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The classical mean value result of Murty and Murty (1991) and Bump, Friedberg, and Hoffstein (1990) on derivatives of modular form L-functions $L(s,f)$ proves (roughly speaking) the existence of infinitely many imaginary quadratic fields $K$ for which the Heegner hypothesis holds, ie., every prime dividing the level of a given modular form $f$ splits in $K$, such that a $L(s,f)$ has a simple zero at $s=k$, with $2k$ being weight of $f$, and such that $L(k,f,\chi)\neq 0$ for the twist $\chi$ associated to $K$ (or vice versa).

I am looking to see if this result has been generalized to (any) fields $K$ in which the Heegner hypothesis fails to hold. Searching through the literature has not proved to be a simple task.

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    $\begingroup$ You probably wanted to write "simple zero" instead of "simple pole". $\endgroup$
    – GH from MO
    Commented Jun 16, 2017 at 20:49

2 Answers 2

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Let $K/\mathbb{Q}$ be a number field, and let $\pi$ be a cuspidal automorphic representation of $\mathrm{GL}_2(\mathbb{A}_K)$, where $\mathbb{A}_K$ is the ring of adeles over $K$. Consider the standard $L$-function $L(s,\pi)$ associated to $\pi$. The proof of Theorem B (part 2) of Friedberg and Hoffstein (Nonvanishing theorems for automorphic L-functions on GL(2), 1995, Ann. of Math.) yields (under suitable hypotheses) a mean value theorem for $L'(\frac{1}{2},\pi\otimes\chi)$ as $\chi$ varies over the quadratic Hecke characters. This gives a field-uniform version of Murty and Murty (1991) and Bump, Friedberg, and Hoffstein (1990). Perhaps this is a good place to look. Another paper which is perhaps related to your question is https://arxiv.org/abs/0802.4027.

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Here is what the Friedberg-Hoffstein result says: In non-technical language, suppose you are given a GL(2) automorphic L-series with the property that there exists some quadratic twist such that the functional equation of the twisted L-series has a negative sign. Then there must exist infinitely many distinct quadratic twists of this L-series with the property that the twisted L-series has a simple zero at the center of the critical strip.

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  • $\begingroup$ +1 Thank you! I would accept both answers if I could. $\endgroup$
    – Tian An
    Commented Oct 24, 2017 at 14:32

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