14
$\begingroup$

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.

Question: Is there some form of comparison theorem between the Beilinson regulator and the Dirichlet regulator?

$\endgroup$
2
  • $\begingroup$ By the Bass-Milnor-Serre theorem, $K_1(O_F)=O_F^*$; so the logarithmic embedding of $O_F^*$/(roots of unity in $F$) can be seen as a $K$-theory regulator map, generalized by Beilinson and Borel. $\endgroup$ Commented Aug 7, 2010 at 6:29
  • $\begingroup$ Hi Robin, sure, both the Beilinson regulator and the Dirichlet regulator are defined on $K_1(O_F)$, but the former is defined in terms of Gillet's general theory of Chern classes, and the latter in terms of the explicit logarithm formula found in any algebraic number theory textbook, and the question is if both definitions agree, or if they differ by some constant factor. $\endgroup$ Commented Aug 7, 2010 at 15:05

1 Answer 1

2
$\begingroup$

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 $\log |\; |:K_1(\mathbb C)\to \mathbb R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.