# Lecture 13 

:::{.remark}
For $\sigma\in \Field(K, \CC)$, we define an equivalence relation by $\sigma \sim (z\mapsto \bar z)\circ \sigma$, and get equivalence classes of size exactly 1 or 2.
The point of this: such embeddings induce the same norm.
The equivalence classes are the **infinite places** and we sometimes write $v\divides \infty$.
We saw that
\[
K_\infty \da \bigoplus _{v\divides \infty} K\complete v \cong K\tensor_\QQ \RR
.\]
:::

:::{.example title="?"}
An example of the above isomorphism, in the simplest example for which we know nearly nothing about the GLC for $\GL_2(K)$, namely $K = \QQ(2^{1\over 3})$.
For elliptic curves over CM or totally real fields, one can hope to prove potential modularity and there are statement for $\GL_n$ over such fields, but here $K$ is neither CM nor totally real.
Given a random elliptic curve $E/K$ here, the chances of proving it is modular are virtually zero!

In this case, note that $x^3-2$ is irreducible over $\QQ$ by Eisenstein, or by passing to the local extensions $\QQ_2(2^{1\over 3} )$ has degree at least three since the valuation increases by a factor of 3.
Write $x^3-2 = (x-\alpha)(x-w)(x-\bar w)$, we can then compute
\[
K_\infty 
&= {\QQ[x] \over \gens{x^3-2}} \tensor_\QQ \RR \\
&\cong {\RR[x] \over \gens{x^3-2}} \\
&\cong {\RR[x] \over \gens{x-\alpha}} \oplus {\RR[x] \over x^2+\alpha x + \alpha^2} \\
&\cong \RR \oplus \CC \\
&= K_{v} \oplus K_{[v']}
,\]
where we've used that the factors are irreducible and coprime over $\RR$.
Note that there are two isomorphisms to $\CC$ here, and
\[
v: K\to \RR \\
\sqrt{2} &\mapsto \alpha \\ \\
v': K\to \RR \\
\sqrt{2} &\mapsto w \text{ or } \bar w
.\]
:::

:::{.definition title="Finite Adeles"}
Let $K\in \NF$, so that $\Places(K) \cong \mspec \OO_K$ consists of finitely many places.
For $p$ a finite place, we can complete to obtain $K\complete p, \OO_{k, \hat p}$, generalizing $\QQpadic$ and $\ZZpadic$.
Define the **finite adeles of $K$** as
\[
\AA_{K, \Fin} 
&\da\resprod_{p\in \Places(K)} K\complete\mfp \\
&\da \ts{(x_p) \in \prod_{p\in \Places(K)} K\complete p \st x_p \in \OO_{K, \hat p} \text{ for all but finitely many }p} \\
&\subseteq \prod_{p\in \Places(K)} K\complete \mfp
.\]

Equivalently,
\[
\AA_{K, \Fin} \da \ts{(x_p) \in \prod_p K\complete p \st \exist S \text{ finite where } x_p\in \OO_{K, \hat p} \,\forall p\not\in S }
,\]
where one thinks of $S$ as a set of bad primes depending on $(x_p)$.
This forms a ring under pointwise operations, but is topologized in the following way: use that $\OO_{K, \complete p}$ is compact and thus the subring $R\da \prod_p \OO_{K, \complete p}$ is compact by Tychonoff, and declare $R$ to be an open neighborhood of zero.
This yields a basis of opens about zero, which can now be translated.
:::

:::{.remark}
Note that the naive product is not locally compact, so does not admit a good theory of Haar measures
The motivation for the restricted product: an element of $K$ has only finitely many primes involved in the denominator, so only maps into finitely many factors.
There is also a diagonal embedding
\[
\Delta: K &\to \AA_{K, f} \\
\lambda = {a\over b} &\mapsto \cdots, S\da \ts{p\in \Places(K) \st p\divides b} = \ts{p \in \Places(K) \st v_p( \lambda) < 0}
,\]
which is always a finite set.
:::

:::{.definition title="Adeles"}
The full ring of **adeles of $K$** is defined as
\[
\AA_K \da \AA_{K,\Fin} \times K_{\infty} = \resprod_{v\in\Places(K)} K\complete v
.\]
:::

:::{.remark}
Note that $K$ is a 1-dimensional ring and contains $\OO_K$, and in analogy there is a ring of meromorphic functions $\CC(t) \contains \CC[t]$ and one could take a Laurent expansion about any point $v$ to get an element of $\prod_{v\in \CC} \CC\fls{t-v}$.
This contains a subring $\prod_{v\in \CC} \CC\fps{t-v}$ of holomorphic functions, so
\[
\CC(t) \embeds \prod_{v\in \CC} \CC\fls{t-v} \contains \prod_{v\in \CC} \CC\fps{t-v}
.\]

Idea: $\CC(t)$ embeds into the restricted product, since a meromorphic function has only finitely many poles, so this product encodes informations about meromorphic functions in a way such that modifying the function at one place won't change its behavior at another.
:::

:::{.remark}
More generally, the GLC would apply to global fields such as $K = \FF_q(t)$, and has essentially been proved by the Lafforgues using moduli spaces of shtukas.
This are sort of like elliptic curves... but also not.
:::

:::{.exercise title="?"}
Note that $K_\infty \cong\tensor_\QQ \RR \cong K\tensor_\QQ \QQ_\infty$, and show
\[
\AA_{K, \Fin} &\cong K\tensor_\QQ \AA_{\QQ, \Fin} \\
\AA_K &\cong K\tensor_\QQ \AA_\QQ
.\]
:::

:::{.lemma title="?"}
\[
\AA_{\QQ, \Fin} \cong \QQ \oplus \prod_{p} \ZZpadic
,\]
i.e. for all $x\da (x_p) \in \AA_{\QQ,\Fin}$ (so $x_p \in \QQpadic$ for all primes and $x_p\in \ZZpadic$ for all but finitely many $p$) there exists a $\lambda \in \QQ$ such that $x = \lambda + \mu$ for some $\mu \in \prod_p \ZZpadic$.
:::

:::{.remark}
The analogy: for a fixed meromorphic function, one can find a rational function to subtract off to remove all of the poles.
So an adele can be modified by a globally meromorphic function to produce an entire function, which is a Riemann-Roch type of statement (associated to local functions to some global function with the same poles).
:::

:::{.proof title="of lemma"}
By induction on $\size S$, where for the base case when $S = \emptyset$ one can take $\mu = x$.
Let $S$ be finite such that $x_p\in \ZZpadic$ for all $p\not\in S$.
For a general $x$, choose $p\in S$ with $x_p \in \QQpadic$ and take a \(p\dash \)adic expansion as a Laurent series with a finite tail:
\[
x_p = \lambda + f \da \qty{a_{-n}p^{-n} + \cdots + a_{-1}p\inv} + \qty{a_0 + \cdots}, \qquad a_i\in \ts{0,\cdots, p-1}
,\]
so that $\lambda$ is the finite tail and satisfies $\lambda = N/p^{-n}$ for some $N$.
Thus has a single pole at $p$, or more rigorously $v_r( \lambda) \geq 0$ for all $r\neq p$.
Now write $x-\lambda = (y_p)$, where the bad set for $(y_p)$ is $S\sm \ts{p}$, so we're done by induction.
:::

:::{.exercise title="Challenging"}
Show that
\[
\AA_K \cong K + \prod_{p} \OO_{K, \hat p}
.\]
The above proof doesn't quite work if $p$ is a prime that is not principal, so one needs input from class groups.
:::

## Ideles

:::{.remark}
We're interested in $\AA_K\units$, and more generally $\GL_n(\AA_K)$.
:::

:::{.definition title="Ideles"}
One can check that
\[
\AA_{K, \Fin}\units = \resprod_{p} K\complete{p}\units \da \ts{(x_p) \in \prod_p K_\complete{p}\units \st x_p\in \OO_{K, \hat p}\units \text{ for almost all } p}
,\]
and $\AA_K\units = \resprod_v K\complete{v}\units$.
The topology on $\AA_{K, \Fin}\units$ is such that $\prod_p \OO_{K, \hat p}\units$ is open; this is not the subspace topology from $\AA_{K, \Fin}$.
:::

:::{.warnings}
$\AA_K\units \embeds \AA_K$ but the topology is not the subspace topology.
:::

:::{.example title="?"}
An element in $\AA_K\sm\AA_K\units$: take $x_p = p$ and set $x = (x_p) \in \AA_{\QQ, \Fin}$ and in fact $x\in \prod_p \ZZpadic$, however $1/x\in \prod_p \QQpadic$ but $1/x\not\in \AA_{\QQ, \Fin}$ since there are problems at infinitely many primes.
:::

## Global Class Field Theory

:::{.remark}
Recall that $G_K/G_K^c = G_K^\ab$ is the maximal Hausdorff abelian quotient of $G_K$, where $G_K^c$ is the topological closure of the commutator subgroup.
This yields a factorization:

\begin{tikzcd}
	{\bar K} \\
	\\
	&& {K^\ab} \\
	\\
	K
	\arrow["{G_K}", from=5-1, to=1-1]
	\arrow["{G_K^\ab}"', from=5-1, to=3-3]
	\arrow[from=3-3, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGJhciBLIl0sWzIsMiwiS15cXGFiIl0sWzAsNCwiSyJdLFsyLDAsIkdfSyJdLFsyLDEsIkdfS15cXGFiIiwyXSxbMSwwXV0=)

After choosing $\bar K \contains K$, one can form the infinite extension
\[
K^\ab = \Union_S L, \qquad S \da \ts{L \st K \leq L \leq \bar K, L/K \text{ finite }, G(L/K) \in \Ab\Grp}
.\]

:::

:::{.remark}
Note that for $K = \QQ$, Kronecker-Weber yields
\[
\QQ^\ab = \Union_{N\geq 1} \QQ(\zeta_N)
.\]
Class field theory tells you what Galois group for $K^\ab/K$ is, but not necessarily what $K^\ab$ is itself.
It exists for $\QQ$ by Kronecker-Weber, and for imaginary quadratic fields for reasons involving $j\dash$invariants of elliptic curves, but for real quadratic fields this is completely open.
:::

:::{.question title="An open problem"}
What is $K^\ab$ for $K = \QQ(2^{1\over 3})$?
:::


:::{.theorem title="GCFT"}
There is a surjection $r_K$, the **global Artin map**, which is a continuous group morphism:

\begin{tikzcd}
	{\dcosetl{K\units}{\AA_K\units}} && {G(K^\ab/K)} && 0
	\arrow["{r_K}", two heads, from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGRjb3NldGx7S1xcdW5pdHN9e1xcQUFfS1xcdW5pdHN9Il0sWzIsMCwiRyhLXlxcYWIvSykiXSxbNCwwLCIwIl0sWzAsMSwicl9LIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzEsMl1d)

:::

:::{.remark}
The Artin map can't be an isomorphism since the RHS is profinite, but the LHS contains $K_\infty\units \cong (\RR\units)^{r_1} \times (\CC\units)^{r_2}$, so any connected component must be in the kernel of $r_K$ since the RHS is totally disconnected and continuous images of connected sets are connected.
This happens to be the entire kernel when $K$ has no units, $K=\QQ$, and imaginary quadratic number fields.
Since the identity is closed in the image, the kernel must be closed in the source by continuity.
:::

:::{.theorem title="from CFT"}
If $C_K$ is defined as the image of $(K_\infty\units)^0)$ in $\dcosetl{K\units}{\AA_K\units}$, then 
\[
\ker r_K = \cl_\top(C_K)
,\]
the topological closure.
:::

:::{.remark}
It turns out that for $K=\QQ$ or an imaginary quadratic field, $C_K$ is already closed, but there are examples where it is not.
Thus $G(K^\ab/K)\cong \dcoset{K\units}{\AA_K\units}{\cl_\top(C_K)}$.
:::


:::{.remark}
Some properties of $r_K$:

\begin{tikzcd}
	{\tv{1,1,\cdots, 1, x_p, 1, \cdots }} & {\dcosetl{K\units}{\AA_K\units}} && {G(K^\ab/K)} \\
	\\
	{x_p} & {K\complete{p}\units} && {G_{K\complete p}^\ab}
	\arrow[from=3-4, to=1-4]
	\arrow["{r_K \text{ local Artin}}", from=3-2, to=3-4]
	\arrow["{r_K \text{ global Artin}}", from=1-2, to=1-4]
	\arrow[maps to, from=3-1, to=1-1]
	\arrow[from=3-2, to=1-2]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwwLCJcXGRjb3NldGx7S1xcdW5pdHN9e1xcQUFfS1xcdW5pdHN9Il0sWzMsMCwiRyhLXlxcYWIvSykiXSxbMywyLCJHX3tLXFxjb21wbGV0ZSBwfV5cXGFiIl0sWzEsMiwiS1xcY29tcGxldGV7cH1cXHVuaXRzIl0sWzAsMCwiXFx0dnsxLDEsXFxjZG90cywgMSwgeF9wLCAxLCBcXGNkb3RzIH0iXSxbMCwyLCJ4X3AiXSxbMiwxXSxbMywyLCJyX0sgXFx0ZXh0eyBsb2NhbCBBcnRpbn0iXSxbMCwxLCJyX0sgXFx0ZXh0eyBnbG9iYWwgQXJ0aW59Il0sWzUsNCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dLFszLDBdXQ==)

So the global Artin maps glues all of the local Artin maps.

If $L/K$ is finite, $G_L \injects G_K$ and there is an induced map $G(L^\ab/L)\to G(K^\ab/K)$ which need not be injective.
There is a commuting square involving norms and transfers:

\begin{tikzcd}
	{\dcosetl{L\units}{\AA_K\units}} && {G(L^\ab/L)} \\
	\\
	{\dcosetl{K\units}{\AA_K\units}} && {G(K^\ab/K)}
	\arrow[from=1-3, to=3-3]
	\arrow["{r_K}", from=3-1, to=3-3]
	\arrow["{r_L}", from=1-1, to=1-3]
	\arrow["{\Norm_{L/K}}", from=1-1, to=3-1]
	\arrow["{\text{Transfer}}"', color={rgb,255:red,92;green,92;blue,214}, curve={height=18pt}, dashed, from=3-3, to=1-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-24pt}, dashed, hook, from=3-1, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

:::