The conductor of an Artin representation is a positive integer that measures its ramification. It can be expressed as a product of local conductors.
Let $K/\Q$ be a Galois extension and $\rho:\Gal(K/\Q)\to\GL(V)$ an Artin representation. Then the conductor of $\rho$ is $ \prod_p p^{f(\rho,p)} $ for non-negative integers $f(\rho,p)$, where the product is taken over prime numbers $p$.
To define the exponents $f(\rho,p)$, fix a prime $\mathfrak{p}$ of $K$ above $p$ and consider the corresponding extension of local fields $K_{\mathfrak{p}}/\Q_p$ with Galois group $G$. Then $G$ has a filtration of higher ramification groups in lower numbering $G_i$, as defined in Chapter IV of Serre's Local Fields [MR:0554237, 10.1007/978-1-4757-5673-9]. In particular, $G_{-1}=G$, $G_0$ is the inertia group of $K_\mathfrak{p}/\Q_p$, and $G_1$ is the wild inertia group, which is a finite $p$-group.
Let $g_i = |G_i|$. Then \[ f(\rho, p) = \sum_{i\geq 0} \frac{g_i}{g_0} (\dim(V) - \dim(V^{G_i}))\] where $V^{G_i}$ is the subspace of $V$ fixed by $G_i$.
Note that if $p$ is unramified in $K$, then $f(\rho,p)=0$ and conversely, if $\rho$ is faithful and $p$ is ramified in $K$, then $f(\rho,p)>0$.
- Review status: reviewed
- Last edited by John Jones on 2023-04-26 10:22:33
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