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$\DeclareMathOperator\Mp{Mp}$Let $F$ be a number field and $\pi$ be an irreducible cuspidal automorphic representation of $\operatorname{PGL}_2(\mathbb{A}_F)$.

Then we can think a submodule $L_{\pi}^2$ of $L_{\mathrm{disc}}^2(\Mp_2)$, the discrete spectrum of automorphic functions on $\Mp_2(F) \backslash \Mp_2(\mathbb{A})$, as

$L_{\pi}^2:=\sum_{a \in F^{\times} \backslash F^{\times^2}} \Theta(\pi \otimes \chi_a)$, where $\chi_a$ is the quadratic character of $\mathbb{A}^{\times}/F^{\times}$ associated to the quadratic extension $F(\sqrt{a})/F$ by global class field theory and $\Theta$ is a theta lift from $\operatorname{SO}_3 \simeq \operatorname{PGL}_2$ to $\Mp_2$.

Then Shimura—Waldspurger correspondence asserts that $L_{\pi}^2$ is the full near equivalence class in $L_{\mathrm{disc}}^2(\Mp_2)$ such that each irreducible summand is in the global Waldspurger packet $A_{\pi}$ of $\pi$ and the number of irreducible summand of $L_{\pi}^2$ is half the number of $A_{\pi}$.

Since $\pi$ is irreducible cuspidal of $\operatorname{PGL} _2$, the number of $A_{\pi}$ should be finite. Therefore, I think there are only finitely many Hecke quadratic characters $\chi$'s such that $L(\frac{1}{2},\pi \times \chi) \ne 0$ because $\Theta(\pi \otimes \chi) \ne 0$ is equal to $L(\frac{1}{2},\pi \times \chi) \ne 0$.

However, the paper of Friedberg and Hoffstein (Theorem B in https://www.jstor.org/stable/2118638) claims that there are infinitely many quadratic characters $\chi$ such that $L(\frac{1}{2},\pi \times \chi) \ne 0$. So I think it contradicts to $\sharp A_{\pi} <\infty$.

What is wrong in this reasoning?

Any comments are welcome!

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    $\begingroup$ Why should the number of A_pi be finite? Doesn't this argument prove that it can't be? $\endgroup$
    – David Loeffler
    Commented Jan 5, 2022 at 12:43
  • $\begingroup$ @DavidLoeffler, The definition of $A_{\pi}$ is the restricted tensor product of local Waldspurger packets $A_{\pi_v}$'s. (Here, $A_{\pi_v}=\{\sigma_v^{+}, \sigma_v^{-}\}$ and $\sigma_v^{+}$ is the theta lift of $\pi_v$ as an irreducible representation of $SO(3)=PGL_2$ to $Mp_2$ and $\sigma_v^{-}$ is the theta lift of $JL(\pi_v)$ to $Mp_2$, where $JL(\pi_v)$ is the J-L transfer of $\pi_v$ to $D^{\times}$, where $D$ is a quaternion division algebra over $F_v$. When there is no J-L transfer of $\pi_v$, define $\sigma^{-}$ as 0. $\endgroup$
    – Andrew
    Commented Jan 5, 2022 at 12:59
  • $\begingroup$ @DavidLoeffler, Since the J-L transfer exists only when $\pi_v$ is discrete series and $\pi_v$ can't be discrete series at unramified places, I thought $A_{\pi}$ should be finite such that $|A_{\pi}|=2^{S_{\pi}}$, where $S_{\pi}$ is the set of places $v$ such that $\pi_v$ is discrete series. If there is something wrong in this argument, I would appreciate if you point it out. $\endgroup$
    – Andrew
    Commented Jan 5, 2022 at 13:03

1 Answer 1

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Revised. The global Waldspurger packet of $\pi$ is indeed finite, as you say in your comment. It's elements are metaplectic representations which are in bijection with the Vogan packet of $\pi$, i.e., me the set of all cuspidal representations $\pi_D$ of a quaternion algebra $D/F$ such that $\pi_D$ corresponds to $\pi$ in the sense of Jacquet-Langlands. There are finitely many relevant $D$'s, which are determined by the finite set of places at which $\pi$ is discrete series.

However this finiteness is unrelated to the question of how many twists of $\pi$ will have nonvanishing central $L$-value. Let $\chi_E$ be the quadratic character associated to a quadratic extension $E/F$. Waldspurger relates $L(1/2, \pi)L(1/2, \pi \otimes \chi_E)$ to a period $P_E$ of some $\pi_D$ in the Vogan packet. As $E$ varies, the period changes, as does the "right" choice for the $D$ on which to choose the $\pi_D$. In particular, infinitely many periods (and thus $L$-values) may vanish and infinitely may be nonzero on the same $\pi_D$.

So there is no contradiction between finiteness of the Vogan packet and the nonvanishing of infinitely many twists. I think the confusion is arising from an assumption that the theta lifts $\Theta(\pi \otimes \chi_a)$ give different automorphic representations for different $a \in F^\times/F^{(2)}$, which is not true.

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  • $\begingroup$ thanks for such kind reply! But due to my stupidness, it is not still clear to me. For the definition of global Waldspurger packet of $\pi$ I mean is the set of irreducible representation of $Mp_2$ associated to $\pi$ though there is a correspondence between irreducible representation of $Mp_2$ and the set of all irreducible cuspidal representation of $D^{\times}$ corresponding to $\pi$ via JL correspondence. It is known that the global $A$-packet of $Mp_2$ associated to $\pi$ consists $\otimes \sigma_v^{\epsilon_v}$ such that $\prod_v \epsilon_v=\epsilon(\frac{1}{2},\pi)$. $\endgroup$
    – Andrew
    Commented Jan 5, 2022 at 15:18
  • $\begingroup$ Since $\epsilon_v=1$ for almost $v$, then $L_{\pi}^2$, the full near equivalent class associated to $\pi$, should be the finite sum of such automorphic representations of $Mp_2$. On the other hands, it is known that $L_{\pi}^2:=\sum_{a \in F^{\times} \backslash F^{\times^2}} \Theta(\pi \otimes \chi_a)$. $\endgroup$
    – Andrew
    Commented Jan 5, 2022 at 15:29
  • $\begingroup$ If we choose $\pi$ as good $\pi_D$ such that $\pi_D$ has infinitely many quadratic twists whose central $L$-values are non-zero, then from the above decompostion, $L_{\pi}^2$ has infinitely many irreducible summands because if $L(\frac{1}{2},\pi \times \chi) \ne 0$ for some quadratic character $\chi$, then the theta lifting $\Theta(\pi \otimes \chi)$ is also nonzero and are all distinguished for different $\chi$'s. I am afraid if these two facts I know about the decomposition of $L_{\pi}^2$ is wrong. $\endgroup$
    – Andrew
    Commented Jan 5, 2022 at 15:30
  • $\begingroup$ @Andrew Ah, I misread your comment. See revised answer. $\endgroup$
    – Kimball
    Commented Jan 5, 2022 at 17:08
  • $\begingroup$ oh, thank you very much! The theta lift $\Theta(\pi \otimes \chi_a)$ is with respect to additive character $\psi_a$ which is related to $a$. I confused that all theta lifts are with respect to the same additive character $\psi$. If it is the case, all theta lift should be different. However, since additive characters in the theta liftings also vary as quadratic characters $\chi$‘s vary, the set of all theta lifts should be finite though $\pi \otimes \chi_a$ all all different. Thank you very much for pointing out some important points. $\endgroup$
    – Andrew
    Commented Jan 6, 2022 at 13:16

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