# How does one understand geometric CFT in terms of modularity?

I have recently asked a question in a similar vein: What makes Geometric CFT easier than CFT?

but I'm afraid I wasn't quite ripe to ask it yet. I have since consulted with the following sources:

As well as some sources refreshing my memory on classic CFT:

My motivation for studying geometric class field theory was first and foremost to solidify my understanding of classic class field theory. I thought that perhaps the geometric intuition will shed some light on the rather massive apparatus of CFT.

In order to be concrete, I will specify a version of geometric CFT:

Theorem A: Let $C$ be a smooth projective, geometrically irreducible curve over a finite field $k$, and let $K$ be its function field. Then the (Artin reciprocity) map $\Phi_K:Div(C)\rightarrow \pi_1^{ab}(C)$ given by $p\mapsto Frob_p$ factors through $Pic(C)$, and induces an isomorphism between the profinite completion of $Pic(C)$ and $\pi_1^{ab}(C)$.

This appears in Toth's master thesis (linked above) as Theorem 1.1.4. In order to understand this as analogous to classic CFT, one need only notice that $K^{\times}\backslash \mathbb{I}_K/ \prod _{p \mbox{ closed point in } C}\hat{O_p}$ (where $\mathbb{I}_K$ is the ideles, and $\hat{O_p}$ is the completion of the stalk at the closed point $p$) is isomorphic to $Pic(C)$.

In other words the theorem above can be viewed as analogous to the adelic point of view of CFT (i.e. the isomorphism between the profinite completions of $K^{\times}\backslash \mathbb{I}_K/\prod_v O_v^{\times}$ with $Gal(K^{ab}/K)$, where $K$ is a number field). I am interested in understanding what the analogous picture in Geometric CFT to modular formulations (rather than adelic ones) of classic CFT.

### Question

How does one understand geometric CFT (as described in the theorem above) in terms of modularity results? In other words, what is the analogous statement to the fact that (up to finitely many primes) the splitting of primes in an abelian extension of $\mathbb{Q}$ is determined by what those primes are conjugate to modulo some conductor? Is there a geometric intuition behind the analogous statement in geometric CFT?

### EDIT

After reading the comments carefully, and going back to my old notes on Class Field Theory to refresh my mind about a few things that I got wrong in the comments, I have come to the conclusion that the following is the question that I really wanted to ask:

We have:

Theorem B: or $K$ a number field, we have that $Gal(K^{ab}/K)$ is isomorphic to the profinite completion of $K^{\times}\backslash \mathbb{I}_K/D_K$ where $D_K$ is the connected component of $1$.

Theorem A above is not analogous to Theorem B because, as Felipe pointed out, Theorem A is only about abelian extensions that are unramified, whereas Theorem B allows ramification everywhere. My question is: what is the analogous statement to Theorem A in the Number Theory case?

I am in particular confused by the subtlety regarding the distinction between $D_K$ and the product $\prod _{p \mbox{ closed point in } C}\hat{O_p}$. Are they the same?

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$Gal(K^{ab,un}/K) = \pi_1^{ab}(C)$ because $C$ is projective. – Felipe Voloch Jun 30 '12 at 0:55
The fundamental group only captures unramified covers, by definition. $\pi_1({\mathbb{P}}^1)$ is trivial. – Felipe Voloch Jun 30 '12 at 2:07
The product $\prod_v O_v^{\times}$ does not include the archimedian places in the number field case. In the function field case, if you exclude a finite set of places you get a bigger group. Read Serre "Groupes algebriques et corps de classes" – Felipe Voloch Jun 30 '12 at 13:14
@Makhalan: It is not true that $K^\times \backslash I_K / \prod_\nu O_\nu^\times$ is isomorphic (after taking profinite completions) to $Gal(K^{ab}/K)$, or even $Gal(K^{ab,un}/K)$. Something that is true: $I_K/K^\times$ is a topological group. Call its connected component $D_K$. Then there is an isomorphism $K^\times \backslash I_K/ D_K \cong Gal(K^{ab}/K)$. For example, the group $\mathbb{Q}^\times \backslash I_\mathbb{Q} / \prod_p \mathbb{Z}_p^\times$ is connected, and dividing by itself we can recover the classical fact that the rational field has no unramified extensions. – Dror Speiser Jun 30 '12 at 20:04
It is not that $D_K$ is the product of completions. $D_K$ is some product over the finitely many infinite places. For example, for the rationals, there is an isomorphism $\mathbb{Q}^\times \backslash I_\mathbb{Q} \cong \mathbb{R}_+^\times \times \prod_p \mathbb{Z}_p^\times$, and the connected component of 1 is $\mathbb{R}_+^\times \times 1$. So $Gal(\mathbb{Q}^{ab}/\mathbb{Q})\cong \prod_p \mathbb{Z}_p^\times$. To get the unramified abelian extensions of $\mathbb{Q}$, we further divide by $\prod_p \mathbb{Z}_p^\times$, to get $\mathbb{Q}^{ab,un}=\mathbb{Q}$. – Dror Speiser Jul 2 '12 at 0:55