# 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 $\chi_{\infty}$?

## 1 Answer

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).