Conductors of non-abelian number fields?

Is there a definition out there of the notion of conductor of a non-abelian number field (i.e. a finite extension of Q whose Galois group is non-abelian)? If not, is there anyone you know of working on it? The definition for abelian number fields uses class field theory; it comes out of Artin reciprocity (see page 525 of Neukirch's Algebraic number theory).

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Sorry, does non-abelian mean "with non-abelian Galois group"? –  Ben Webster Oct 31 '09 at 16:49
Yeah, sorry. Similarly, abelian means with abelian Galois group. And the Galois group I mean is that over Q. You can also define the conductor of an extension of number fields L/K with abelian Galois group (again using the Artin reciprocity map). –  Rob Harron Oct 31 '09 at 17:02

I think that a good notion of "conductor" isn't going to be intrinsic to the extension K/Q; rather, you might choose some finite-dimensional complex representation rho of Gal(K/Q) and then use the Artin conductor of the resulting Galois representation. When K/Q is abelian, there aren't so many interesting choices of rho.

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I don't know the precise definition of the conductor of a nonabelian Galois extension of Q, but see page 10 of Langlands's expository article "Representation Theory: Its Rise and Its Role in Number Theory" sunsite.ubc.ca/DigitalMathArchive/Langlands/pdf/gibbs-ps.pdf . Presumably the thesis of Joe Buhler referenced therein gives a precise definition, or a reference to one.

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Buhler is discussing: given a Galois extension K/Q with Galois group A_5, what is the smallest Artin conductor of a two-dimensional Galois representation factoring through Gal(K/Q) whose image in PGL(2,C) is A_5? Buhler finds it's 800, and occurs for the quintic polynomial Langlands mentions. Langlands (and Buhler 2 or 3 times) refers to this as the "conductor of K", but Buhler generally refers to this as the minimal conductor of a corresponding projective representation. He makes no assertion that this is a definition of the conductor of K. It certainly offers a possibility though. Thanks. –  Rob Harron Oct 31 '09 at 19:45