\input{"preamble.tex"}

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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    MSRI Summer School: Automorphic Forms and the Langlands Program
  }
    \\ {\normalsize MSRI Summer School 2017, Lectures by Kevin
Buzzard} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes based on
Kevin's video lectures, which are not quite a transcription or an
attempt to capture all of the information, but rather to jot down some
reference material for myself to refer back to later! } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2022-03-14 }
\end{flushleft}


\newpage

% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage

\hypertarget{lecture-1-overview-and-what-questions-were-interested-in}{%
\section{Lecture 1: Overview and what questions we're interested
in}\label{lecture-1-overview-and-what-questions-were-interested-in}}

\begin{remark}

References:

\begin{itemize}
\tightlist
\item
  Kevin's website: \url{https://www.ma.imperial.ac.uk/~buzzard/MSRI/}
\item
  Youtube Playlist:
  \url{https://www.youtube.com/watch?v=Rv59aRUMfio\&list=PLhsb6tmzSpiysoRR0bZozub-MM0k3mdFR}
\end{itemize}

\end{remark}

\begin{remark}

This course: adapted from handwritten notes from a course by Richard
Taylor in 1992 at Caltech. The original course focused on
\(\operatorname{GL}_2\), we'll discuss \(\operatorname{GL}_n\). About a
year before Wiles-Taylor!

Ways to learn:

\begin{itemize}
\tightlist
\item
  Engage with the material
\item
  Type it up.\footnote{DZG: Like me!} Better than hours of videos!
\item
  Complete the exercises and fill in the gaps. See also the problem set
  on the \href{https://www.ma.imperial.ac.uk/~buzzard/MSRI/}{course
  website}.
\item
  Talk to the experts at the talks
\item
  Complete the project on the website on the abelian \(p{\hbox{-}}\)adic
  Langlands correspondence. This mimics the classical correspondence,
  and is very much in its infancy at the moment.
\end{itemize}

In all cases, it's useful to work with other people, communicate,
interact, etc.

\end{remark}

\begin{remark}

The story begins with Dirichlet characters -- let
\(N\in {\mathbb{Z}}_{\geq 1}\) and consider a character
\begin{align*}
\chi: C_n^{\times}\to {\mathbb{C}}&& \in {\mathsf{Grp}}
.\end{align*}
Attached to \(\chi\) is a Galois representation
\begin{align*}
\rho_\chi: G \coloneqq{ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu/{\mathbb{Q}}) \to \operatorname{GL}_1({\mathbb{C}})
.\end{align*}
Note that \(G\) is an infinite group. This arises as the following
composition:

\begin{center}
\begin{tikzcd}
    G &&&& {\operatorname{GL}_1({\mathbb{C}})} \\
    \\
    {{ \mathsf{Gal}} ({\mathbb{Q}}(\zeta_N)/{\mathbb{Q}})} &&&& {C_N^{\times}}
    \arrow["{=}", from=3-1, to=3-5]
    \arrow["\chi"', from=3-5, to=1-5]
    \arrow[two heads, from=1-1, to=3-1]
    \arrow["{\rho_\chi}", from=1-1, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJHIl0sWzQsMCwiXFxHTF8xKFxcQ0MpIl0sWzQsMiwiQ19uXFx1bml0cyJdLFswLDIsIlxcR2FsKFxcUVEoXFx6ZXRhX24pL1xcUVEpIl0sWzMsMiwiPSJdLFsyLDEsIlxcY2hpIiwyXSxbMCwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMCwxLCJcXHJob19cXGNoaSJdXQ==}{Link
to Diagram}
\end{quote}

Here the equality denotes a canonical isomorphism, and is given by the
map \(n\mapsto (\zeta_N \mapsto \zeta_N^n)\).

\end{remark}

\begin{remark}

There is a version of this for \(\operatorname{GL}_2\). Let \(f\) be a
cuspidal modular form which is an eigenform for the Hecke operators
\(T_p\), which are endomorphisms of the space of modular forms. So
\(T_p f = \lambda f\) for some \(\lambda \in {\mathbb{C}}\), and it
turns out that the subfield
\(\left\langle{\lambda_p}\right\rangle {\mathbb{C}}\) (which could be
infinitely generated) is a number field -- it is equipped with an
embedding into \({\mathbb{C}}\) and has finite degree over
\({\mathbb{Q}}\).

\end{remark}

\begin{theorem}[Deligne, 60s/70s]

Let \(\ell \in {\mathbb{Z}}\) is prime and \(\lambda \divides \ell\) is
a prime of \(E_f\), i.e.~a nonzero prime ideal of
\({\mathcal{O}}_{E_f}\). Following an observation due to Serre, Deligne
constructs a map
\begin{align*}
\rho_f: G \to \operatorname{GL}_2({ \operatorname{cl}} _\mathsf{Alg}(E_f { {}_{ \widehat{\lambda} } } ) )
,\end{align*}
where the RHS is completing \(E_f\) at \(\lambda\) and taking an
algebraic closure. This is an \(\ell{\hbox{-}}\)adic representation, and
formally resembles the \({\mathbb{C}}{\hbox{-}}\)representation above in
the sense that \(\rho_f\) is ``attached'' to \(f\). Deligne's
construction uses étale cohomology.

\end{theorem}

\begin{remark}

For \(f\) a modular form, there is a level \(N\geq 1\), a weight
\(k\geq 1\), and a Dirichlet character \(\chi\). It turns out that
\(\rho_f\) is unramified away from \(N\ell\) -- note that there is no
analog of \(\ell\) for the \(\operatorname{GL}_1\) case. If \(p\) is a
prime not dividing \(N\ell\), then \(\rho_f(\operatorname{Frob}_p)\) has
characteristic polynomial
\begin{align*}
x^2 + \lambda_p x + p^{k-1}\chi(p)
.\end{align*}
Since \(p\notdivides N\ell\), \(p\notdivides N\) and thus
\(\chi(p) \neq 0\). Note that this is the trace of the representation
\(\rho\), and it turns out that the conjugacy classes
\(\operatorname{Frob}_p\) are dense in \(G\). By \textbf{Chebotarev
density}, there is at most one semisimple \(\rho_f\) with this property,
and Deligne's theorem is that there is \emph{at least} one. These days,
\(N{\hbox{-}}\)dimensional \(\ell{\hbox{-}}\)adic Galois representations
are common precisely because we now know they are the right things to
look at. Historically, number theorists may not have considered modular
forms number-theoretic objects -- instead, they were considered objects
of harmonic analysis, and number theory likely focused on things like
class numbers and Iwasawa's main conjecture.

\end{remark}

\begin{remark}

A word on Deligne's construction -- how does he find a 2-dimensional
\(\ell{\hbox{-}}\)adic representation of \(G\)? He constructs \(\rho_f\)
using étale cohomology with nontrivial coefficients. A modern take is
that it would come from a motive, and étale cohomology produces motives,
although one would normally take trivial coefficients. There is a slight
issue in constructing \(\ell{\hbox{-}}\)adic sheaves on the modular
curve. Deligne later uses trivial coefficients and takes cohomology on
some power of a universal elliptic curve. Thus Deligne's proof is a
partial proof that the \(\rho_f\) are motivic. This all assumes
\(k\geq 2\), and the \(k=1\) case was handled by Deligne-Serre in the
70s.

\begin{quote}
\#todo What is \(k\)
\end{quote}

\end{remark}

\begin{question}

Some questions arising from Deligne's construction:

\begin{itemize}
\item
  What do the representations locally look like at ramified primes
  \(p\), i.e.~\(p\divides N\ell\)? There is a formal meaning here: the
  global Galois group contains the local one, and \(G\) is embedded
  canonically and only ambiguous up to conjugacy. One can restrict the
  global representation to this local representation In the unramified
  case, we know the characteristic polynomial, and this essentially
  determines the local behavior, although there is a semisimplicity
  issue involving the Tate conjecture -- specifying the characteristic
  polynomial of a matrix doesn't uniquely determine it, due to possible
  multiplicity in eigenvalues.

  \begin{itemize}
  \item
    Case 1: if \(p\divides N\) and \(p\neq \ell\), the answer is given
    by the conjectured \textbf{local Langlands correspondence}. These
    are theorems for \(\operatorname{GL}_2({ {\mathbb{Q}}_p })\) and
    \(\operatorname{GL}_n(K)\) for any
    \(K\in\mathsf{Local}\mathsf{Field}\). This correspondence aims to
    relate (possibly infinite dimensional) representations of groups
    like \(\operatorname{GL}_n({ {\mathbb{Q}}_p })\) to Weil-Deligne
    representations, which are similar to Galois representations. We'll
    soon explain what this has to do with modular forms.
  \item
    Case 2: if \(p=\ell\), so we have a \(p{\hbox{-}}\)adic
    representation of \(G\), there should be a
    \textbf{\(p{\hbox{-}}\)adic local Langlands correspondence}. This is
    essentially the boundary of what we currently know, and in a sense
    we don't even know what the right question should be. This is a
    theorem for \(\operatorname{GL}_2\), but unknown for
    \(\operatorname{GL}_3\) and above. Ask Rebecca about this!
  \end{itemize}
\end{itemize}

\end{question}

\begin{remark}

An easier variant: instead of asking for \(\rho_f\), ask for
\(\mkern 1.5mu\overline{\mkern-1.5mu\rho\mkern-1.5mu}\mkern 1.5mu_f: { \mathsf{Gal}} ({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }/{\mathbb{Q}})\to \operatorname{GL}_2(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_\ell)\),
i.e.~reduce \(\operatorname{mod}\ell\). Identify
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_\ell\)
with the residue field of
\(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu_\lambda\)
at \(\lambda\). So reduce Deligne's \(\ell{\hbox{-}}\)adic
representation to get a \(\operatorname{mod}\ell\) representation. These
reps are easier to understand since there are tricks, e.g.~looking for
\(\mkern 1.5mu\overline{\mkern-1.5mu\rho\mkern-1.5mu}\mkern 1.5mu_f\) in
\(A[\ell]\) for some \(A\in {\mathsf{Ab}}{\mathsf{Var}}\). Note that
\(\rho_f\) involves étale cohomology.

\end{remark}

\begin{question}

Questions Taylor asked at the time:

Are \(\rho_\chi, \rho_f\) special cases of a general story? Note that
this relates Dirichlet characters to modular forms.

Answer: yes, kind of. There is the following theorem from 2013, which
was reproved by Scholze using perfectoid magic.

\end{question}

\begin{theorem}[?]

Let \(E\) be a totally real (so all embeddings
\(E\xhookrightarrow{\sigma} {\mathbb{C}}\) have
\(\sigma(E) \subseteq {\mathbb{R}}\)) or CM number field (totally
imaginary extension of a totally real number field) and let \(\pi\) be a
\emph{cuspidal automorphic representation} of
\(\operatorname{GL}_n({\mathbb{A}}_E)\). Suppose that \(\pi\) is
``cohomological'', which is a condition on the weights and the PDEs that
the automorphic forms come from and is a strong algebraicity condition.
Then there is a representation
\(\rho_\pi: { \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu/E)\to \operatorname{GL}_n({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }_\ell)\)
attached to \(\pi\) in a canonical way, which is the analog of
\(\operatorname{ch}\mathrm{poly}\rho_f(\operatorname{Frob}_p)\)

\end{theorem}

\begin{remark}

Automorphic forms will be solutions to elliptic PDEs Unrelated, but see
\href{https://www.galoisrepresentations.com/about/}{Frank Calegari's
blog}.

\end{remark}

\begin{remark}

Meta-theorem: Galois representations come from étale cohomology groups
and their \(p{\hbox{-}}\)adic deformations, say of an algebraic variety
defined over a number field e.g.~a Shimura variety. General idea: given
an algebraic or analytic object like \(\chi, f, \pi\) (or more generally
\emph{motivic} objects), some technical machinery produces
representations of Galois groups. Can we classify the image of this
correspondence? Which Galois representations come from such things? I.e.
given \(\rho: G\to \operatorname{GL}_n(k)\) for \emph{some}
\(k\in \mathsf{Field}\), is \(\rho\) isomorphic to some \(\rho'\) coming
from an algebraic variety?

Dimension 1: let \(K/{\mathbb{Q}}\) be finite Galois and
\(\rho: { \mathsf{Gal}} (K/{\mathbb{Q}})\to \operatorname{GL}_1({\mathbb{C}})\).
Is \(\rho\cong \rho_\chi\) for \(\chi\) a Dirichlet character? We can
take an epi-mono factor any group morphism and Galois theory works
better with epis, so replace \(K\) with a subfield \(L \leq K\) to make
\(\rho\) \emph{injective}. This is because
\(\rho: { \mathsf{Gal}} (K/{\mathbb{Q}}) \twoheadrightarrow{ \mathsf{Gal}} (L/{\mathbb{Q}}) \hookrightarrow\operatorname{GL}_1({\mathbb{C}})\).
So we assume
\begin{align*}
\rho: { \mathsf{Gal}} (K/{\mathbb{Q}}) \hookrightarrow{\mathbb{C}}^{\times}
,\end{align*}
hence
\({ \mathsf{Gal}} (K/{\mathbb{Q}})\in {\mathsf{Ab}}{\mathsf{Grp}}\). A
reminder of what a \(\rho_\chi\) will look like:

\begin{center}
\begin{tikzcd}
    {C_N ^{\times}} && {\mathbb{C}}^{\times}\\
    \\
    {{ \mathsf{Gal}} ({\mathbb{Q}}(\zeta_N)/{\mathbb{Q}})}
    \arrow[Rightarrow, no head, from=3-1, to=1-1]
    \arrow["\chi", from=1-1, to=1-3]
    \arrow["{\rho_\chi}"', from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJDX04gXFx1bml0cyJdLFsyLDAsIlxcQ0NcXHVuaXRzIl0sWzAsMiwiXFxHYWwoXFxRUShcXHpldGFfTikvXFxRUSkiXSxbMiwwLCIiLDAseyJsZXZlbCI6Miwic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFswLDEsIlxcY2hpIl0sWzIsMSwiXFxyaG9fXFxjaGkiLDJdXQ==}{Link
to Diagram}
\end{quote}

By the same trick, we can factor \(\rho_\chi\) to assume it is injective
from some \(L \leq {\mathbb{Q}}(\zeta_N)\) a subfield of a cyclotomic
field. So the question reduces to the following:

\end{remark}

\begin{question}

If \(K\in \mathsf{Number}\mathsf{Field}\) is Galois over
\({\mathbb{Q}}\) with
\({ \mathsf{Gal}} (K/{\mathbb{Q}})\in {\mathsf{Ab}}{\mathsf{Grp}}\),
does there exist an \(N\geq 1\) with
\(K \hookrightarrow{\mathbb{Q}}(\zeta_N)\).

\end{question}

\begin{remark}

Answer: yes, but this takes some work and has a name, the
Kronecker-Weber theorem (an explicit special case of global CFT). Note
that the converse is clear: subfields of cyclotomic fields will yield
Galois groups which are subgroups of a cyclic group and hence abelian.

So for all
\(\rho: { \mathsf{Gal}} ({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }/{\mathbb{Q}}) \to \operatorname{GL}_1({\mathbb{C}})\)
(so the image is finite order), there is a
\(\chi: C_N^{\times}\to {\mathbb{C}}^{\times}\) with
\(\rho\cong \rho_\chi\). Thus is the proposed correspondence, the image
is everything, and the proof is class field theory.

\end{remark}

\begin{remark}

What about \(\operatorname{GL}_2\)? Let \(f\) be a cuspidal modular
eigenform as before, there is a Galois representation
\(\rho_f: { \mathsf{Gal}} ({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }/{\mathbb{Q}}) \to \operatorname{GL}_2({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }_\ell)\)
such that

\begin{itemize}
\tightlist
\item
  \(\rho_f\) is absolutely irreducible
\item
  \(\rho_f\) is ``odd'', i.e.~writing
  \({\overline{{\in }}}{ \mathsf{Gal}} ({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }/{\mathbb{Q}})\)
  for complex conjugation,
  \(\operatorname{det}\rho_f({\overline{{)}}} = -1\).
\item
  \(\rho_f\)is unramified away from a finite set of primes and carries
  some \(p{\hbox{-}}\)adic Hodge theory and is \emph{potentially
  semistable}.
\end{itemize}

Étale cohomology always has the third property, which is sometimes
called ``being geometric'', and the conjecture is that geometric
representations come from geometry.

\end{remark}

\begin{conjecture}

Fontaine-Mazur asked in the 1990s if all \(\rho\) satisfying these
properties are of the form \(\rho_f\) for some \(f\), which became the
Fontaine-Mazur conjecture. This is basically known now, see Emerton and
Kisin.

\end{conjecture}

\begin{remark}

Similarly, we can ask that if
\(\rho: { \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu/{\mathbf{E}}) \to \operatorname{GL}_n({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }_\ell)\)
satisfies some assumptions, is it the case that \(\rho \cong \rho_\pi\)
for some \(\pi\). State of the art: BLGGT, proves this in many cases
using very technical \(p{\hbox{-}}\)adic Hodge theory. See also the
10-author paper.

\end{remark}

\hypertarget{lecture-2-part-1-classical-local-langlands}{%
\section{Lecture 2: Part 1, Classical Local
Langlands}\label{lecture-2-part-1-classical-local-langlands}}

References: Serre's \emph{Local Fields}

\begin{quote}
\url{https://www.youtube.com/watch?v=1mhSFioQInU\&list=PLhsb6tmzSpiysoRR0bZozub-MM0k3mdFR\&index=2}
\end{quote}

\begin{remark}

Vaguely stated local Langlands for \(\operatorname{GL}_n\) over
\(K/{ {\mathbb{Q}}_p }\) a finite extension: there is a bijective
correspondence
\begin{align*}
\left\{{\substack{
  \text{Certain $\infty{\hbox{-}}$dimensional irreducible} \\
  \text{reps of }\operatorname{GL}_n \text{ over } {\mathbb{C}}
}}\right\}
&\rightleftharpoons
\left\{{\substack{
  \text{Certain $n{\hbox{-}}$dimensional reps of a group} \\
  A \text{ related to }{ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu /K)
}}\right\}
.\end{align*}
LHS: complicated reps of a simple group, vs RHS: relatively simple reps
of a complicated group. For \(n=1\), this recovers local CFT. This is an
entirely local statement, and is in fact a theorem now but the proof is
global. For \(n\geq 2\), local Langlands for \(\operatorname{GL}_n\)
over \(K\) is a theorem due to Harris-Taylor, and the proofs are again
global (i.e.~working with number fields). Oddly there isn't quite a
``local'' statement of the above bijection. This statement really should
be categorified, which has happened on the function field side via the
geometric Langlands correspondence.

\end{remark}

\begin{remark}

Working toward infinite Galois groups, recall the finite case: let
\(L/K\) be a finite extension, then \(L\) is Galois iff normal and
separable. Recall that separable only needs to be checked in
characteristic \(p\), and normal means being a splitting field. Then
\({ \mathsf{Gal}} (L/K) \in {\mathsf{Fin}}{\mathsf{Grp}}\) (which is a
theorem) are the auts of \(L\) fixing \(K\) pointwise, and there is a
correspondence
\begin{align*}
{ \mathsf{Gal}} (L/{-}):\, 
\left\{{\substack{
  \text{Subfields } K \leq M \leq L
}}\right\}
&\rightleftharpoons
\left\{{\substack{
  \text{Subgroups } 0 \leq { \mathsf{Gal}} (L/M) \leq { \mathsf{Gal}} (L/K)
}}\right\} \\
M &\mapsto \left\{{\sigma\in { \mathsf{Gal}} (L/M) {~\mathrel{\Big\vert}~}{ \left.{{\sigma}} \right|_{{M}} } = \operatorname{id}_M}\right\}
.\end{align*}

\end{remark}

\begin{definition}[Galois Extensions]

For the infinite case, let \(K\in \mathsf{Field}\) and let \(L/K\) be
algebraic, possibly with \([L:K] = \infty\). We say \(L\) is again
Galois iff normal and separable.

\end{definition}

\begin{warnings}

Note that the cardinalities need not match up here: \(L\) can be a
countably infinite extension with an uncountable Galois group.

\end{warnings}

\begin{remark}

For a fixed element \(\lambda \in L\) and
\(\sigma\in { \mathsf{Gal}} (L/K)\), one doesn't need to actually work
in the infinite group: there exists a finite \(M\) with
\(K\leq M\leq L\) and \(M\) finite Galois over \(L\) with
\(\lambda \in M\) -- e.g.~one can take \(M\) to be the splitting field
of \(\min_{\lambda, L}(x)\in K[x]\). Moreover there is a canonical map
\(\mathop{\mathrm{Res}}_{L/M}: { \mathsf{Gal}} (L/K)\to { \mathsf{Gal}} (M/K)\)
and \(\sigma(\lambda)\) is determined by \(\tilde\sigma(\lambda)\) for
\(\tilde \sigma = \mathop{\mathrm{Res}}_{L/M}(\sigma)\). Note that if
\(L/K\) is separable then \(M/K\) is separable -- normality may not
descend this way, but will if \(M\) is a splitting field. This argument
shows that there is an injection
\begin{align*}
{ \mathsf{Gal}} (L/K) \hookrightarrow\prod_{\substack{ K \subseteq M \subseteq L \\ M/K \text{ finite Galois }}} { \mathsf{Gal}} (M/K) \implies
{ \mathsf{Gal}} (L/K) \cong \cocolim_{\substack{K \subseteq M \subseteq L \\ M/K \text{ finite Galois}}} M \in {\mathsf{Grp}}
,\end{align*}
and it turns out that \({ \mathsf{Gal}} (L/K)\) is a closed subspace of
\(\prod { \mathsf{Gal}} (M/K)\) in the subspace topology.

\end{remark}

\begin{warnings}

Each \({ \mathsf{Gal}} (M/K)\) has the discrete topology, but
\({ \mathsf{Gal}} (L/K)\) need not be discrete in the product topology!
The basic opens for the product topology on \(\prod X_i\) are of the
form \(\prod U_i\) with only finitely many \(U_i\neq X_i\).

\end{warnings}

\begin{proposition}[FTGT]

If \(L/K\) is Galois, i.e.~algebraic normal and separable, with
\({ \mathsf{Gal}} (L/K)\) given the projective limit topology, there is
a bijection
\begin{align*}
{ \mathsf{Gal}} (L/{-}):\,
\left\{{\substack{
  \text{Subfields } K \leq M \leq L
}}\right\}
&\rightleftharpoons
\left\{{\substack{
  \text{Closed subgroups of }{ \mathsf{Gal}} (L/K)
}}\right\}
\end{align*}
One might convince themselves that ``restricting to the identity on
\(M\)'' is a closed condition. Note that everything is discrete in the
finite case, so all subgroups were closed!

\end{proposition}

\begin{example}[?]

Take \(K={\mathbb{Q}}\) and \(L = \displaystyle\bigcup_{n\geq 1} L_n\)
where \(L_n\coloneqq{\mathbb{Q}}(\zeta_{p^n}) \subseteq {\mathbb{C}}\).
There is a canonical isomorphism
\({ \mathsf{Gal}} (L_n/{\mathbb{Q}}) = C_{p^n}^{\times}\), so
\(G_L \hookrightarrow\prod_{n\geq 1} C_{p^n}^{\times}\). Note that this
is not a surjection in general: note that the \(L_i\) are filtered, and
we can restrict \({ \mathsf{Gal}} (L/K)\to { \mathsf{Gal}} (L_n/K)\) by
just restricting automorphisms. So writing
\(\phi_n \coloneqq{ \left.{{\phi}} \right|_{{L_n}} }\) for any
\(\phi \in { \mathsf{Gal}} (L/K)\), knowing \(\phi_n\) implies knowing
\(\phi_{\leq n}\) (by restriction). Under the canonical identification,
\(\phi_n \in C_{p^n}^{\times}\), and for \(m\leq n\) we have
\(\phi_m = \phi_n \operatorname{mod}p^m\). In this case, there is a
homeomorphism
\begin{align*}
{ \mathsf{Gal}} (L/{\mathbb{Q}}) = \cocolim C_{p^n}^{\times}\coloneqq{ {\mathbb{Z}}_p }^{\times}\hookrightarrow{ {\mathbb{Z}}_p }
,\end{align*}
noting that
\({ {\mathbb{Z}}_p }^{\times}\hookrightarrow{ {\mathbb{Z}}_p }\) is a
topological group in a topological ring, and the subspace topology works
here since inversion is continuous.

\end{example}

\begin{example}[?]

Let \(K\) be finite, say \({\sharp}K = q\), and
\(L = \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu\).
Noting that
\(L_n \coloneqq{\mathbb{F}}_{q^n} \subseteq L_m \coloneqq{\mathbb{F}}_{q^m} \iff n\divides m\),
we can form the filtered colimit
\begin{align*}
L = \colim_n L_n = \displaystyle\bigcup_{n\geq 1} L_n
.\end{align*}
Recall that
\({ \mathsf{Gal}} (L_n/{\mathbb{F}}_q) = C_n = \left\langle{\operatorname{Frob}_q}\right\rangle\)
where \(\operatorname{Frob}_q(x) \coloneqq x^q\) has order \(n\). So
\({ \mathsf{Gal}} (L/K) \hookrightarrow\prod C_n\); which subset is it?
If \(g_n\) is in the image, then \(g_n \operatorname{mod}m = g_m\) for
all \(m\divides n\), so
\begin{align*}
{ \mathsf{Gal}} (L/K) = \cocolim_n C_n = \widehat{{\mathbb{Z}}}
,\end{align*}
the profinite integers.

\end{example}

\begin{warnings}

\begin{align*}
{\mathbb{F}}_q \subseteq {\mathbb{F}}_{q^2} \not\subseteq {\mathbb{F}}_{q^3}
\end{align*}
for dimension reasons, so one can't form the usual directed system here.

\end{warnings}

\hypertarget{local-fields}{%
\subsection{Local Fields}\label{local-fields}}

\begin{remark}

For us, the examples of local fields will be \(K/{ {\mathbb{Q}}_p }\)
finite extensions. To be happy with \({ {\mathbb{Q}}_p }\), see
exercises in Cassels' book on elliptic curves. Choose an algebraic
closure
\(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K\),
which is unique but not up to unique isomorphism. Slight issue with
isomorphisms here: picking two algebraic closures
\(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu'\)
and two random isomorphisms
\(i: \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu \to \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu'\)
and
\(j: \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu'\to \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\),
it may not be that \(ij = \operatorname{id}\). On
\({ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\),
this induces an inner automorphism, so we should avoid trying to work
with explicit elements of this group (which in some sense are not
well-defined).

\end{remark}

\begin{remark}

Recall that there is a (normalized) valuation
\(v: { {\mathbb{Q}}_p }\to {\mathbb{Z}}\cup\left\{{\infty}\right\}\)
where \(v(p^n u) = n\) for \(u\in { {\mathbb{Z}}_p }^{\times}\).
Similarly \(K \supseteq{\mathcal{O}}_K \supseteq{\mathfrak{p}}_K\) its
unique maximal/prime ideal, which is principal and generated by a
uniformizer \({\mathfrak{p}}_K = \left\langle{\pi_K}\right\rangle\).
Thus for every \(k\in K\) we can write \(k = \pi_K^n u\), and we have a
valuation
\begin{align*}
v: K^{\times}&\to {\mathbb{Z}}\\
\pi_K &\mapsto 1 \\
\pi_K^n u &\mapsto n 
.\end{align*}

\end{remark}

\begin{definition}[Unramified extensions]

Let \(L/K\) be algebraic, possibly infinite. The valuation \(v_K\)
extends to \(L^{\times}\to {\mathbb{Q}}\), and
\(L \supseteq{\mathcal{O}}_L = \left\{{0}\right\} \cup\left\{{\lambda \in L {~\mathrel{\Big\vert}~}v(\lambda) \geq 0}\right\}\).
Note \({\mathcal{O}}_L \supseteq{\mathfrak{p}}_L\) its unique maximal
ideal, which need not be principal here. Write
\(\kappa_L = {\mathcal{O}}_L/{\mathfrak{p}}_L\), which is an algebraic
extension of \(\kappa_K \coloneqq{\mathcal{O}}_K/{\mathfrak{p}}_K\),
which here is a finite field. If \(L/K\) is Galois, there is a
surjective map
\begin{align*}
{ \mathsf{Gal}} (L/K)\twoheadrightarrow{ \mathsf{Gal}} (\kappa_L/\kappa_K)
.\end{align*}
This need not be injective in general, so we say \(L/K\) is
\textbf{unramified} iff this is a bijection.

\end{definition}

\begin{proposition}[?]

Let \(K/{ {\mathbb{Q}}_p }\) be a finite extension and
\({\mathfrak{p}}_K \in \operatorname{Spec}{\mathcal{O}}_K\) with
\({\mathfrak{p}}_K = \left\langle{\pi_K}\right\rangle\) generated by a
uniformizer. TFAE:

\begin{itemize}
\tightlist
\item
  \({\mathfrak{p}}_L = \pi_K {\mathcal{O}}_L\), so the same element
  generates the maximal ideal of \(L\),
\item
  \(v_K: L^{\times}\to {\mathbb{Q}}\) has image \({\mathbb{Z}}\).
\item
  \(L/K\) is unramified.
\end{itemize}

\end{proposition}

\begin{remark}

The compositum of two unramified extensions of \(K\) is again
unramified, and if \(L/K\) there is a unique unramified subextension
\(M\) with \(L \supseteq M \supseteq K\). Moreover
\({ \mathsf{Gal}} (M/K) = { \mathsf{Gal}} (\kappa_M/\kappa_L)\), since
all unramified extensions are Galois, and this will be a procyclic
group.

\end{remark}

\hypertarget{lecture-3}{%
\section{Lecture 3}\label{lecture-3}}

\textless!--\textgreater{}
\href{https://www.youtube.com/watch?v=_4di81CzUt0\&list=PLhsb6tmzSpiysoRR0bZozub-MM0k3mdFR\&index=3}{Video}--\textgreater{}

\begin{remark}

Plans: group theoretic properties of
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\)
for \(K\) a \(p{\hbox{-}}\)adic field, tamely ramified extensions, and
the theorems of local CFT.

\end{remark}

\begin{remark}

Setup: \(K/{ {\mathbb{Q}}_p }\) finite with \(L/K\) algebraic, normal,
and separable (automatic if we assume \(\operatorname{ch}k = 0\)), so
Galois. The goal is to understand
\(G_K\coloneqq{ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\),
we'll first show it surjects onto an easier group and investigate the
kernel:

\end{remark}

\begin{definition}[Inertia subgroup]

The \textbf{inertia subgroup} is defined as the kernel of the reduction
map to residue fields:

\begin{center}
\begin{tikzcd}
    0 & \textcolor{rgb,255:red,214;green,92;blue,92}{I_{L/K} \coloneqq\ker \kappa} \\
    \\
    & {G(L/K)} \\
    \\
    0 & {G(\kappa_L, \kappa_K) \cong { \mathsf{Gal}} (L/K)/I_{L/K}}
    \arrow["\kappa", from=3-2, to=5-2]
    \arrow[from=1-2, to=3-2]
    \arrow[from=1-1, to=1-2]
    \arrow[from=5-2, to=5-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMSwyLCJcXEdhbChML0spIl0sWzEsNCwiXFxHYWwoXFxrYXBwYV9MLCBcXGthcHBhX0spIFxcY29uZyBcXEdhbChML0spL0lfe0wvS30iXSxbMSwwLCJJX3tML0t9IFxcZGEgXFxrZXIgXFxrYXBwYSIsWzAsNjAsNjAsMV1dLFswLDAsIjAiXSxbMCw0LCIwIl0sWzAsMSwiXFxrYXBwYSJdLFsyLDBdLFszLDJdLFsxLDRdXQ==}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{remark}

Note that \(\kappa_L\) is finite, so \(G(\kappa_L/\kappa_L)\) is an
easier group to understand, and in fact is procyclic (topologically
monogenic). It's clear that \(I_{L/K} \leq G_{L/K}\) is a closed
subgroup, and thus corresponds to some \(M\) with
\(K \subseteq M \subseteq L\) where \(G(L/M) = I_{L/K}\) and
\({ \mathsf{Gal}} (M/K) \cong { \mathsf{Gal}} (\kappa_L/\kappa_K)\). In
fact, \(M\) will be the union of all subfields of \(L\) containing \(K\)
which are unramified over \(K\).

\end{remark}

\begin{remark}

A special interesting case is when
\(L = \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu\),
then \(M = K^{\scriptscriptstyle\mathrm{un}}\) will be the maximal
unramified extension:

\begin{center}
\begin{tikzcd}
    {\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu} \\
    & {} \\
    {K^{\scriptscriptstyle\mathrm{un}}} \\
    & {} \\
    K
    \arrow[from=5-1, to=3-1]
    \arrow[from=3-1, to=1-1]
    \arrow["{I_{\mkern 1.5mu\overline{\mkern-1.5muL\mkern-1.5mu}\mkern 1.5mu/K}}"', shift right=5, curve={height=30pt}, dashed, no head, from=3-1, to=1-1]
    \arrow["{\widehat{{\mathbb{Z}}}}"', shift right=5, curve={height=30pt}, dashed, no head, from=5-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJcXGJhcntLfSJdLFswLDIsIkteXFx1bnJhbSJdLFswLDQsIksiXSxbMSwxXSxbMSwzXSxbMiwxXSxbMSwwXSxbMSwwLCJJX3tcXGJhciBML0t9IiwyLHsib2Zmc2V0Ijo1LCJjdXJ2ZSI6NSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn0sImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwxLCJcXGhhdHtcXFpafSIsMix7Im9mZnNldCI6NSwiY3VydmUiOjUsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9LCJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link
to Diagram}
\end{quote}

Note that
\begin{align*}
G(K^{\scriptscriptstyle\mathrm{un}}/K){ \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }{ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5mu\kappa\mkern-1.5mu}\mkern 1.5mu_K/\kappa_K) { \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }\widehat{{\mathbb{Z}}}
.\end{align*}

\end{remark}

\begin{question}

Two stories in CFT: what are the groups, and what are the corresponding
fields?

\end{question}

\begin{remark}

These are two genuinely different stories, e.g.~for
\(K\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\), for
\(K^{\operatorname{ab}}\) the maximal abelian extension it's precisely
known what the Galois group is, but what is the actual extension (the
Hilbert class field)? If \(K = { {\mathbb{Q}}_p }\), then
\begin{align*}
K^{\scriptscriptstyle\mathrm{un}}= \displaystyle\bigcup_{m\geq 1, p\notdivides m} { {\mathbb{Q}}_p }(\zeta_m)
,\end{align*}
being careful because \({ {\mathbb{Q}}_p }(\zeta_p)\) is ramified. This
might also be true for other fields \(K\).

\end{remark}

\begin{remark}

Setup: let \(L/K\) be Galois, but now assume
\(I_{L/K}\in{\mathsf{Fin}}{\mathsf{Grp}}\), e.g.~when \(L/K\) is a
finite extension. We know \(I_{L/K} {~\trianglelefteq~}G(L/K)\) is
normal since it is a kernel, and if \(L/K\) is finite then the quotient
is finite cyclic. Put a filtration on the inertia group and we'll look
at the filtered pieces: note that if \(\sigma\in I_{L/K}\), then
\(\sigma: L\to L\) descends to a local morphism
\(\sigma: {\mathcal{O}}_L\to{\mathcal{O}}_L\) preserving
\({\mathfrak{p}}_L\).

\begin{question}

How much does \(\sigma\) disrupt these local pieces?

\end{question}

The claim is that since \(I_{L/K}\) finite implies
\({\mathfrak{p}}_L = \left\langle{\pi_L}\right\rangle\) is principal,
where the idea is that the uniformizer in the base extends all the way
to \(L\):

\begin{center}
\begin{tikzcd}
    L && {{\mathfrak{p}}_L} \\
    \\
    M && {{\mathfrak{p}}_M = \left\langle{\pi_K}\right\rangle \in \operatorname{Spec}{\mathcal{O}}_M} \\
    \\
    K && {{\mathfrak{p}}_K = \left\langle{\pi_K}\right\rangle \in \operatorname{Spec}{\mathcal{O}}_K}
    \arrow[from=5-1, to=3-1]
    \arrow[from=3-1, to=1-1]
    \arrow["{\text{unram}}", shift left=4, curve={height=-30pt}, dashed, no head, from=5-1, to=3-1]
    \arrow["{I_{L/K}, \text{finite}}", shift left=5, curve={height=-30pt}, dashed, no head, from=3-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Moreover the discrete valuation \(v_L: L\twoheadrightarrow{\mathbb{Z}}\)
satisfies \(v_L = {\sharp}I_{L/K} v_K\) on \(K^{\times}\). So write
\({\mathfrak{p}}_L = \left\langle{\pi_L}\right\rangle\) for some
uniformizer. Define a filtration by
\begin{align*}
I_{L/K, i} = \left\{{\sigma\in I_{L/K} {~\mathrel{\Big\vert}~}{ \sigma(\pi_L)\over \pi_L}\in 1 + {\mathfrak{p}}_L^i  }\right\}, \qquad I_{L/K, 0} \coloneqq I_{L/K}
,\end{align*}
where we note that \(\sigma(\pi_L)\) will still be a uniformizer, so the
quotient is a unit, and we are measuring how far it is from 1. All of
the subquotients turn out to be abelian.

\begin{quote}
Setting \(i=1\) should recover the Sylow subgroup.
\end{quote}

One can check that this defines a series of normal subgroups of
\(G(L/K)\):
\begin{align*}
I_{L/K} = I_{L/K, 0} \supseteq I_{L/K, 1} \supseteq I_{L/K, 2}\supseteq\cdots
.\end{align*}
Note that since \(L = M\left\langle{\pi_L}\right\rangle\), so if
\(\sigma\) fixes \(\pi_L\) it fixes all of \(L\) and must be the
identity. Thus if \(\sigma\neq \operatorname{id}\), then \(\sigma\)
moves \(\pi_L\) and \(\sigma(\pi_L)/\pi_L \neq 1\), so for \(i\gg 1\)
one has
\(\sigma(\pi_L)/\pi_L\neq 1\operatorname{mod}{\mathfrak{p}}_L^i\). The
conclusion is that \(I_{L/K, i} = 1\) for \(i\gg 1\), making this a
finite filtration.

We can write down an embedding
\begin{align*}
I_{L/K}/ I_{L/K, 1} &\hookrightarrow\kappa_L^{\times}\\
\sigma &\mapsto \sigma(\pi_L)/\pi_L
.\end{align*}

\begin{quote}
Jackie is now happy about this.
\end{quote}

In particular, the domain is cyclic of order prime to \(p\). So given
the full Galois group, we factored out the cyclic unramified part and
were left with inertia, and now we've factored something simple out of
the remaining inertia part. Bad news: \(I_{L/K, 1}\) is complicated!

\end{remark}

\begin{definition}[Lower Numbering]

For \(i\geq 1\), define an embedding
\begin{align*}
I_{L/K, i} / I_{L/K, i+1} &\hookrightarrow{\mathfrak{p}}_L^i/{\mathfrak{p}}_L^{i+1} \overset{{\mathsf{Ab}}{\mathsf{Grp}}}{{ { \, \xrightarrow{\sim}\, }}} {\mathbb{G}}_a(\kappa_L) \\
\sigma &\mapsto {\sigma(\pi_L) \over \pi_L} - 1
.\end{align*}
Note that the RHS could be an infinite field, but turns out to be a
finite-dimension \({\mathbb{F}}_p{\hbox{-}}\)vector space and thus the
domain is isomorphic as a group to \(C_p^n\) for some \(n\). We call
this the filtration the \textbf{lower numbering}. In particular, its
order is a power of \(p\).

\end{definition}

\begin{remark}

Upshot: \(I_{L/K, 1}\) has \(p{\hbox{-}}\)power order, is a normal
subgroup, and the quotient is cyclic of order prime to \(p\), making it
the unique Sylow \(p{\hbox{-}}\)subgroup of \(I_{L/K}\). In particular,
this is a \(p{\hbox{-}}\)group, hence solvable.

\end{remark}

\begin{definition}[Tame and wild ramification]

We say \(L/K\) is \textbf{tamely ramified} iff \(I_{L/K, 1} = 1\).
Otherwise, we say \(L/K\) is \textbf{wildly ramified}.

\end{definition}

\begin{warnings}

This seems to mean ``ramified but not too badly'', but unramified
extensions are also tamely ramified! However, wildly ramified does imply
ramified.

\end{warnings}

\begin{remark}

We're interested in
\(L = \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu\)
where \(I_{L/K}\) is not finite, so we'll need to glue using a limit.
Note that the lower numbering doesn't behave well with respect to
extensions of \(L\) since extensions \(L'/L\) change the uniformizer.
More precisely, if \(L'/L/K\) with \(L/K\) and \(L'/K\) Galois with
\(I_{L'/K}\) finite, there is a map
\(I_{L'/K}\twoheadrightarrow I_{L/K}\) but \(I_{L'/K, i}\) does not get
identified with \(I_{L/K, i}\) -- the issue is that \(\pi_L\) may have
nothing to do with \(\pi_{L'}\), and one is trying to control things
\(\operatorname{mod}\pi_L^i\) and \(\operatorname{mod}\pi_{L'}^i\).

The fix: introduce a relabeling.

\end{remark}

\begin{definition}[Upper numbering]

Set \(g_i = {\sharp}I_{L/K, i}\) so
\(g_0 \geq g_1 \geq \cdots \geq g_M = 1\) for some \(M\gg 0\). Define a
PL continuous function \(\phi\), a scaling factor. Some facts about it:

\begin{itemize}
\tightlist
\item
  \(\phi\) will be linear on \((i, i+1)\) for all \(i\)
\item
  \(\phi(0) = 0\)
\item
  The slope of \(\phi\) on \((i, i+1)\) will be \({g_{i+1} \over g_0}\),
\item
  \(\phi: [0, \infty)\to [0,\infty)\) is a strictly increasing
  bijection.
\end{itemize}

Note that the slopes are decreasing, and limit to a constant slope
\(g_M/g_0\). For \(v\in {\mathbb{R}}_{\geq 0}\), define
\begin{align*}
I_{L/K, v} \coloneqq I_{L/K, {\left\lceil v \right\rceil}}
.\end{align*}
Then define the renormalized \textbf{upper numbering}
\begin{align*}
I_{L/K}^u \coloneqq I_{L/K, \phi^{-1}(u)}
.\end{align*}

\end{definition}

\begin{remark}

The point: the upper numbering will now lift to infinite extensions.

\end{remark}

\begin{proposition}[?]

If \(L'/L/K\) are all Galois with \(I_{L'/K}\) finite, then
\begin{align*}
I_{L/K}^u = \operatorname{im}I_{L'/K}^u
.\end{align*}

\end{proposition}

\begin{proof}[?]

Omitted, see Serre's \emph{Local Fields}, or Cassels-Froehlich.

\end{proof}

\begin{remark}

If one graphs \(v\mapsto {\sharp}I_{L/K, v}\), there are jump
discontinuities at random \({\mathbb{Z}}{\hbox{-}}\)points. After the
rescaling, the graph of \(u\mapsto {\sharp}I_{L/K, i}^u\) still jumps,
but now at \({\mathbb{Q}}{\hbox{-}}\)points instead of on
\({\mathbb{Z}}\). The denominators of all of the jumps will all divide
\(g_0\).

\end{remark}

\begin{theorem}[Hasse-Arf]

If \(L/K\) is abelian, then the jumps in \(I_{L/K}^u\) are in fact in
\({\mathbb{Z}}\).

\end{theorem}

\begin{remark}

If \(L/K\) is any Galois extension, define \(I_{L/K}^u\) by gluing
\(I_{M/K}^u\) for all \(M/K\) algebraic and \(I_{M/K}\) finite.

\end{remark}

\begin{remark}

It turns out that \(L/K\) is tamely ramified
\(\iff I_{L/K, {\varepsilon}} = 1\) for all
\({\varepsilon}> 0 \iff I_{L/K}^\delta = 1\) for all \(\delta > 0\), and
this last condition makes sense for any (possibly infinite) Galois
extensions.

Note that compositing tame extensions is still tame, so \(L/K\) contains
a maximal tamely ramified extensions. So we've split off some easier
parts of the full extension:

\begin{center}
\begin{tikzcd}
    L \\
    \\
    {K^2} \\
    & {} \\
    {K^1} \\
    & {} \\
    K
    \arrow[from=7-1, to=5-1]
    \arrow[from=5-1, to=3-1]
    \arrow["{\text{maximal unramified, cyclic}}"', shift right=5, curve={height=30pt}, dashed, no head, from=7-1, to=5-1]
    \arrow[from=3-1, to=1-1]
    \arrow["{\text{maximal tamely ramified, cyclic}}", shift left=5, curve={height=-30pt}, dashed, no head, from=7-1, to=3-1]
    \arrow["{\text{wildly ramified, pro-}p}"', shift right=5, curve={height=30pt}, dashed, no head, from=5-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Recall that \(G(K_2/K_1) \cong I_{L/K} / I_{L/K, 1}\) was cyclic of
order \(m\) with \(p\notdivides m\). By Kummer theory, one can generally
write down cyclic extensions of order \(m\) over some field as long as
the field contains \(m\)th roots of unity.

\begin{quote}
See Birch's article in Cassels-Froehlich.
\end{quote}

Since \(p\divides m\),
\(K^{\scriptscriptstyle\mathrm{un}}\supseteq\mu_m\) the \(m\)th roots of
unity in
\(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu\). So if
\(K^2/K^{\scriptscriptstyle\mathrm{un}}\) is Galois with
\(G(K^2/K^{\scriptscriptstyle\mathrm{un}})\cong C_m\), then by Kummer
theory, this extension arises by taking an \(m\)th root, so
\(K^2 \cong {\mathsf{K}}^{\scriptscriptstyle\mathrm{un}}(a^{1\over m})\)
for some \(a\in K^{\scriptscriptstyle\mathrm{un}}\). E.g. since
\({\mathbb{Q}}\) contains square roots of unity \((\pm 1)\), every
quadratic extension \(K/{\mathbb{Q}}\) is obtained as
\(K = {\mathbb{Q}}(\sqrt a)\). Here
\(K^{\scriptscriptstyle\mathrm{un}}\) only has one prime, so
\(K^2 = K^{\scriptscriptstyle\mathrm{un}}(\pi_K^{1\over m})\) since
\(m\)th roots of units give unramified extensions and
\(a = \pi_K^\ell u\) for some unit.

Thus the tamely ramified extension
\(K^2 = K^t\coloneqq\displaystyle\bigcup_{m\geq 1, p\notdivides m} K^{\scriptscriptstyle\mathrm{un}}(\pi_K^{1\over m})\).
Moreover there is a canonical isomorphism
\begin{align*}
G(K^{\scriptscriptstyle\mathrm{un}}(\sqrt[m]{\pi_K})/K^{\scriptscriptstyle\mathrm{un}}) &{ \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }\mu_m \\
\sigma &\mapsto \sigma(\sqrt[m]{\pi_K})/\sqrt[m]{\pi_K}
.\end{align*}
Thus
\begin{align*}
G(K^t/K^{\scriptscriptstyle\mathrm{un}}) = \cocolim \mu_m \cong \cocolim \left\{{ C_m {~\mathrel{\Big\vert}~}{p\notdivides m}}\right\} = \prod_{\ell \neq p} { {\mathbb{Z}}_\ell }
.\end{align*}

\end{remark}

\begin{remark}

Our decomposition of the absolute Galois group is now:

\begin{center}
\begin{tikzcd}
    {\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu} \\
    \\
    {K^t} \\
    \\
    {K^{\scriptscriptstyle\mathrm{un}}} \\
    \\
    K
    \arrow[from=7-1, to=5-1]
    \arrow[from=5-1, to=3-1]
    \arrow[from=3-1, to=1-1]
    \arrow["{=\widehat{{\mathbb{Z}}} = \prod_{\ell} { {\mathbb{Z}}_\ell }}"', curve={height=30pt}, dashed, no head, from=7-1, to=5-1]
    \arrow["{\cong \prod_{\ell\neq p}{ {\mathbb{Z}}_\ell }}"', curve={height=30pt}, dashed, no head, from=5-1, to=3-1]
    \arrow["{\text{pro{\hbox{-}}}p}"', curve={height=30pt}, dashed, no head, from=3-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Note that if the top field is a finite extension, this is just factoring
the full Galois group as

\begin{itemize}
\tightlist
\item
  a cyclic group with a canonical generator,
\item
  a cyclic group with with a non-canonical generator, and
\item
  a finite \(p{\hbox{-}}\)group.
\end{itemize}

The story thus far packages all of this together for all finite
extensions at once by taking direct limits.

\end{remark}

\begin{remark}

Note that \(G(K^t/K)\) is unknown at this point, and will in general be
a semidirect product of the two (pro)cyclic group and its (pro)cyclic
quotient. Take a canonical generator to write
\begin{align*}
\left\langle{\operatorname{Frob}}\right\rangle = G(K^{\scriptscriptstyle\mathrm{un}}/K) = { \mathsf{Gal}} (\kappa(K^{\scriptscriptstyle\mathrm{un}})/\kappa(K))
\end{align*}
which contains the map \(x\mapsto x^q\) where
\(q \coloneqq{\sharp}\kappa(K)\), and \(\operatorname{Frob}\) is defined
by pulling this back to
\({ \mathsf{Gal}} (K^{\scriptscriptstyle\mathrm{un}}/K)\) along the
canonical isomorphisms. We can lift \(\operatorname{Frob}\) to
\(G(K^t/K)\) (which we want to understand), then it acts on the normal
subgroup \({ \mathsf{Gal}} (K^t/K^{\scriptscriptstyle\mathrm{un}})\) by
conjugation,
i.e.~\(\sigma \mapsto \operatorname{Frob}\circ \sigma\circ \operatorname{Frob}^{-1}\).
This is independent of choice of lift, since another lift is of the form
\(\psi \operatorname{Frob}\) where \(\psi\) is in the abelian part and
thus commutes with \(\sigma\). So we need to specify this action to say
what \(G(K^t/K)\) is. We have a canonical isomorphism
\begin{align*}
G(K^t/K^{\scriptscriptstyle\mathrm{un}}) { \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }\cocolim_m \mu_M(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu)
,\end{align*}
so it suffices to give an automorphism of this group. The following
exercise yields the glue:

\begin{exercise}[?]

Note that Frobenius acts on the RHS, and \(x\mapsto x^q\) makes sense
here since \(q\) is a power of \(p\) and \(m\) is prime to \(p\). Check
that the map induced by \(\operatorname{Frob}\) is
\(\zeta \mapsto \zeta^q\).

\end{exercise}

\end{remark}

\begin{remark}

Next: statements of local CFT.

\end{remark}

\hypertarget{lecture-4-statements-of-local-class-field-theory}{%
\section{Lecture 4: Statements of Local Class Field
Theory}\label{lecture-4-statements-of-local-class-field-theory}}

\begin{remark}

Idea: LCFT offers an interpretation of
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)^{\operatorname{ab}}\),
and a vast number of group cohomology computations.

\end{remark}

\begin{remark}

Just the statements today, the proofs require clever tricks like
Lubin-Tate groups.

We've reduced studying
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\) to
looking at the inertia group \(I\). The obstacle: the Sylow
\(p{\hbox{-}}\)subgroup for \(I\) is complicated. We'll try to
understand instead
\({\mathsf{Ab}}(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu, K))\),
its abelianization.

\end{remark}

\begin{definition}[Weil Group]

Let \(K/{ {\mathbb{Q}}_p }\) be finite, so we have a SES
\begin{align*}
1 \to I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K} \to G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K) \to \widehat{{\mathbb{Z}}} = \left\langle{\operatorname{Frob}}\right\rangle \to 1
.\end{align*}
Note that \(\operatorname{Frob}\) isn't quite a well-defined element in
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\) --
any two lifts differ by inertia. Consider
\begin{align*}
\operatorname{Frob}^{\mathbb{Z}}\coloneqq\left\{{\cdots, \operatorname{Frob}^{-2}, \operatorname{Frob}^{-1}, \operatorname{id}, \operatorname{Frob}, \operatorname{Frob}^2, \cdots}\right\} \cong {\mathbb{Z}}\subseteq \widehat{{\mathbb{Z}}}
.\end{align*}
Define the \textbf{Weil group} \(W_K\) by the following diagram:

\begin{center}
\begin{tikzcd}
    1 && {I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}} && {W_K} && {\operatorname{Frob}^{\mathbb{Z}}} && 1 \\
    \\
    1 && {I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}} && {G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)} && {\widehat{{\mathbb{Z}}}} && 1
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-7, to=1-9]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=3-7]
    \arrow[from=3-7, to=3-9]
    \arrow[Rightarrow, no head, from=1-3, to=3-3]
    \arrow[hook, from=1-5, to=3-5]
    \arrow[hook, from=1-7, to=3-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

More precisely,
\begin{align*}
W_K \coloneqq\left\{{g\in G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K) {~\mathrel{\Big\vert}~}\operatorname{im}(g) \in \widehat{{\mathbb{Z}}}= { \mathsf{Gal}} (K^{\scriptscriptstyle\mathrm{un}}/K) \text{ is in } \operatorname{Frob}^{\mathbb{Z}}}\right\}
.\end{align*}

Topologize \(W_K\) in the following way:
\(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K} \subseteq W_K\)
is open, but not an open subgroup of
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\)
since \(0\) is closed in \(\widehat{{\mathbb{Z}}}\) and preimages send
closed sets to closed sets, making
\(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}\)
closed in
\({ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\).
Thus
\(W_K/I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K} = {\mathbb{Z}}^{\operatorname{disc}}\).

Using that the embedding
\({\mathbb{Z}}^{\operatorname{disc}}\to \widehat{{\mathbb{Z}}}\) is
continuous, we can equivalently define \(W_K\) by a pullback in
\({\mathsf{Top}}{\mathsf{Grp}}\):

\begin{center}
\begin{tikzcd}
    {W_K} && {{\mathbb{Z}}^{\operatorname{disc}}} \\
    \\
    {G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)} && {\widehat{{\mathbb{Z}}}}
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJXX0siXSxbMiwwLCJcXFpaXlxcZGlzYyJdLFsyLDIsIlxcaGF0e1xcWlp9Il0sWzAsMiwiXFxHYWwoXFxiYXIgSy9LKSJdLFszLDJdLFsxLDJdLFswLDFdLFswLDNdLFswLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{warnings}

\(W_K\) does \emph{not} use the subspace topology.

\end{warnings}

\begin{remark}

We haven't lost much by passing to \(W_K\), since its image is dense in
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\).
Note that at every finite stage \(L/K\),
\(W_K\twoheadrightarrow G(L/K)\).

\end{remark}

\begin{remark}

For \(G\in {\mathsf{Top}}{\mathsf{Grp}}\), define
\(G^c \in {\mathsf{Top}}{\mathsf{Ab}}{\mathsf{Grp}}\) to be the
(topological) closure
\({ \operatorname{cl}} _G \left\langle{\left\{{ghg^{-1}h^{-1}{~\mathrel{\Big\vert}~}g, h\in G}\right\}}\right\rangle\).
Note that \(G^{\operatorname{ab}}\coloneqq G/G^c\) is the maximal
abelian Hausdorff quotient of \(G\). Why take the topological closure:
if \(\operatorname{id}\) is not closed, then
\({ \operatorname{cl}} _G(\operatorname{id}) {~\trianglelefteq~}G\) is
closed, as are all of its cosets, so quotienting by this makes points
closed and yields a Hausdorff space.

\end{remark}

\begin{theorem}[Main Theorem of LCFT]

If \(K/{ {\mathbb{Q}}_p }\) is finite, then there is a canonical
isomorphism, \textbf{the Artin map}:
\begin{align*}
r_K: K^{\times}{ \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }W_K^{\operatorname{ab}}
.\end{align*}

\end{theorem}

\begin{remark}

Note that \(W_K\) is not a profinite group, and is mapping to a discrete
group \({\mathbb{Z}}\). This is what \(K^{\times}\) looks like: it
contains \({\mathcal{O}}_{K}^{\times}\), which is profinite, and the
quotient is \({\mathbb{Z}}^{\operatorname{disc}}\). So \(K^{\times}\)
also has a discrete and profinite part.

\end{remark}

\hypertarget{properties-of-the-artin-map}{%
\subsection{Properties of the Artin
Map}\label{properties-of-the-artin-map}}

\begin{remark}

See Serre's article in Cassels-Froehlich for proofs!

\end{remark}

\hypertarget{the-canonical-isomorphism}{%
\subsubsection{The Canonical
Isomorphism}\label{the-canonical-isomorphism}}

\begin{proposition}[Properties of the Artin map]

\begin{itemize}
\tightlist
\item
  \(r_K\) restricts to maps:
\end{itemize}

\begin{center}
\begin{tikzcd}
    K^{\times}&& {W_K^{\operatorname{ab}}} \\
    \\
    {{\mathcal{O}}_K^{\times}} && {\operatorname{im}(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K})} \\
    \\
    {1 + {\mathfrak{p}}_K^i,\, i\geq 1} && {\operatorname{im}(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}^i)}
    \arrow["{\sim}", from=1-1, to=1-3]
    \arrow["{\sim}", from=3-1, to=3-3]
    \arrow["{\sim}", from=5-1, to=5-3]
    \arrow[hook, from=5-1, to=3-1]
    \arrow[hook, from=5-3, to=3-3]
    \arrow[hook, from=3-3, to=1-3]
    \arrow[hook, from=3-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\begin{itemize}
\tightlist
\item
  Note that only the upper numbering makes sense here, since these are
  infinite extensions. This is weird because of the difference in jumps
  for the filtrations, but is explained by the \textbf{Hasse-Arf
  theorem}.
\item
  \(r_k(\pi_K) \in \operatorname{Frob}^{-1}\cdot \operatorname{im}(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K})\),
  which is a coset in a quotient.

  \begin{itemize}
  \tightlist
  \item
    Note that \(\pi_K\) is a \emph{choice} of uniformizer, which
    e.g.~for \({ {\mathbb{Q}}_p }\) is \(p\), but a random extension one
    can take any root of an Eisenstein polynomial. So there is some
    ambiguity on both sides of this map, but we use the one that sends
    \(\pi_K\) to a \(\operatorname{Frob}^{-1}\).
  \end{itemize}
\end{itemize}

\end{proposition}

\begin{remark}

Confusing comment: if
\(\phi:A{ \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }B\)
is a canonical isomorphism in \({\mathsf{Ab}}{\mathsf{Grp}}\), then
\(\phi \circ \iota\) where \(\iota(x) = x^{-1}\) is just as canonical!
This occurs e.g.~for the Weil pairing on an elliptic curve
\(E[n]{ {}^{ \scriptscriptstyle\times^{2} } }\to \mu_n\), which are maps
between abelian groups. This is an issue here since there are really two
canonical isomorphisms we could call \(r_K\), by composing with
inversion on \(K^{\times}\), but the last isomorphism
\(r_k(\pi_K) \in \operatorname{Frob}^{-1}\Im(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K})\)
singles out \emph{which} isomorphism it is since if \(\pi_K\) is a
uniformizer, \(\pi_K^{-1}\) is not. So we can tell the difference: one
isomorphism sends Frobenius to a uniformizer, the other to the
\emph{inverse} of a uniformizer.

\end{remark}

\begin{definition}[Geometric and Arithmetic Frobenius]

The \textbf{arithmetic Frobenius} is \(\operatorname{Frob}\), and
\textbf{geometric Frobenius} is \(\operatorname{Frob}^{-1}\).

\end{definition}

\begin{remark}

Deligne's convention: associate the uniformizer with \emph{geometric}
Frobenius.

\end{remark}

\hypertarget{abelian-extensions}{%
\subsubsection{Abelian Extensions}\label{abelian-extensions}}

\begin{remark}

If \(L/K\) is finite, then
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/L)\hookrightarrow G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\)
is a subgroup, and there is a map \(W_L \hookrightarrow W_K\). These are
injections by TFTGT, and there is a diagram:

\begin{center}
\begin{tikzcd}
    {G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/L)} && {{ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)} \\
    \\
    {W_L} & {} & {W_K} \\
    \\
    {W_L^{\operatorname{ab}}} && {W_K^{\operatorname{ab}}}
    \arrow[hook, from=1-1, to=1-3]
    \arrow[hook, from=3-1, to=3-3]
    \arrow[from=5-1, to=5-3]
    \arrow[two heads, from=3-1, to=5-1]
    \arrow[hook, from=3-1, to=1-1]
    \arrow[two heads, from=3-3, to=5-3]
    \arrow[hook, from=3-3, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{warnings}

Note that (topological) abelianization may not preserve monomorphism!
E.g. consider \(G\) with \(G^{\operatorname{ab}}= 1\) and a map
\(C_n\hookrightarrow G\) from a cyclic group, then
\(C_n^{\operatorname{ab}}= C_n\twoheadrightarrow 1\).

\end{warnings}

\begin{remark}

One can fill in the following diagram with the norm:

\begin{center}
\begin{tikzcd}
    L^{\times}&& {W_L^{\operatorname{ab}}} \\
    \\
    K^{\times}&& {W_K^{\operatorname{ab}}}
    \arrow["{r_L}", from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
    \arrow["{r_K}"', from=3-1, to=3-3]
    \arrow["{\operatorname{Nm}_{L/K}}"', from=1-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJMXFx1bml0cyJdLFswLDIsIktcXHVuaXRzIl0sWzIsMCwiV19MXlxcYWIiXSxbMiwyLCJXX0teXFxhYiJdLFswLDIsInJfTCJdLFsyLDNdLFsxLDMsInJfSyIsMl0sWzAsMSwiXFxObV97TC9LfSIsMl1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{remark}

For \(H \leq G\) a finite index subgroup, there is a \textbf{transfer}
map on the Weil groups, the \emph{Verlagerung transfer}
\begin{align*}
V: G^{\operatorname{ab}}\to ^{\operatorname{ab}}
g &\mapsto \prod_{1\leq i\leq n} \gamma_i g\gamma_i^{-1}, && \gamma_i \in G/H \text{ coset representatives}
.\end{align*}
which behaves like a norm. It fits into a diagram:

\begin{center}
\begin{tikzcd}
    L^{\times}&& {W_L^{\operatorname{ab}}} \\
    \\
    K^{\times}&& {W_K^{\operatorname{ab}}}
    \arrow["{r_K}", from=3-1, to=3-3]
    \arrow["{r_L}", from=1-1, to=1-3]
    \arrow["\operatorname{transfer}"', from=3-3, to=1-3]
    \arrow[hook, from=3-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJLXFx1bml0cyJdLFswLDAsIkxcXHVuaXRzIl0sWzIsMCwiV19MXlxcYWIiXSxbMiwyLCJXX0teXFxhYiJdLFswLDMsInJfSyJdLFsxLDIsInJfTCJdLFszLDIsIlxcdHJhbnNmZXIiLDJdLFswLDEsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{remark}

If \(L/K\) is finite Galois, noting that \(W_L \leq W_K\) is a finite
index normal subgroup with \(W_K/W_L = G(L/K)\). Recall that \(W_L^c\)
is the closure of \([W_L, W_L]\) and is a characteristic subgroup, so
\(W_L^c{~\trianglelefteq~}W_K\). Define the quotient
\begin{align*}
W_{L/K} \coloneqq W_K/W_L^c
.\end{align*}

\end{remark}

\begin{definition}[Fundamental Class in Galois Cohomology]

There is a canonical SES

\begin{center}
\begin{tikzcd}
    &&&& {} \\
    \\
    1 && {W_L^{\operatorname{ab}}= L^{\times}} \\
    \\
    && {W_L/K} \\
    \\
    1 && {G(L/K)}
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=5-3]
    \arrow[from=5-3, to=7-3]
    \arrow[from=7-3, to=7-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbNCwwXSxbMiwyLCJXX0xeXFxhYiA9IExcXHVuaXRzIl0sWzIsNCwiV19ML0siXSxbMiw2LCJcXEdhbChML0spIl0sWzAsNiwiMSJdLFswLDIsIjEiXSxbNSwxXSxbMSwyXSxbMiwzXSxbMyw0XV0=}{Link
to Diagram}
\end{quote}

This extension gives a \textbf{fundamental class}
\begin{align*}
\alpha_{L/K} \in H^2(G(L/K); L^{\times})
,\end{align*}
and is in fact a nontrivial element (so not a semidirect product). It
turns out that this \(H^2\) is cyclic of order \(n=[L:K]\) and generated
by \(\alpha\), and \(({-}) \smile\alpha_{L/K}\) is a map between
cohomology groups and is often an isomorphism.

\end{definition}

\begin{remark}

Say \(L/K\) is an abelian extension iff it is Galois with abelian Galois
group. There is a maximal abelian extension \(K^{\operatorname{ab}}\),
which exists since compositing preserves abelian extensions. We now
essentially understand \(G(K^{\operatorname{ab}}/K)\):

\begin{center}
\begin{tikzcd}
    && {\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu} \\
    \\
    && {K^{\operatorname{ab}}} \\
    {K^{\scriptscriptstyle\mathrm{un}}} \\
    && K
    \arrow[hook, from=5-3, to=3-3]
    \arrow[hook, from=3-3, to=1-3]
    \arrow["{?}"', curve={height=30pt}, dashed, no head, from=3-3, to=1-3]
    \arrow["{G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)^{\operatorname{ab}}}"', curve={height=30pt}, dashed, no head, from=5-3, to=3-3]
    \arrow[from=5-3, to=4-1]
    \arrow[from=4-1, to=3-3]
    \arrow["{G(K^{\scriptscriptstyle\mathrm{un}}/K) \cong \widehat{{\mathbb{Z}}}}", curve={height=-30pt}, dashed, no head, from=5-3, to=4-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

We now have a diagram:

\begin{center}
\begin{tikzcd}
    {I_{K^{\operatorname{ab}}/K}} && {G(K^{\operatorname{ab}}/K)} && {{ \mathsf{Gal}} (K^{\scriptscriptstyle\mathrm{un}}/K) = \widehat{{\mathbb{Z}}}} \\
    \\
    {I_{K^{\operatorname{ab}}/K}} && {W_K^{\operatorname{ab}}\underset{CFT}{{ { \, \xrightarrow{\sim}\, }}} K^{\times}} && {\mathbb{Z}}
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[Rightarrow, no head, from=1-1, to=3-1]
    \arrow[hook, from=3-3, to=1-3]
    \arrow[hook, from=3-5, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJJX3tLXlxcYWIvS30iXSxbMCwyLCJJX3tLXlxcYWIvS30iXSxbMiwyLCJXX0teXFxhYiBcXHVuZGVyc2V0e0NGVH17XFxpc299IEtcXHVuaXRzIl0sWzIsMCwiXFxHYWwoS15cXGFiL0spIl0sWzQsMCwiXFxHYWwoS15cXHVucmFtL0spID0gXFxoYXR7XFxaWn0iXSxbNCwyLCJcXFpaIl0sWzEsMl0sWzIsNV0sWzAsM10sWzMsNF0sWzAsMSwiIiwxLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFs1LDQsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
to Diagram}
\end{quote}

Note that \(G(K^{\operatorname{ab}}/K)\) is a profinite topological
group, so can't quite be \(K^{\times}\), but is some kind of completion.

\end{remark}

\begin{remark}

Conclusion: after choosing a uniformizer/Frobenius there is a
non-canonical isomorphism
\begin{align*}
G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)^{\operatorname{ab}}= { \mathsf{Gal}} (K^{\operatorname{ab}}/K) { { \, \xrightarrow{\sim}\, }}{\mathcal{O}}_K^{\times}\times \widehat{{\mathbb{Z}}}\cong I_{K^{\operatorname{ab}}/K} \times { \mathsf{Gal}} (K^{\scriptscriptstyle\mathrm{un}}/K)
.\end{align*}
We know
\(G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)\) is
solvable, since it factors as cyclic and pro\({\hbox{-}}p\) parts, but
the difficulty lies in how they extend. CFT is enough to understand the
abelianization, but then we were stuck for 50 years! Next: Langlands
ideas generalizing this story.

\end{remark}

\hypertarget{lecture-5}{%
\section{Lecture 5}\label{lecture-5}}

\begin{remark}[A neat trick]

A neat trick for pre/post-multiplying by diagonal matrices:
\begin{align*}
\begin{bmatrix}
\mathrel{\Big|}& \mathrel{\Big|}\\
\mathbf{v}_{1} & \mathbf{v}_{2} \\
\mathrel{\Big|}& \mathrel{\Big|}
\end{bmatrix}
\cdot
\begin{bmatrix}
d_1 & 0 
\\
0 & d_2
\end{bmatrix}
=
\left[\begin{array}{cc}
\mathrel{\Big|}& \mathrel{\Big|}\\
d_1\mathbf{v}_{1} & d_2\mathbf{v}_{2} \\
\mathrel{\Big|}& \mathrel{\Big|}
\end{array}\right]
.\end{align*}

Similarly
\begin{align*}
\begin{bmatrix}
d_1 & 0 
\\
0 & d_2
\end{bmatrix}
\cdot
\begin{bmatrix}
-  & \mathbf{v}_1 & - \\  
-  & \mathbf{v}_2 & - \\  
\end{bmatrix}
=
\begin{bmatrix}
-  & d_1 \mathbf{v}_1 & - \\  
-  & d_2 \mathbf{v}_2 & - \\  
\end{bmatrix}
.\end{align*}

I.e. post-multiplication by a diagonal matrix acts by scaling the
columns, and pre-multiplying scales the rows. A useful consequence:
\begin{align*}
{
\begin{bmatrix}
  {l_1} & {0} 
\\
  {0} & {l_2}
\end{bmatrix}
} \cdot {
\begin{bmatrix}
  {a} & {b} 
\\
  {c} & {d}
\end{bmatrix}
} \cdot {
\begin{bmatrix}
  {r_1} & {0} 
\\
  {0} & {r_2}
\end{bmatrix}
} = {
\begin{bmatrix}
  {l_1 r_1 a} & {l_1 r_2 b} 
\\
  {l_2 r_1 c} & {l_2 r_2 d}
\end{bmatrix}
}
,\end{align*}
i.e.~entry \(a_{ij}\) gets hit by the nonzero entries in the \(i\)th row
on the left and \(j\)th column on the right.

\end{remark}

\hypertarget{conductor}{%
\subsection{Conductor}\label{conductor}}

\begin{remark}

Next goal: statements of local Langlands conjectures (LLC). For
\(\operatorname{GL}_n/K\) with \(K/{ {\mathbb{Q}}_p }\) finite, there
will be some correspondence between representations of some ``Galois
group'' and representations of \(\operatorname{GL}_n(K)\). Here the
``Galois group'' is the Weil-Deligne group, which is related to the Weil
group, so we'll discuss \(n{\hbox{-}}\)dimensional representations of
\(W_K\).

\end{remark}

\begin{remark}

Recall we have a diagram:

\begin{center}
\begin{tikzcd}
    1 && {I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}} && {{ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K)} && {\widehat{{\mathbb{Z}}}} && 1 \\
    \\
    1 && {I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}} && {W_K} && {\mathbb{Z}}&& 1
    \arrow[Rightarrow, no head, from=3-3, to=1-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-7, to=1-9]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=3-7]
    \arrow[from=3-7, to=3-9]
    \arrow[hook, from=3-5, to=1-5]
    \arrow[hook, from=3-7, to=1-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

We're really interested in reps of \(G_K\) here, where e.g.~étale
cohomology or the Tate module of an elliptic curve induce
(\(\ell{\hbox{-}}\)adic) representations. Note that if one \emph{starts}
with a \(G_K\) representation on \(\widehat{{\mathbb{Z}}}\), one gets a
\(W_K\) representation on \({\mathbb{Z}}\), but the converse is not true
since reps may not lift, despite the fact that
\({\mathbb{Z}}\hookrightarrow\widehat{{\mathbb{Z}}}\) is dense.

\end{remark}

\begin{remark}

Let \(E\) be a field with the discrete topology and put the discrete
topology on \(\operatorname{GL}_n(E)\). Consider continuous
representations \(\rho: W_K\to \operatorname{GL}_n(E)\). Note that
\begin{align*}
\rho \text{ is continuous } \iff \ker \rho \leq W_K \text{ is open}
.\end{align*}
Note that
\(\rho( I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K} ) \subseteq \operatorname{GL}_n(E)\)
is the continuous image of a compact space, thus a compact subset of a
discrete group, and thus finite since \(\operatorname{GL}_n(E)\) is
covered by its open points and this must admit a finite subcover. So
we've reduced the hard part, inertia, to something more finite. We'll
now try to attach numerical invariants by pushing the upper/lower
filtrations into the image of \(\rho\). Since
\(\rho(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K})\)
is a finite quotient, we have a factoring of Galois groups:

\begin{center}
\begin{tikzcd}
    {\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu} \\
    && {L = L(\rho)} \\
    {K^{\scriptscriptstyle\mathrm{un}}}
    \arrow["{I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}}", hook, from=3-1, to=1-1]
    \arrow["{\rho(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K})}"', hook, from=3-1, to=2-3]
    \arrow[hook, from=2-3, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGJhciBLIl0sWzAsMiwiS15cXHVucmFtIl0sWzIsMSwiTCA9IEwoXFxyaG8pIl0sWzEsMCwiSV97XFxiYXIgSy9LfSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMiwiXFxyaG8oSV97XFxiYXIgSy9LfSkiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDAsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
to Diagram}
\end{quote}

So write
\(\rho(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}) = I_{L/K} \supseteq I_{L/K, 1}\supseteq\cdots\).

\end{remark}

\begin{definition}[Conductor]

Let \(\rho: W_K\to \operatorname{GL}_N(E) = \mathop{\mathrm{Aut}}_E(V)\)
be a representation as above where \(V = E^n\) as a vector space, and
define the \textbf{conductor of \(\rho\)} as
\begin{align*}
f(\rho) \coloneqq\sum_{i\geq 0} {1\over [I_{L/K}: I_{L/K,i} ]} \dim( V/V^{I_{L/K, i}} ) \in {\mathbb{Q}}_{\geq 0}
,\end{align*}
where for \(H\to \mathop{\mathrm{Aut}}(V)\) we define the fixed points
as
\(V^H \coloneqq\left\{{v\in V{~\mathrel{\Big\vert}~}hv = v {~\mathrel{\Big\vert}~}\forall h\in H}\right\}\).
Note that this sum is finite, since
\(i\gg 0 \implies I_{L/K, i} = \left\{{1}\right\}\) and the dimension
appearing is zero.

\end{definition}

\begin{remark}

It should be easy to show that \(f(\rho) = 0 \iff \rho\) is unramified
\(\iff \rho(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}) = \left\{{1}\right\}\).
One can also get a trivial lower bound by examining the first term,
which is \(\dim(V/V^{I_{L/K}})\).

\end{remark}

\begin{example}[?]

Representations of \(K^{\times}\) can be used to get reps of \(W_K\):

\begin{center}
\begin{tikzcd}
    && {W_K} \\
    \\
    K^{\times}&& {W_K^{\operatorname{ab}}} \\
    \\
    && {}
    \arrow[two heads, from=1-3, to=3-3]
    \arrow["\sim"', from=3-1, to=3-3]
    \arrow["\exists"', dashed, two heads, from=1-3, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJLXFx1bml0cyJdLFsyLDIsIldfS15cXGFiIl0sWzIsNF0sWzIsMCwiV19LIl0sWzMsMSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzAsMSwiXFxzaW0iLDJdLFszLDAsIlxcZXhpc3RzIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn0sImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

We have a valuation \(v_K: K^{\times}\to {\mathbb{Z}}\) where
\(v(\pi_K) = 1\), so define a norm
\({\left\lVert {\lambda} \right\rVert} \coloneqq{\varepsilon}^{-v(\lambda)}\)
for some \({\varepsilon}\in (0, 1)\), e.g.~taking
\({\varepsilon}\coloneqq 1/p\) for \(K= { {\mathbb{Q}}_p }\). Note that
changing \({\varepsilon}\) just yields an equivalent norm, and e.g.~in
non-Archimedean geometry we only use equivalence classes of norms
anyway!

\begin{remark}

If \(K\in \mathsf{Field}\) is complete with respect to a nontrivial
non-Archimedean norm, there is a good theory of \emph{rigid geometry}.
E.g. for \(K = {\mathbb{C}}{\left[\left[ t \right]\right] }\), one can
take \(v(t^n + \cdots) \coloneqq n\). Note that
\({\mathbb{C}}{\left[\left[ t \right]\right] }\) is not locally compact,
so doesn't admit a Haar measure.

\end{remark}

If \(K/{ {\mathbb{Q}}_p }\) is finite, there is a canonical choice for
this norm. Since \(\kappa(K)\) is a finite field, \(K\) is locally
compact since \({\mathcal{O}}_K \cong { {\mathbb{Z}}_p }^d\) which is a
compact open neighborhood of the identity. So \(K\) is a locally compact
Hausdorff abelian group, and thus admits an additive
(translation-invariant) Haar measure
\(\mu: K\to {\mathbb{R}}_{\geq 0} \cup\left\{{\infty}\right\}\). We'll
have \(\mu({\mathcal{O}}_K) < \infty\), so we can normalize to
\(\mu({\mathcal{O}}_K) = 1\). Note that \(\mu(X) = \mu(X+a)\) for any
\(a\in K\) and any subset \(X \subseteq K\), and
\(X \cap Y = \emptyset \implies \mu(X \cup Y) = \mu(X) + \mu(Y)\). We
can then write
\begin{align*}
{\mathcal{O}}_K = {\textstyle\coprod}_{\lambda \in \kappa(K)} \tilde \lambda + {\mathfrak{p}}_K
\end{align*}
where \(\tilde \lambda\) is a lift of
\(\lambda \in \kappa(K) = {\mathcal{O}}_K/{\mathfrak{p}}_K\) to
\({\mathcal{O}}_K\). Thus
\begin{align*}
\mu({\mathcal{O}}_K) = q\mu({\mathfrak{p}}_K) \implies \mu({\mathfrak{p}}_K) = 1/q,\qquad q\coloneqq{\sharp}\kappa(K)
.\end{align*}

\end{example}

\hypertarget{norms}{%
\subsection{Norms}\label{norms}}

\begin{remark}

If \(a\in K^{\times}\), one can define
\({\left\lVert {a} \right\rVert}\) as the factor by which multiplication
scales the Haar measure. E.g. for \(X = [0, 1]\) and the Lebesgue
measure, for \(\lambda \in {\mathbb{R}}^{\times}\) this yields
\({\left\lVert {\lambda} \right\rVert} = \mu(\lambda X)\). Noting that
\(\lambda [0, 1] = [0, \lambda]\) for \(\lambda > 0\), this recovers
\({\left\lVert {\lambda} \right\rVert} = {\left\lvert {\lambda} \right\rvert}\),
the usual absolute value. In general, one can define
\begin{align*}
{\left\lVert {\lambda} \right\rVert} \coloneqq{\mu(\lambda X) \over \mu(X)}
.\end{align*}

\end{remark}

\begin{remark}

If we want to set \(v(\pi_K)= 1\) and
\({\left\lVert {\pi_K} \right\rVert} = {\varepsilon}^1\), what should
one choose for \({\varepsilon}\)? Using the Haar measure trick above, if
\(\mu({\mathcal{O}}_K) = 1\),
\begin{align*}
{\left\lVert {\pi_K} \right\rVert} = \mu(\pi_K {\mathcal{O}}_K) = \mu({\mathfrak{p}}_K) = 1/q
,\end{align*}
so take \({\varepsilon}\coloneqq 1/q\). Note that this needs to be
chosen carefully to get a global product formula on adeles later.

\end{remark}

\begin{definition}[Norm]

So for \(v_K: K^{\times}\to {\mathbb{Z}}\) and \(\lambda \neq 0\), we
define
\begin{align*}
{\left\lVert {{-}} \right\rVert}: K\ &\to {\mathbb{Q}}_{\geq 0} \\
{\left\lVert {\lambda} \right\rVert} &\coloneqq\qty{1\over q}^{v_K(\lambda)} = q^{-v_K(\lambda)}
,\end{align*}
which restricts to
\begin{align*}
{\left\lVert {{-}} \right\rVert}: K^{\times}\to {\mathbb{Q}}_{>0}
.\end{align*}
We also set \({\left\lVert {0} \right\rVert} = 0\).

\end{definition}

\begin{remark}

Under the identification, for \(E={\mathbb{Q}}\) or any field of
characteristic zero, we now have an induced representation
\begin{align*}
{\left\lVert {{-}} \right\rVert}: W_K\to \operatorname{GL}_1(E)
.\end{align*}
Moreover, \({\left\lVert {{-}} \right\rVert}^m\) for
\(m\in {\mathbb{Z}}_{\geq 1}\) all produce representations of \(W_K\).

\end{remark}

\begin{exercise}[?]

Show that
\begin{align*}
f({\left\lVert {{-}} \right\rVert}^m) = 0
.\end{align*}

\end{exercise}

\begin{remark}

What is
\({\left\lVert {\widetilde{ \operatorname{Frob}} } \right\rVert}\),
where \(\widetilde{\operatorname{Frob}} \in W_K\) is a lift of
\(\operatorname{Frob}\in {\mathbb{Z}}\) along
\(W_K\twoheadrightarrow{\mathbb{Z}}\)? Use that
\(r_K: K^{\times}\to W_K^{\operatorname{ab}}\) sends
\({\mathcal{O}}_K^{\times}\pi_K\) to
\(\operatorname{Frob}^{-1}I_{K^{\operatorname{ab}}/K}\), and
\({\left\lVert {\pi_K} \right\rVert} = 1/q\), so
\({\left\lVert {\operatorname{Frob}^{-1}} \right\rVert} = 1/q\) and thus
\({\left\lVert {\operatorname{Frob}} \right\rVert} = q\).

\begin{quote}
See Tate's article, ``Number Theoretic Background''
\end{quote}

\end{remark}

\hypertarget{weil-deligne-representations}{%
\subsection{Weil-Deligne
Representations}\label{weil-deligne-representations}}

\begin{remark}

We'll assume \(\operatorname{ch}E = 0\) since we'll want to take norms
of elements in \(W_K\) and regard them as elements in \(E\), or really
elements of \({\mathbb{Q}}_{>0}\).

\end{remark}

\begin{definition}[Weil-Deligne representations]

A \textbf{Weil-Deligne} representation is a pair \((\rho_0, N)\) where
\begin{align*}
\rho_0: W_K \to \mathop{\mathrm{Aut}}_E(V) = \operatorname{GL}_n(E)
\end{align*}
is a continuous representation where \(V\cong E^n\), and
\(N \in \mathop{\mathrm{End}}_E(V)\) is nilpotent. We also require a
compatibility condition:
\begin{align*}
\rho_0(\sigma) N \rho_0(\sigma)^{-1}= {\left\lVert {\sigma} \right\rVert} N \qquad \forall \sigma\in W_K
.\end{align*}

\end{definition}

\begin{remark}

Note that this recovers the previous notion of representations by
setting \(N=0\). For \(N\neq 0\), taking
\(\sigma = \widetilde{\operatorname{Frob}}\) this equation reduces to
saying that \(N\) is conjugate to \(qN\) for \(q\) a scalar. Since
\(\operatorname{Spec}qN = \left\{{q\lambda {~\mathrel{\Big\vert}~}\lambda \in \operatorname{Spec}N}\right\}\),
so if there is any nonzero eigenvalue there are infinitely many, so the
only eigenvalue is zero and this forces \(N\) to be nilpotent.

\end{remark}

\begin{example}[?]

Are there examples where \(N\neq 0\) is nilpotent? We need dimension at
least 2, so let \(E = {\mathbb{Q}}\) and
\(V= \mathop{\mathrm{span}}_{\mathbb{Q}}\left\{{e_1, e_0}\right\}\).
Then
\begin{align*}
\rho_0(\sigma) e_1 &= {\left\lVert {\sigma} \right\rVert}^1 e_1 \\
\rho_0(\sigma) e_0 &= {\left\lVert {\sigma} \right\rVert}^0 e_0 \\ \\
\rho_0(\sigma) &= {
\begin{bmatrix}
  {{\left\lVert {\sigma} \right\rVert}} & {0} 
\\
  {0} & {1}
\end{bmatrix}
}
,\end{align*}
which is a direct sum of two 1-dimensional representation. Here
\(Ne_0 = e_1\) and \(Ne_1 = 0\), so
\(N = { \begin{bmatrix}  {0} & {1} \\  {0} & {0} \end{bmatrix} }\).
Checking the equation:
\begin{align*}
{
\begin{bmatrix}
  {{\left\lVert {\sigma} \right\rVert}} & {0} 
\\
  {0} & {1}
\end{bmatrix}
} \cdot {
\begin{bmatrix}
  {0} & {1} 
\\
  {0} & {0}
\end{bmatrix}
} \cdot {
\begin{bmatrix}
  {{\left\lVert {\sigma} \right\rVert}^{-1}} & {0} 
\\
  {0} & {1}
\end{bmatrix}
} 
= {
\begin{bmatrix}
  {0} & {{\left\lVert {\sigma} \right\rVert}} 
\\
  {0} & {0}
\end{bmatrix}
} = {\left\lVert {\sigma} \right\rVert} N
.\end{align*}

\end{example}

\begin{example}[The Steinberg representation]

Let
\(V = {\mathbb{Q}}^n = \mathop{\mathrm{span}}_{\mathbb{Q}}\left\{{e_0, \cdots, e_{n-1}}\right\}\)
and define
\(\rho_0(\sigma)e_i = {\left\lVert {\sigma} \right\rVert}^i e_i\) and
\(Ne_i = e_{i+1}\) with \(N e_{n-1} = 0\) yields a Weil-Deligne
representation. This is the most basic indecomposable example, and the
rest arise from tensoring with reps of \(W_K\) and taking direct sums.

\end{example}

\begin{definition}[Semisimple Weil-Deligne Representations]

A Weil-Deligne representation \((\rho_0, N)\) is
\textbf{\(F{\hbox{-}}\)semisimple} if
\(\rho_0(\widetilde{\operatorname{Frob}})\) is a semisimple matrix
(i.e.~diagonalizable over
\(\mkern 1.5mu\overline{\mkern-1.5muE\mkern-1.5mu}\mkern 1.5mu\)),
independent of the choice of lift of \(\operatorname{Frob}\).

\end{definition}

\hypertarget{lecture-6-representations-of-operatornamegl_nk-and-the-local-langlands-conjecture}{%
\section{\texorpdfstring{Lecture 6: Representations of
\(\operatorname{GL}_n(k)\) and the Local Langlands
Conjecture}{Lecture 6: Representations of \textbackslash operatorname\{GL\}\_n(k) and the Local Langlands Conjecture}}\label{lecture-6-representations-of-operatornamegl_nk-and-the-local-langlands-conjecture}}

\begin{remark}

The LLC will take the following form:
\begin{align*}
\left\{{\substack{
  \text{$n{\hbox{-}}$dimensional $F{\hbox{-}}$semisimple } \\
  \text{Weil-Deligne representations of } W_k
}}\right\}/\sim
&\rightleftharpoons
\left\{{\substack{
  \text{Irreducible admissible reps} \\
  \text{ of } \operatorname{GL}_n(K)
}}\right\}
,\end{align*}
where we're working toward describing the RHS.

\end{remark}

\begin{warnings}

Most representations here will be infinite dimensional.

\end{warnings}

\begin{remark}

Setup:

\begin{itemize}
\item
  \(E\in {\mathsf{Top}}\mathsf{Field}\) with the discrete topology and
  any characteristic,
\item
  \(V\in {\mathsf{E}{\hbox{-}}\mathsf{Mod}}\) an infinite-dimensional
  \(E{\hbox{-}}\)vector space,
\item
  \(K/{ {\mathbb{Q}}_p }\) a finite extension,
\item
  \(\operatorname{GL}_n(K)\) has the \(p{\hbox{-}}\)adic topology,
  described by a bases of open neighborhoods of the identity:
  \begin{align*}
  {\mathcal{B}}\coloneqq\left\{{M \coloneqq{
  \begin{bmatrix}
    {a} & {b} 
  \\
    {c} & {d}
  \end{bmatrix}
  } \in \operatorname{GL}_n({\mathcal{O}}_K) {~\mathrel{\Big\vert}~}M \equiv I_n \operatorname{mod}{\mathfrak{p}}_K^m\, \forall m\in {\mathbb{Z}}_{\geq 1}}\right\}
  .\end{align*}
\item
  \(\pi\in {\mathsf{Grp}}(\operatorname{GL}_n(K), \mathop{\mathrm{Aut}}_E(V))\),
  where we're not using the topology on \(E\) yet.
\end{itemize}

We want a sensible notion of continuity for \(\pi\).

\end{remark}

\begin{definition}[Smooth representations]

A representations \(\pi\) is \textbf{smooth} iff stabilizers are open,
i.e.~
\begin{align*}
\forall v\in V, \, {\operatorname{Stab}}_\pi(v) \coloneqq\left\{{g\in \operatorname{GL}_n(K) {~\mathrel{\Big\vert}~}g.v \coloneqq\pi(g)(v) = v}\right\} \text{ is open in } \operatorname{GL}_n(K)
.\end{align*}

\end{definition}

\begin{definition}[Admissible representations]

A smooth representation \(\pi\) is \textbf{admissible} iff for all
\(U \leq \operatorname{GL}_n(K)\) open subgroups, the fixed set \(V^U\)
is finite dimensional.

\end{definition}

\begin{remark}

On terminology: admissible will always imply smooth.

\end{remark}

\begin{example}[?]

If \(\dim V = 1\) and \(\pi(g) = 1\), then \(\pi\) is admissible. If
\(\dim V = \infty\) instead, this is smooth but not admissible, and in
some sense is an infinite direct sum of trivial 1-dimensional reps.

\end{example}

\begin{fact}

Irreducible and smooth implies admissible.

\end{fact}

\begin{definition}[Irreducible representations]

A representation \(\pi: G\to \operatorname{GL}(V)\) for
\(G = \operatorname{GL}_n(K)\) is \textbf{irreducible} iff there are
only 2 \(G{\hbox{-}}\)invariant subspaces: \(0\) and \(V\) itself. Note
that this still holds in infinite-dimensions in this case, i.e.~we don't
have to require the invariant subspaces to be closed.

\end{definition}

\begin{remark}

By LCFT, we know
\(K^{\times}{ { \, \xrightarrow{\sim}\, }}W_k^{\operatorname{ab}}\), but
how do we understand \(W_K\)? Langlands' insight: reinterpreting this
isomorphism in terms of representations. If the groups are isomorphic,
their reps are isomorphic, so
\begin{align*}
\left\{{\substack{
  \text{Irreducible 1-dimension reps} \\
  \text{of }K^{\times}= \operatorname{GL}_1(K)
}}\right\}/\sim
&\rightleftharpoons
\left\{{\substack{
  \text{Irreducible 1-dimensional reps} \\
  \text{ of } W_K
}}\right\}/\sim
\end{align*}

So the conjecture is that there should be a canonical bijection, say of
sets, of the following form:
\begin{align*}
{\mathsf{Irr}}{\mathsf{Rep}}^{\dim = n} \operatorname{GL}_1(K) \rightleftharpoons{\mathsf{Irr}}{\mathsf{Rep}}^{\dim = n} W_k
.\end{align*}
Next time we'll see how to generalize this to \(n=2\), where the \(n=1\)
case is class field theory, and we'll see how to match things up.

\end{remark}

\hypertarget{lecture-7-local-langlands}{%
\section{Lecture 7: Local Langlands}\label{lecture-7-local-langlands}}

\hypertarget{statement-of-llc}{%
\subsection{Statement of LLC}\label{statement-of-llc}}

\begin{conjecture}[The Local Langlands Correspondence for $\GL_n$]

There is a canonical bijective correspondence of sets
\begin{align*}
\displaystyle\coprod_{n\geq 1}
F{\hbox{-}}\ss\mathsf{WD}{\mathsf{Rep}}{_{\scriptstyle / \sim} }
&\rightleftharpoons
\displaystyle\coprod_{n\geq 1} {\mathsf{sm}}{\mathsf{Irr}}{\mathsf{Adm}}{\mathsf{Rep}}(\operatorname{GL}_n(K)){_{\scriptstyle / \sim} }\\
\rho &\rightleftharpoons\pi
\end{align*}
The objects on the left are \(F{\hbox{-}}\)semisimple Weil-Deligne reps,
on the right are \emph{admissible} reps. Note that we've not yet seen
interesting objects showing up naturally on either side! Both sides are
somewhat pathological at this point.

Compare to the Taniyama-Shimura conjecture, where elliptic curves come
from modular forms, and both show up naturally. See also the BSD
conjecture -- if you don't know an elliptic curve is modular, you can't
meromorphically extend the \(L\) function.

\end{conjecture}

\begin{remark}

Here ``canonical'' means that the bijection satisfies a long list of
properties. Some examples of what should match up:

\begin{itemize}
\item
  Duality. A representation \(V\) will have a left
  \(G{\hbox{-}}\)action, so \(V {}^{ \vee }\) will have a right
  \(G{\hbox{-}}\)action, and composing with inversion makes it a left
  action again -- typically this yields a different representation.
\item
  Conductors. The conductor on the RHS is harder to define.
\item
  \(L{\hbox{-}}\)functions.
\item
  \({\varepsilon}\) factors.
\item
  \({\varepsilon}\) factors of pairs -- these turn out to be crucial!
\item
  \(\cdots\)
\end{itemize}

This list of properties became so long that there is now a theorem
showing that there is at most one bijection satisfying all of them.
Another theorem shows that there is \emph{at least} one bijection.

\begin{itemize}
\tightlist
\item
  In the function field case, this was established in the 80s (Laumm,
  Rapoport, Stuhler).
\item
  In the \(p{\hbox{-}}\)adic field case, this was proved in 2000
  (Harris-Taylor)
\end{itemize}

All of the proofs are global, i.e.~use number fields, despite the fact
that Galois groups of number fields are more complicated than Galois
groups of local fields.

\end{remark}

\begin{remark}

Philosophy: this gives a direction to go after class field theory, and
highlights that \(G_K\) isn't the fundamental object, but rather the
category \({\mathsf{Rep}}(G_K)\). In analogy, note that
\(\pi_1 X \in {\mathsf{Grp}}\) is not canonically defined, but
\(\Pi_1 X\in {\mathsf{Grpd}}\) is. Defining the group depends on choice
of a base point, and a path \(x_0\to x_1\) produces an isomorphism
\(\pi_1(X, x_0)\to \pi_1(X, x_1)\), but e.g.~on a torus the paths can
wind around genus so that two choices of path need not be homotopic.
However, there is a theorem that says
\({\mathsf{Rep}}(\pi_1 X) { { \, \xrightarrow{\sim}\, }}\mathsf{LocSys}(X)\).
In this analogy, choosing a point for \(\pi_1\) is like choosing an
algebraic closure
\(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K\).

\end{remark}

\begin{remark}

Next goal: finding natural examples of the LHS of the correspondence,
and at least some global occurrences on the RHS, and show that the LLC
is useful. We'll consider

\begin{itemize}
\tightlist
\item
  The \(n=1\) case, recovering CFT,
\item
  Weil-Deligne representations showing up naturally,
\item
  Examples of smooth irreducible admissible reps \(\pi\), e.g.~in the
  cohomology of Shimura varieties.
\end{itemize}

\end{remark}

\hypertarget{the-n1-case}{%
\subsection{\texorpdfstring{The \(n=1\)
case}{The n=1 case}}\label{the-n1-case}}

\begin{conjecture}

If \(G\in{\mathsf{Alg}}{\mathsf{Grp}}_{/ {K}}\) is connected and
reductive, i.e.~over
\(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu\), \(G\)
is isomorphic to one of
\begin{align*}
\operatorname{GL}_n, {\operatorname{SL}}_n, \operatorname{PGL}_n, {\operatorname{O}}_n, {\mathsf{Sp}}_{2g}, E_6, E_7, E_8, F_4, G_2
,\end{align*}
then there is a LLC for \(G\): there is a canonical surjection

\begin{align*}
F \text{ finite fibers} \hookrightarrow
\left\{{\substack{
  \text{Certain Weil-Deligne reps} (\rho, N): W_K \to {}^L G({\mathbb{C}})
}}\right\}
&\twoheadrightarrow
{\mathsf{sm}}{\mathsf{Irr}}{\mathsf{Rep}}( G(K) )
\end{align*}
where \({}^LG\) is an \(L{\hbox{-}}\)group for \(G\) and the fibers are
referred to as \textbf{\(L{\hbox{-}}\)packets}. This surjection is
similarly supposed to satisfy a big list of properties, but it is not
known if these uniquely characterize the surjection.

\end{conjecture}

\begin{remark}

Consider \(n=1\) dimensional reps, so on the LHS we have pairs
\((\rho, N): W_K\to \operatorname{GL}_1({\mathbb{C}})\) where \(N\)
being \(1\times 1\) nilpotent forces \(N=0\). Note that \(\ker \rho_0\)
is closed and the quotient is Hausdorff and abelian, so \(\rho_0\)
factors through \(W_K^{\operatorname{ab}}\). So the LHS reduces to
1-dimensional continuous reps of \(W_K^{\operatorname{ab}}\) over
\({\mathbb{C}}\).

A coincidence in dimension 1: the RHS reads smooth admissible
irreducible reps of \(K^{\times}\), and one can show that admissible and
irreducible implies \(\dim \pi < \infty\) and further \(\dim \pi = 1\).
This needs that we have compact open normal subgroups, and this fails
quite seriously for \(\operatorname{GL}_2\). In the \(n=1\) case,
continuity is equivalent to smooth and admissible, so we're considering
\begin{align*}
\pi \in {\mathsf{Top}}{\mathsf{Grp}}(K^{\times}, {\mathbb{C}}^{\times}), \qquad
\rho \in {\mathsf{Top}}{\mathsf{Grp}}(W_L^{\operatorname{ab}}, {\mathbb{C}}^{\times})
.\end{align*}
and these are equal by local class field theory.

\end{remark}

\begin{remark}

A source of Weil-Deligne reps: \textbf{\(\ell{\hbox{-}}\)adic
representations}. Let \(K/{ {\mathbb{Q}}_p }\) be finite, and suppose
\(\rho\in {\mathsf{Top}}{\mathsf{Grp}}(G_K, \operatorname{GL}_n({ {\mathbb{Q}}_\ell }))\)
where \(G_K\) is a profinite topological group with the profinite
topology, and \({ {\mathbb{Q}}_\ell }\) has \(\ell{\hbox{-}}\)adic
topology. Assume also \(\ell \neq p\).

Note that the target is now \emph{not} discrete, so there won't be
finite inertia. The wild part of inertia will be pro\({\hbox{-}}p\) and
end up being finite, while the tame part will have a large
\(\ell{\hbox{-}}\)adic component. In the discrete case, we knew the
image of inertia was finite by a compactness argument, but that may not
hold here.

\end{remark}

\begin{example}[of where Weil-Deligne reps show up]

These show up in nature, e.g.~

\begin{itemize}
\tightlist
\item
  In the \(\ell{\hbox{-}}\)adic Tate module \(T_\ell(E/K)\) of an
  elliptic curve \(E/K\).
\item
  In \(\ell{\hbox{-}}\)adic etale cohomology
  \(H^i_\text{ét}(X_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}; { {\mathbb{Q}}_\ell })\).
  The vast majority of Weil-Deligne representations come from here.
\item
  In \(\ell{\hbox{-}}\)adic deformations of the above, i.e.~taking an
  \(\ell{\hbox{-}}\)adic representation, reducing
  \(\operatorname{mod}\ell\), and try to deform back into characteristic
  zero. Some families come reps of \(\operatorname{GL}_n(R)\) for \(R\)
  an affinoid object. See also eigencurves.
\end{itemize}

These are slightly more difficult because we can no longer control
inertia.

\end{example}

\begin{remark}

Given a \(\rho\), one can construct a WD rep. One example of where
\(T_\ell(E/K)\) is easy to compute: if \(E\) is an elliptic curve with
split (bad) multiplicative reduction, then \(E\) can be uniformized to
obtain
\begin{align*}
E(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu) \cong \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu^{\times}/ q^{\mathbb{Z}}\qquad q\in K, {\left\lvert {q} \right\rvert} < 1
.\end{align*}

This
\begin{align*}
{ \left.{{T_\ell E}} \right|_{{E_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu /K}}} } = {
\begin{bmatrix}
  {\chi} & {*} 
\\
  {0} & {\operatorname{id}}
\end{bmatrix}
}
,\end{align*}
where \(\chi\) can be a nontrivial infinite cyclotomic character. Note
that this yields infinite inertia.

\end{remark}

\begin{remark}

Recall that if \(\rho\) is an \(\ell{\hbox{-}}\)adic representation as
above,
\(\rho(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K})\)
can be infinite but can't be too bad,
e.g.~\(\rho(I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu /K}^{\varepsilon})\)
must be finite for \({\varepsilon}>0\) since it is pro\({\hbox{-}}p\).
The tame inertia isn't so bad: we have
\begin{align*}
{ \mathsf{Gal}} (K^t/K^{\scriptscriptstyle\mathrm{un}}) \cong \prod_{r\neq p\text{ prime}} {\mathbb{Z}}_r
,\end{align*}
and we should worry about the \({ {\mathbb{Z}}_\ell }\) component. To
isolate this part, fix
\(t\in { \mathsf{Gal}} (K^t/K^{\scriptscriptstyle\mathrm{un}}) \twoheadrightarrow{ {\mathbb{Z}}_\ell }\)
and \(\phi\in G_K\) lifting
\(\operatorname{Frob}\in G_{K^{\scriptscriptstyle\mathrm{un}}}\). The
full \(G_K\) now breaks into three stages:

\begin{itemize}
\tightlist
\item
  Unramified: controlled by what happens to \(\phi\),
\item
  Tamely ramified: controlled by \(t\),
\item
  Wildly ramified: pro\({\hbox{-}}p\) and hence finite.
\end{itemize}

\end{remark}

\begin{proposition}[Grothendieck]

If \(\rho\in {\mathsf{Top}}{\mathsf{Grp}}(G_K, \operatorname{GL}_n(E))\)
with \(E/{ {\mathbb{Q}}_\ell }\) is finite, then there exists a unique
Weil-Deligne representation
\((\rho, N) \in {\mathsf{Top}}{\mathsf{Grp}}(W_K, \operatorname{GL}_n(E)^{\operatorname{disc}})\).
This satisfies the following: if
\(\sigma\in I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}\),
for every \(m\in {\mathbb{Z}}_{\geq 0}\),
\begin{align*}
\rho(\phi^m \sigma) = \rho_0(\phi^m \sigma) \exp\qty{N\cdot t(\sigma)}
,\end{align*}
where nilpotence of \(N\) guarantees that the series expansion for
\(\exp({-})\) is finite.

\end{proposition}

\begin{remark}

Note that we can always restrict \(\rho\) using that \(W_K \leq G_K\),
but this may not be continuous. Since we dropped the topology on
\(\operatorname{GL}_n(E)\), it makes it harder for \(\rho_0\) to be
continuous.

\end{remark}

\begin{remark}

Not all WD reps show up this way, since \(\rho(\phi)\) has restrictions
on its eigenvalues -- if \((\rho_0, N)\) arises this way, eigenvalues of
\(\rho(\phi)\) will be in \({ {\mathbb{Q}}_\ell }^{\times}\). However,
this is essentially the only obstruction: given \((\rho_0, N)\) with
\(\lambda_i \in { {\mathbb{Q}}_\ell }^{\times}\), then it comes from a
\(\rho\).

\end{remark}

\hypertarget{smooth-admissible-reps-pi-of-operatornamegl_nk}{%
\subsection{\texorpdfstring{Smooth admissible reps \(\pi\) of
\(\operatorname{GL}_n(K)\)}{Smooth admissible reps \textbackslash pi of \textbackslash operatorname\{GL\}\_n(K)}}\label{smooth-admissible-reps-pi-of-operatornamegl_nk}}

\begin{remark}[on conductors]

Given \(\pi: K^{\times}\to {\mathbb{C}}^{\times}\) smooth admissible
irreducible, define
\begin{align*}
f(\pi) = 
\begin{cases}
0 & { \left.{{\pi}} \right|_{{{\mathcal{O}}_K^{\times}}} } = 1 
\\
r & \text{ if $r \in {\mathbb{Z}}_{>0}$ is minimal such that } { \left.{{\pi}} \right|_{{1+ {\mathfrak{p}}_K^r {\mathcal{O}}_K}} }=1.
\end{cases}
.\end{align*}
As \(r\) increases, this gives a basis of neighborhoods of the identity.
In the LLC for \(n=1\), if
\(\rho_0 = (\rho_0, N=0) \rightleftharpoons\pi\), it turns out that
\(f(\rho_0) = f(\pi)\).

\end{remark}

\hypertarget{the-n2-case}{%
\subsection{\texorpdfstring{The \(n=2\)
case}{The n=2 case}}\label{the-n2-case}}

\begin{remark}

Constructing a \(\pi\): given a pair of characters, one can construct a
representation of \(\operatorname{GL}_2(K)\). Let
\(\chi_1, \chi_2\in {\mathsf{Top}}{\mathsf{Grp}}(K^{\times}\to{\mathbb{C}}^{\times})\)
and define
\begin{align*}
I(\chi_1, \chi_2) \coloneqq\left\{{\phi: \operatorname{GL}_2(K) \to {\mathbb{C}}{~\mathrel{\Big\vert}~}
\phi \text{ is locally constant}, \phi\qty{ {
\begin{bmatrix}
  {a} & {b} 
\\
  {0} & {d}
\end{bmatrix}
} g } = \chi_1(a) \chi_2(d) {\left\lVert {a\over d} \right\rVert}^{1\over 2} \phi(g)
}\right\}
.\end{align*}

\begin{remark}

The norm term is a fudge factor!

\end{remark}

Note that since the source is totally disconnected, there can be many
locally constant but non-constant functions,
e.g.~\(\chi_{{ {\mathbb{Z}}_p }}(x)\) on \({ {\mathbb{Q}}_p }\). This is
a vector space under pointwise operations in the target, and is supposed
to look like \(\operatorname{Ind}_B^G V\) for \(V\) a 1-dimensional
representation of
\(B = { \begin{bmatrix}  {*} & {*} \\  {0} & {*} \end{bmatrix} }\). Now
define
\begin{align*}
\pi: \operatorname{GL}_2(K) &\to \mathop{\mathrm{Aut}}_{{\mathsf{{\mathbb{C}}}{\hbox{-}}\mathsf{Mod}}} (I(\chi_1, \chi_2)) \\
g &\mapsto \phi\mapsto (h\mapsto \phi(hg))
,\end{align*}
so that \((\pi(g)\phi)(h) = \phi(hg)\) for all
\(g,h\in \operatorname{GL}_2(K)\). One can check that this defines an
action -- write this as \(g.\phi(h) = \phi(hg)\), then one needs to
check that \((g_1 g_2)\phi = g_1(g_2\phi)\). Consider instead writing
\((x) f \coloneqq f(x)\), then the equation reads
\((h)(g\phi) = (hg)\phi\) and the check is
\((h)(g_1 g_2(\phi)) = (hg_1)(g_2\phi)\), and one can rearrange the
brackets by definition. Thus this yields a representation of
\(\operatorname{GL}_2(K)\).

\end{remark}

\hypertarget{lecture-8}{%
\section{Lecture 8}\label{lecture-8}}

\begin{question}

Is this representation \(\pi\) smooth, admissible, and irreducible?

\end{question}

\begin{lemma}[?]

Let
\(B(K) \coloneqq\left\{{{ \begin{bmatrix}  {*} & {*} \\  {0} & {*} \end{bmatrix} } \in \operatorname{GL}_2(K)}\right\} \leq \operatorname{GL}_2(K)\),
then there is a decomposition into upper triangular and integer
matrices:
\begin{align*}
\operatorname{GL}_2(K) = B(K) \cdot \operatorname{GL}_2({\mathcal{O}}_K) \coloneqq\left\{{ bg {~\mathrel{\Big\vert}~}b\in B(K), g\in \operatorname{GL}_2({\mathcal{O}}_K)}\right\}
.\end{align*}

\end{lemma}

\begin{remark}

Why this is useful: \(\operatorname{GL}_2({\mathcal{O}}_K)\) is compact,
so locally constant functions
\(\phi: \operatorname{GL}_2({\mathcal{O}}_K)\to {\mathbb{C}}\) will only
take finitely many values (using that continuous images of compact sets
are compact and compact subsets of discrete spaces are finite). So
\(\phi(\operatorname{GL}_2({\mathcal{O}}_K))\) is a finite set, and
\(\phi(B(K))\) is controlled by \(I(\chi_1, \chi_2)\).

\end{remark}

\begin{proof}[Specializing a proof for $\GL_n$]

Let
\(\gamma \coloneqq{ \begin{bmatrix}  {a} & {b} \\  {c} & {d} \end{bmatrix} }\in \operatorname{GL}_2(K)\),
then we want to produce \(\beta, \kappa\) with \(\gamma = \beta\kappa\).

\begin{itemize}
\tightlist
\item
  Without loss of generality \(\gamma\in {\operatorname{SL}}_2(K)\) by
  left-multiplying by
  \({ \begin{bmatrix}  { (\operatorname{det}\gamma)^{-1}} & {0} \\  {0} & {1} \end{bmatrix} }\in B(K)\).
\item
  Without loss of generality, \(c, d\in {\mathcal{O}}_K\) and at least
  one is a unit: To scale \(c,d\), choose \(\alpha\in K^{\times}\) so
  that \(ac,ad\in {\mathcal{O}}_K\) and at least one is in
  \({\mathcal{O}}_K^{\times}\) since not both of \(c,d\) are zero.
  Left-multiply by
  \({ \begin{bmatrix}  {\alpha^{-1}} & {0} \\  {0} & {\alpha} \end{bmatrix} }\in B(K)\)
  to send \(c\to \alpha c, d\to\alpha d\).
\item
  Without loss of generality, \(c\in {\mathcal{O}}_K^{\times}\). If
  \(d \in {\mathcal{O}}_K^{\times}\) and
  \(c\not\in {\mathcal{O}}_K^{\times}\), right-multiply by
  \({ \begin{bmatrix}  {0} & {-1} \\  {1} & {0} \end{bmatrix} }\in {\operatorname{SL}}_2({\mathcal{O}}_K)\)
  to swap \(c, d\)
\item
  Check
  \({ \begin{bmatrix}  {1} & {-a/c} \\  {0} & {1} \end{bmatrix} } \cdot { \begin{bmatrix}  {a} & {b} \\  {c} & {d} \end{bmatrix} } = { \begin{bmatrix}  {0} & {-1/c} \\  {c} & {d} \end{bmatrix} } \in \operatorname{GL}_2({\mathcal{O}}_K)\),
  noting that the first matrix is in \(B(K)\).
\end{itemize}

\end{proof}

\begin{remark}

More generally, for \(\operatorname{GL}_n\) the Weyl group \(S_n\) is
involved.

\end{remark}

\begin{exercise}[A really good one.]

Show that \(I(\chi_1, \chi_2)\) is smooth and admissible.

\end{exercise}

\begin{observation}

The norm term in \(I(\chi_1, \chi_2)\) is a fudge factor, and
\begin{align*}
{\left\lVert {a\over d} \right\rVert}^{1\over 2} = {{\left\lVert {a} \right\rVert} ^{1\over 2}\over {\left\lVert {d} \right\rVert} ^{1\over 2}}
.\end{align*}
One could redefine
\begin{align*}
\tilde\chi_1(x) &\coloneqq\chi_1(x) {\left\lVert {x} \right\rVert}^{-{ 1\over 2} } \\
\tilde\chi_2(x) &\coloneqq\chi_2(x) {\left\lVert {x} \right\rVert}^{1\over 2}
\end{align*}
to make the formula read
\begin{align*}
\phi {
\begin{bmatrix}
  {a} & {b} 
\\
  {c} & {d}
\end{bmatrix}
} g = \tilde \chi_1(a) \tilde \chi_2(d)\phi(g)
.\end{align*}

\end{observation}

\begin{warnings}

The fudge factor is needed here: taking the trivial characters will
yield a subspace of constant functions in \(I(\chi_1, \chi_2)\) when the
\(\chi_i\) are trivial reps. When \(\chi_1 = \chi_2\), there is an
invariant 1-dimensional subrepresentation, and is thus not irreducible.

\end{warnings}

\begin{remark}[on group representations]

Recall that for \(H\leq G\in {\mathsf{Grp}}\) with \([G:H] < \infty\),
one can induce \(H{\hbox{-}}\)representations to
\(G{\hbox{-}}\)representations. If \(\chi\) is a character of \(H\),
then \(\operatorname{Ind}_H^G \chi\) is a character of \(G\). If
\(\chi\) is trivial, its induction will be unlikely to be irreducible
since it contains the 1-dimensional trivial representation.

\end{remark}

\begin{exercise}[?]

Show that if
\(\chi_1/\chi_2 = {\left\lVert {{-}} \right\rVert}_K^{\pm 2}\), then the
naive definition of \(I(\chi_1, \chi_2)\) admits a 1-dimensional
quotient.

\end{exercise}

\begin{remark}

What's really going on with \(I(\chi_1, \chi_2)\): the pair
\((\chi_1, \chi_2)\) is a 1-dim representation \(\psi\) of \(B\), and
we're writing down an explicit model for
\(\operatorname{Ind}_B^{\operatorname{GL}_2} \psi\). Moreover there is a
pairing involving an integral on \(G\) and \(B\) with respect to a Haar
measure:
\begin{align*}
I(\chi_1, \chi_2) \times I(\chi_1^{-1}, \chi_2^{-1}) \to {\mathbb{C}}
.\end{align*}
Where the fudge factor comes from: at some point one changes a left Haar
measure to a right one, and there is a fudge factor of
\({\left\lVert {a/d} \right\rVert}\), so one splits it between the two
terms to yield a duality
\(I(\chi_1, \chi_2) {}^{ \vee }\cong I(\chi_1^{-1}, \chi_2^{-1})\).

\end{remark}

\begin{fact}

\(I(\chi_1, \chi_2)\) is irreducible if
\(\chi_1/\chi_2 \neq {\left\lVert {{-}} \right\rVert}^{\pm 1}\).

\end{fact}

\begin{exercise}[?]

If \(\chi_1/\chi_2 = {\left\lVert {{-}} \right\rVert}^{- 1}\), show that
there is a 1-dimensional representation \(\Psi\) of
\(\operatorname{GL}_2(K)\) given by
\((\chi_1 \cdot {\left\lVert {{-}} \right\rVert}^{1\over 2})\circ \operatorname{det}\)
fitting into a SES
\begin{align*}
0\to \psi \to I(\chi_1, \chi_2) \to S(\chi_1, \chi_2) \to 0
,\end{align*}
where \(S\) is irreducible. If instead
\(\chi_1/\chi_2 = {\left\lVert {{-}} \right\rVert}^{1}\), then there is
a SES
\begin{align*}
0 \to S(\chi_1, \chi_2) \to I(\chi_1, \chi_2)\to (\chi_2 \cdot {\left\lVert {{-}} \right\rVert}^{1\over 2})\circ \operatorname{det}\to 0
.\end{align*}

\end{exercise}

\begin{remark}

We've induced characters to get reps of \(\operatorname{GL}_2(K)\), how
do we get more? An observation due to Weil: if \(K\) is any field,
\begin{align*}
\left\langle{ T \coloneqq{
\begin{bmatrix}
  {t} & {0} 
\\
  {0} & {t^{-1}}
\end{bmatrix}
}, U\coloneqq{
\begin{bmatrix}
  {1} & {u} 
\\
  {0} & {1}
\end{bmatrix}
}, {
\begin{bmatrix}
  {0} & {-1} 
\\
  {1} & {0}
\end{bmatrix}
} {~\mathrel{\Big\vert}~}\cdots}\right\rangle \cong {\operatorname{SL}}_2(K)
,\end{align*}
where there are some explicit relations. So we can cook up
\({\operatorname{SL}}_2(K)\) reps by specifying these generators and
checking that the relations hold. Note that \(\operatorname{GL}_2(K)\)
is an extension of \({\operatorname{SL}}_2(K)\) by an abelian group, so
this can be used to construct \(\operatorname{GL}_2\) reps.

\end{remark}

\begin{remark}

A large source of \({\operatorname{SL}}_2(K)\) and hence
\(\operatorname{GL}_2(K)\) reps: let
\({\operatorname{SL}}_2(K) \curvearrowright L^2(K; {\mathbb{C}})\) by
\begin{align*}
(Tf)(x) &= f(tx) \\
(Uf)(x) &= f(u+x) \\
(Wf)(x) &= \widehat{f}(x)
,\end{align*}
i.e.~\(W\) acts by the Fourier transform and is given by an explicit
integral.

\end{remark}

\begin{fact}

From Jacquet-Langlands: if

\begin{itemize}
\tightlist
\item
  \(L/K\) is a quadratic extension,
\item
  \(\chi: L^{\times}\to{\mathbb{C}}^{\times}\) is admissible,
\item
  \(\chi \neq \chi \circ \sigma\) for any nontrivial
  \(\sigma\in{ \mathsf{Gal}} (L/K)\) (i.e.~\(\chi\) is not its own
  Galois conjugate),
\end{itemize}

Jacquet-Langlands construct an irreducible infinite-dimensional
representation of \(\operatorname{GL}_2(K)\) using \(L^2(L)\) and
Fourier transforms. We'll call this representation
\(\mathrm{BC}_L^K(\psi)\), for \emph{base change}.

\end{fact}

\begin{fact}

We have a collection of infinite-dimensional smooth admissible
irreducible representations of \(\operatorname{GL}_2(K)\).

\begin{itemize}
\tightlist
\item
  \(I(\chi_1, \chi_2) \cong I(\chi_2, \chi_1)\) which is irreducible
  when \(\chi_1/\chi_2 \neq {\left\lVert {{-}} \right\rVert}^{\pm 1}\)
\item
  \(S(\chi, \chi\cdot {\left\lVert {{-}} \right\rVert})\)
\item
  \(\mathrm{BC}_L^K(\psi)\).
\end{itemize}

Moreover, for \(\operatorname{ch}\kappa_K = p > 2\), these exhaust all
such representations. For \(p=2\), there are exceptional
representations. So we know all possibilities for \(\pi\).

\end{fact}

\begin{warnings}

The obvious representations of \(\operatorname{GL}_2(K)\) like the
2-dimensional representation
\(K_{\operatorname{disc}}{ {}^{ \scriptscriptstyle\times^{2} } }\) work,
due to the \(p{\hbox{-}}\)adic topology -- the usual action has
stabilizers which are closed but not open. The only finite-dimensional
representations are of the form \(\chi \circ \operatorname{det}\) for
\(\chi: K^{\times}\to{\mathbb{C}}^{\times}\) a character.

\end{warnings}

\hypertarget{lecture-9}{%
\section{Lecture 9}\label{lecture-9}}

\begin{remark}

Recall that we've seen the following reps of \(\operatorname{GL}_2(K)\)
for \(K\in \mathsf{Field}_{p{\hbox{-}}\text{adic}}\):

\begin{itemize}
\tightlist
\item
  \(I(\chi_1, \chi_2) \subseteq {\mathsf{Grp}}(\operatorname{GL}_2(K), {\mathbb{C}}_{\operatorname{disc}})\)
  with \(\chi_1/\chi_2 \neq {\left\lVert {{-}} \right\rVert}^{\pm 1}\),
\item
  \(S(\chi, \chi\times {\left\lVert {{-}} \right\rVert})\), a
  subset/quotient of \(I\),
\item
  \(\mathrm{BC}_L^K(\psi) \in {\mathsf{Grp}}(L^{\times}, {\mathbb{C}}^{\times})\)
  admissible with \(\psi \neq \psi \circ \sigma\),
\item
  \(\chi \circ \operatorname{det}\), a subset/quotient of \(I\) which is
  1-dimensional.
\end{itemize}

For \(p>2\), these are all of the smooth admissible irreducible reps.

\end{remark}

\begin{exercise}[?]

Prove that \(I\) is smooth and admissible.

\end{exercise}

\hypertarget{conductors}{%
\subsection{Conductors}\label{conductors}}

\begin{remark}

Let \(\pi\) be a smooth admissible irreducible representation of
\(\operatorname{GL}_2(K)\) of infinite dimension. Note that these
definitions generalize to \(G\) any connected reductive algebraic group,
and there's a notion of genericity for \(\pi\) -- it turns out that for
\(\operatorname{GL}_2\), \(\pi\) is generic iff \(\dim \pi = \infty\).

\end{remark}

\begin{theorem}[Casselman]

For \(n\geq 0\), define
\begin{align*}
G \coloneqq U_1({\mathfrak{p}}_K^n) = \left\{{M\coloneqq{
\begin{bmatrix}
  {a} & {b} 
\\
  {c} & {d}
\end{bmatrix}
} \in \operatorname{GL}_2({\mathcal{O}}_K) {~\mathrel{\Big\vert}~}M \equiv {
\begin{bmatrix}
  {*} & {*} 
\\
  {0} & {1}
\end{bmatrix}
} \operatorname{mod}{\mathfrak{p}}^n}\right\} \leq \operatorname{GL}_2(K)
,\end{align*}
which is a local analog of \(\Gamma_1(p^n)\) for modular forms. These
are all compact and open, and therefore by admissibility
\begin{align*}
d(\pi, n) \coloneqq\dim \pi^G < \infty
,\end{align*}
where the RHS denotes the \(G{\hbox{-}}\)invariants. There is a
\emph{conductor} \({\mathfrak{f}}(\pi)\in {\mathbb{Z}}_{\geq 0}\) such
that
\begin{align*}
d(\pi, n) = \max(0, 1+n-{\mathfrak{f}}(\pi)) \qquad n\geq 0
,\end{align*}
i.e.~the sequence is zeros for a fixed number of \(n\) depending on
\(\pi\) and increases linearly beyond that.

\end{theorem}

\begin{warnings}

Conductors are delicate!

\end{warnings}

\begin{remark}

For a complicated representation, one would expect the first few of
these invariants to be zero, since this is asking for invariants under
the action of a maximal compact subgroup in the case \(n=0\) (which
recover \(\operatorname{GL}_2({\mathcal{O}}_K)\)), but the groups get
smaller as \(n\) increases so \(d(\pi, n)\) generally increases.

\end{remark}

\begin{remark}

A glimpse at the proof: for interesting reps of
\(\operatorname{GL}_2(K)\), one can use \emph{Whittaker models} that
realize \(\pi\) as a set of functions on \(K\).

\end{remark}

\begin{exercise}[?]

Compute the conductors for \(I\) and \(S\):

\begin{itemize}
\tightlist
\item
  \({\mathfrak{f}}(I(\chi_1, \chi_2)) = {\mathfrak{f}}(\chi_1) + {\mathfrak{f}}(\chi_2)\)
\item
  \({\mathfrak{f}}(S(\chi, \chi {\left\lVert {} \right\rVert})) = 1\) if
  \({\mathfrak{f}}(\chi) = 0\) (the unramified case) and is
  \(2{\mathfrak{f}}(\chi)\) if \({\mathfrak{f}}(\chi) > 0\) (the
  ramified case).
\end{itemize}

\end{exercise}

\hypertarget{central-characters}{%
\subsection{Central Characters}\label{central-characters}}

\begin{exercise}[?]

Show that Schur's lemma holds: if \(\pi\) is an irreducible admissible
representation of \(\operatorname{GL}_2(K)\) then there is an admissible
character \(\chi_\pi: K^{\times}\to {\mathbb{C}}^{\times}\) such that
for all \(\lambda \in K^{\times}= Z(\operatorname{GL}_n(K))\), using the
diagonal embedding
\begin{align*}
K^{\times}&\to \operatorname{GL}_n(K) \\
\lambda &\mapsto \operatorname{diag}(\lambda, \lambda,\cdots, \lambda)
,\end{align*}
the action is given by
\begin{align*}
\lambda . \pi = \chi(\pi)\lambda
\end{align*}
where \(\chi(\pi)\) is a scalar called the \textbf{central character}.

\end{exercise}

\begin{exercise}[?]

Show that

\begin{itemize}
\tightlist
\item
  \(\chi_{I(\chi_1, \chi_2)} = \chi_1 \chi_2\).
\item
  \(\chi_{S(\chi, \chi {\left\lVert {} \right\rVert})} = \chi_1 \chi_2\).
\item
  \(\chi_{\phi \circ \operatorname{det}} = \phi^2\).
\end{itemize}

\end{exercise}

\hypertarget{local-langlands-for-operatornamegl_2}{%
\subsection{\texorpdfstring{Local Langlands for
\(\operatorname{GL}_2\)}{Local Langlands for \textbackslash operatorname\{GL\}\_2}}\label{local-langlands-for-operatornamegl_2}}

\begin{remark}

We'll assume LLC for \(\operatorname{GL}_1\). Some notation: for
\(\chi_i \in {\mathsf{Grp}}(K^{\times},{\mathbb{C}}^{\times})\), attach
\(\rho_i\in {\mathsf{Grp}}(W_K, {\mathbb{C}}^{\times})\). Note that
\(I(\chi_1,\cdots, \chi_n)\) makes sense for \(\operatorname{GL}_n(K)\),
and the fudge factor here is \(\rho\), the half-sum of positive weights.
There would be a corresponding semisimple Galois representation given by
\(\rho_1\oplus\cdots\rho_n\).

\end{remark}

\begin{remark}

We'll match up infinite dimensional reps of \(\operatorname{GL}_2(K)\)
with 2-dimensional reps of \(G_K\):

\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.68}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.31}}@{}}
\toprule
Weil-Deligne Representations & Galois Representations \\
\midrule
\endhead
\(\pi\) & \(\rho\) \\
\(I(\chi_1, \chi_2)\) & \((\rho_0 = \rho_1 \oplus \rho_2, N=0)\) \\
\(S(\chi_1, \chi_1\cdot {\left\lVert {{-}} \right\rVert})\) &
\(\qty{ \rho_0 = { \begin{bmatrix} {{\left\lVert {{-}} \right\rVert}\rho_2} & {0} \\ {0} & {\rho_1} \end{bmatrix} }, N = { \begin{bmatrix} {0} & {1} \\ {0} & {0} \end{bmatrix} }}\) \\
\(\chi_1 \circ \operatorname{det}\) &
\(\qty{ \rho_0 = { \begin{bmatrix} {\rho_1 \cdot {\left\lVert {{-}} \right\rVert}^{1\over 2}} & {0} \\ {0} & {\rho_1 \cdot {\left\lVert {{-}} \right\rVert}^{-{ 1\over 2} } } \end{bmatrix} }, N=0 }\) \\
\(\mathrm{BC}_L^K(\psi), \psi \in {\mathsf{Grp}}(L^{\times}, {\mathbb{C}}^{\times}) \leadsto_{\text{CFT }} \sigma \in {\mathsf{Grp}}(W_L, {\mathbb{C}}^{\times})\)
& \(\qty{\rho_0 = \operatorname{Ind}_{W_L}^{W_K} \sigma, N=0}\) \\
Conductors \({\mathfrak{f}}(\pi)\) & \({\mathfrak{f}}(\rho_0, N)\) \\
Central characters \(\chi_\pi\) & \(\operatorname{det}\rho_0\) \\
\bottomrule
\end{longtable}

Note that many of these are reducible, and an irreducible \(\pi\) can be
matched with a reducible \(\rho\). In general, irreducible WD reps get
matched with the most complicated \(\pi\) reps, the
\textbf{supercuspidal} \(\rho\) (here \(\mathrm{BC}\)).

For \(p=2\), there are more things on both sides -- on the right, it
turns out there is an \(S_4\) extension of \({\mathbb{Q}}_2\). This
can't happen for \({ {\mathbb{Q}}_p }\) for \(p>2\):

\begin{itemize}
\tightlist
\item
  Wild inertia must be trivial, since it is pro-\(p\),
\item
  Tame inertia must be cyclic
\item
  The unramified extension must also be cyclic
\end{itemize}

So this would force \(S_4\) to have a cyclic subgroup with a cyclic
quotient, which doesn't exist. Thus no extension of a
\(p{\hbox{-}}\)adic field for \(p\geq 3\) can be an \(S_4\) extension.

Using such an \(S_4\), one can find a lift of the following form:

\begin{center}
\begin{tikzcd}
    &&&& {\operatorname{GL}_2({\mathbb{C}})} \\
    \\
    {W_{{\mathbb{Q}}_2}} && {S_4} && {\operatorname{PGL}_2({\mathbb{C}})}
    \arrow[two heads, from=3-1, to=3-3]
    \arrow[hook, from=3-3, to=3-5]
    \arrow["\exists"', dashed, from=3-1, to=1-5]
    \arrow[two heads, from=1-5, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJXX3tcXFFRXzJ9Il0sWzIsMiwiU180Il0sWzQsMiwiXFxQR0xfMihcXENDKSJdLFs0LDAsIlxcR0xfMihcXENDKSJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMywiXFxleGlzdHMiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMywyLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XV0=}{Link
to Diagram}
\end{quote}

The indicated lift exists by a theorem of Tate, and is measured by an
element of \(H^2(W_{{\mathbb{Q}}_2}; {\mathbb{C}}^{\times}) = 1\). Using
this, one gets a 2-dimensional representation
\(W_{{\mathbb{Q}}_2} \to \operatorname{GL}_2({\mathbb{C}})\) which will
be irreducible due to the \(S_4\), and won't be on the list above since
the induced representation will be of generalized dihedral type,
i.e.~the image will have an index 2 normal (and thus abelian) subgroup,
while \(S_4\) has no such subgroup.

\end{remark}

\begin{remark}

A good way of getting irreducible reps is inducing up.

\end{remark}

\begin{remark}

The \(\rho\) side is definitely the less complicated story! So far, we
haven't seen any natural \(\pi\) reps -- the ones we'll build will come
from modular forms.

\end{remark}

\begin{definition}[Conductors and determinants of a Weil-Deligne representation]

The \textbf{conductor} of a Weil-Deligne representation \((\rho_0, N)\)
is defined as
\begin{align*}
{\mathfrak{f}}(\rho_0, N) = {\mathfrak{f}}(\rho_0) + \dim\qty{V^{S} \over (\ker N)^S},\, S \coloneqq I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}
,\end{align*}
where the second term is \(0\) when \(N=0\). The \textbf{determinant} is
the 1-dimensional representation
\begin{align*}
\operatorname{det}(\rho_0) \in {\mathsf{Grp}}(W_K, {\mathbb{C}}^{\times})
.\end{align*}

\end{definition}

\begin{exercise}[A computation]

Check that if the LLC for \(\operatorname{GL}_2\) hold, there are
matchings

\begin{itemize}
\tightlist
\item
  \(\pi \rightleftharpoons(\rho_0, N)\)
\item
  \({\mathfrak{f}}(\pi) \rightleftharpoons f(\rho_0, N)\)
\item
  \(\chi_\pi \rightleftharpoons\operatorname{det}\rho\) by LCFT
\end{itemize}

\end{exercise}

\begin{exercise}[?]

Check that if \(p > 2\), all of the \(F{\hbox{-}}\)semisimple
2-dimensional WD reps of \(W_K\) are listed above. Use Tate's ``Number
Theoretic Background'' 2.2.5.2, which says that all irreducible
representations of \(W_K\) that aren't inductions must have dimension
\(p^k\) for some \(k\).

\end{exercise}

\begin{remark}

Conductors measure to what extent something is ramified. Here
\({\mathfrak{f}}(\pi)\) controls a dimension.

\end{remark}

:::\{.remark ``Title on the unramified case''\} Consider the
\({\mathfrak{f}}(\pi) = 0\) unramified case. This forces

\begin{itemize}
\item
  \(\pi = I(\chi_1, \chi_2)\) (these are called \textbf{principal series
  representations}), and the characters factor:
  \(\chi_i: K^{\times}\to K^{\times}/{\mathcal{O}}_K^{\times}\to {\mathbb{C}}^{\times}\).
  The \(\chi_i\) are 1-dimensional and unramified.
\item
  \(\pi = \operatorname{det}\circ \chi\) where \(\chi\) factors as
  above. On the \(W_K\) side, this forces
  \(\rho = \rho_1 \oplus \rho_2\) and
  \(\rho_i: W_K\to W_K/I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}\)
  and \(N=0\).
\end{itemize}

:::

\begin{remark}

Supposing \(\dim \pi = \infty\) and \({\mathfrak{f}}(\pi) = 0\), the
invariants \(\pi^G\) for \(G=\operatorname{GL}_2({\mathcal{O}}_K)\)
satisfy \(\dim \pi^G = 1\). So from an infinite-dimensional rep, we
isolate a nontrivial 1-dimensional subspace which carries information
about \(\pi\).

This construction works extremely generally: let
\(G \in {\mathsf{Alg}}{\mathsf{Grp}}_{/ {K}}\) be connected reductive
and unramified, e.g.~\(G = \operatorname{GL}_n\), and let \(\pi\) be a
smooth admissible representation of \(G(K)\). We say \(\pi\) \textbf{is
unramified} if there exists a hyperspecial maximal compact subgroup
\(H\leq G(K)\) (e.g.~\(H = \operatorname{GL}_n({\mathcal{O}}_K)\)) with
\(\pi^H\neq 0\).

Note that \(G\) being unramified means there is a model over
\({\mathcal{O}}_K\), so a group scheme over
\(\operatorname{Spec}{\mathcal{O}}_K\), and \(G({\mathcal{O}}_K)\) is a
hyperspecial maximal compact subgroup.

\end{remark}

\begin{remark}

The invariants \(\pi^H\) here will be the finite piece we pick out of an
infinite dimensional representation -- unfortunately it is not
\(\operatorname{GL}_2(K)\) invariant The trick: use Hecke operators. Let
\(G = \operatorname{GL}_2(K)\), or any locally compact totally
disconnected topological group and \(\pi\) be an admissible
representation of \(G\). If \(U, V\leq G\) are compact open subgroups,
e.g.~\(U = U_1({\mathfrak{p}}_K^n)\) or
\(\operatorname{GL}_2({\mathcal{O}}_K)\), and if \(g\in G\) then there
exists a Hecke operator
\begin{align*}
[UgV] \in {\mathsf{{\mathbb{C}}}{\hbox{-}}\mathsf{Mod}}(\pi^V &\to \pi^U)\\
x&\mapsto \sum_{1\leq i\leq r} g_i x
,\end{align*}
where we take a coset decomposition
\(UgV = {\textstyle\coprod}_{1\leq i\leq r} g_i V\), which must be
finite using compactness and that \(V\) is open. This is a type of
averaging process or a trace.

\end{remark}

\begin{exercise}[?]

Show \(UgV \in \pi^U\) and is independent of the choices of \(g_i\).

\end{exercise}

\begin{remark}

Now just consider \(\operatorname{GL}_2(K)\) with
\({\mathfrak{f}}(\pi) = 0\) and
\(U=V=\operatorname{GL}_2({\mathcal{O}}_K)\). Define a map
\begin{align*}
T \coloneqq[UgV]\in {\mathsf{{\mathbb{C}}}{\hbox{-}}\mathsf{Mod}}(\pi^U \to \pi^U), \qquad g \coloneqq{
\begin{bmatrix}
  {\pi_K} & {0} 
\\
  {0} & {1}
\end{bmatrix}
}
S \coloneqq[UhV]\in {\mathsf{{\mathbb{C}}}{\hbox{-}}\mathsf{Mod}}(\pi^U \to \pi^U), \qquad h \coloneqq{
\begin{bmatrix}
  {\pi_K} & {0} 
\\
  {0} & {\pi_K}
\end{bmatrix}
}
.\end{align*}
Since \(\dim \pi^U = 1\), \(T\) and \(S\) act by scalars
\(t,s\in {\mathbb{C}}\).

\end{remark}

\begin{exercise}[Forces unraveling definitions]

If \(\pi = I( \lambda_1, \lambda_2)\),
\(\chi_1/\chi_2 = {\left\lVert {{-}} \right\rVert}^{\pm 1}\), and
\({\mathfrak{f}}(\pi) = 0\), find the values of \(t\) and \(s\).
Possible solutions:

\begin{itemize}
\tightlist
\item
  \(t = \sqrt{q_K} \cdot (\alpha+ \beta)\) where \(q_K\) is the size of
  the residue field
\item
  \(s = \chi_\pi(\pi_K) = \alpha \beta\) where
  \(\alpha = \chi_1(\pi_K), \beta = \chi_2(\pi_K)\).
\end{itemize}

Also do the 1-dimensional unramified case. As a consequence, show that
if \(\pi\) is an admissible irrep of \(\operatorname{GL}_2(K)\) and
\({\mathfrak{f}}(\pi) = 0\), then \(\pi\) is 1-dimensional or
\(\pi = I(\chi_1, \chi_2)\). You can use that \(\mathrm{BC}\) don't have
conductor zero. By LLC, \(\pi\) should match up with \(N=0\),
\(\rho_0: W_K\to W_K/I_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K} \cong {\mathbb{Z}}\),
which contains \(\operatorname{Frob}\), and we can further embed
\({\mathbb{Z}}\hookrightarrow\operatorname{GL}_2({\mathbb{C}})\). Then
\(\rho_0(\operatorname{Frob})\) has characteristic polynomial
\begin{align*}
x^2 - - {t\over \sqrt{q_K}}x + s
.\end{align*}

\end{exercise}

\begin{remark}

What's the point? In the unramified case, we can construct the LLC in a
more conceptual way: isolated a 1-dim space on which Hecke operators
act, choose them carefully and look at their eigenvalues \(t,s\), and
some magic machine produces the punchline above.

\end{remark}

\begin{exercise}[?]

More generally for \(G = \operatorname{GL}_n(K)\), one can take
\(T_i = U \operatorname{diag}(\pi_K, \cdots, \pi_K, 1, \cdots, 1) U\)
with \(i\) copies of \(\pi_K\) where
\(U \coloneqq\operatorname{GL}_n({\mathcal{O}}_K)\). Then the \(T_i\)
eigenvalue is \(t_i\), what is
\({\mathrm{charpoly}}_{\rho_0(\operatorname{Frob})}(x)\)?

\end{exercise}

\begin{remark}

If \(G = G(K)\) with \(K\) unramified with \(\pi\) an unramified
representation of \(G\), the Langlands reinterpretation of the
\textbf{Satake correspondence} associates to \(\pi\) a semisimple
conjugacy class \(C\) in \({}^L G({\mathbb{C}})\). In this case, we'll
have \(\pi \to \rho_0 \cdot \rho_0(\operatorname{Frob}) = C\).

So even though the LLC for a general \(G\) is not well-defined, there is
a conjectural map from \(\pi\) to \(\rho\), and in the simplest case
where everything is unramified, LLC sends an unramified \(\pi\) to an
unramified \(\rho\), which is the same data as specifying the image of
Frobenius, i.e.~a choice of conjugacy class.

\end{remark}

\hypertarget{lecture-10-part-2-global-langlands}{%
\section{Lecture 10: Part 2, Global
Langlands}\label{lecture-10-part-2-global-langlands}}

\begin{remark}

Global means number fields, which are harder than local fields but
perhaps more familiar. From now on,
\(K\in \mathsf{Number}\mathsf{Field}\), so \(K/{\mathbb{Q}}\) is finite.
We'll study the global groups \({ \mathsf{Gal}} (L/K)\) and its relation
to local Galois groups -- taking limits will produce structure on
\(G_K\). There is an analog of the Weil group in the global case, but
it's much more complicated -- similar to how we needed to replace local
Weil groups with WD reps, we'll need to replace global Weil groups with
``global Langlands groups'', which won't quite be defined.

The same machines for producing \(\ell{\hbox{-}}\)adic representations
of \(G_K\) in the local setting will work here:

\begin{itemize}
\tightlist
\item
  \(T_\ell X\) the \(\ell{\hbox{-}}\)adic Tate module of \(X\) an
  elliptic curve or abelian variety,
\item
  \(H_*^\text{ét}(X; { {\mathbb{Q}}_\ell })\) for
  \(X\in {\mathsf{sm}}\mathop{\mathrm{Proj}}{\mathsf{Var}}_{/ {K}}\),
\end{itemize}

yielding a \(\rho\) side. The \(\pi\) side will be \textbf{automorphic
representations}, which we'll define.

\end{remark}

\begin{conjecture}[Global Langlands]

Every automorphic representation of \(\operatorname{GL}_2(K)\)
corresponds to a 2-dimensional representation of the global Langlands
group.

\end{conjecture}

\begin{remark}

Locally, given an \(\ell{\hbox{-}}\)adic representation of \(G_K\), a
proposition of Grothendieck let us construct a WD rep, but we don't get
all such WD reps this way since there was an \(\ell{\hbox{-}}\)adic unit
issue. Here, they'll give us a source of automorphic reps, but perhaps
not all of them. However, reps that are ``algebraic'' (so those with a
good \(p{\hbox{-}}\)adic Hodge theory) should match on both sides.

\end{remark}

\begin{conjecture}

There is a correspondence between automorphic reps \(\pi\) of
\(\operatorname{GL}_n(K)\) and motivic \(n{\hbox{-}}\)dimensional
representations
\(\rho: G_K\to \operatorname{GL}_n(\mkern 1.5mu\overline{\mkern-1.5mu{ {\mathbb{Q}}_\ell }\mkern-1.5mu}\mkern 1.5mu)\)
which are unramified away from a finite set of places which support a
good de Rham and \(p{\hbox{-}}\)adic Hodge theory. Both should come from
motives, and this should be compatible with the LLC in the following
sense: global \(\ell{\hbox{-}}\)adic reps \(\rho\) of \(G_K\) should
restrict to local \(\ell{\hbox{-}}\)adic representations, and global
algebraic automorphic representations \(\pi\) should be related to local
\(\pi\)s.

\end{conjecture}

\begin{remark}

Note that the global Langlands conjecture above is essentially
uncheckable, except perhaps for \(\operatorname{GL}_1\) where it
suffices to understand the abelianization of the global Langlands group
\((?)^{\operatorname{ab}}\cong \dcosetl{K^{\times}}{{\mathbb{A}}_K^{\times}}\).
This can be handled with global class field theory.

\end{remark}

\hypertarget{galois-groups}{%
\subsection{Galois Groups}\label{galois-groups}}

\begin{remark}

Let \(K/{\mathbb{Q}}\) be a finite Galois extension, then
\(K \supseteq{\mathcal{O}}_K\) -- note that when \(K\) is local,
\({\mathcal{O}}_K\) is a DVR and has a unique prime ideal, but e.g.~if
\(K={\mathbb{Q}}\) then \({\mathcal{O}}_K = {\mathbb{Z}}\) has
infinitely many primes. Pick
\({\mathfrak{p}}\in \operatorname{Spec}K \subseteq \operatorname{mSpec}K\)
nonzero, so
\(\kappa({\mathfrak{p}}) \coloneqq{\mathcal{O}}_K/{\mathfrak{p}}\) is a
finite field. Take the \({\mathfrak{p}}{\hbox{-}}\)adic completion of
\(K\):
\begin{align*}
{\mathcal{O}}{ {}_{ \widehat{K, {\mathfrak{p}}} } } \coloneqq\cocolim_n {\mathcal{O}}_K/{\mathfrak{p}}^n, \qquad K{ {}_{ \widehat{{\mathfrak{p}}} } } \coloneqq\operatorname{ff}({\mathcal{O}}_{K, {\mathfrak{p}}})
.\end{align*}
Equivalently, pick \(\lambda \in K^{\times}\) and consider the
fractional ideal \(\lambda {\mathcal{O}}_K\) it generates, which is a
finitely generated \({\mathcal{O}}_K{\hbox{-}}\)submodule of
\({\mathcal{O}}_K\). It factors into principal fractional ideals:
\begin{align*}
\lambda {\mathcal{O}}_K = {\mathfrak{p}}^{v_{\mathfrak{p}}(\lambda)} \times \prod_{{\mathfrak{q}}_i \neq {\mathfrak{p}}}{\mathfrak{q}}_i^{e_i}
.\end{align*}
where \(v_{\mathfrak{p}}: K^{\times}\twoheadrightarrow{\mathbb{Z}}\) and
\({\left\lVert {\lambda} \right\rVert}_{{\mathfrak{p}}} \coloneqq{\varepsilon}^{-v_{\mathfrak{p}}( \lambda)}\)
where
\({\varepsilon}\coloneqq q_{\mathfrak{p}}\coloneqq{\sharp}\kappa({\mathfrak{p}})\).
This norm induces a metric
\(d(x,y) \coloneqq{\left\lVert {x-y} \right\rVert}_{\mathfrak{p}}\), and
one can take the Cauchy completion to define the local field
\(K{ {}_{ \widehat{{\mathfrak{p}}} } }\). Note that
\(K{ {}_{ \widehat{{\mathfrak{p}}} } }/{ {\mathbb{Q}}_p }\) is a finite
extension where
\({\mathfrak{p}}\cap{\mathbb{Z}}=\left\langle{p}\right\rangle\).

\end{remark}

\begin{remark}

Let \(L/K\) be a finite Galois extension of number fields and
\({\mathfrak{p}}\in \operatorname{Spec}{\mathcal{O}}_K\) nonzero as
above. Then \({ \mathsf{Gal}} (L/K) \in{\mathsf{Fin}}{\mathsf{Grp}}\) is
the global Galois group we'll study. We can then extend
\({\mathfrak{p}}\):

\begin{center}
\begin{tikzcd}
    L && {{\mathcal{O}}_L} && {{\mathfrak{p}}{\mathcal{O}}_L} \\
    \\
    K && {{\mathcal{O}}_K} && {\mathfrak{p}}
    \arrow[hook', from=3-1, to=1-1]
    \arrow[hook', from=3-3, to=1-3]
    \arrow[dashed, from=3-5, to=1-5]
    \arrow[hook', from=3-3, to=3-1]
    \arrow[hook', from=1-3, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\begin{warnings}

The ideal \({\mathfrak{p}}{\mathcal{O}}_L \in \operatorname{Id}(L)\)
need not be prime! E.g. take
\(L/K = {\mathbb{Q}}(\sqrt p)/{\mathbb{Q}}\), then \(p\) becomes a
square in \(L\).

\end{warnings}

We can factor
\begin{align*}
{\mathfrak{p}}{\mathcal{O}}_L = \prod_{1\leq i\leq g} P_i^{e_i}, \qquad P_i\in \operatorname{Spec}{\mathcal{O}}_L
.\end{align*}

Since \({ \mathsf{Gal}} (L/K)\curvearrowright L\), it acts on
\({\mathcal{O}}_L\) and fixes
\({\mathfrak{p}}\subseteq {\mathcal{O}}_K \subseteq K\). This action
fixes \({\mathfrak{p}}{\mathcal{O}}_L\) as a set and permutes its
elements, and permutes the \(P_i\in \operatorname{Spec}{\mathcal{O}}_L\)
appearing in the factorization above since \(\sigma(P_i)\) is again
prime and divides \({\mathfrak{p}}\). This is clearly true by transport
of structure, since \(\sigma: L\to L\) is an isomorphism of fields.

\end{remark}

\begin{fact}

Galois acts transitively, so there is only one orbit and thus all of the
\(e_i\) are preserved by \(\sigma\) and all of the \(P_i\) are
isomorphic to \(\sigma(P_i)\). As a corollary, all of the completions
are isomorphic:
\begin{align*}
L{ {}_{ \widehat{P_1} } } \cong L{ {}_{ \widehat{P_2} } } \cong \cdots L{ {}_{ \widehat{P_g} } }
.\end{align*}

\end{fact}

\begin{remark}

Let \(P \coloneqq P_1\in \operatorname{Spec}L\) be a fixed choice of a
prime above \({\mathfrak{p}}\). Note that if
\({\mathcal{P}}\coloneqq\left\{{P_1, P_2,\cdots, P_g}\right\}\) then by
Orbit-Stabilizer,
\begin{align*}
{ \mathsf{Gal}} (L/K)/ {\operatorname{Stab}}(P_1) { { \, \xrightarrow{\sim}\, }}{\mathrm{Orb}}({ \mathsf{Gal}} (L/K) \curvearrowright{\mathcal{P}}) = {\mathcal{P}}
.\end{align*}
Define the \textbf{decomposition group}
\begin{align*}
D_P \coloneqq\left\{{\sigma\in { \mathsf{Gal}} (L/K) {~\mathrel{\Big\vert}~}\sigma(P) = P}\right\} \implies { \mathsf{Gal}} (L/K)/D_P \cong {\mathcal{P}}
.\end{align*}
Note that this need not be a normal subgroup. Now \(\sigma: L\to L\)
descends to a map
\(\sigma: ({\mathcal{O}}_L, P) \to ({\mathcal{O}}_L, P)\) and thus to
the completions
\(\sigma: L{ {}_{ \widehat{P} } } \to L{ {}_{ \widehat{P} } }\) by
transport of structure.

\end{remark}

\begin{fact}

\(L{ {}_{ \widehat{P} } } / \kappa({\mathfrak{p}})\) is Galois and
\(D_P \cong { \mathsf{Gal}} (L{ {}_{ \widehat{P} } }/\kappa({\mathfrak{p}}))\),
which is the local Galois group, and
\(D_P \hookrightarrow{ \mathsf{Gal}} (L/K)\).

\end{fact}

\begin{remark}

So given \({ \mathsf{Gal}} (L/K)\), choose
\({\mathfrak{p}}\in \operatorname{Spec}{\mathcal{O}}_K\) and
\(P\divides {\mathfrak{p}}{\mathcal{O}}_L\) with
\(P\in \operatorname{Spec}{\mathcal{O}}_L\) to produce
\(D_P \subseteq { \mathsf{Gal}} (L/K)\) with
\(D_P\cong { \mathsf{Gal}} (L{ {}_{ \widehat{P} } }/\kappa({\mathfrak{p}}))\).
Note that there is an inertia subgroup
\(I \subseteq { \mathsf{Gal}} (L{ {}_{ \widehat{P} } }/\kappa({\mathfrak{p}}))\),
and the quotient by \(I\) is generated by \(\operatorname{Frob}\). It
turns out that \(I\) is almost always trivial:
\begin{align*}
{\mathfrak{p}}\notdivides {\operatorname{disc}}(L/K) \implies I = 1
,\end{align*}
so \(L{ {}_{ \widehat{P} } }\) is an unramified extension of
\(\kappa({\mathfrak{p}})\) and there is a distinguished element
\(\operatorname{Frob}_P \in D_P \subseteq { \mathsf{Gal}} (L/K)\). Note
that changing \(L\) might change \(\operatorname{Frob}_P\), so this
depends on \({\mathfrak{p}}\), the extension \(L\), and
\(P\divides {\mathfrak{p}}{\mathcal{O}}_L\). Consider choosing a
different \(P'\): since Galois acts transitively, there is some
\(\sigma\in { \mathsf{Gal}} (L/K)\) with \(\sigma(P) = P'\) and thus
\begin{align*}
D_{P'} = \sigma D_P \sigma^{-1}, \qquad
\operatorname{Frob}_{P'} = \sigma \operatorname{Frob}_P \sigma^{-1}
\end{align*}
by transport of structure. So this yields a well-defined \emph{conjugacy
class}
\begin{align*}
\operatorname{Frob}_{\mathfrak{p}}\coloneqq\left\{{\operatorname{Frob}_{P'} {~\mathrel{\Big\vert}~}P'\divides {\mathfrak{p}}{\mathcal{O}}_L}\right\}
,\end{align*}
which works for all
\({\mathfrak{p}}\notdivides {\operatorname{disc}}(L/K)\).

\end{remark}

\hypertarget{lecture-11}{%
\section{Lecture 11}\label{lecture-11}}

\begin{quote}
Look up Fontaine's \(B_{\mathrm{dR}}\)!
\end{quote}

\hypertarget{the-frobenius-machine}{%
\subsection{The Frobenius Machine}\label{the-frobenius-machine}}

\begin{remark}

Last time: global Galois groups, i.e.~Galois groups of number fields.
The setup:

\begin{itemize}
\tightlist
\item
  \(L/K\) finite Galois global,
\item
  \({\mathfrak{p}}\in \operatorname{Spec}{\mathcal{O}}_K\) a nonzero
  prime,
\item
  \({\mathfrak{p}}{\mathcal{O}}_L\in \operatorname{Id}({\mathcal{O}}_L)\),
  probably not in \(\operatorname{Spec}{\mathcal{O}}_L\),
\item
  A fixed prime \(P\divides {\mathfrak{p}}\) and a factorization
  \begin{align*}
  {\mathfrak{p}}{\mathcal{O}}_L = P^e \times \prod_{{\mathfrak{q}}\divides {\mathfrak{p}}}{\mathfrak{q}}_i^{e_i}
  .\end{align*}
\item
  The decomposition group.
  \begin{align*}
  G(L/K) \supseteq D_{P/{\mathfrak{p}}} \coloneqq\left\{{\sigma \in G(L/K) {~\mathrel{\Big\vert}~}\sigma P = P}\right\}
  .\end{align*}
\end{itemize}

\end{remark}

\begin{fact}

If \(\sigma\in D_{P/{\mathfrak{p}}}\), then \(\sigma: L\to L\) descends
to \(\sigma: {\mathcal{O}}_L \to {\mathcal{O}}_L\) restricts to the
identity on \(K\), \({\mathcal{O}}_L\), and \({\mathfrak{p}}\). Moreover
\(\sigma(P) = P\), so there is an induced map
\({\mathcal{O}}_L/{\mathfrak{p}}^n{\circlearrowleft}\) and thus a map on
the completion \(\sigma: L{ {}_{ \widehat{P} } } {\circlearrowleft}\)
fixing \(K{ {}_{ \widehat{{\mathfrak{p}}} } }\), yielding an element
\begin{align*}
\sigma\in G( L{ {}_{ \widehat{P} } }, K{ {}_{ \widehat{{\mathfrak{p}}} } } )
= D_{P/{\mathfrak{p}}} \supseteq I_{P/{\mathfrak{p}}} 
= I_{L{ {}_{ \widehat{P} } } / K{ {}_{ \widehat{{\mathfrak{p}}} } }} 
.\end{align*}
So the global group contains the local group.

For global fields, we have a discriminant
\(\Delta = {\operatorname{disc}}(L/K) \in \operatorname{Id}({\mathcal{O}}_K)\),
and \(p\notdivides \Delta\) then \(I_{P/{\mathfrak{p}}} = 1\) for all
\(P\divides {\mathfrak{p}}\), making \({\mathfrak{p}}\) unramified. Note
that changing \(P\) yields isomorphic inertia groups. Common situation:
things in NT tend to be unramified outside a finite set of primes \(S\);
here
\(S = \left\{{p\in \operatorname{Spec}{\mathcal{O}}_K {~\mathrel{\Big\vert}~}p\notdivides \Delta}\right\}\).
For all \(P\divides {\mathfrak{p}}\), we get a cyclic group
\begin{align*}
D_{P/{\mathfrak{p}}}/I_{P/{\mathfrak{p}}} = D_{P/{\mathfrak{p}}} = \left\langle{\operatorname{Frob}_P}\right\rangle = G(\kappa(P)/ \kappa({\mathfrak{p}}))
\end{align*}
since this is a local Galois group where the inertia is trivial. So each
\(P\) yields an element \(\operatorname{Frob}_P\in G(L/K)\), and varying
\(P\) yields a conjugacy class
\(\operatorname{Frob}_{\mathfrak{p}}\coloneqq\left\{{\operatorname{Frob}_P {~\mathrel{\Big\vert}~}P\divides {\mathfrak{p}}}\right\}\).

\end{fact}

\begin{theorem}[Chebotarev]

Every conjugacy class \(C \subseteq G(L/K)\) is of the form
\(C = \operatorname{Frob}_{\mathfrak{p}}\) for infinitely many
\({\mathfrak{p}}\in \operatorname{Spec}{\mathcal{O}}_K\), which is a
weak form of \textbf{Chebotarev density}. Moreover the density of such
\({\mathfrak{p}}\) is exactly \({\sharp}C/{\sharp}G(L/K)\).

\end{theorem}

\begin{remark}

A useful consequence: it suffices to specify representations on
\(\operatorname{Frob}_{\mathfrak{p}}\) for all \({\mathfrak{p}}\). There
is a variant for infinite extensions, which we'll always assume to be
cofinitely unramified. Fix an algebraic closure
\(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu \coloneqq{ \operatorname{cl}} _\mathsf{Alg}(K)/K\);
then \(G_K\) is ramified at every prime
\({\mathfrak{p}}\in\operatorname{Spec}K\), so the reduction to
\(\operatorname{Frob}\) above no longer works.

Let \(S \subseteq \operatorname{mSpec}{\mathcal{O}}_K\) be a finite set
of primes, we'll show that \(\operatorname{Frob}_{\mathfrak{p}}\) makes
sense for \({\mathfrak{p}}\not\in S\). Suppose
\(K \subseteq L_1, L_2 \subseteq K\) with \(L_i/K\) finite and
unramified outside of \(S\), then the compositum \(L_1L_2\) is again
unramified outside of \(S\). So define
\begin{align*}
K_S \coloneqq\displaystyle\bigcup_{L\in \tilde S} L, \qquad \tilde S \coloneqq\left\{{ {L/K {~\mathrel{\Big\vert}~}L \text{ is finite Galois unramified outside of } S } }\right\}
.\end{align*}
Note that every number field other than \({\mathbb{Q}}\) is ramified at
some prime.

\end{remark}

\begin{question}

Consider \(K={\mathbb{Q}}\) and let \(S=\left\{{p}\right\}\) be a fixed
prime. What is \(K^S\)? Note that
\({\mathbb{Q}}(\zeta_{p^n}) \subseteq K^S\) for all \(n\geq 1\).

\end{question}

\begin{example}[?]

Like \(K={\mathbb{Q}}\) and fix \(N\in {\mathbb{Z}}_{\geq 1}\), and take
\(S = \left\{{p\in \operatorname{Spec}{\mathbb{Z}}{~\mathrel{\Big\vert}~}p\divides N}\right\}\).
Then \({\mathbb{Q}}(\zeta_N)/{\mathbb{Q}}\) is unramified outside of
\(S\) and has Galois group \(C_N^{\times}\). Thus if
\(p\in {\mathbb{Z}}\) is prime and \(p\not\in S\), so
\(p\notdivides N\), there is a canonical conjugacy class
\(\operatorname{Frob}_p \in C_N^{\times}\) -- since this group is
abelian, this is in fact an element. It could be \(p\) or \(p^{-1}\),
and one can work out that it must be \(p\).

\end{example}

\begin{corollary}[?]

There are infinitely many primes in any arithmetic progression.

\end{corollary}

\begin{remark}

Fixing \(K={\mathbb{Q}}\) and a model of
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\),
for \(S = \left\{{p}\right\}\) one has
\(K^S\supseteq{\mathbb{Q}}(\zeta_{p^n})\) for all \(n\) and thus their
union
\begin{align*}
{\mathbb{Q}}(\zeta_{p^\infty}) \coloneqq\displaystyle\bigcup_{n\geq 1} {\mathbb{Q}}(\zeta_{p^n}) \subseteq { \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }
.\end{align*}
. Therefore there is a surjection
\begin{align*}
G(K^S/K) 
&\twoheadrightarrow G({\mathbb{Q}}(\zeta_{p^\infty})/{\mathbb{Q}})\\
&{ { \, \xrightarrow{\sim}\, }}\cocolim_n G({\mathbb{Q}}(\zeta_{p^n})/{\mathbb{Q}})\\
&{ \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }\cocolim_n C_n^{\times}\\
&{ { \, \xrightarrow{\sim}\, }}{ {\mathbb{Z}}_p }^{\times}
.\end{align*}

We can get Frobenius elements in such infinite extensions: if
\(r\neq p\) is prime, since the groups are abelian we have elements
\(\operatorname{Frob}_r\in C_{p^n}^{\times}= G({\mathbb{Q}}(\zeta_{p^n})/{\mathbb{Q}})\)
which gets identified with \(r\) as a residue class. Taking the inverse
limit yields a Frobenius element
\(\operatorname{Frob}_r \in { {\mathbb{Z}}_p }^{\times}\).

\end{remark}

\begin{remark}

Upshot: for
\(K \in \mathsf{Number}\mathsf{Field}, S\subseteq {\operatorname{Places}}(K)\)
finite,
\begin{align*}
G(K^S/K) = \cocolim \left\{{ G(L/K) {~\mathrel{\Big\vert}~}L/K \text{ finite, Galois, unramified outside of } S }\right\}
\end{align*}
and for all \({\mathfrak{p}}\not\in S\), we get conjugacy classes
\(\operatorname{Frob}_{{\mathfrak{p}}, L/K} \subseteq G(L/K)\) which
glue in the limit to a conjugacy class
\begin{align*}
\operatorname{Frob}_{{\mathfrak{p}}} = \operatorname{Frob}_{{\mathfrak{p}}, K^S/K} \subseteq G(K^S/K)
.\end{align*}
So we have names for conjugacy classes in this group, although not for
elements. By Chebotarev density, the following map is surjective
\begin{align*}
\left\{{{\mathfrak{p}}\in \operatorname{mSpec}K {~\mathrel{\Big\vert}~}{\mathfrak{p}}\not\in S}\right\} &\twoheadrightarrow G(L/K)/\mathop{\mathrm{Inn}}G(L/K) \\
{\mathfrak{p}}&\mapsto \operatorname{Frob}_{\mathfrak{p}}
,\end{align*}
where \(G\curvearrowright G\) by inner automorphisms, and the RHS
denotes conjugacy classes.

\end{remark}

\begin{corollary}[Density of Frobenii]

If \(L/K\) is infinite Galois and unramified outside of
\(S \subseteq {\operatorname{Places}}(K)\), then
\begin{align*}
\left\{{\operatorname{Frob}_{\mathfrak{p}}{~\mathrel{\Big\vert}~}{\mathfrak{p}}\not\in S}\right\} { \underset{ {\scriptscriptstyle\mathrm{dense}} }{\hookrightarrow} }G(L/K)
.\end{align*}

\end{corollary}

\begin{corollary}[?]

If \(F: G(L/K)\to X\) is continuous and factors through
\(G(L/K)/\mathop{\mathrm{Inn}}G(L/K)\), so is constant on conjugacy
classes, it may be possible to recover \(F\) from
\(\left\{{ F(\operatorname{Frob}{\mathfrak{p}}) {~\mathrel{\Big\vert}~}{\mathfrak{p}}\in \operatorname{mSpec}K\setminus S}\right\}\).
Upshot: if \(F\) is the characteristic polynomial of a representation,
then it may suffice to know the characteristic polynomials of Frobenius.

\end{corollary}

\hypertarget{brauer-nesbitt}{%
\subsection{Brauer-Nesbitt}\label{brauer-nesbitt}}

\begin{theorem}[Brauer-Nesbitt]

Let \(G\in {\mathsf{Grp}}\) and \(E\in \mathsf{Field}\) be arbitrary.
Recall that \(\rho: G\to \operatorname{GL}_n(E)\) is semisimple iff
\(\rho\) is a direct sum of irreducible representations. Note that
irreducibles and direct sums of irreducible are semisimple. If
\(\rho_1, \rho_2: G\to \operatorname{GL}_n(E)\) are two semisimple
representations, then
\begin{align*}
{\mathrm{charpoly}}\rho_1(g) = {\mathrm{charpoly}}\rho_2(g)\, \forall g\in G \implies \rho_1 \cong \rho_2
.\end{align*}

\end{theorem}

\begin{warnings}

If one replaces \(GL_n(E)\) with an arbitrary group and asks for the
\(\rho_i\) just to be conjugate, the analogous theorem will no longer be
true. So automorphic representations may not be determined by their
local components.

\end{warnings}

\begin{remark}

If \(\operatorname{ch}K = 0\), then
\({\mathrm{tr}}\rho_1 = {\mathrm{tr}}\rho_2 \implies \rho_1 \cong \rho_2\)
for semisimple \(\rho\), so one doesn't need the entire characteristic
polynomial -- one can compute the entire polynomial from \(\rho(g^i)\)
for \(1\leq i \leq n\), but this uses division by \(n!\).

\end{remark}

\begin{exercise}[Non-example]

Let \(G = C_3\),
\(E = \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_2\),
and \(n=2\). Find non-isomorphic \(\rho_i\) semisimple and reducible
with identical traces. The issue will be the \(2!\) is not invertible in
\(E\).

\end{exercise}

\begin{remark}

Upshot: if
\(\rho\in {\mathsf{Top}}{\mathsf{Grp}}(G(K^S/K)\to \operatorname{GL}_n(E))\)
is a continuous semisimple irrep, say where \(E/{ {\mathbb{Q}}_p }\),
and one knows
\(F_{\mathfrak{p}}\coloneqq{\mathrm{charpoly}}\rho(\operatorname{Frob}_{\mathfrak{p}})\in E[x]\)
for all \({\mathfrak{p}}\not\in S\), then this determines \(\rho\)
uniquely. This is because the function
\(g\mapsto {\mathrm{charpoly}}{\mathrm{charpoly}}\rho(g)\) will be
continuous, then Brauer-Nesbitt says this determines \(\rho\). If we
know \({\mathrm{charpoly}}\operatorname{Frob}_{\mathfrak{p}}\) (as is
often the case in NT), by continuity there will be at most one way to
extend this to a map, so at most one \(\rho\).

\end{remark}

\begin{example}[?]

Let \(K = {\mathbb{Q}}\) and \(S = \left\{{p}\right\}\) for a fixed
prime. Let \(L = {\mathbb{Q}}(\zeta_{p^\infty})\), then
\begin{align*}
G(L/K) = \cocolim_n C_{p^n}^{\times}= { {\mathbb{Z}}_p }^{\times}= \operatorname{GL}_1({ {\mathbb{Z}}_p }) \subseteq \operatorname{GL}_1({ {\mathbb{Q}}_p })
.\end{align*}
We can then define a representation
\begin{align*}
\rho: G_{{\mathbb{Q}}} \twoheadrightarrow G({\mathbb{Q}}(\zeta_{p^\infty})/{\mathbb{Q}}) = { {\mathbb{Z}}_p }^{\times}\to \operatorname{GL}_1({ {\mathbb{Q}}_p })
.\end{align*}

Note that this factors:

\begin{center}
\begin{tikzcd}
    & {G({\mathbb{Q}}^S/{\mathbb{Q}})} \\
    \\
    {G({ \mkern 1.5mu\overline{\mkern-1.5mu \mathbb{Q} \mkern-1.5mu}\mkern 1.5mu }/{\mathbb{Q}})} && {G({\mathbb{Q}}(\zeta_{p^\infty})/{\mathbb{Q}})} && { {\mathbb{Z}}_p }^{\times}&& {\operatorname{GL}_1({ {\mathbb{Q}}_p })}
    \arrow[two heads, from=3-1, to=3-3]
    \arrow[two heads, from=1-2, to=3-3]
    \arrow[from=3-1, to=1-2]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=3-7]
    \arrow["\rho"', curve={height=30pt}, dashed, from=3-1, to=3-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwyLCJHKFxcUVFiYXIvXFxRUSkiXSxbMiwyLCJHKFxcUVEoXFx6ZXRhX3twXlxcaW5mdHl9KS9cXFFRKSJdLFs0LDIsIlxcWlpwYWRpY1xcdW5pdHMiXSxbNiwyLCJcXEdMXzEoXFxRUXBhZGljKSJdLFsxLDAsIkcoXFxRUV5TL1xcUVEpIl0sWzAsMSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzQsMSwiIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzAsNF0sWzEsMl0sWzIsM10sWzAsMywiXFxyaG8iLDIseyJjdXJ2ZSI6NSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to diagram}
\end{quote}

So there is a map
\(G_{{\mathbb{Q}}} \to G({\mathbb{Q}}^S/{\mathbb{Q}})\), which is still
complicated but contains conjugacy classes \(\operatorname{Frob}_r\) for
\(r\not\in S\). This composite is the \(p{\hbox{-}}\)adic
\textbf{cyclotomic character}, which we'll notate \(\omega_p\) (noting
that it's often notated \(\chi_p\)). By Brauer-Nesbitt and Chebotarev,
\(\rho\) is determined by the fact that
\(\rho(\operatorname{Frob}_r) = r\) for all \(r\neq p\).

\end{example}

\begin{warnings}

A confusing fact: let \(p, \ell\) be two different primes and set
\(S= \left\{{p, \ell}\right\}\). We can then produce two Galois
representations that factor through \(G({\mathbb{Q}}^S/{\mathbb{Q}})\):

\begin{center}
\begin{tikzcd}
    && {} \\
    &&&& {G({\mathbb{Q}}(\zeta_{p^\infty})/{\mathbb{Q}})} && {\operatorname{GL}_1({ {\mathbb{Z}}_p }) = { {\mathbb{Z}}_p }^{\times}} \\
    {G_{\mathbb{Q}}} && {G({\mathbb{Q}}^S/{\mathbb{Q}})} \\
    &&&& {G({\mathbb{Q}}(\zeta_{\ell^\infty})/{\mathbb{Q}})} && {\operatorname{GL}_1({ {\mathbb{Z}}_\ell }) = { {\mathbb{Z}}_\ell }^{\times}}
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=2-5]
    \arrow[from=3-3, to=4-5]
    \arrow[from=4-5, to=4-7]
    \arrow[from=2-5, to=2-7]
    \arrow["{\omega_p}", curve={height=-30pt}, dashed, from=3-1, to=2-7]
    \arrow["{\omega_\ell}"', curve={height=30pt}, dashed, from=3-1, to=4-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Note that
\(\omega_p(\operatorname{Frob}_r) = \omega_\ell(\operatorname{Frob}_r) = r\)
for all \(r\not\in S\) and
\begin{align*}
\left\{{\operatorname{Frob}_r {~\mathrel{\Big\vert}~}r\not\in S}\right\} { \underset{ {\scriptscriptstyle\mathrm{dense}} }{\hookrightarrow} }G({\mathbb{Q}}^S/{\mathbb{Q}})
,\end{align*}
so why aren't these representations equal since they have the same
traces? The catch is that Brauer-Nesbitt doesn't apply, since it
requires to representations to \(\operatorname{GL}_n(E)\) where \(E\) is
a single, fixed field. In fact, these two reps are extremely different:
\(\ker \omega_p = {\mathbb{Q}}(\zeta_{p^\infty})\) which is unramified
away from \(p\) and totally ramified at \(p\), while
\(\ker \omega_\ell = {\mathbb{Q}}(\zeta_{\ell^\infty})\). Moreover these
are two completely disjoint extensions of \({\mathbb{Q}}\), since
anything in both fields generates an extension that's unramified away
from \(p\) and away from \(\ell\), and thus unramified everywhere and in
\({\mathbb{Q}}\).

\end{warnings}

\hypertarget{lecture-12-ellhbox-adic-representations}{%
\section{\texorpdfstring{Lecture 12, \(\ell{\hbox{-}}\)adic
Representations}{Lecture 12, \textbackslash ell\{\textbackslash hbox\{-\}\}adic Representations}}\label{lecture-12-ellhbox-adic-representations}}

\hypertarget{families-of-ellhbox-adic-reps}{%
\subsection{\texorpdfstring{Families of \(\ell{\hbox{-}}\)adic
reps}{Families of \textbackslash ell\{\textbackslash hbox\{-\}\}adic reps}}\label{families-of-ellhbox-adic-reps}}

\begin{remark}

See Néron-Shafarevich criterion, Tate modules of AVs.

\end{remark}

\begin{example}[?]

Let \(K\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\), say \(K={\mathbb{Q}}\),
and let \(E/K\) be an elliptic curve. Let \(S_0\) be a finite set of
places of \(K\), so points in \(\operatorname{mSpec}{\mathcal{O}}_K\),
where \(E\) has bad reduction. For \(\ell\) a prime, consider the
\(\ell{\hbox{-}}\)adic Tate module: since
\(E[\ell^n](\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu)\)
receives an action of \(G_K\), setting
\(T_\ell E \coloneqq\cocolim_n E[\ell^n]\) produces a representation
\begin{align*}
\rho_{E, \ell}: G_K\to \operatorname{GL}_2({ {\mathbb{Z}}_\ell })
,\end{align*}
which is well-defined up to conjugation that factors through
\(G(K^{\tilde S_0}/K)\) where
\(\tilde S_0 \coloneqq S_0 \cup\left\{{p{~\mathrel{\Big\vert}~}p\divides \ell}\right\}\).

If \(p\not\in S_0\) and \(p\notdivides \ell\),
\begin{align*}
{\mathrm{charpoly}}\rho_{E, \ell} \operatorname{Frob}_p = x^2 - a_p x - \operatorname{Nm}(p) \in {\mathbb{Q}}[x] \hookrightarrow{ {\mathbb{Q}}_\ell }[x], \qquad a_p \coloneqq 1 + \operatorname{Nm}(p) - {\sharp}E(\kappa(p))
.\end{align*}
Note that this no longer depends on \(\ell\), so
\begin{align*}
\operatorname{Tr}\rho_{E, \ell} \operatorname{Frob}_p = a_p
,\end{align*}
independent of \(\ell\). However,
\(\rho_{E, \ell} \not\cong \rho_{E. p}\) for \(\ell\neq p\), since
\(\rho_{E, \ell}\) is infinitely ramified at \(\ell\) and has wild
inertia, while \(\rho_{E, p}\) will have finite wild inertia at
\(\ell\).

\end{example}

\begin{definition}[$\ell\dash$adic representations of absolute Galois groups]

Setup:

\begin{itemize}
\tightlist
\item
  \(K\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\),
\item
  \(E/{ {\mathbb{Q}}_\ell }\) a finite extension,
\item
  \(S\) a finite set of places in
  \(\operatorname{mSpec}{\mathcal{O}}_K\)
\end{itemize}

Then a morphism of topological groups
\begin{align*}
\rho\in {\mathsf{Top}}{\mathsf{Grp}}(G(K^S/K) \to \operatorname{GL}_n(E))
\end{align*}
where the LHS has the profinite topology and the RHS has the
\(\ell{\hbox{-}}\)adic topology is an \textbf{\(\ell{\hbox{-}}\)adic
representation} of \(G_K\). In this situation we say \(\rho\) is
\emph{unramified outside of \(S\)}.

\end{definition}

\begin{remark}

Note that we regard this as \(G_K\) representation despite having
\(G(K^S/K)\) as the source, since we implicitly require reps to be
unramified away from \(S\) since \(G_K\twoheadrightarrow G(K^S/K)\).

\end{remark}

\begin{definition}[Rational representations]

We say \(\rho\) is \textbf{rational over \(E_0\)} for
\(E_0 \subseteq E\) if
\({\mathrm{charpoly}}\rho \operatorname{Frob}_p \in E[x]\) in fact lies
in \(E_0[x]\).

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  For \(\rho\) the cyclotomic character
  \(\rho: G_{\mathbb{Q}}\to \operatorname{GL}_1({ {\mathbb{Q}}_\ell })\)
  satisfies \(\rho \operatorname{Frob}_r = r\) for all \(r\neq \ell\)
  and is thus rational over \({\mathbb{Q}}\).
\item
  \(T_\ell E\) for \(E\) an elliptic curve is rational over
  \({\mathbb{Q}}\).
\item
  \(\rho = H^1_\text{ét}(X_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}; { {\mathbb{Q}}_\ell })\)
  the \(\ell{\hbox{-}}\)adic etale cohomology of
  \(X\in {\mathsf{sm}}{\mathsf{Alg}}{\mathsf{Var}}_{/ {K}}\) with \(X\)
  proper is rational over \({\mathbb{Q}}\).
\end{itemize}

\end{example}

\begin{definition}[Pure reps and their weights]

The representation \(\rho\) is \textbf{pure of weight \(w\)} iff
\(\rho\) is rational over some
\(E_0 \in \mathsf{Field}_{/ {{\mathbb{Q}}}}\) and for all embeddings
\(i\in \mathsf{Field}(E_0, {\mathbb{C}})\) and for all eigenvalues
\(\alpha\) of \(\rho\operatorname{Frob}_p\), the magnitude satisfies
\begin{align*}
{\left\lvert {i(\alpha)} \right\rvert} = q_p^{-{w\over 2}}, \qquad q_p \coloneqq{\sharp}{\mathcal{O}}_K/p
.\end{align*}

\end{definition}

\begin{theorem}[Deligne]

For \(X\in {\mathsf{sm}}\mathop{\mathrm{proj}}{\mathsf{Var}}_{/ {K}}\)
proper,
\(H^i(X_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}; { {\mathbb{Q}}_\ell })\)
is pure of weight \(i\).

\end{theorem}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  \(H^2({\mathbb{P}}^1_{/ {K}} ; { {\mathbb{Q}}_\ell }) = \omega_\ell^{-1}\)
  is the inverse of the cyclotomic character, making it pure of weight
  \(-2\).
\item
  \(T_\ell E\) is pure of weight -1, so the roots of
  \(x^2-a_px + \operatorname{Nm}(p)\) are complex conjugates, which
  yields the Hasse bound
  \begin{align*}
  {\left\lvert {a_p} \right\rvert} \leq 2\sqrt{\operatorname{Nm}(p)}
  .\end{align*}
\end{itemize}

\end{example}

\begin{remark}

We'll now let \(\ell\) vary, and formalize the notion that the
cyclotomic characters for \(\ell\neq p\) are distinguished.

\end{remark}

\begin{definition}[Compatible Systems]

Setup:

\begin{itemize}
\tightlist
\item
  \(K\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\),
\item
  \(E_0\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\) (e.g.~\({\mathbb{Q}}\)),
\item
  \(S_0 \subseteq {\operatorname{Places}}(K)\) a finite set of places,
\item
  For all \(p\not\in S_0\), a polynomial \(F_p(x)\in E_0[x]\)
  (e.g.~primes of good reduction and
  \(F_p(x) = x-\operatorname{Nm}(p)\)),
\item
  For all finite places \(\lambda \in {\mathcal{O}}_{E_0}\), an
  \(\ell{\hbox{-}}\)adic representation
  \begin{align*}
  \rho_\lambda: G(K^{\tilde S_0}/K) \to \operatorname{GL}_n\qty{ { \operatorname{cl}} _\mathsf{Alg}\qty{ (E_0){ {}_{ \widehat{\lambda} } }} }, \qquad \tilde S_0 \coloneqq S_0 \cup\left\{{p\in {\operatorname{Places}}(K) {~\mathrel{\Big\vert}~}p\divides \ell}\right\}
  .\end{align*}
\end{itemize}

We'll say \(\rho_\lambda\) is a \textbf{compatible system} of
\(\lambda{\hbox{-}}\)adic representations iff for all
\(\lambda\divides \ell\) and for all \(p\not\in \tilde S_0\), so
\(p\not\in S_0\) and \(p\notdivides \ell\),
\begin{align*}
{\mathrm{charpoly}}\rho_\lambda \operatorname{Frob}_p
\end{align*}
is independent of \(\lambda\).

\end{definition}

\begin{remark}

Note that \((E_0){ {}_{ \widehat{\lambda} } }/{ {\mathbb{Q}}_\ell }\) is
a finite extension for any \(\lambda \divides \ell\). If one assumes
\(E_0 = {\mathbb{Q}}\), the above is the data of an
\(\ell{\hbox{-}}\)adic representation for every rational prime \(\ell\)
whose traces all agree.

\end{remark}

\begin{example}[?]

Examples that are known to be compatible systems:

\begin{itemize}
\tightlist
\item
  The cyclotomic characters \(F_p(x) = x-p\),
\item
  \(T_\ell E\) for all \(\ell\), where \(E_0 = {\mathbb{Q}}\) and
  \(F_p(x) = x^2 - a_p x - \operatorname{Nm}(p)\), where \(S_0\) are the
  primes of bad reduction,
\item
  \(H^i_\text{ét}(X; { {\mathbb{Q}}_\ell })\).
\end{itemize}

\end{example}

\begin{remark}

Instead of asking for traces to agree, one can apply LLC: the
\(\rho_\lambda\) are \textbf{strongly compatible} iff for all
\(p\in S_0\) with \(p\divides P\), for all \(\lambda \in E_0\) with
\(\lambda\not\divides P\), consider the restricted representation
\({ \left.{{\rho_\lambda}} \right|_{{G(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu_p/ K_P) }} }\)
yields a WD representation by Grothendieck, and by LLC a representation
\(\pi\) of \(\operatorname{GL}_n(K_p)\) on some infinite dimensional
vector space. One could then define them to be \emph{strongly
compatible} iff \(\pi\) only depends on \(\lambda\). This is unknown for
\(H^i_\text{ét}\).

\begin{quote}
See the \textbf{Weight-Monodromy conjecture}.
\end{quote}

\end{remark}

\hypertarget{adeles-and-global-cft}{%
\subsection{Adeles and Global CFT}\label{adeles-and-global-cft}}

\begin{question}

Motivating question: what is \(G_K^{\operatorname{ab}}\)?

\end{question}

\begin{definition}[Infinite Places]

Let \(K\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\) with
\([K:{\mathbb{Q}}] = d\), so there are \(d\) field embeddings
\(K \xrightarrow{\sigma} {\mathbb{C}}\). These split into two types:

\begin{itemize}
\tightlist
\item
  Let \(r_1\) be the number of \(\sigma\) such that
  \(\sigma(K) \subseteq {\mathbb{R}}\), i.e.~the number of totally real
  embeddings.
\item
  If \(\sigma(K) \not\subseteq {\mathbb{R}}\), writing
  \(\cong(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\)
  for conjugation in \({\mathbb{C}}\), \(\sigma\) and
  \({\overline{{\circ }}}\sigma\) are distinct embeddings that induce
  the same norm since
  \({\left\lvert {z} \right\rvert} = {\left\lvert {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} \right\rvert}\).
  These non-real embeddings come in pairs, so let \(2r_2\) be the number
  of such embeddings.
\end{itemize}

Then
\begin{align*}
r_1 + 2r_2 = d
.\end{align*}
An \textbf{infinite place} \(v\) of \(K\) is either

\begin{itemize}
\tightlist
\item
  A real place \(v = \sigma\in \mathsf{Field}(K\to {\mathbb{R}})\), or
\item
  A complex place, which is the pair of maps
  \(\left\{{\sigma, (z\mapsto \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\circ \sigma}\right\}\in \mathsf{Field}(K\to {\mathbb{C}})\)
  whose image isn't totally real.
\end{itemize}

Further define
\begin{align*}
K_\infty \coloneqq\prod_{v \divides \infty } K{ {}_{ \widehat{v} } }
,\end{align*}
the product of all infinite places, noting that
\(K{ {}_{ \widehat{v} } } \cong {\mathbb{R}}\) or \({\mathbb{C}}\) so
that \(K \hookrightarrow K{ {}_{ \widehat{v} } }\).

\end{definition}

\begin{example}[?]

For \(K = {\mathbb{Q}}(2^{1\over 3})\), note that

\begin{itemize}
\tightlist
\item
  \(r_1 = 1\)
\item
  \(r_2 = 1\)
\end{itemize}

\end{example}

\begin{remark}

It turns out that
\begin{align*}
K_\infty \cong {\mathbb{R}}^{r_1} \times {\mathbb{C}}^{r_2} = K\otimes_{\mathbb{Q}}{\mathbb{R}},
\quad
K_\infty^{\times}\cong ({\mathbb{R}}^{\times})^{r_1} \times ({\mathbb{C}}^{\times})^{r_2}
,\end{align*}
and \(K_\infty^{\times}\) is generally disconnected since
\({\mathbb{R}}^{\times}\) is disconnected. The connected component of
the identity satisfies
\begin{align*}
(K_\infty^{\times})^0 \cong ({\mathbb{R}}_{\geq 0})^{r_1} \times ({\mathbb{C}}^{\times})^{r_2}
.\end{align*}

We'll define the \textbf{adeles} as a restricted product in the category
of topological rings:
\begin{align*}
{\mathbb{A}}_L = \prod_{p<\infty} K{ {}_{ \widehat{p} } } \times K_\infty
,\end{align*}
and the \textbf{ideles} as \({\mathbb{A}}_L^{\times}\). It will be
locally compact, has an explicit compact subspace, and is related to
class groups and unit groups.

\end{remark}

\hypertarget{lecture-13}{%
\section{Lecture 13}\label{lecture-13}}

\begin{remark}

For \(\sigma\in \mathsf{Field}(K, {\mathbb{C}})\), we define an
equivalence relation by
\(\sigma \sim (z\mapsto \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\circ \sigma\),
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
\textbf{infinite places} and we sometimes write \(v\divides \infty\). We
saw that
\begin{align*}
K_\infty \coloneqq\bigoplus _{v\divides \infty} K{ {}_{ \widehat{v} } } \cong K\otimes_{\mathbb{Q}}{\mathbb{R}}
.\end{align*}

\end{remark}

\begin{example}[?]

An example of the above isomorphism, in the simplest example for which
we know nearly nothing about the GLC for \(\operatorname{GL}_2(K)\),
namely \(K = {\mathbb{Q}}(2^{1\over 3})\). For elliptic curves over CM
or totally real fields, one can hope to prove potential modularity and
there are statement for \(\operatorname{GL}_n\) 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 \({\mathbb{Q}}\)
by Eisenstein, or by passing to the local extensions
\({\mathbb{Q}}_2(2^{1\over 3} )\) has degree at least three since the
valuation increases by a factor of 3. Write
\(x^3-2 = (x-\alpha)(x-w)(x-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu)\),
we can then compute
\begin{align*}
K_\infty 
&= {{\mathbb{Q}}[x] \over \left\langle{x^3-2}\right\rangle} \otimes_{\mathbb{Q}}{\mathbb{R}}\\
&\cong {{\mathbb{R}}[x] \over \left\langle{x^3-2}\right\rangle} \\
&\cong {{\mathbb{R}}[x] \over \left\langle{x-\alpha}\right\rangle} \oplus {{\mathbb{R}}[x] \over x^2+\alpha x + \alpha^2} \\
&\cong {\mathbb{R}}\oplus {\mathbb{C}}\\
&= K_{v} \oplus K_{[v']}
,\end{align*}
where we've used that the factors are irreducible and coprime over
\({\mathbb{R}}\). Note that there are two isomorphisms to
\({\mathbb{C}}\) here, and
\begin{align*}
v: K\to {\mathbb{R}}\\
\sqrt{2} &\mapsto \alpha \\ \\
v': K\to {\mathbb{R}}\\
\sqrt{2} &\mapsto w \text{ or } \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu
.\end{align*}

\end{example}

\begin{definition}[Finite Adeles]

Let \(K\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\), so that
\({\operatorname{Places}}(K) \cong \operatorname{mSpec}{\mathcal{O}}_K\)
consists of finitely many places. For \(p\) a finite place, we can
complete to obtain
\(K{ {}_{ \widehat{p} } }, {\mathcal{O}}_{k, \widehat{p}}\),
generalizing \({ {\mathbb{Q}}_p }\) and \({ {\mathbb{Z}}_p }\). Define
the \textbf{finite adeles of \(K\)} as
\begin{align*}
{\mathbb{A}}_{K, {\mathsf{Fin}}} 
&\coloneqq\prod^{\operatorname{res}}_{p\in {\operatorname{Places}}(K)} K{ {}_{ \widehat{{\mathfrak{p}}} } }\\
&\coloneqq\left\{{(x_p) \in \prod_{p\in {\operatorname{Places}}(K)} K{ {}_{ \widehat{p} } } {~\mathrel{\Big\vert}~}x_p \in {\mathcal{O}}_{K, \widehat{p}} \text{ for all but finitely many }p}\right\} \\
&\subseteq \prod_{p\in {\operatorname{Places}}(K)} K{ {}_{ \widehat{{\mathfrak{p}}} } }
.\end{align*}

Equivalently,
\begin{align*}
{\mathbb{A}}_{K, {\mathsf{Fin}}} \coloneqq\left\{{(x_p) \in \prod_p K{ {}_{ \widehat{p} } } {~\mathrel{\Big\vert}~}{\exists}S \text{ finite where } x_p\in {\mathcal{O}}_{K, \widehat{p}} \,\forall p\not\in S }\right\}
,\end{align*}
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 \({\mathcal{O}}_{K, { {}_{ \widehat{p} } }}\) is
compact and thus the subring
\(R\coloneqq\prod_p {\mathcal{O}}_{K, { {}_{ \widehat{p} } }}\) 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.

\end{definition}

\begin{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
\begin{align*}
\Delta: K &\to {\mathbb{A}}_{K, f} \\
\lambda = {a\over b} &\mapsto \cdots, S\coloneqq\left\{{p\in {\operatorname{Places}}(K) {~\mathrel{\Big\vert}~}p\divides b}\right\} = \left\{{p \in {\operatorname{Places}}(K) {~\mathrel{\Big\vert}~}v_p( \lambda) < 0}\right\}
,\end{align*}
which is always a finite set.

\end{remark}

\begin{definition}[Adeles]

The full ring of \textbf{adeles of \(K\)} is defined as
\begin{align*}
{\mathbb{A}}_K \coloneqq{\mathbb{A}}_{K,{\mathsf{Fin}}} \times K_{\infty} = \prod^{\operatorname{res}}_{v\in{\operatorname{Places}}(K)} K{ {}_{ \widehat{v} } }
.\end{align*}

\end{definition}

\begin{remark}

Note that \(K\) is a 1-dimensional ring and contains
\({\mathcal{O}}_K\), and in analogy there is a ring of meromorphic
functions \({\mathbb{C}}(t) \supseteq{\mathbb{C}}[t]\) and one could
take a Laurent expansion about any point \(v\) to get an element of
\(\prod_{v\in {\mathbb{C}}} {\mathbb{C}}{\left(\left( t-v \right)\right) }\).
This contains a subring
\(\prod_{v\in {\mathbb{C}}} {\mathbb{C}}{\left[\left[ t-v \right]\right] }\)
of holomorphic functions, so
\begin{align*}
{\mathbb{C}}(t) \hookrightarrow\prod_{v\in {\mathbb{C}}} {\mathbb{C}}{\left(\left( t-v \right)\right)  } \supseteq\prod_{v\in {\mathbb{C}}} {\mathbb{C}}{\left[\left[ t-v \right]\right]  }
.\end{align*}

Idea: \({\mathbb{C}}(t)\) 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.

\end{remark}

\begin{remark}

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

\end{remark}

\begin{exercise}[?]

Note that
\(K_\infty \cong\otimes_{\mathbb{Q}}{\mathbb{R}}\cong K\otimes_{\mathbb{Q}}{\mathbb{Q}}_\infty\),
and show
\begin{align*}
{\mathbb{A}}_{K, {\mathsf{Fin}}} &\cong K\otimes_{\mathbb{Q}}{\mathbb{A}}_{{\mathbb{Q}}, {\mathsf{Fin}}} \\
{\mathbb{A}}_K &\cong K\otimes_{\mathbb{Q}}{\mathbb{A}}_{\mathbb{Q}}
.\end{align*}

\end{exercise}

\begin{lemma}[?]

\begin{align*}
{\mathbb{A}}_{{\mathbb{Q}}, {\mathsf{Fin}}} \cong {\mathbb{Q}}\oplus \prod_{p} { {\mathbb{Z}}_p }
,\end{align*}
i.e.~for all
\(x\coloneqq(x_p) \in {\mathbb{A}}_{{\mathbb{Q}},{\mathsf{Fin}}}\) (so
\(x_p \in { {\mathbb{Q}}_p }\) for all primes and
\(x_p\in { {\mathbb{Z}}_p }\) for all but finitely many \(p\)) there
exists a \(\lambda \in {\mathbb{Q}}\) such that \(x = \lambda + \mu\)
for some \(\mu \in \prod_p { {\mathbb{Z}}_p }\).

\end{lemma}

\begin{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).

\end{remark}

\begin{proof}[of lemma]

By induction on \({\sharp}S\), where for the base case when
\(S = \emptyset\) one can take \(\mu = x\). Let \(S\) be finite such
that \(x_p\in { {\mathbb{Z}}_p }\) for all \(p\not\in S\). For a general
\(x\), choose \(p\in S\) with \(x_p \in { {\mathbb{Q}}_p }\) and take a
\(p{\hbox{-}}\)adic expansion as a Laurent series with a finite tail:
\begin{align*}
x_p = \lambda + f \coloneqq\qty{a_{-n}p^{-n} + \cdots + a_{-1}p^{-1}} + \qty{a_0 + \cdots}, \qquad a_i\in \left\{{0,\cdots, p-1}\right\}
,\end{align*}
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
\(S\setminus\left\{{p}\right\}\), so we're done by induction.

\end{proof}

\begin{exercise}[Challenging]

Show that
\begin{align*}
{\mathbb{A}}_K \cong K + \prod_{p} {\mathcal{O}}_{K, \widehat{p}}
.\end{align*}
The above proof doesn't quite work if \(p\) is a prime that is not
principal, so one needs input from class groups.

\end{exercise}

\hypertarget{ideles}{%
\subsection{Ideles}\label{ideles}}

\begin{remark}

We're interested in \({\mathbb{A}}_K^{\times}\), and more generally
\(\operatorname{GL}_n({\mathbb{A}}_K)\).

\end{remark}

\begin{definition}[Ideles]

One can check that
\begin{align*}
{\mathbb{A}}_{K, {\mathsf{Fin}}}^{\times}= \prod^{\operatorname{res}}_{p} K{ {}_{ \widehat{p} } }^{\times}\coloneqq\left\{{(x_p) \in \prod_p K_{ {}_{ \widehat{p} } }^{\times}{~\mathrel{\Big\vert}~}x_p\in {\mathcal{O}}_{K, \widehat{p}}^{\times}\text{ for almost all } p}\right\}
,\end{align*}
and
\({\mathbb{A}}_K^{\times}= \prod^{\operatorname{res}}_v K{ {}_{ \widehat{v} } }^{\times}\).
The topology on \({\mathbb{A}}_{K, {\mathsf{Fin}}}^{\times}\) is such
that \(\prod_p {\mathcal{O}}_{K, \widehat{p}}^{\times}\) is open; this
is not the subspace topology from \({\mathbb{A}}_{K, {\mathsf{Fin}}}\).

\end{definition}

\begin{warnings}

\({\mathbb{A}}_K^{\times}\hookrightarrow{\mathbb{A}}_K\) but the
topology is not the subspace topology.

\end{warnings}

\begin{example}[?]

An element in \({\mathbb{A}}_K\setminus{\mathbb{A}}_K^{\times}\): take
\(x_p = p\) and set
\(x = (x_p) \in {\mathbb{A}}_{{\mathbb{Q}}, {\mathsf{Fin}}}\) and in
fact \(x\in \prod_p { {\mathbb{Z}}_p }\), however
\(1/x\in \prod_p { {\mathbb{Q}}_p }\) but
\(1/x\not\in {\mathbb{A}}_{{\mathbb{Q}}, {\mathsf{Fin}}}\) since there
are problems at infinitely many primes.

\end{example}

\hypertarget{global-class-field-theory}{%
\subsection{Global Class Field Theory}\label{global-class-field-theory}}

\begin{remark}

Recall that \(G_K/G_K^c = G_K^{\operatorname{ab}}\) 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{center}
\begin{tikzcd}
    {\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu} \\
    \\
    && {K^{\operatorname{ab}}} \\
    \\
    K
    \arrow["{G_K}", from=5-1, to=1-1]
    \arrow["{G_K^{\operatorname{ab}}}"', from=5-1, to=3-3]
    \arrow[from=3-3, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGJhciBLIl0sWzIsMiwiS15cXGFiIl0sWzAsNCwiSyJdLFsyLDAsIkdfSyJdLFsyLDEsIkdfS15cXGFiIiwyXSxbMSwwXV0=}{Link
to Diagram}
\end{quote}

After choosing
\(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu \supseteq K\),
one can form the infinite extension
\begin{align*}
K^{\operatorname{ab}}= \displaystyle\bigcup_S L, \qquad S \coloneqq\left\{{L {~\mathrel{\Big\vert}~}K \leq L \leq \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu, L/K \text{ finite }, G(L/K) \in {\mathsf{Ab}}{\mathsf{Grp}}}\right\}
.\end{align*}

\end{remark}

\begin{remark}

Note that for \(K = {\mathbb{Q}}\), Kronecker-Weber yields
\begin{align*}
{\mathbb{Q}}^{\operatorname{ab}}= \displaystyle\bigcup_{N\geq 1} {\mathbb{Q}}(\zeta_N)
.\end{align*}
Class field theory tells you what Galois group for
\(K^{\operatorname{ab}}/K\) is, but not necessarily what
\(K^{\operatorname{ab}}\) is itself. It exists for \({\mathbb{Q}}\) by
Kronecker-Weber, and for imaginary quadratic fields for reasons
involving \(j{\hbox{-}}\)invariants of elliptic curves, but for real
quadratic fields this is completely open.

\end{remark}

\begin{question}[An open problem]

What is \(K^{\operatorname{ab}}\) for
\(K = {\mathbb{Q}}(2^{1\over 3})\)?

\end{question}

\begin{theorem}[GCFT]

There is a surjection \(r_K\), the \textbf{global Artin map}, which is a
continuous group morphism:

\begin{center}
\begin{tikzcd}
    {\dcosetl{K^{\times}}{{\mathbb{A}}_K^{\times}}} && {G(K^{\operatorname{ab}}/K)} && 0
    \arrow["{r_K}", two heads, from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGRjb3NldGx7S1xcdW5pdHN9e1xcQUFfS1xcdW5pdHN9Il0sWzIsMCwiRyhLXlxcYWIvSykiXSxbNCwwLCIwIl0sWzAsMSwicl9LIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzEsMl1d}{Link
to Diagram}
\end{quote}

\end{theorem}

\begin{remark}

The Artin map can't be an isomorphism since the RHS is profinite, but
the LHS contains
\(K_\infty^{\times}\cong ({\mathbb{R}}^{\times})^{r_1} \times ({\mathbb{C}}^{\times})^{r_2}\),
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,
\(K={\mathbb{Q}}\), and imaginary quadratic number fields. Since the
identity is closed in the image, the kernel must be closed in the source
by continuity.

\end{remark}

\begin{theorem}[from CFT]

If \(C_K\) is defined as the image of \((K_\infty^{\times})^0)\) in
\(\dcosetl{K^{\times}}{{\mathbb{A}}_K^{\times}}\), then
\begin{align*}
\ker r_K = { \operatorname{cl}} _\top(C_K)
,\end{align*}
the topological closure.

\end{theorem}

\begin{remark}

It turns out that for \(K={\mathbb{Q}}\) or an imaginary quadratic
field, \(C_K\) is already closed, but there are examples where it is
not. Thus
\(G(K^{\operatorname{ab}}/K)\cong \dcoset{K^{\times}}{{\mathbb{A}}_K^{\times}}{{ \operatorname{cl}} _\top(C_K)}\).

\end{remark}

\begin{remark}

Some properties of \(r_K\):

\begin{center}
\begin{tikzcd}
    {{\left[ {1,1,\cdots, 1, x_p, 1, \cdots } \right]}} & {\dcosetl{K^{\times}}{{\mathbb{A}}_K^{\times}}} && {G(K^{\operatorname{ab}}/K)} \\
    \\
    {x_p} & {K{ {}_{ \widehat{p} } }^{\times}} && {G_{K{ {}_{ \widehat{p} } }}^{\operatorname{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}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMSwwLCJcXGRjb3NldGx7S1xcdW5pdHN9e1xcQUFfS1xcdW5pdHN9Il0sWzMsMCwiRyhLXlxcYWIvSykiXSxbMywyLCJHX3tLXFxjb21wbGV0ZSBwfV5cXGFiIl0sWzEsMiwiS1xcY29tcGxldGV7cH1cXHVuaXRzIl0sWzAsMCwiXFx0dnsxLDEsXFxjZG90cywgMSwgeF9wLCAxLCBcXGNkb3RzIH0iXSxbMCwyLCJ4X3AiXSxbMiwxXSxbMywyLCJyX0sgXFx0ZXh0eyBsb2NhbCBBcnRpbn0iXSxbMCwxLCJyX0sgXFx0ZXh0eyBnbG9iYWwgQXJ0aW59Il0sWzUsNCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dLFszLDBdXQ==}{Link
to Diagram}
\end{quote}

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

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

\begin{center}
\begin{tikzcd}
    {\dcosetl{L^{\times}}{{\mathbb{A}}_K^{\times}}} && {G(L^{\operatorname{ab}}/L)} \\
    \\
    {\dcosetl{K^{\times}}{{\mathbb{A}}_K^{\times}}} && {G(K^{\operatorname{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["{\operatorname{Nm}_{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}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

\end{remark}

\hypertarget{lecture-14}{%
\section{Lecture 14}\label{lecture-14}}

\hypertarget{lecture-15}{%
\section{Lecture 15}\label{lecture-15}}

\hypertarget{lecture-16}{%
\section{Lecture 16}\label{lecture-16}}

\hypertarget{lecture-17}{%
\section{Lecture 17}\label{lecture-17}}

\hypertarget{lecture-18}{%
\section{Lecture 18}\label{lecture-18}}

\hypertarget{lecture-19}{%
\section{Lecture 19}\label{lecture-19}}

\hypertarget{lecture-20}{%
\section{Lecture 20}\label{lecture-20}}

\hypertarget{intro}{%
\section{Intro}\label{intro}}

\begin{remark}

Some additional resources:

\begin{itemize}
\tightlist
\item
  \url{https://math.stackexchange.com/questions/48981/the-langlands-program-for-beginners}
\end{itemize}

\end{remark}

\begin{remark}

Some vague definitions:

\begin{itemize}
\item
  The congruence subgroups \(\Gamma(N), \Gamma_0(N), \Gamma_1(N)\).
  \begin{align*}
  \Gamma(N) &\coloneqq\left\{{M\in {\operatorname{SL}}_2({\mathbb{Z}}) {~\mathrel{\Big\vert}~}M\equiv I \operatorname{mod}N}\right\} \\
  \Gamma_0(N) &\coloneqq\left\{{M\in {\operatorname{SL}}_2({\mathbb{Z}}) {~\mathrel{\Big\vert}~}M\equiv {
  \begin{bmatrix}
    {*} & {*} 
  \\
    {0} & {*}
  \end{bmatrix}
  } \operatorname{mod}N}\right\}
  .\end{align*}

  \begin{itemize}
  \tightlist
  \item
    A \textbf{congruence subgroup of level \(N\)} is any
    \(H \supseteq\Gamma(N)\)
  \item
    The \textbf{level} is the smallest \(N\) such that
    \(H \supseteq\Gamma(N)\).
  \item
    Recovers \({\operatorname{SL}}_2({\mathbb{Z}}) = \Gamma_0(1)\).
  \item
    Principal congruence subgroups of \textbf{level} \(N\):
    \begin{align*}
    1\to \Gamma(N) \to {\operatorname{SL}}_2({\mathbb{Z}}) \xrightarrow{\operatorname{mod}N} {\operatorname{SL}}_2({\mathbb{Z}}/N{\mathbb{Z}})\to 1
    .\end{align*}
  \end{itemize}
\item
  Fuchsian groups: discrete subgroups of
  \({\operatorname{SL}}_2({\mathbb{R}})\).
\item
  \(Y(\Gamma)\): modular curves of the form
  \(\dcosetl{\Gamma}{{\mathbb{H}}}\) for \(\Gamma\) a Fuchsian group of
  the first kind. For congruence subgroups, abbreviated \(Y(N)\).
\item
  \(X(\Gamma)\): the compactification of \(Y(\Gamma)\) obtained by
  adding cusps. For congruence subgroups, abbreviated \(X(N)\).
\item
  Shimura-Taniyama-Weil theorem: for \(E\) an elliptic curve, there is a
  cover \(X_0(N) \to E\) where \(N\) is the conductor of \(E\).
\item
  Bad reduction for an elliptic curve: primes \(p\) for which the
  equation reduces \(\operatorname{mod}p\) to a singular curve.
\item
  Factors of automorphy: for
  \(\gamma = { \begin{bmatrix}  {a} & {b} \\  {c} & {d} \end{bmatrix} }\in \Gamma\),
  \(j(\gamma, \tau) \coloneqq(c\tau + d)\).
\item
  Slash operators:
  \(f\mathrel{\Big|}[\gamma]_k \coloneqq f(\gamma({-}))\cdot j(\gamma,{-})^{-k}\)
\item
  Classical automorphic forms of weight \(k\) and level \(N\):

  \begin{itemize}
  \tightlist
  \item
    \(f\in \operatorname{Mero}({\mathbb{H}}, {\mathbb{C}})\),
  \item
    \(f\mathrel{\Big|}[\gamma]_k = f\) for all \(\gamma \in \Gamma\) a
    congruence subgroup of level \(N\),
  \item
    Meromorphic at cusps, so the corresponding Fourier expansion at the
    cusps has a finite tail.
  \item
    Note that these conditions guarantee the corresponding \(L\)
    function will be meromorphic with known poles, or holomorphic for
    cusp forms.
  \end{itemize}
\item
  Classical modular forms as automorphic forms:
\item
  \(f\in \mathop{\mathrm{Hol}}({\mathbb{H}}, {\mathbb{C}})\)
\item
  \(f\mathrel{\Big|}[\gamma]_k = f\) for all \(\gamma\),
\item
  Holomorphic at cusps, so Fourier expansion as no negative terms.
\item
  \textbf{Cusp form}: modular forms with vanishing constant Fourier
  coefficient at every cusp.
\item
  Automorphic forms \(f\) of weight \(2k\) correspond to meromorphic
  differential forms
  \(\omega = f(z) (dz){ {}^{ \scriptstyle\otimes_{{\mathbb{C}}}^{k} } }\).
\item
  Weakly modular forms of \textbf{weight} \(k\) for
  \({\operatorname{SL}}_2({\mathbb{Z}})\):
  \(f(\gamma(\tau)) = (c\tau + d)^k f(\tau)\) for all
  \(\gamma\in \Gamma\) and \(\tau \in {\mathbb{H}}\).

  \begin{itemize}
  \item
    Automorphic forms: meromorphic at \(\infty\), form spaces
    \({\mathcal{A}}_k(\Gamma)\)
  \item
    Modular forms: holomorphic on
    \({\mathbb{H}}\cup\left\{{\infty}\right\}\), form spaces
    \({\mathcal{M}}_k(\Gamma)\)
  \item
    Cusp forms: vanishing at cusp points, i.e.~\(f(\infty) = 0\), form
    spaces \({\mathcal{S}}_k(\Gamma)\).
  \item
    The containment:
    \begin{align*}
    {\mathcal{A}}_k(\Gamma) \supseteq {\mathcal{M}}_k(\Gamma) \supseteq {\mathcal{S}}_k(\Gamma)
    .\end{align*}
  \item
    The Eisenstein space:
    \begin{align*}
    0 \to {\mathcal{S}}_K(\Gamma) \to {\mathfrak{m}}_k(\Gamma) \to {\mathcal{E}}_k(\Gamma) \to 0
    .\end{align*}
  \end{itemize}
\item
  Eisenstein series:
  \begin{align*}
  G_{k}(\tau):=\sum_{(c, d) \in \mathbb{Z}^{2}}^{\prime} \frac{1}{(c \tau+d)^{k}}
  ,\end{align*}
  where \(G_k(\infty) = 2\zeta(k)\).

  \begin{itemize}
  \tightlist
  \item
    Normalization:
    \begin{align*}
    E_{k}\coloneqq\frac{G_{k}}{2 \zeta(k)} \in {\mathcal{M}}_{k}(\Gamma)
    .\end{align*}
  \end{itemize}
\item
  The discriminant form:
  \begin{align*}
  \Delta: {\mathbb{H}}\to {\mathbb{C}}\\
  \Delta &= g_2^3-27 g_3^2, \quad g_2 \coloneqq 60 G_4,\quad g_3 \coloneqq 14- G_6
  .\end{align*}

  \begin{itemize}
  \tightlist
  \item
    Facts:
    \(\Delta/(2\pi)^{12}\in {\mathcal{M}}_{12}({\operatorname{SL}}_2({\mathbb{Z}}))\)
    and has a Fourier expansion \(\sum_k \tau(k) q^k\) for
    \(q=\exp(2\pi i \tau)\) where \(\tau\) is the Ramanujan \(\tau\)
    function. Moreover
    \(\Delta \in {\mathcal{S}}_{12}({\operatorname{SL}}_2({\mathbb{Z}}))\)
  \end{itemize}
\item
  Modular curves \(Y(N)\): ?

  \begin{itemize}
  \tightlist
  \item
    \(X(N)\): ?
  \end{itemize}
\item
  Geometric interpretations:

  \begin{itemize}
  \tightlist
  \item
    Modular forms of weight \(k\) and level \(N\): meromorphic
    differential forms on \(X_0(N)\) which are multiples of a certain
    divisor, i.e.~holomorphic sections of lines bundles on modular
    curves, so
    \({\mathcal{M}}_k(\Gamma) \cong H^0(X; {\mathcal{L}}_{\Gamma, k})\)
    for some line bundle.
  \item
    Automorphic form: for \(G\) an algebraic group, a function
    \(f: G\to {\mathbb{C}}\) which is invariant with respect to some
    subgroup \(\Gamma \leq G\)
  \end{itemize}
\item
  Petersson inner product: for \(f,g\) modular forms of weight \(k\) for
  \(\Gamma\), at least one cuspidal, and \(F\) a fundamental domain for
  \(\Gamma\),
  \begin{align*}
  {\left\langle {f},~{g} \right\rangle} \coloneqq\iint_F f\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu y^k \, {\,dx\,dy\over y^2}
  .\end{align*}
\item
  Mellin transform:
  \begin{align*}
  M(f)(s) \coloneqq\int_{{\mathbb{R}}\geq 0} f(y) y^s {\,dy\over y}
  .\end{align*}

  \begin{itemize}
  \tightlist
  \item
    Note that \(M(e^{-x})(s) = \Gamma(s)\)
  \end{itemize}
\item
  Riemann zeta is the L function associated to the Jacobi theta
  function, which is modular of weight \(1/2\) with respect to
  \(\left\langle{z\mapsto z+2,{ \begin{bmatrix}  {0} & {-1} \\  {1} & {0} \end{bmatrix} } }\right\rangle\).
\item
  First instance of Langlands: Taniyama-Shimura-Weil theorem. For
  \(E_{/ {{\mathbb{Q}}}}\) an elliptic curve of conductor \(N\), there
  is a weight 2 cusp form which is a Hecke eigenform on \(\Gamma_0(N)\)
  with \(L(E, s) = L(f, s)\). More generally, all \(L\) functions
  attached to algebraic varieties should arise as \(L\) functions coming
  from automorphic forms.
\end{itemize}

\end{remark}

\addsec{ToDos}
\listoftodos[List of Todos]
\cleardoublepage

% Hook into amsthm environments to list them.
\addsec{Definitions}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]
\cleardoublepage

\addsec{Theorems}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]
\cleardoublepage

\addsec{Exercises}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]
\cleardoublepage

\addsec{Figures}
\listoffigures
\cleardoublepage


\printbibliography[title=Bibliography]


\end{document}
