6
$\begingroup$

Doing class field theory from the point of view of class formations, it is my understanding that to get the artin reciprocity map, one does this by inverting an isomorphism which is given by the Tate-Nakayama lemma, which goes as follows:

Let $G$ be a finite group, and let $C$ be a $G$-module and $u \in H^2(G,C)$. If for every subgroup $H$ of $G$ we have that (1) $H^{1}(H,C)=0$, (2) the size of $H^{2}(H,C)$ is the same as the size of $H$ and is generated by $Res(u)$

Then for any $G$-module $M$ such that $Tor_{1}^{\mathbb{Z}}(M,C)=0$, cup-product with $u$ will give an isomorphism $$\widehat{H}^{i}(G,M) \longrightarrow \widehat{H}^{i+2}(G,M \otimes_{\mathbb{Z}} C), \qquad \text{ for all } i \in \mathbb{Z}$$

If one then applies this to number fields $K,F$ with $K/F$, $G=Gal(K/F)$, $C=C_K$ (idele class group) and $M=\mathbb{Z}$, then the case $i=-2$ gives an isomorphism whose inverse is the artin reciprocity map.

My question is what else do all the other isomorphisms tell us if anything?

Thank you.

$\endgroup$

1 Answer 1

1
$\begingroup$

I don't know the answer to your question in all generality. However, if one continues with $M=\mathbb{Z}$ and $C=C_K$ then one can ask what happens as we vary $i$.

  • for $i=-3$ we identify $H^{-3}(C, \mathbb{Z})$ with elements of $C_K$ of norm $1$ modulo ``principal elements'' (those of the form $I_GC_K$ where $I_G$ is the augmentation ideal). This isomorphism is used in section 2 of Peter Roquette's article in Cassels and Frohlich,
  • for $i=-1$ the groups are trivial,
  • for $i=0$ we identify $H^2(G, C_k)$ with $\mathbb{Z}/n\mathbb{Z}$,
  • for $i=1$ the groups are trivial,
  • the $i=2$ case is described in the exercises on page 157 of

Artin, Emil; Tate, John, Class field theory, Providence, RI: AMS Chelsea Publishing (ISBN 978-0-8218-4426-7/hbk). vii, 192 p. (2009). ZBL1179.11040.

For some applications of these isomorphisms where $M$ is not $\mathbb{Z}$, you can look at the following paper by Kottwitz (in particular see section 4): https://arxiv.org/abs/1401.5728

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .