# Explicit examples of algebraic Hecke characters with infinite image?

Jerry Shurman has a lovely set of notes explaining the classical definition of Hecke characters, the idelic definition of Hecke characters, their relationship, and the classification of algebraic Hecke characters for $\mathbb{Q}$ as Dirichlet characters. He also gives a single family of examples of algebraic Hecke characters with infinite order, namely $\displaystyle \chi: \mathbb{Z}[i] \to \mathbb{C}^{\times}$ given by $\displaystyle \chi(z) = \left(\frac{z}{|z|}\right)^{4n}$ for integers $n$.

It's clear that one has essentially the same family for imaginary quadratic number fields with class number $1$. But what about imaginary quadratic fields with higher class number? I imagine that one has one family analogous to the one above for each ideal class, but I don't know what they should look like...

What do infinite image algebraic Hecke characters for real quadratic fields look like? Because the unit group is infinite, one can't kill the unit group as above, by putting a $4$ in the exponent...

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The "most obvious" algebraic Hecke characters of a field $K$ are the characters of the ideal class group of $K$, which have trivial infinity-type and trivial conductor. There might be no non-trivial examples (as in the case of Q). But you can get more examples by:

• beefing up the conductor (which gets you Dirichlet characters, or more generally characters of ray class groups, but never anything of infinite order)

• changing the infinity-type (the restriction of $\chi$ to the connected component of the identity in $(K \otimes \mathbb{R})^\times \subseteq \mathbb{A}_K^\times$).

Over $K=\mathbb{Q}$, the infinity-types are pretty restricted: the only algebraic characters of the group of positive reals are the maps $x \mapsto x^k$, so you just get powers of the "norm" character (the character of the ideles whose restriction to $\mathbb{R}_+$ is the identity, and which sends a uniformizer at a prime $p$ to $1/p$).

Over a number field the game is more subtle. Let's first suppose $K$ is totally real of degree $n$. The infinity-type of a character looks like $z \mapsto z_1^{k_1} \dots z_n^{k_n}$ for integers $k_1, ..., k_n$, where $z_1, ..., z_n$ are the embeddings into $\mathbb{R}$, but there is a constraint that the infinity-type needs to vanish on a finite-index subgroup of the global units, and this forces the vector $k_1, ..., k_n$ to be orthogonal to the lattice in $\mathbb{R}^n$ generated by the vectors

$$\{ (\log |u_1|, ..., \log |u_n|) : u \in \mathcal{O}_K^\times\}.$$

By Dirichlet's unit theorem this lattice has rank $r_1 + r_2 - 1 = n-1$, so its orthogonal complement has rank at most 1 -- it's spanned by $(1, ..., 1)$. This tells us that every algebraic Hecke character is just a finite-order character times a power of the norm character, which is a bit boring.

For non-totally-real fields the game gets more interesting because there are not so many units. If $K$ is a CM field of even degree $2d$, then the unit group has rank $d-1$, and you can show that the weights of algebraic Hecke characters span a lattice of rank $d + 1$ (spanned by the norm character and characters of the form $x \mapsto \sigma_i(x) / \overline{\sigma_i(x)}$ for each embedding $\sigma_i: K \hookrightarrow \mathbb{C}$).

If $K$ is not either totally real or CM, things are more interesting still: the lattice spanned by the logs of the units has rank $r_1 + r_2 - 1$, so its orthogonal complement has dimension $1 + r_2$ over $\mathbb{R}$, but you can't find enough integer vectors in the orthogonal complement. For instance, if $K = \mathbb{Q}(\sqrt[3]{2})$ then the only possibility is the norm character, again (it is a fun exercise to check this by hand in this case). This is an instance of a theorem of [edit: Emil Artin and] Andre Weil: for any number field $K$, and any algebraic Groessencharacter of $K$, the infinity-type of the character must factor through the norm map to the maximal CM subfield of $K$ [edit: or to $\mathbb{Q}$ if there is no such subfield].

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Dear David: for the theorem you mention at the end, it is probably more appropriate to attribute it to both Artin and Weil (in view of comments Weil makes about this result in his paper where the result is used). It may also be worth noting that when $K$ contains no CM subfield then "maximal CM subfield" means $\mathbf{Q}$ (for the purpose of the statement of this theorem). –  user27056 Nov 8 '12 at 22:54
@xbnv: Thanks, I have edited my answer in the light of your comments. –  David Loeffler Nov 9 '12 at 15:47
For complex CM fields, the your possible weights $\sigma_i/\overline\sigma_i$ are not the whole story: you gave a sublattice of index $2^d$. It's more that, for each CM type $\Phi$, i.e. a choice of one element of each $\{\sigma_i,\overline\sigma_i\}$, you can add the weight $\sum_{\sigma \in \Phi} \sigma$. (Thus your example is the difference of a CM type and its conjugate, or twice a CM type minus the norm.) Note that the Hecke characters attached to CM abelian varieties need precisely the weights $\Phi$. –  Jay Dec 24 '12 at 16:36

Let $n$ be the maximal order of an element of the class group. Take an ideal, take the $n$th power, take a generator, square it, divide by its conjugate.

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