# On Weil's characters of type (A)

In Weil's paper

"On a certain type of characters of the idele-class group of an algebraic number field",

Weil introduces a class of characters on the Idele class group (of not necessarily finite order) which take algebraic values. He calls them, characters of type (A) (the (A) probably stands for algebraic, I guess). On the last 3 lines of p. 3, he says that Artin pointed out to him how to use Minkowski's theorem on units (in absolutely normal number fields) to show essentially that all such characters come from CM fields.

Basically, this question boils down to understand the following: For which number fields $K$; collection of integers $f_{\iota}$; and a suitable choice of an integer $m$, is it possible to have $$\prod_{\iota\in S_{\infty} }\left(\frac{\iota(\epsilon)}{\overline{\iota(\epsilon)}}\right)^{mf_{\iota}}=1$$ for all $\epsilon\in \mathcal{O}_K^{\times}$? Here $S_{\infty}$ denotes the set of infinite places of $K$.

Q1: Is there a reference in the literature where I can find a proof (or a sketch of a proof) of this result?

Q2: If no such reference exists, then how does one prove it from Minkowski's theorem?

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See Laurent Fargues's great article 'Motives and automorphic forms: the potentially abelian case', available here: www-irma.u-strasbg.fr/~fargues/Motifs_abeliens.pdf It's Proposition 1.12. –  Keerthi Madapusi Pera Jan 30 '13 at 14:49
Also, do you mean Dirichlet's theorem, rather than Minkowski's? –  Keerthi Madapusi Pera Jan 30 '13 at 14:49
Fargues' article is indeed great in many ways but I don't find his proof of Proposition 1.12 all that clear -- he leaves a lot of checking to the reader. –  David Loeffler Jan 30 '13 at 20:16
see Patrikis' 2012 thesis: math.ias.edu/~patrikis/variationsoct2012.pdf (in particular, 5.1.3). the first proof given is the standard one (see Keerthi's answer below), but it also indicates a second proof. –  fherzig Jan 31 '13 at 19:37

Now that I've had the chance to look at Fargues's proof carefully, it seems incomplete for what it claims to prove, but it might answer the question under consideration.

Suppose that you have a collection of integers $f_{\iota}$ indexed by embeddings $\iota:K\hookrightarrow\mathbb{C}$ and satisfying for some integer $m$ (this is slightly different from the OP's identity):

$$\prod_{\iota}\iota(\epsilon)^{mf_{\iota}}=1,\text{ for all \epsilon\in\mathcal{O}_K^\times}.$$

Then I claim that there exists a CM sub-field $L\subset K$ such that $f_{\iota}$ only depends on $\iota\vert_L$ (for the purposes of this statement, totally real fields are CM).

The assertion that allows us to reduce this to Dirichlet's unit theorem is the following: It suffices to show that, for all $\sigma\in Aut(\mathbb{C})$, $f_{\sigma\iota}+f_{\sigma\overline{\iota}}$ is independent of $\iota$ and $\sigma$.

(EDIT: It was incorrectly stated in the original version that it was enough to require independence from $\iota$ alone).

To see this, we reinterpret this condition as follows: After fixing an embedding $\bar{\mathbb{Q}}\hookrightarrow\mathbb{C}$, we can view the collection $(f_{\iota})$ as an element $f$ of the $G_{\mathbb{Q}}$-module of maps $Hom(K,\bar{\mathbb{Q}})\to\mathbb{Z}$. The Galois action is given by $(\sigma(f))_{\iota}=f_{\sigma^{-1}\iota}$. From this point of view, our claim amounts to: For every complex conjugation $c$ of $\bar{\mathbb{Q}}$, $f_{\iota}+f_{c(\iota)}$ is independent of $c$ and $\iota$. In particular, for all $\sigma\in G_{\mathbb{Q}}$, and all complex conjugations $c$, we have:

$$f_{\sigma^{-1}\iota}+f_{c(\sigma^{-1}\iota)}=f_{\iota}+f_{c\iota}.$$

Another way to write this is, for all $\iota$:

$$\sigma(f)_{\iota}+\sigma(c(f))_{\iota}=f_{\iota}+c(f)_{\iota}.$$

So $\sigma$ stabilizes $f$ if and only if it stabilizes $c(f)$, for all complex conjugations $c$. Let $L$ be the fixed field of the stabilizer of $f$ in $G_{\mathbb{Q}}$. We have shown that $L$ is stabilized by all complex conjugations. To show that $L$ is CM we need to now need to know that all complex conjugations have the same action on $L$. This amounts to showing that $f_{c(\iota)}$ does not depend on the choice of $c$. But, since $f_{\iota}+f_{c(\iota)}$ does not depend on $c$ by hypothesis, this is clear.

Let us return to the first identity above. Taking the logarithm of its absolute value gives us:

$$\sum_{\iota}f_{\iota}\vert \iota(\epsilon)\vert=0.$$

But consider the map

$$\ell:\mathcal{O}_K^\times\otimes\mathbb{R}\rightarrow\oplus_{v\vert\infty}\mathbb{R}.$$ given by $\ell(\epsilon\otimes 1)_v=\log\vert \epsilon\vert_v$ if $v$ is real and by $\ell(\epsilon\otimes 1)_v=2\log\vert \epsilon\vert_v$ if $v$ is complex. The indexing set here is the set of inequivalent archimedean norms on $K$. Then Dirichlet's unit theorem says that $\ell$ is an isomorphism onto the hyperplane $H$ where the sum of the co-ordinates is identically $0$.

This shows that, for all $(b_v)\in H$,
$$\sum_{v}\frac{f_{\iota(v)}+f_{\overline{\iota}(v)}}{2}b_v=0.$$ Here, if $v$ is complex, $\iota(v)$ and $\overline{\iota}(v)$ are the two complex embeddings inducing $v$. If $v$ is real, they're the same embedding. If one uses the usual basis of $H$ consisting of differences of the basis vectors for the ambient space, one finds that $f_{\iota}+f_{\overline{\iota}}$ is independent of $\iota$. Applying the same argument to the character $\epsilon\mapsto\prod_{\iota}\sigma^{-1}(\iota(\epsilon))^{f_{\iota}}=\prod_\iota\iota(\epsilon)^{f_{\sigma\iota}}$, one sees that $f_{\sigma\iota}+f_{\sigma\overline{\iota}}$ is independent of $\iota$ as well.

(EDIT: The remainder of the proof has been changed to reflect the correction in the criterion above.)

Set $w(\sigma)=f_{\sigma\iota}+f_{\sigma\bar{\iota}}$: this doesn't depend on $\iota$. To finish the proof, we must show that it is independent of $\sigma$ as well. First, let $L$ be as above. From what we have proven already, we find that $L$ is stable under all complex conjugations. Assume that $L$ is totally complex; this implies that $K$ is also totally complex. Then, if $n=[K:\mathbb{Q}]$, we have:

$$2nw(1)=\sum_{\iota}(f_{\iota}+f_{\bar{\iota}})=\sum_{\iota}(f_{\sigma\iota}+f_{\sigma\bar{\iota}})=2nw(\sigma).$$

So, in this case, the independence is clear.

To finish, it is enough to show the following: If $L$ has one real place, then $L=\mathbb{Q}$. But the hypothesis implies that there is a complex conjugation $c$ that acts trivially on $L$, and so $2f_{\iota}=f_{\iota}+f_{c(\iota)}$ is independent of $\iota$. In this case, the Hecke character must be the twist of a finite order character by a power of the norm character.

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Hi Keerthi, I'm confused about one point. If $Stab(f)=H\leq G_{\mathbf{Q}}$, then you are showing that for all complex conjugation $c\in G_{\mathbf{Q}}$, $cHc^{-1}=H$. But this does not seem to imply that $L:=\bar{\mathbf{Q}}^{H}$ is a CM field. For example, if $L$ is Galois over $\mathbf{Q}$ and not CM, then the condition $cHc^{-1}=H$ is trivially satisfied since $H$ is normal in $G_{\mathbf{Q}}$. Am I right? –  Hugo Chapdelaine Jan 31 '13 at 17:10
Hi Hugo, you're right! I was confused about why I seemed to be getting post facto that $f_{\sigma\iota}+f_{\bar{\sigma}\iota}$ was independent also of $\sigma$. But of course I need to show this a priori! I'll see if I can fix the argument. –  Keerthi Madapusi Pera Jan 31 '13 at 20:25
Edits made. I think the proof is now correct. –  Keerthi Madapusi Pera Jan 31 '13 at 21:13
Also, as Florian points out in the comments above, pretty much the same proof appears in Patrikis's thesis. –  Keerthi Madapusi Pera Jan 31 '13 at 21:23
Thanks a lot Keerthi! Your proof seems to be correct. –  Hugo Chapdelaine Jan 31 '13 at 23:05

Schappacher, Periods of Hecke characters, Chapter Zero, involves some of this, though he quotes SGA 4.5 section 5 for some parts, which could also be useful. I can't track a specific results like what you want, but I think he covers it, in an essence.