Dirichlet's regulator vs Beilinson's regulator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:29:15Z http://mathoverflow.net/feeds/question/34801 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34801/dirichlets-regulator-vs-beilinsons-regulator Dirichlet's regulator vs Beilinson's regulator Andreas Holmstrom 2010-08-06T22:35:32Z 2011-01-06T23:48:43Z <p>Consider a number field $F$ with ring of integers $O_F$. The Beilinson regulator can in this particular setting be viewed as a map from $K_n(O_F)$ to a suitable real vector space. Here $n$ is any positive odd integer. For $n \geq 3$, there is also another regulator map, defined by Borel, and Burgos has proved that the Borel regulator (suitably normalized) is twice Beilinson's regulator. For $n=1$, the Borel regulator is not defined (as far as I understand, correct me if I'm wrong), but we do have the original and most basic example of a regulator, which is that of Dirichlet.</p> <p>Question: Is there some form of comparison theorem between the Beilinson regulator and the Dirichlet regulator?</p> http://mathoverflow.net/questions/34801/dirichlets-regulator-vs-beilinsons-regulator/51357#51357 Answer by profilesdroxford54 for Dirichlet's regulator vs Beilinson's regulator profilesdroxford54 2011-01-06T23:48:43Z 2011-01-06T23:48:43Z <p>As you comment, the Beilinson regulator is defined using Chern classes for Deligne cohomology. In particular, for $K_1(\mathbb C)$, the first Chern class induces the identity from $K_1(\mathbb C)\simeq \mathbb C^*$ to $H^1_{Deligne}(pt, \mathbb Z(1))\simeq \mathbb C^*$. One passes to $H^1_{Deligne}(pt, \mathbb R(1))\simeq \mathbb R$ by taking the logarithm. So the Belinson regulator for $K_1$ is induced by the map $\log |\; |:K_1(\mathbb C)\to \mathbb R$.</p>