f\in S_2(\Gamma_0(N))$, $n\geq 1$. Let $X_0(N)$ be a smooth projective model over $\mathbb Q$ of the modular curve associated to $\Gamma_0(N)$ and let $$(f,g)=\int_{X_0(N)}f(z)\overline{g(z)}dxdy,$$ $z=x+iy$, be the Petersson inner product of $f,g\in S_2(\Gamma_0(N))$.

Question (a) Does there exist a constant $c>0$, depending at most on $\Gamma_0(N)$ (or $X_0(N)$), with the following property: Suppose $f\in S_2(\Gamma_0(N))$ is an eigenform with $a_1(f)=1$ which lies in the new part, $g\in S_2(\Gamma_0(N))$ and $a_n(f),a_n(g)\in\mathbb Z, n\geq 1$. If $(f,g)>0$, then $$(f,g)\geq c.$$

(b) Can one find such a constant $c>0$ with an explicit dependence on $N$.

(c) Can one find such a constant $c>0$ which is absolute.

My main interest is when $f$ in the above question is in addition a newform for $\Gamma_0(N)$ with $a_1(f)=1$, but I don't know if this assumption simplifies things. Further, David Loeffler mentions below that $(f,g)$ is related to a residue of a certain $L$-series at $s=2$.

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I am a bit confused about your last remark. If you take two different newforms $f$ and $g$ satisfying your normalization, then $(f,g)=0$ by multiplicity one. –  GH from MO Apr 9 '13 at 16:06
Dear GH, thanks for the remark. The assumptions of the question still hold: There I assume that (f,g)>0. I will edit the question to make this more clear. –  ranicl Apr 9 '13 at 16:51
Given $f$, the possible $(f,g)$ form a subgroup of ${\bf R}$, which is either discrete or dense. Once the space of cuspforms has dimension at least $2$ one would expect it to be dense unless $f = 0$ (why should two or more "random" Petersson products be proportional?), and thus to contain arbitrarily small positive elements. Is there a further missing assumption? –  Noam D. Elkies Apr 9 '13 at 17:24
If you assume that $f$ is a newform with integer coefficients, then the orthogonal of $f$ with respect to the Petersson scalar product admits a basis consisting of forms with integral coefficients, thus the image of the linear map on $S_2(\Gamma_0(N),\mathbf{Z})$ given by $g \mapsto (f,g)$ is a lattice of $\mathbf{R}$. Thus $c$ exists in this case, but it is not clear how to compute a lower bound in terms of $N$ because of the possible congruences between $f$ and other newforms as David explained. –  François Brunault Apr 9 '13 at 21:39
Ah, I saw the "integer coefficients" part but didn't appreciate the significance of "newform" (implying not just in the "new" space but an actual Hecke eigenform). –  Noam D. Elkies Apr 10 '13 at 2:50
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2 Answers

up vote 1 down vote accepted

It's not hard to see that the answer to (a) is yes. There is a basis of $S_2^{\textrm{new}}(\Gamma_0(N))$ consisting of newforms. These newforms come into Galois orbits $\{f^\sigma\}_{\sigma}$. Here $\sigma$ runs through the embeddings of $K_f$ into $\mathbf{C}$, where $K_f$ is the field generated by the Fourier coefficients of $f$. A basic fact is that the $\mathbf{C}$-span of a given Galois orbit is generated by cusp forms with integral coefficients. This follows from considering the forms $\sum_{\sigma} \sigma(a) f^\sigma$ where $a$ runs through a $\mathbf{Z}$-basis of the ring of integers of $K_f$.

It follows that if the newform $f$ has integral coefficients, then its orthogonal complement $f^\perp$ is generated by cusp forms with integral coefficients. Now consider the linear map $\lambda_f : S_2(\Gamma_0(N),\mathbf{Z}) \to \mathbf{R}$ given by $\lambda_f(g)=(f,g)$. By the previous remark, the kernel of $\lambda_f$ has rank one less than the rank of $S_2(\Gamma_0(N),\mathbf{Z})$, which implies that the image of $\lambda_f$ is of the form $c_f \mathbf{Z}$ for some $c_f >0$. Thus we can take for $c$ the minimum of all the $c_f$. In fact $c_f = (f,f)/m_f$ for some integer $m_f$ measuring the congruences of $f$ with other cusp forms.

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Dear Francois Brunault, thats a fantastic answer. Thank you very much!!! –  ranicl Apr 11 '13 at 11:06
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I typed a comment but the formatting wouldn't come out right, so here it is as an answer!

I cannot work out why you expect the "Plancherel or Parseval type" formula to work. Does it not bother you a little that $a_n(f)$ and $a_n(g)$ are perfectly capable of being integers for all $n$, so your series is obviously divergent?

Much better is to consider the series $$ L(f, g, s) = \sum_{n \ge 1} a_n(f) \overline{a_n(g)} n^{-s},$$ which converges for $Re(s) > 2$ (this is not so easy to see, but it is easy to show that it converges for $Re(s) \gg 0$). This has meromorphic continuation to all $s \in \mathbb{C}$ with a pole at $s = 2$ at which the residue is (maybe up to a normalizing constant depending on $N$ that I have forgotten) the Petersson product $\langle f, g \rangle$.

But this does not help you to get lower bounds on $\langle f, g \rangle$ as far as I can see. Some quite grotty things can happen, e.g. if $f$ is an eigenform and there is another newform $f'$ with $f = f'$ modulo some integer $N$, then one can take $g = (f - f')/N$, and this will be integral but its Petersson product with $f$ will be $\langle f, f \rangle / N$. So the issue of bounding Petersson products below is quite closely related to the issue of congruences between eigenforms.

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Dear David Loeffler, thanks a lot for the helpful comment. I will edit my question according to your comment. –  ranicl Apr 9 '13 at 13:09
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