$D_v$ is division if and only if $\ve(1/2, \pi'_{E_0, v})=-1$

-   $\ve(1/2, \pi\otimes\eta_{E_0})=1$,

where $\pi'_{E_0}$ denotes the base change of $\pi'=\operatorname{JL}(\pi)$ to $\GL_2(\A_{E_0})$ and $\eta_{E_0}$ is the qudratic character of $\A^\times$ attached to $E_0$. Suppose we have such an $E_0$. Theorem \[thm:FH\] asserts that there are infinitely many quadratic extension $E$ of $F$ such that $E_v=E_{0, v}$ for any $v\in S(\pi)$ and $L(1/2, \pi\otimes\eta_{E})\neq0$. From Theorem \[thm:W2\] (2), we see that $\pi$ is $E^\times$-distinguished for each such $E$.

Now we assume we have $E_0$ which satisfies (a). For $v\in S(\pi)$, $\pi_v$ is generic unramified and has trivial central character, hence for any quadratic étale algebra $K_v$ over $F_v$, we have $\ve(\pi_{K_v})=1$. Combining this with (a), we get $$1=\ve(1/2, \pi'_{E_0})=\ve(1/2, \pi)\ve(1/2, \pi\otimes\eta_{E_0})\eta_{E_0}(-1)=\ve(1/2, \pi\otimes\eta_{E_0}).$$ Thus (b) follows from (a). The only remaining problem is to find $E_0$ which satisfies (a). This can be done if there is a quadratic étale algebra $\cE_v$ over $F_v$ at each $v\in S(\pi)$ so that we have $\ve(1/2, \pi'_{\cE_v})=-1$ if and only if $D_v$ is division. Once we have such a family $\{\cE_v\}$, we can take a quadratic extension $E_0$ so that $E_{0, v}=K_v$ for each $v\in S(\pi)$ by the strong approximation theorem. However, it seems not 