Importance of the compactness of idele group

Question

Si $k$ is a number field and $J_k$ is the group of $V_k^*$ of invertible elements of the adele ring $V_k$ with the induced topology given by the morphism $V_k\to V_k\times V_K,\ x\mapsto(x,x^{-1})$.

In addition, $k^*$ is a topological subgroup of $J_k$.

Now, let $d:J_k\to \mathbb R, \ \alpha=\{\alpha_v\}\mapsto \displaystyle d(\alpha)=\prod_{all v}\vert \alpha_v\vert_v$

We denote $Ker d=J_k^1$.

By the product formula $k^*\subseteq J_k^1$.

A relevant result is the compactness of the topological group $J_k^1/k^*$ in the quotient topology.

I don't know of this topic, but in some books say's that the compacteness is importante on Class Field Theory, for example in Cassels-Frohlich A.N.T.

I know a little of C.F.T but not enough for see the importance of the compactness.

They could give me some important results on C.F.T. involving the compactness of this group.

Thanks for all.

Answer 1

In a compact topological group, every open subgroup is necessarily of finite index, because the cosets form an open cover of the space.

One of the major results of global class field theory is that there is a bijection between finite abelian extensions $K$ of $k$, and open subgroups $W$ of $J_k$ satisfying $k^{\ast} \subseteq W \subseteq J_k$. This bijection satisfies $[J_k : W] = [K : k]$, and in particular, every open subgroup of $J_k$ is of finite index if it contains $k^{\ast}$. To give an explicit description of $W$, it is the subgroup generated by $k^{\ast}$ and $N_{K/k}(J_K)$, where $N_{K/k}$ is the idele norm.

Question: without any class field theory, how can we deduce that any open subgroup $W$ of $J_k$ containing $k^{\ast}$ is of finite index in $J_k$ in the first place?

Without too much work, we can see that this follows from the compactness of $J_k^1/k^{\ast}$. Such a $W$ corresponds to an open subgroup of $J_k/k^{\ast}$. Now $J_k/k^{\ast}$ is not compact, but is isomorphic (as topological groups) to the cartesian product $$J_k^1/k^{\ast} \times (0, \infty)$$ where $J_k^1/k^{\ast}$ is compact. You can see that an open subgroup of the above product must be of the form $$H \times (0, \infty)$$ where $H$ is an open subgroup of $J_k^1/k^{\ast}$. Since $H$ is necessarily of finite index in $J_k^1/k^{\ast}$, $W$ must be of finite index in $J_k$.

Answer 2

This compactness implies finiteness of generalized class groups, and implies the generalized Dirichlet units theorem. (It can also be proven from those results, but a much simpler proof of the compactness is more direct, using Haar measure.)

This compactness also has a large impact on the harmonic analysis involved in the Iwasawa-Tate approach to zeta and L-functions for $GL(1)$, by restricting the types of Hecke characters that can appear.

In fact, I would tend to say that this compactness is "prior" to classfield theory, in that global classfield theory (formulated in a modern way) would not make much sense without this fact.