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Edits as suggested by xbnv
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David Loeffler
David Loeffler
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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].

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 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 $K$.

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].

Source Link
David Loeffler
David Loeffler
  • 37k
  • 3
  • 89
  • 194

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 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 $K$.