There are two concepts of self dual character, one is for global and another is for local.
Let $K$ be an imaginary quadratic number field, and a Hecke character $\chi : \mathbb{A}_K^{\times}/K^{\times} \to \mathbb{C}^{\times} $ is called self-dual if for any $x \in \mathbb{A}_K^{\times}$ $$ \chi(x \cdot \bar{x}) = |x|_{\mathbb{A}_K}$$ where the bar means conjugate and $|x |_{\mathbb{A}_K} = \prod_v|x_v|_v$ .
Fix a prime $p$. For any $K_v,\ v|p$, we konw that $K_v/Q_p$ is a quadratic extension. Let $\omega : K_v^{\times} \to \mathbb{C}^{\times} $ be a character, then it's called self-dual if $\omega|_{Q_p^\times}$ is the quadratic character of $Q_p^{\times}$ associated to the quadratic extension $K_v/Q_p$.
My question is, if a global Hecke character is self-dual, can we say it's every local component is self-dual?
Form the name I think that's true, but I have problem in proving this satement.
I consider the case that $K_v/Q_p$ is ramified, and let denote the quadratic character of $Q_p^{\times}$ induced from quadratic extension $K_v/Q_p$ by $\omega$. Then because in such a case $p \in Nm_{K_v/Q_p}K_v$, we know that $\omega(p) = 1$. Then we ask will a self-dual character of $K$, say $\chi$, always give $\chi_v(p) = 1?$
And I think that can't be true at all ! Since we can embed $p$ into $\mathbb{A}_K$ by $x=(1,1,...,1,p,1,...)$, only the $v$-place is $p$ and other places be $1$. Then for a self-dual Hecke character $\chi$, we have $$ \chi(x) = |x \cdot \bar{x}|_{\mathbb{A}_K} = |p^2|_v = p^{-4}$$ And we also have $$ \chi(x) = \prod_w \chi_w(x_w) = \chi_v(p) $$ Thus we always have $\chi_v(p) = p^{-4}$, which means $\chi_v$ can't be self-dual !
Did I make some mistakes?
