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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 $\rho:\Gal(K/\Q)\to\GL_n(\C)$ 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 $f(\rho, p) = \sum_{i\geq 0} \frac{g_i}{g_0} (\chi(1) - \chi(G_i))$ where $\chi$ is the character of the representation $\rho$ and $\chi(G_i)$ is the average value of $\chi$ on $G_i$. Since $\rho$ maps to $\GL_n(\C)$, $\chi(1)=n$.

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 $f(\rho,p)>0$.

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• Last edited by John Jones on 2020-10-25 00:24:04
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