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

Question

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.

Answer 1

Dear Daniel, the answer is yes.

An easy key lemma:

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Lemma: Let $\alpha: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ be a continuous character, and $n \geq 1$ an integer. Then there exists one and only one character $\beta: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \alpha$.

Proof: Using the isom $\mathbb R^\ast_+ \simeq \mathbb R$, and the polar decomposition, we are reduce to prove that if $\alpha: \mathbb R \rightarrow \mathbb R$ (resp. $\alpha: \mathbb R \rightarrow \mathbb R / \mathbb Z$) then there exists a unique $\beta$ of the same type such that $\alpha=n \beta$. But any character $\alpha$ as above is of the form $\alpha(x)=ax$ (resp. $\alpha(x)= ax \pmod{\mathbb Z}$) for a unique $a \in \mathbb R$. It is thus clear that taking $\beta(x) = (a/n) x$ works and is the unique possible choice for $\beta$. QED

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Now consider the composed map: $i: \mathbb R^\ast_+ \hookrightarrow I_k^\infty \hookrightarrow I_k \rightarrow I_k/k^\ast$, where the first map is the diagonal embedding of $\mathbb R^\ast$ in each of the component at infinity of $I_k$ (the product of which I call $I_k^\infty$). This map is clearly injective. Now call a Hecke character $\chi: I_k/k^\ast \rightarrow \mathbb C^\ast$ good if it is trivial on the image of $i$.

Prop: for any Hecke character $\chi$, there are exactly one good character in its equivalence class.

Proof: Note that $|i(x)|=x^n$ if $n=[k:\mathbb Q]$ (recall that the $|\ |$ on the component $\mathbb C$ is the square of the complex modulus).

If $\chi$ and $\chi'$ are good characters in the same class, then $\psi = \chi (\chi')^{-1}$ is trivial on $I_k^1$ and on $i(\mathbb R^\ast)$, hence on $I_k$ since any element $x$ on $I_k$ can be written as $(x/ i(y)) i(y)$, with $y = |x|^{1/n} \in \mathbb R^\ast_+$, hence is in $I_k^1 i(\mathbb R_+^\ast)$. Hence the uniqueness.

For the existence, by the lemma there exists $\beta: \mathbb R_\ast^+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \chi \circ i$, and consider $\chi' = \chi\ \ \beta^{-1}(|\bullet|)$.