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Fix integers $k\geq2$ and $N>1$, and let $S(k,N)$ denote the normalized new Hecke eigenforms in $S_k(\Gamma_1(N))$. [If it makes my question easier to answer, feel free to replace this with $\Gamma_0(N)$].

For $f \in S(k,N)$, one can define an associated adjoint L-function $L(s,\mathrm{ad}\ f)$. Of particular interest is the special value at $s=1$, which is related to congruences between modular forms. There is a period $\Omega_f$ such that $\frac{L(1,\mathrm{ad}\ f)}{\Omega_f}$ is algebraic.

I am interested in the finite sum

$\displaystyle\sum_{f \in S(k,N)}\frac{L(1,\mathrm{ad}\ f)}{\Omega_f} $.

What, if anything, can be said about this sum? Can we bound it?

There appears to be a vast amount of literature on related topics (e.g. the paper http://www.sci.ccny.cuny.edu/~bfeigon/Average_L-values.pdf seems relevant), but I am unqualified to determine what is known for the sum I wrote above.

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Here is a naive way I think you can get some bounds. Warning: I have not thought through this too carefully, and the bounds from such an approach might not be as good as you want in any case.

The adjoint $L$-value is essentially the Petersson norm $(f,f)$. If $f$ is a new form, then it is $\frac{2^{k}}{N} (f,f)$ with normalization as in that Feigon-Whitehouse paper. So you want to bound $(f,f)$. Well, the Petersson norm is equivalent to the sup norm (of $y^{k/2}|f|$) on $S(k,N)$. Asymptotic bounds for the sup norm have been studied analytically, for instance by Blomer-Holowinsky (which also contains a summary of the relation between the sup norm and Petersson norm) in the case of square free level and recently by Saha in greater generality. This gives you bounds for individual newforms, and a slightly cruder bound for a general cusp form. Then you can use the dimension formula to at least get a rough bound on this sum. One can maybe do better by trying to filter forms in $S(k,N)$ according to their "true level."

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  • $\begingroup$ Thank you! This is probably as good an answer as I'm going to get, but I'm going to wait a day or two before accepting this just in case. $\endgroup$ – Jeff H Apr 11 '15 at 12:59
  • $\begingroup$ @JeffH Well, I can imagine an analytic number theorist might have a better answer. In any case, if at some point you discover something nice, let me know. I have some vague ideas for a different approach, which I may think about if I ever have time. $\endgroup$ – Kimball Apr 11 '15 at 18:03

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