Let $K$ be a number field and let $G$ be the group of automorphisms of $K$ over $\mathbf Q$. The group $G$ acts in a natural way on the ideal class group of $K$. I would like to know if there are any results giving a formula for the number of orbits of this action (or equivalently a formula for the number of ideal classes that are fixed by some element of $G$). In particular, I would like to compare the number of orbits to the class number of $K$.
[Math] Ideal classes fixed by the Galois group
algebraic-number-theoryclass-field-theorynt.number-theory
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In principle, this follows from Borel and Serre's compactification of arithmetic orbifolds. Let $K$ be a field with $r$ real places and $s$ complex places, and $H_{r,s}=(\mathbb{H}^2)^r\times(\mathbb{H}^3)^s$. Then $SL_2(K)\leq PSL_2(\mathbb{R})^r\times PSL_2(\mathbb{C})^s$ by taking the product of the various Galois embeddings, and acts on $H_{r,s}$. Then via this embedding, $SL_2(\mathcal{O}K)$ acts discretely on $H_{r,s}$, with finite covolume. There are finitely many cusps of this orbifold $H_{r,s}/SL_2(\mathcal{O}_K)$, corresponding to the orbits of $PSL_2(\mathcal{O}_K)$ acting on $\mathbb{P}^1(K)$, which Borel and Serre provide a compactification for. When $K=\mathbb{Q}$, this compactifies $\mathbb{H}^2/PSL_2(\mathbb{Z})$ by a circle, and for $K=\mathbb{Q}(\sqrt{-D}), D\in \mathbb{N}$, $\mathbb{H}^3/PSL_2(\mathcal{O}_K)$ is compactified by Euclidean 2-orbifolds. In the real quadratic case, the compactification is by solv 3-orbifolds.
One may also deduce this from the fact that $H_{r,s}/SL_2(\mathcal{O}_K)$ is finite volume and from the Margulis lemma, which describes the structure of the cusps. I'm not sure who originally proved this, but Borel gave explicit formulae for the volume (although these formulae involve the class number).
This answer is not meant to indicate that this is how one should prove that the class group is finite, but to show how it fits into a certain mathematical context.
Dear Tim,
As you're probably aware, this is part of the 'anabelian' etcetera.
It suffices to recover all intertia subgroups $I_v\subset H$, because their union will then be a normal subgroup $N$ such that $H/N$ is the Galois group of the maximal extension of $K$ unramified everywhere. We can get the ideal class group then by (topological) abelianization. The fact that we can get all the decomposition groups $D_v\subset G$ is Neukirch's theorem (together with Artin-Schreier at infinity). This says the maximal subgroups isomorphic to a local Galois group are exactly the decomposition groups. If you want to make this purely group-theoretic for the finite places, you invoke the theorem of Jannsen-Wingberg that lays out a presentation for all local Galois groups and consider maximal elements in the lattice of subgroups isomorphic to such an explicit presentation. Once you have the $D_v$, there is a standard group-theoretic recipe for $I_v$, which escapes me for the moment. But I'll get back to you with it, if you don't figure it out in the meanwhile.
Added:
OK, so here is the easy part. Now let $F$ be a finite extension of $\mathbb{Q}_p$ and $D=Gal(\bar{F}/F)$. We know that $D^{ab}$ fits into an exact sequence $$0\rightarrow U_F\rightarrow D^{ab}\rightarrow \hat{\mathbb{Z}}\rightarrow 0,$$ so we recover $p$ as the unique prime such that the topological $\mathbb{Z}_p$-rank of $D^{ab}$ is $r_D\geq 2$. The order $q_D$ of the residue field is 1 greater than the order of the prime-to-$p$ torsion subgroup of $D^{ab}$. Also, we know $r_D=1+[F:\mathbb{Q}_p]$. Now we apply the same reasoning to the subgroups of finite index in $D$ to figure out those corresponding to unramified extensions. That is, consider the subgroups $E$ of finite index such that $q^{r_D-1}_E=q_D^{r_E-1}$. Then the inertia subgroup of $D$ is the intersection of all of these.
Best Answer
I assume you want $K$ to be Galois over $\mathbb{Q}$. More generally, let $L/K$ be a Galois extension of number fields. The the class group $C_K$ of $K$ maps to $C_L^{G_{L/K}}$, the part of $C_L$ fixed by the Galois group of $L/K$, and you seem to be asking what the quotient $C_L^{G_{L/K}}/C_K$ looks like.
Taking cohomology of the exact sequences $$ 1\to R_L^*\to L^*\to L^*/R_L*\to1 \quad\text{and}\quad 1\to L^*/R_L* \to I_L \to C_L \to 1 $$ gives (if I'm not mistaken) exact sequences $$ 0 \to H^1(G_{L/K},L^*/R_L*) \to H^2(G_{L/K},R_L^*) \to \text{Br}(L/K) $$ and $$ 0 \to C_K \to C_L^{G_{L/K}} \to H^1(G_{L/K},L^*/R_L*), $$ so the quotient that you're interested in naturally injects $$ C_L^{G_{L/K}}/C_K \hookrightarrow \text{Ker}\Bigl(H^2(G_{L/K},R_L^*) \to \text{Br}(L/K)\Bigr). $$ The Galois structure of unit groups has been much studied. You might look at some of Ted Chinburg's papers (http://www.math.upenn.edu/~ted/CVPubs9-10-07.html)