Hecke characters of imaginary quadratic fields Let $K$ be a number field and let $\mathfrak m$ be an integral ideal. Let $I(\mathfrak m)$ be the group of fractional ideals of $K$ coprime with $\mathfrak m$. Let $P(\mathfrak m)$ be the group of principal fractional ideals coprime with $\mathfrak m$. Finally, let $K_{\mathfrak m}$ be the set of elements of $K$ which are coprime with $\mathfrak m$. Recall that there is a map $K\to \mathbb R\otimes K\simeq (\mathbb R)^{r_1}\times (\mathbb C)^{r_2}$ which sends $\alpha\mapsto 1\otimes\alpha$ where $r_1$ (resp. $r_2$) is the number of real (resp. complex) places of $K$.
A Hecke character of $K$ of conductor $\mathfrak m$ and infinity type $\chi_{\infty}$ is a homomorphism
$$\chi\colon I(\mathfrak m)\to \mathbb C^*$$
such that there exists a continuous homomorphism
$$\chi_{\infty}\colon (\mathbb R^*)^{r_1}\times (\mathbb C^*)^{r_2}\to \mathbb C^*$$
with the property that
$$\chi((\alpha))=\chi_{\infty}(1\otimes \alpha)^{-1} \text{ for all }\alpha \in K_{\mathfrak m}$$
Now my question is the following: assume that $K$ is an imaginary quadratic field. How do I know that there exist nontrivial Hecke characters on $K$? When $K$ has class number $1$ everything is easy, for example when $K=\mathbb Q(i)\subseteq \mathbb C$ one can just take $\mathfrak m=\mathcal O_K$ and $\chi_n(\alpha)=\left(\frac{\alpha}{|\alpha|}\right)^{4n}$ for $n\in \mathbb Z$. Then $\chi_{n,\infty}(\alpha)=(\alpha/|\alpha|)^{-4n}$ and we're ok. But what happens in the general case? Is it possible to write down explicitely a nontrivial Hecke character? And is it true that given an infinity type $\chi_{\infty}$ there always exist a Hecke character with infinity type $\chi_{\infty}$?
 A: Regarding the question on the general case and explicit constructions: two steps are necessary to get an (algebraic) Hecke $\chi$ of a given infinity type.

*

*Define $\chi$ on the principal ideals in $P(\mathfrak{m})$.

*Extend $\chi$ from $P(\mathfrak{m})$ to all fractional ideals in $I(\mathfrak{m})$.

Suppose $K$ is an imaginary quadratic field. For simplicity, let’s exclude the cases $K=\mathbb{Q}(i)$ and  $K=\mathbb{Q}(\sqrt{-3})$. Then we have $\mathcal{O}_{K}^{\times}=\{1,-1\}$.
For the first step, we choose two integers $n$ and $m$ and set $$\chi_{\infty} (a) = a^n \bar{a}^m . $$
If the weight $w=n+m$ is even then $\chi_{\infty}(-1)=1$ and $\chi_{\infty}$ yields a character on $P(\mathfrak{m})$. We can choose any ideal $\mathfrak{m}$ with even character $\chi_{fin}$ on $(\mathcal{O}_K/\mathfrak{m})^{\times}$ and set
$$ \chi( (a) ) = \chi_{fin} (a) \chi_{\infty}(a) .$$
This also works for $\mathfrak{m} = (1)$, where $\chi((a))=\chi_{\infty}(a)$.
If $w$ is odd, we need to choose a non-trivial ideal $\mathfrak{m}$ and an odd character $\chi_{fin}$, i.e. $\chi_{fin}(-1 + \mathfrak{m} )=-1$. Then $\chi$ can be defined as above.
Now let’s look at the second step. If the class number of $K$ is not $1$, then we need to extend $\chi$ from the principal ideals $(a) \in P(\mathfrak{m})$ to all fractional ideals in $I(\mathfrak{m})$.
We start with an explicit construction. Decompose the ray class group $I(\mathfrak{m}) / P(\mathfrak{m})$ into a sum of cyclic groups. Choose a generator of each component. Say $\mathfrak{a}$ is one of them and $\mathfrak{a}$ has order $n$. Then $\mathfrak{a}^n=(b)$ is principal and we can define $\chi(\mathfrak{a})$ by taking any $n$-th root of $\chi( (b) )$. This defines $\chi$ on  $I(\mathfrak{m})$. Note that the construction depends on choices and two such extensions differ by a character of the ray class group $I(\mathfrak{m}) / P(\mathfrak{m})$.
Another way to extend the character from $P(\mathfrak{m})$ to $I(\mathfrak{m})$ is to use the induced representation. This is a sum of characters,  and we choose of one them as $\chi$. Again,  two such characters differ by a ray class character.
A: The infinity type of a Hecke character is constrained: the units in $ \mathcal{O}_{K}^{\times} \cap K_{\mathfrak{m}}$ must lie within its kernel. So, one may not get any Hecke characters for a given infinity type unless the conductor $ \mathfrak{m}$ is replaced by a smaller ideal within $ \mathfrak{m}$.
            Similarly, if one insists on a fixed conductor, one may not find any Hecke characters until the infinity type is modified. For instance, $ \chi_{\infty}(\alpha)= (\alpha/|\alpha|)$ won't pass for a Hecke character in the example you quoted. We need to pass to its fourth power (as $\mathcal{O}_{K}^{\times}$ has exponent four in this case).
