# Special value of $L$-function

Let $p$ be a prime number. Let $f$ be a newform of weight 2 on $Γ_0(p)$, and $E_f$ denote the associated newform quotient of $J_0 (N)$ over $\mathbb{Q}$. Is there a way to express the algebraic part of the special $L$-value of $E_f$ over $K$, a totally real quadratic field, in terms of the conjugacy classes of maximal orders in the definite quaternion algebra over $\mathbb{Q}$ ramified at $p$ and $\infty$?. p.s: when $K$ is a quadratic imaginary field, there is such an expression.

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No. $\;\;\;\;\;$ –  David Hansen Aug 27 '12 at 21:48

Elaborating David H's hilariously laconic comment... :)

Lacking any other answer: work of Shimura and Waldspurger from 20+ years ago may be the relevant stuff to see what is true.

My not-so-well-informed perception of the situation is that whatever one is likely to prove (or is known by this date, or at least not-so-long-ago) goes by way of a theta correspondence.

But/and a theta correspondence from an orthogonal group to $SL_2(\mathbb R)$ (or a two-fold cover, if necessary) maps at archimedean places to holomorphic discrete series essentially only when the source repn of the orthogonal group is the trivial repn on a definite orthogonal group, such as $SO(2)$. In contrast, at archimedean places $SO(1,1)$ produces principal series, so such an orthogonal group's afms map to waveforms, not holomorphic things.

That is, if things go by way of Weil (-Segal-Shale) repns, then hitting holomorphic things is a significant constraint, whatever is going on at finite primes.

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Let me elaborate on my rudely terse comment. Formulas of the type I think the questioner has in mind express $L(1/2, f \otimes \chi)$ as a period integral of a certain vector in a Jacquet-Langlands correspondant $\pi_f^{JL}$ of $\pi_f$ against $\chi$, where $\chi$ is a character of $\mathbf{A}_K^\times$. Since you need $K^\times \hookrightarrow B^\times$ for this to work where $B$ is the quaternion algebra on which $\pi_f^{JL}$ lives, $B$ will always be split at $\infty$ when $K$ is real quadratic. –  David Hansen Sep 24 '12 at 12:48