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 $t \in \mathbb{R}$.



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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.



Does every equivalence class of Hecke characters contain a distinguished element?
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 $\chi' = \chi\ \ \beta^{-1}(|\bullet|)$.


