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$\DeclareMathOperator\Res{Res}\DeclareMathOperator\Gal{Gal}$I am trying to understand lemma 3.1 of "Abelian Varieties over \mathbb Q and modular forms" of Ribet. ArXiv link

Just so everyone is on the same page.

-Let $E$ be a number field, $\lambda$ a place over $l$ and $E_{\lambda}$ the corresponding complete local field.
-Let $\mathbb I$ be the group of ideles of $K$ and $\mathbb I_{\infty}$ the subgroup of those ideles whose only nontrivial components are on the infinite places.
-$T=\Res_{K/\mathbb Q}(G_{m/K})$ is the torus obtained by restrincting scalar from $K$ to $\mathbb Q$ of the multiplicative group $\mathbb G_{m/K}$.

On this lemma Ribet says that a locally algebraic character $\delta_{\lambda}:\Gal(\overline{\mathbb Q}/\mathbb Q)\longrightarrow E_{\lambda}$ is associated to a Grossencharacter of type $A_0$ or in modern lenguage an algebraic Hecke character.

By that I understand that what he means is that $\delta_{\lambda}\circ i:\mathbb I_{K}\longrightarrow E_{\lambda}$ is and algebraic Hecke character in particular ir takes values in $E$. My definition of algebraic Hecke character is a character of the idele class group $\chi:\mathbb I_{K} \longrightarrow \mathbb C$ such that when restricted to the infinite places it is of the form $\chi(x_{\infty})=\prod_{\nu \in S_{\infty}} x_{\nu}^{n_{\nu}}$ with $n_{\nu}\in \mathbb Z$.

For the proof of this fact Ribet cite page 761 of *"Galois action on division points of abelian varieties with real multiplications" JSTOR link behind paywall.

What says there is that all locally algebraic representations come from linear representations of a certain algebraic group $S_{\mathfrak m}$ called the Serre torus. Later on this same article it is stated that characters of the serre torus are given by a pair of homomorphisms $\chi:\mathbb I_{K}\longrightarrow \overline{\mathbb Q}$ and $f:T\longrightarrow S_{\mathfrak m}$ such that
1-$\chi(x)=1$ when $x\in U_{\mathfrak m}$
2-$\chi(x)=f(x)$ when $x\in K^*$

I suppose that using this two properties one can show that $\chi$ is an algebraic Hecke character but the things I have tried did not work and I can not found the proof anywhere. I am also wondering if in the case of a representation of dimension 1 we can skip the use of the Serre torus $S_{\mathfrak m}$ since the condition of being locally algebraic seem to me very close to conditions 1 and 2. I am not very used to work with ideles so do not worry about being too explicit.

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For $K =\mathbb Q$, class field theory (or, in this case, Kronecker-Weber) says that the abelianization of the Galois group of $\mathbb Q$ is $\mathbb I/\mathbb I_\infty$. So a character $\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q) \to E_\lambda$, which necessarily factors through the abelianization, is the same as a character $\mathbb I/\mathbb I_\infty \to E_\lambda$.

Thus we have a character $\chi\colon \mathbb I \to E_\lambda$ that is trivial on $\mathbb I_\infty$. Since the trivial representation is algebraic, why isn't $\chi$ already a Grossencharacter? The reason is Grossencharacters are required to have the image of the units of each non-archimidean local field be finite, i.e. in this case we want the image of $\mathbb Z_p^\times$ to be finite for all $p$. This is true for all $p\neq l $ but may not be true for $l$.

Instead, the restriction of $\chi$ to $\mathbb Z_l^\times$ is locally algebraic, i.e. there is some open subgroup $U$ and some integer $n$ such that $x\in U$ is sent by $\chi\mid_{\mathbb Z_\ell^\times}$ to $x^n$.

All we have to do to get a Grossencharacter is to modify the character $\chi$ by dividing $\chi(x)$ for an idele $x$ by the $n$th power of the $\mathbb Q_l$-component of $x$ and then multiplying by the $n$th power of the $\mathbb R$-component of $x$. For $x\in \mathbb Q$ the $\mathbb Q_p$ and $\mathbb R$ component are the same rational number so this doesn't do anything.

This preserves all local properties of the character at places other than $l$ and $\infty$. At $l$ we have made the character trivial on the open subgroup $U$, giving the desired finite image of the units. At $\infty$ we have made the character look like $x \to x^n$ and hence be algebraic.

We can send Grossencharacters to locally algebraic Galois characters by reversing this process.

For other fields $K$ it's not much more complicated, we just replace the formula $x^n$ by some product of powers of images of $x$ under different embeddings of $K$.

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  • $\begingroup$ Why do we know that the image is inside $\mathbb C$ and not in $E_{\lambda}$? Because otherwise I cannot multiply by the $\mathbb R$ component. $\endgroup$ Commented May 23 at 17:47
  • $\begingroup$ @JoseCanseco Well, one can just embed $E_\lambda$ into $\mathbb C$. $\endgroup$
    – Will Sawin
    Commented May 23 at 17:59
  • $\begingroup$ Can I do it in a continuous way? $\endgroup$ Commented May 23 at 18:13
  • $\begingroup$ @JoseCanseco One can check indirectly that the final character ends up being continuous. The point is the original character is finitely ramified so after passing to an open subgroup of the ideles it agrees with the $n$th power of the $\mathbb Z_\ell$ component. So when you multiply and divide, after restricting to the open subgroup you get the $n$th power of the $\mathbb R$ component, which is continuous. Since it's continuous on an open subgroup it's continuous on every coset of that open subgroup and thus continuous everywhere. $\endgroup$
    – Will Sawin
    Commented May 23 at 20:43

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