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Daniel Loughran
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Does every equivalence class of Hecke characters contain a distinguished element?

Let $k$ be a number field and let $I_k$ denote the idele group of $k$. Let $$|\cdot|: (x_v) \mapsto \prod_{v \in \Omega_k} |x_v|,$$ denote the adelic norm map. If $I_k^1$ denotes the kernel of this map, then we have a short exact sequence $$1 \to I^1_k \to I_k \to \mathbb{R}_{>0} \to 1. \qquad (*)$$ Next, recall that a Hecke character for $k$ is simply a continuous character

$$\chi:I_k \to S^1 \subset \mathbb{C}^*,$$

which is trivial on $k^* \subset I_k$.

We say that two Hecke characters are equivalent if their restrictions to $I^1_k$ are equal. It follows easily from the sequence $(*)$ that every Hecke character equivalent to a fixed Hecke charater $\chi$ has the form $\chi|\cdot|^{it}$, for some $t \in \mathbb{R}$.

Does every equivalence class of Hecke characters contain a distinguished element?

I won't deny that this question is slightly vague; what I want is something like a canonically defined Hecke character in each equivalence class.

If my calculations are correct, then if an equivalence class of Hecke characters contains a character $\chi$ of finite order, then $\chi$ is the unique character of finite order in its class. I certainly count such a character as being distinguished. The problem is therefore with Hecke characters of infinite order, which I have to say I don't understand that well. Perhaps there is a character in the class whose L-function has certain special properties? Many of the references I have come across about Hecke characters choose a splitting of the exact sequence $(*)$ in order to decompose Hecke characters. I certainly don't view this as canonical as there is no canonical choice of splitting.

Daniel Loughran
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