# Lecture 11 

> Look up Fontaine's $B_{\dR}$!

## The Frobenius Machine

:::{.remark}
Last time: global Galois groups, i.e. Galois groups of number fields.
The setup:

- $L/K$ finite Galois global,
- $\mfp\in \spec \OO_K$ a nonzero prime,
- $\mfp \OO_L\in \Id(\OO_L)$, probably not in $\spec \OO_L$,
- A fixed prime $P\divides \mfp$ and a factorization
\[
\mfp \OO_L = P^e \times \prod_{\mfq\divides \mfp }\mfq_i^{e_i}
.\]
- The decomposition group.
\[
G(L/K) \contains D_{P/\mfp} \da \ts{\sigma \in G(L/K) \st \sigma P = P}
.\]

:::

:::{.fact}
If $\sigma\in D_{P/\mfp}$, then $\sigma: L\to L$ descends to $\sigma: \OO_L \to \OO_L$ restricts to the identity on $K$, $\OO_L$, and $\mfp$.
Moreover $\sigma(P) = P$, so there is an induced map $\OO_L/\mfp^n\selfmap$ and thus a map on the completion $\sigma: L\complete P \selfmap$ fixing $K\complete\mfp$, yielding an element \[
\sigma\in G( L\complete{P}, K\complete{\mfp} )
= D_{P/\mfp} \contains I_{P/\mfp} 
= I_{L\complete P / K\complete\mfp} 
.\]
So the global group contains the local group.

For global fields, we have a discriminant $\Delta = \disc(L/K) \in \Id(\OO_K)$, and $p\notdivides \Delta$ then $I_{P/\mfp} = 1$ for all $P\divides \mfp$, making $\mfp$ 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 $S = \ts{p\in \spec \OO_K \st p\notdivides \Delta}$.
For all $P\divides \mfp$, we get a cyclic group 
\[
D_{P/\mfp}/I_{P/\mfp} = D_{P/\mfp} = \gens{\Frob_P} = G(\kappa(P)/ \kappa(\mfp))
\]
since this is a local Galois group where the inertia is trivial.
So each $P$ yields an element $\Frob_P\in G(L/K)$, and varying $P$ yields a conjugacy class $\Frob_\mfp \da \ts{\Frob_P \st P\divides \mfp}$.
:::

:::{.theorem title="Chebotarev"}
Every conjugacy class $C \subseteq G(L/K)$ is of the form $C = \Frob_\mfp$ for infinitely many $\mfp\in \spec \OO_K$, which is a weak form of **Chebotarev density**.
Moreover the density of such $\mfp$ is exactly $\size C/\size G(L/K)$.
:::

:::{.remark}
A useful consequence: it suffices to specify representations on $\Frob_\mfp$ for all $\mfp$.
There is a variant for infinite extensions, which we'll always assume to be cofinitely unramified.
Fix an algebraic closure $\bar K \da \cl_\alg(K)/K$; then $G_K$ is ramified at every prime $\mfp\in\spec K$, so the reduction to $\Frob$ above no longer works.

Let $S \subseteq \mspec \OO_K$ be a finite set of primes, we'll show that $\Frob_\mfp$ makes sense for $\mfp\not\in S$.
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 
\[
K_S \da \Union_{L\in \tilde S} L, \qquad \tilde S \da \ts{ {L/K \st L \text{ is finite Galois unramified outside of } S } }
.\]
Note that every number field other than $\QQ$ is ramified at some prime.
:::

:::{.question}
Consider $K=\QQ$ and let $S=\ts{p}$ be a fixed prime.
What is $K^S$?
Note that $\QQ(\zeta_{p^n}) \subseteq K^S$ for all $n\geq 1$.
:::

:::{.example title="?"}
Like $K=\QQ$ and fix $N\in \ZZ_{\geq 1}$, and take $S = \ts{p\in \spec \ZZ \st p\divides N}$.
Then $\QQ(\zeta_N)/\QQ$ is unramified outside of $S$ and has Galois group $C_N\units$.
Thus if $p\in \ZZ$ is prime and $p\not\in S$, so $p\notdivides N$, there is a canonical conjugacy class $\Frob_p \in C_N\units$ -- since this group is abelian, this is in fact an element.
It could be $p$ or $p\inv$, and one can work out that it must be $p$.
:::

:::{.corollary title="?"}
There are infinitely many primes in any arithmetic progression.
:::

:::{.remark}
Fixing $K=\QQ$ and a model of $\bar\QQ$, for $S = \ts{p}$ one has $K^S\contains \QQ(\zeta_{p^n})$ for all $n$ and thus their union 
\[
\QQ(\zeta_{p^\infty}) \da \Union_{n\geq 1} \QQ(\zeta_{p^n}) \subseteq \QQbar
.\].
Therefore there is a surjection
\[
G(K^S/K) 
&\surjects G(\QQ(\zeta_{p^\infty})/\QQ)\\
&\iso \cocolim_n G(\QQ(\zeta_{p^n})/\QQ )\\
&\caniso \cocolim_n C_n\units \\
&\iso \ZZpadic\units
.\]

We can get Frobenius elements in such infinite extensions:
if $r\neq p$ is prime, since the groups are abelian we have elements $\Frob_r\in C_{p^n}\units = G(\QQ(\zeta_{p^n})/\QQ)$ which gets identified with $r$ as a residue class.
Taking the inverse limit yields a Frobenius element $\Frob_r \in \ZZpadic\units$.
:::

:::{.remark}
Upshot: for $K \in \Number\Field, S\subseteq \Places(K)$ finite,
\[
G(K^S/K) = \cocolim \ts{ G(L/K) \st L/K \text{ finite, Galois, unramified outside of } S }
\]
and for all $\mfp\not\in S$, we get conjugacy classes $\Frob_{\mfp, L/K} \subseteq G(L/K)$ which glue in the limit to a conjugacy class
\[
\Frob_{\mfp} = \Frob_{\mfp, K^S/K} \subseteq G(K^S/K)
.\]
So we have names for conjugacy classes in this group, although not for elements.
By Chebotarev density, the following map is surjective
\[
\ts{\mfp\in \mspec K \st \mfp\not\in S} &\surjects G(L/K)/\Inn G(L/K) \\
\mfp &\mapsto \Frob_\mfp
,\]
where $G\actson G$ by inner automorphisms, and the RHS denotes conjugacy classes.
:::

:::{.corollary title="Density of Frobenii"}
If $L/K$ is infinite Galois and unramified outside of $S \subseteq \Places(K)$, then
\[
\ts{\Frob_\mfp\st \mfp\not\in S} \embedsdense G(L/K)
.\]
:::

:::{.corollary title="?"}
If $F: G(L/K)\to X$ is continuous and factors through $G(L/K)/\Inn G(L/K)$, so is constant on conjugacy classes, it may be possible to recover $F$ from $\ts{ F(\Frob \mfp) \st \mfp\in \mspec K\sm S}$.
Upshot: if $F$ is the characteristic polynomial of a representation, then it may suffice to know the characteristic polynomials of Frobenius.
:::

## Brauer-Nesbitt

:::{.theorem title="Brauer-Nesbitt"}
Let $G\in \Grp$ and $E\in \Field$ be arbitrary.
Recall that $\rho: G\to \GL_n(E)$ is semisimple iff $\rho$ is a direct sum of irreducible representations.
Note that irreducibles and direct sums of irreducible are semisimple.
If $\rho_1, \rho_2: G\to \GL_n(E)$ are two semisimple representations, then
\[
\charpoly \rho_1(g) = \charpoly \rho_2(g)\, \forall g\in G \implies \rho_1 \cong \rho_2
.\]
:::

:::{.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.
:::

:::{.remark}
If $\characteristic K = 0$, then $\tr \rho_1 = \tr \rho_2 \implies \rho_1 \cong \rho_2$ 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 $n!$.
:::

:::{.exercise title="Non-example"}
Let $G = C_3$, $E = \bar\FF_2$, and $n=2$. 
Find non-isomorphic $\rho_i$ semisimple and reducible with identical traces.
The issue will be the $2!$ is not invertible in $E$.
:::

:::{.remark}
Upshot: if $\rho\in \Top\Grp(G(K^S/K)\to \GL_n(E))$ is a continuous semisimple irrep, say where $E/\QQpadic$, and one knows $F_\mfp \da \charpoly \rho(\Frob_\mfp)\in E[x]$ for all $\mfp\not\in S$, then this determines $\rho$ uniquely.
This is because the function $g\mapsto \charpoly \charpoly \rho(g)$ will be continuous, then Brauer-Nesbitt says this determines $\rho$.
If we know $\charpoly \Frob_\mfp$ (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 $\rho$.
:::

:::{.example title="?"}
Let $K = \QQ$ and $S = \ts{p}$ for a fixed prime.
Let $L = \QQ(\zeta_{p^\infty})$, then
\[
G(L/K) = \cocolim_n C_{p^n}\units = \ZZpadic\units = \GL_1(\ZZpadic) \subseteq \GL_1(\QQpadic)
.\]
We can then define a representation 
\[
\rho: G_{\QQ} \surjects G(\QQ(\zeta_{p^\infty})/\QQ) = \ZZpadic\units \to \GL_1(\QQpadic)
.\]

Note that this factors:

\begin{tikzcd}
	& {G(\QQ^S/\QQ)} \\
	\\
	{G(\QQbar/\QQ)} && {G(\QQ(\zeta_{p^\infty})/\QQ)} && \ZZpadic\units && {\GL_1(\QQpadic)}
	\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}

> [Link to diagram](https://q.uiver.app/?q=WzAsNSxbMCwyLCJHKFxcUVFiYXIvXFxRUSkiXSxbMiwyLCJHKFxcUVEoXFx6ZXRhX3twXlxcaW5mdHl9KS9cXFFRKSJdLFs0LDIsIlxcWlpwYWRpY1xcdW5pdHMiXSxbNiwyLCJcXEdMXzEoXFxRUXBhZGljKSJdLFsxLDAsIkcoXFxRUV5TL1xcUVEpIl0sWzAsMSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzQsMSwiIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzAsNF0sWzEsMl0sWzIsM10sWzAsMywiXFxyaG8iLDIseyJjdXJ2ZSI6NSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d)

So there is a map $G_{\QQ} \to G(\QQ^S/\QQ)$, which is still complicated but contains conjugacy classes $\Frob_r$ for $r\not\in S$.
This composite is the \(p\dash \)adic **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 $\rho(\Frob_r) = r$ for all $r\neq p$.
:::

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

\begin{tikzcd}
	&& {} \\
	&&&& {G(\QQ(\zeta_{p^\infty})/\QQ)} && {\GL_1(\ZZpadic) = \ZZpadic\units} \\
	{G_\QQ} && {G(\QQ^S/\QQ)} \\
	&&&& {G(\QQ(\zeta_{\ell^\infty})/\QQ)} && {\GL_1(\ZZladic) = \ZZladic\units}
	\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}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwyLCJHX1xcUVEiXSxbMiwyLCJcXEdhbChcXFFRXlMvXFxRUSkiXSxbMiwwXSxbNCwxLCJHKFxcUVEoXFx6ZXRhX3twXlxcaW5mdHl9KS9cXFFRKSJdLFs0LDMsIkcoXFxRUShcXHpldGFfe1xcZWxsXlxcaW5mdHl9KS9cXFFRKSJdLFs2LDEsIlxcR0xfMShcXFpacGFkaWMpID0gXFxaWnBhZGljXFx1bml0cyJdLFs2LDMsIlxcR0xfMShcXFpabGFkaWMpID0gXFxaWmxhZGljXFx1bml0cyJdLFswLDFdLFsxLDNdLFsxLDRdLFs0LDZdLFszLDVdLFswLDUsIlxcb21lZ2FfcCIsMCx7ImN1cnZlIjotNSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzAsNiwiXFxvbWVnYV9cXGVsbCIsMix7ImN1cnZlIjo1LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=)

Note that 
$\omega_p(\Frob_r) = \omega_\ell(\Frob_r) = r$ for all $r\not\in S$ and
\[
\ts{\Frob_r \st r\not\in S} \embedsdense G(\QQ^S/\QQ)
,\]
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 $\GL_n(E)$ where $E$ is a single, fixed field.
In fact, these two reps are extremely different: $\ker \omega_p = \QQ(\zeta_{p^\infty})$ which is unramified away from $p$ and totally ramified at $p$, while $\ker \omega_\ell = \QQ(\zeta_{\ell^\infty})$. 
Moreover these are two completely disjoint extensions of $\QQ$, since anything in both fields generates an extension that's unramified away from $p$ and away from $\ell$, and thus unramified everywhere and in $\QQ$.
:::