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.

  • $\begingroup$ Doesn't Weil have some type "A0" of Hecke characters, which are the "algebraic" ones? They include other infinite ones, the Grossencharacters, in CM extensions (see 2nd link below). They do not include those used in XV of Lang (involving a log of a fundamental unit in a real quadratic field, to trivialize units) as in this question mathoverflow.net/questions/65059/… See also mathoverflow.net/questions/111851/… and $\endgroup$
    – v08ltu
    Jun 12, 2013 at 21:15

1 Answer 1


Dear Daniel, the answer is yes.

An easy key lemma:

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

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 $\chi' = \chi\ \ \beta^{-1}(|\bullet|)$.

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    $\begingroup$ The integer $n$ is not specified anywhere. Was a condition left out? $\endgroup$
    – KConrad
    Jun 14, 2013 at 1:54
  • $\begingroup$ Thanks for the answer, but I just want to clarify something: Is the map $\mathbb{R}_{>0} \hookrightarrow I^{\infty}_k$ canonical? Or does it depend on a choice of isomorphism $k \otimes_{\mathbb{Q}} \mathbb{R} \cong \mathbb{R}^{r_1} \times \mathbb{C}^{r_2}$? $\endgroup$ Jun 14, 2013 at 8:59
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    $\begingroup$ @Kconrad: you're right. $n$ is the degree of $k$ over $\mathbb Q$. I have edited my answer to say this. @Daniel: the map $i$ is canonical. The only freedom you have in defining the isomorphism of $\R$-\algebras $k \otimes_{\mathbb Q} \mathbb R \simeq \mathbb R^{r_1} \times \mathbb C^{r_2}$ is that you can let act the complex conjugacy on each of the $\mathbb C$ factors (hence you have $2^{r_2}$ such isomorphisms). But as far as the embedding of $\R$ is concern, this choice does not change anything. $\endgroup$
    – Joël
    Jun 14, 2013 at 14:08
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    $\begingroup$ Right I think I get it now. The map $i$ is "as good as" a splitting of the exact sequence $(*)$. It gives you a distinguished subgroup for you to play with. Thanks again. $\endgroup$ Jun 14, 2013 at 16:17
  • $\begingroup$ Out of interest, I think there is a nice analytic way to think of this in terms of the functional equation of the associated L-function. Namely, $\chi$ is good if and only if the sum of the imaginary parameters of the associated Gamma factors is equal to zero. $\endgroup$ Jun 14, 2013 at 16:33

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