What are non-abelian \$L\$-functions? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:48:27Z http://mathoverflow.net/feeds/question/78928 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78928/what-are-non-abelian-l-functions What are non-abelian \$L\$-functions? James D. Taylor 2011-10-23T21:31:53Z 2011-10-23T21:42:09Z <p>I have heard people discussing the utility of \$L\$-functions, claiming that since they are essentially cohomological entities, they are "abelian" and therefore lack force.</p> <p>From looking around on the web, I see that this idea has a base of believers, and that there is some notion of non-abelian \$L\$-functions.</p> <h3>Question</h3> <p>What is the definition of non-abelian \$L\$-functions? Does it have to do with replacing cohomology with homotopy in some way? How does it relate to the original definition of \$L\$-function (in particular, what is the analogue of the characteristic polynomial?)? What is the context in which it arises?</p> http://mathoverflow.net/questions/78928/what-are-non-abelian-l-functions/78930#78930 Answer by anon for What are non-abelian \$L\$-functions? anon 2011-10-23T21:42:09Z 2011-10-23T21:42:09Z <p>"I have heard people discussing the utility of L-functions, claiming that since they are essentially cohomological entities, they are "abelian" and therefore lack force." This sounds like twaddle to me. Artin L-series are non-abelian; the L-series of representations of the Weil group are non-abelian; automorphic L-series are nonabelian (unless they are on GL1); the L-series of motives are usually nonabelian. Perhaps you could clarify your question.</p>