I understand that in the number field / function field analogy, the ideles $\mathbb I_K$ of a number field $K$ are supposed to be analogous to the Picard group of a function field.
Question: Is this more than an analogy? Is there an actual geometric setting in which the ideles parameterize "line bundles" over a geometric object attached to $K$?
My first inclination was to see if this was the case in Berkovich geometry, in the case $K = \mathbb Q$. It's encouraging to note that ideles correspond to Cech 1-cocycles for reasonable coverings of the Berkovich space $\mathcal M(\mathbb Z)$ with values in $\mathcal O_\mathbb Z^\times$. But nevertheless, every such cocycle is a coboundary, so it seems to not literally be the case that the ideles parameterize line bundles on $\mathcal M(\mathbb Z)$.
Best Answer
As explained in the comments, I disagree with this analogy. Nonetheless, there is a way you can realize the idele class group (not the ideles) as a group of line-bundle-like objects under the tensor product.
First, some context.
Suppose we have a number field $K$ with ring of integers $O$.
We have a surjective map from the idele class group of $K$ to the Arakelov class group, and another surjective map from the Arakelov class group to the ordinary class group.
The ordinary class group is isomorphic to the group of line bundles under the tensor product. The Arakelov class group is isomorphic to the group of Arakelov line bundles under the tensor product (these are described in Neukirch’s Algebraic Number Theory but he calls them "invertible metrized $O$-modules"). An Arakelov line bundle is basically a locally principal $O$-module $M$ with an inner product (aka metric) on $M \otimes_\mathbb{Z} \mathbb{C} $, and forgetting this inner product corresponds to the surjective map from the Arakelov to the ordinary class group.
To associate a sort of line bundle to idele class group elements, we need to add a bit more structure to our Arakelov line bundles. Then, like before, the surjective map from the idele class group to the Arakelov class group corresponds to forgetting this extra structure.
In order to describe it, I first need to give a different perspective on what an Arakelov line bundle is.
Let $A$ be a one-dimensional $K$ vector space. For every place $\nu$ of $K$, choose a norm on the completion $A_\nu = A \otimes_K K_\nu$ which agrees with the corresponding valuation at $\nu$, so that any nonzero element of $A$ has norm $1$ at cofinitely many places, and the norm on $A_\nu$ has the same image as the valuation on $K_\nu$. The resulting structure, of $A$ plus a choice of norm for each place, is equivalent to an Arakelov line bundle (the elements of $A$ with norm at most 1 at every finite place are the $O$-module, and the norms at the infinite places tell you the metric).
We can specify a norm on a completion $A_\nu$ by choosing a nonzero element of $A_\nu$ and requiring it to have norm $1$. This uniquely determines the norm on all other elements of $A_\nu$. So to get an Arakelov bundle, it’s sort of like we picked an element of $A_\nu$ for every place $\nu$, and then forgot everything except for the induced norms.
To get the extra structure I was referring to, we simply refrain from forgetting this extra data. Let’s define an “idelic line bundle” to be a one-dimensional vector space $A$ over $K$, plus a choice of nonzero element $a_\nu$ of $A_\nu$ for each place $\nu$, so that for any nonzero element $a$ of $A$, $|a / a_\nu|_\nu = 1$ at cofinitely many places $\nu$ (note that dividing vectors is ok here since the vector spaces are one-dimensional).
There is an obvious notion of tensor product and of isomorphism of these line bundles. It is straightforward to show that the isomorphism classes form a group under the tensor product, and it is isomorphic to the idele class group.
I’m pretty sure we can also get a sort of “Picard group” isomorphic to a ray class group in a similar way but I will leave that as an exercise.
Also, this construction works just as well for global fields of positive characteristic.