# Are the ideles literally a Picard group?

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 $$\mathcal M(\mathbb Z)$$.

• This might not be relevant, but the idele class group surjects onto the class group $\mathrm{Pic}(\mathrm{Spec}(O_K))$. Nov 28, 2018 at 19:43
• Not answering as I don't really know this stuff. In "The Idèle Class Group" by Lenstra, on page 4, there is this exact sequence:$$1\to\mu\to\prod_{\mathfrak p}\{x\in K_{\mathfrak p}\mid|x|_{\mathfrak p}=1\}\to C_K\to\operatorname{Pic}_K\to1,$$where $\mu$ is the group of roots of 1, $C_K$ is the idèle class group, and $\operatorname{Pic}_K$ is the group of isomorphism classes of invertible $\mathcal O_K$-modules equipped with a metric (see the text). Seems it is this additional metric that signifies passage from ideals to idèles. Nov 28, 2018 at 20:22
• "I understand that in the number field / function field analogy, the ideles IK of a number field K are supposed to be analogous to the Picard group of a function field." This analogy seems a bit wonky to me. The ideles of a number field are analogous to the ideles of a function field (of a curve over a finite field). The number field analogue of the Picard group is the Arakelov Picard group. Anyway, I sort of have an answer to your question and will write it up when I get a chance (hopefully tomorrow). Nov 29, 2018 at 9:06
• @TomPrice I'm sure I've got a number of fundamental confusions going on here. I'd really appreciate anything you can say to set me straight, thanks. Nov 29, 2018 at 22:32

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.

• Thanks! The real lesson for me is to read Neukirch. Dec 4, 2018 at 3:41
• It's a great book so you can't go wrong there. That being said, the later paragraphs, on idelic line bundles and the alternative description of Arakelov line bundles, are my own ideas and not in Neukirch as far as I know. Dec 4, 2018 at 4:24

It seems it would be helpful to discuss idealic/adelic view on line/vector bundles, which is quite simple and intuitive. At the end we might arrive at a kind of asnwer to the question.

Step 1: Bundles are described by transition functions from one chart of the covering to another - in case of line bundles we have invertible elements living on intersection , in vector bundles case - $$GL(O_{intersection})$$. (Looking forward idels are exactly these transition functions for specific covering). Do not forget about factorization $$GL(O_{U1})$$ \ $$GL(O_{intersection})$$ / $$GL(O_{U2})$$.

Step 2: Consider 1d-scheme - i.e. curve (over ANY(!) field ) or Spec(Z). Consider quite specific covering - we take all geometric points $$x$$ of the curve and "general point" (i.e. Spec(FractionField)). And consider "infinitesemal neighbourhood" of every point $$U_x$$ - it is $$Spec( K_x [[t]] )$$ . Fact 1. Infinitesemal neighbourhoods of different points do not intersect. Fact 2. Intersection of $$U_x$$ and "generic point" is $$Spec( K_x ((t)) )$$.

Comibining step 1 and 2 we arrive to idelic/adelic picture of bundles:

Line bundles = K(x) \ $$\prod_x K_x ((t))^*$$ / $$\prod_x K_x [[t]]^*$$

You can recognize idels at the middle: $$\prod_x K_x ((t))^*$$

Vector bundles = GL( FractionField) \ $$\prod_x GL( K_x ((t)) )$$ / $$\prod_x GL( K_x [[t]] )$$

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I am not sure other steps are necessary, but let me discuss.

If I undestand you question correctly you want to exclude factorization over / $$\prod_x K_x [[t]]^*$$ from the factor above - just to have idels. That can be done.

Claim: K(x) \ $$\prod_x K_x ((t))^*$$ describes line bundles over geometiric object - each point $$x$$ is glued to all points in its infinitesemal neighbourhood.

Step 3: Gluing points A,B means we consider functions: f(A)=f(B). Now take $$A = B+ \epsilon$$ for $$\epsilon^n=0$$. If we consider f(A)=f(B) we glue A to its n-neighbourhood. The function field will be $$f^{(k)}(A)=0$$ for all k <= n, that means series of the form $$C+ c_{n+1}t^{n+1}+c_{n+2}t^{n+2}$$. Taking limit n to infinity we kill all the functions. However the fraction field does not change - i.e. intersection with generic point is still the same. In that way idels are preserved but the factor is $$\prod_x K_x [[t]]^*$$ is killed.

PS

Curves with $$f^{(k)}(A)=0$$ considered in Serre book "Algebraic Groups and Class Fields", despite being singular, one can consider their Jacobians and it is related to ramified covering in the same way as usual Jacobion related to unramified coverings.