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Nov 11, 2018 at 15:22 answer added Nodt Greenish timeline score: 3
Apr 11, 2013 at 11:06 vote accept ranicl
Apr 10, 2013 at 17:57 answer added François Brunault timeline score: 2
Apr 10, 2013 at 12:57 history edited ranicl CC BY-SA 3.0
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Apr 10, 2013 at 12:54 comment added ranicl Dear Noam D. Elkies and Francois Brunault, thanks for your helpful comments. I edited the question to make it more reasonable to be true. @ Francois Brunault, as far as I can see your comment answers my edited question (a). Would it be possible that you explain in more detail how to get this constant? (Unfortunately I don't understand your arguments)
Apr 10, 2013 at 2:50 comment added Noam D. Elkies 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).
Apr 9, 2013 at 21:39 comment added François Brunault 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.
Apr 9, 2013 at 17:24 comment added Noam D. Elkies 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?
Apr 9, 2013 at 16:51 history edited ranicl CC BY-SA 3.0
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Apr 9, 2013 at 16:51 comment added ranicl 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.
Apr 9, 2013 at 16:06 comment added GH from MO 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.
Apr 9, 2013 at 13:18 history edited ranicl CC BY-SA 3.0
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Apr 9, 2013 at 12:57 answer added David Loeffler timeline score: 2
Apr 9, 2013 at 11:58 history asked ranicl CC BY-SA 3.0