# Are there partially algebraic Hecke characters?


Let $\chi\colon \mathbb{A}_F^\times/F^\times \to \C^\times$ be a Hecke character with local components $\chi_v$ for each place $v$ of $F$. When $v$ is an archimedean place, then there are $m_v\in\Z$, $\sigma_v,\varphi_v\in\R$ such that for all $x\in F_v^\times$, $$\chi_v(x) = \left(\frac{x}{|x|}\right)^{m_v}|x|^{\sigma_v+i\varphi_v}.$$ Recall that $\chi$ is algebraic if for every archimedean place $v$ the local component $\chi_v$ is a rational function, i.e. if there exists integers $p_v,q_v\in\Z$ such that $$\chi_v(x) = x^{p_v}\bar{x}^{q_v}.$$ We understand well when algebraic characters exists: up to a finite order character, they factor through the norm to the maximal CM subfield of $F$ (or to $\Q$ if there is no CM subfield).

For an algebraic character, all $\varphi_v = 0$, and up to finite index and multiplication by a power of the norm this characterises algebraic characters.

Now let $\Sigma$ be a subset of the set of archimedean places of $F$.

Does there exist Hecke characters as above such that $\varphi_v=0$ if and only if $v\in\Sigma$?

If $|\Sigma|\le 1$ then you can obtain such characters by multiplying any Hecke character by a suitable power of the norm.

If $\Sigma$ is the set of all archimedean places, we know the answer by the usual theory of algebraic characters.

Otherwise, since I do not see an obvious reason why such characters should exist, the "obvious guess" is that they do not. Is it true?

I was able to rule out a couple more cases in an ad-hoc way, and it seems to quickly involve transcendence problems.