# Lecture 4: Statements of Local Class Field Theory :::{.remark} Idea: LCFT offers an interpretation of $G(\bar K/K)^\ab$, and a vast number of group cohomology computations. ::: :::{.remark} Just the statements today, the proofs require clever tricks like Lubin-Tate groups. We've reduced studying $G(\bar K/K)$ to looking at the inertia group $I$. The obstacle: the Sylow $p\dash$subgroup for $I$ is complicated. We'll try to understand instead $\Ab(G(\bar K, K))$, its abelianization. ::: :::{.definition title="Weil Group"} Let $K/\QQpadic$ be finite, so we have a SES \[ 1 \to I_{\bar K/K} \to G(\bar K/K) \to \hat{\ZZ} = \gens{\Frob} \to 1 .\] Note that $\Frob$ isn't quite a well-defined element in $G(\bar K/K)$ -- any two lifts differ by inertia. Consider \[ \Frob^\ZZ \da \ts{\cdots, \Frob^{-2}, \Frob\inv, \id, \Frob, \Frob^2, \cdots} \cong \ZZ \subseteq \hat{\ZZ} .\] Define the **Weil group** $W_K$ by the following diagram: \begin{tikzcd} 1 && {I_{\bar K/K}} && {W_K} && {\Frob^\ZZ} && 1 \\ \\ 1 && {I_{\bar K/K}} && {G(\bar K/K)} && {\hat{\ZZ}} && 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} > [Link to Diagram](https://q.uiver.app/?q=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) More precisely, \[ W_K \da \ts{g\in G(\bar K/K) \st \im(g) \in \hat \ZZ = \Gal(K^\unram/K) \text{ is in } \Frob^\ZZ } .\] Topologize $W_K$ in the following way: $I_{\bar K/K} \subseteq W_K$ is open, but not an open subgroup of $G(\bar K/K)$ since $0$ is closed in $\hat{\ZZ}$ and preimages send closed sets to closed sets, making $I_{\bar K/K}$ closed in $\Gal(\bar K/K)$. Thus $W_K/I_{\bar K/K} = \ZZ^\disc$. Using that the embedding $\ZZ^\disc \to \hat{\ZZ}$ is continuous, we can equivalently define $W_K$ by a pullback in $\Top\Grp$: \begin{tikzcd} {W_K} && {\ZZ^\disc} \\ \\ {G(\bar K/K)} && {\hat{\ZZ}} \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} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJXX0siXSxbMiwwLCJcXFpaXlxcZGlzYyJdLFsyLDIsIlxcaGF0e1xcWlp9Il0sWzAsMiwiXFxHYWwoXFxiYXIgSy9LKSJdLFszLDJdLFsxLDJdLFswLDFdLFswLDNdLFswLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) ::: :::{.warnings} $W_K$ does *not* use the subspace topology. ::: :::{.remark} We haven't lost much by passing to $W_K$, since its image is dense in $G(\bar K/K)$. Note that at every finite stage $L/K$, $W_K\surjects G(L/K)$. ::: :::{.remark} For $G\in \Top\Grp$, define $G^c \in \Top\Ab\Grp$ to be the (topological) closure $\cl_G \gens{\ts{ghg\inv h\inv \st g, h\in G}}$. Note that $G^\ab \da G/G^c$ is the maximal abelian Hausdorff quotient of $G$. Why take the topological closure: if $\id$ is not closed, then $\cl_G(\id) \normal G$ is closed, as are all of its cosets, so quotienting by this makes points closed and yields a Hausdorff space. ::: :::{.theorem title="Main Theorem of LCFT"} If $K/\QQpadic$ is finite, then there is a canonical isomorphism, **the Artin map**: \[ r_K: K\units \caniso W_K^\ab .\] ::: :::{.remark} Note that $W_K$ is not a profinite group, and is mapping to a discrete group $\ZZ$. This is what $K\units$ looks like: it contains $\OO_{K}\units$, which is profinite, and the quotient is $\ZZ^\disc$. So $K\units$ also has a discrete and profinite part. ::: ## Properties of the Artin Map :::{.remark} See Serre's article in Cassels-Froehlich for proofs! ::: ### The Canonical Isomorphism :::{.proposition title="Properties of the Artin map"} - $r_K$ restricts to maps: \begin{tikzcd} K\units && {W_K^\ab} \\ \\ {\OO_K\units} && {\im(I_{\bar K/K})} \\ \\ {1 + \mfp_K^i,\, i\geq 1} && {\im(I_{\bar K/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} > [Link to Diagram](https://q.uiver.app/?q=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) - 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 **Hasse-Arf theorem**. - $r_k(\pi_K) \in \Frob\inv \cdot \im(I_{\bar K/K})$, which is a coset in a quotient. - Note that $\pi_K$ is a *choice* of uniformizer, which e.g. for $\QQpadic$ 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 $\Frob\inv$. ::: :::{.remark} Confusing comment: if $\phi:A\caniso B$ is a canonical isomorphism in $\Ab\Grp$, then $\phi \circ \iota$ where $\iota(x) = x\inv$ is just as canonical! This occurs e.g. for the Weil pairing on an elliptic curve $E[n]\cartpower{2}\to \mu_n$, 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 $K\units$, but the last isomorphism $r_k(\pi_K) \in \Frob\inv \Im(I_{\bar K/K})$ singles out *which* isomorphism it is since if $\pi_K$ is a uniformizer, $\pi_K\inv$ is not. So we can tell the difference: one isomorphism sends Frobenius to a uniformizer, the other to the *inverse* of a uniformizer. ::: :::{.definition title="Geometric and Arithmetic Frobenius"} The **arithmetic Frobenius** is $\Frob$, and **geometric Frobenius** is $\Frob\inv$. ::: :::{.remark} Deligne's convention: associate the uniformizer with *geometric* Frobenius. ::: ### Abelian Extensions :::{.remark} If $L/K$ is finite, then $G(\bar K/L)\injects G(\bar K/K)$ is a subgroup, and there is a map $W_L \injects W_K$. These are injections by TFTGT, and there is a diagram: \begin{tikzcd} {G(\bar K/L)} && {\Gal(\bar K/K)} \\ \\ {W_L} & {} & {W_K} \\ \\ {W_L^\ab} && {W_K^\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} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.warnings} Note that (topological) abelianization may not preserve monomorphism! E.g. consider $G$ with $G^\ab = 1$ and a map $C_n\injects G$ from a cyclic group, then $C_n^\ab = C_n\surjects 1$. ::: :::{.remark} One can fill in the following diagram with the norm: \begin{tikzcd} L\units && {W_L^\ab} \\ \\ K\units && {W_K^\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["{\Norm_{L/K}}"', from=1-1, to=3-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJMXFx1bml0cyJdLFswLDIsIktcXHVuaXRzIl0sWzIsMCwiV19MXlxcYWIiXSxbMiwyLCJXX0teXFxhYiJdLFswLDIsInJfTCJdLFsyLDNdLFsxLDMsInJfSyIsMl0sWzAsMSwiXFxObV97TC9LfSIsMl1d) ::: :::{.remark} For $H \leq G$ a finite index subgroup, there is a **transfer** map on the Weil groups, the *Verlagerung transfer* \[ V: G^\ab\to ^\ab g &\mapsto \prod_{1\leq i\leq n} \gamma_i g\gamma_i\inv, && \gamma_i \in G/H \text{ coset representatives} .\] which behaves like a norm. It fits into a diagram: \begin{tikzcd} L\units && {W_L^\ab} \\ \\ K\units && {W_K^\ab} \arrow["{r_K}", from=3-1, to=3-3] \arrow["{r_L}", from=1-1, to=1-3] \arrow["\transfer"', from=3-3, to=1-3] \arrow[hook, from=3-1, to=1-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJLXFx1bml0cyJdLFswLDAsIkxcXHVuaXRzIl0sWzIsMCwiV19MXlxcYWIiXSxbMiwyLCJXX0teXFxhYiJdLFswLDMsInJfSyJdLFsxLDIsInJfTCJdLFszLDIsIlxcdHJhbnNmZXIiLDJdLFswLDEsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d) ::: :::{.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 $W_L^c\normal W_K$. Define the quotient \[ W_{L/K} \da W_K/W_L^c .\] ::: :::{.definition title="Fundamental Class in Galois Cohomology"} There is a canonical SES \begin{tikzcd} &&&& {} \\ \\ 1 && {W_L^\ab = L\units} \\ \\ && {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} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbNCwwXSxbMiwyLCJXX0xeXFxhYiA9IExcXHVuaXRzIl0sWzIsNCwiV19ML0siXSxbMiw2LCJcXEdhbChML0spIl0sWzAsNiwiMSJdLFswLDIsIjEiXSxbNSwxXSxbMSwyXSxbMiwzXSxbMyw0XV0=) This extension gives a **fundamental class** \[ \alpha_{L/K} \in H^2(G(L/K); L\units) ,\] 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 $(\wait) \cupprod \alpha_{L/K}$ is a map between cohomology groups and is often an isomorphism. ::: :::{.remark} Say $L/K$ is an abelian extension iff it is Galois with abelian Galois group. There is a maximal abelian extension $K^\ab$, which exists since compositing preserves abelian extensions. We now essentially understand $G(K^\ab/K)$: \begin{tikzcd} && {\bar K} \\ \\ && {K^\ab} \\ {K^\unram} \\ && 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(\bar K/K)^\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^\unram/K) \cong \hat\ZZ}", curve={height=-30pt}, dashed, no head, from=5-3, to=4-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) We now have a diagram: \begin{tikzcd} {I_{K^\ab/K}} && {G(K^\ab/K)} && {\Gal(K^\unram/K) = \hat{\ZZ}} \\ \\ {I_{K^\ab/K}} && {W_K^\ab \underset{CFT}{\iso} K\units} && \ZZ \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} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJJX3tLXlxcYWIvS30iXSxbMCwyLCJJX3tLXlxcYWIvS30iXSxbMiwyLCJXX0teXFxhYiBcXHVuZGVyc2V0e0NGVH17XFxpc299IEtcXHVuaXRzIl0sWzIsMCwiXFxHYWwoS15cXGFiL0spIl0sWzQsMCwiXFxHYWwoS15cXHVucmFtL0spID0gXFxoYXR7XFxaWn0iXSxbNCwyLCJcXFpaIl0sWzEsMl0sWzIsNV0sWzAsM10sWzMsNF0sWzAsMSwiIiwxLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFs1LDQsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d) Note that $G(K^\ab/K)$ is a profinite topological group, so can't quite be $K\units$, but is some kind of completion. ::: :::{.remark} Conclusion: after choosing a uniformizer/Frobenius there is a non-canonical isomorphism \[ G(\bar K/K)^\ab = \Gal(K^\ab/K) \iso \OO_K\units \times \hat\ZZ \cong I_{K^\ab/K} \times \Gal(K^\unram/K) .\] We know $G(\bar K/K)$ is solvable, since it factors as cyclic and pro$\dash p$ 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. :::