# Lecture 2: Part 1, Classical Local Langlands References: Serre's *Local Fields* > :::{.remark} Vaguely stated local Langlands for $\GL_n$ over $K/\QQpadic$ a finite extension: there is a bijective correspondence \[ \correspond{ \text{Certain $\infty\dash$dimensional irreducible} \\ \text{reps of }\GL_n \text{ over } \CC } &\mapstofrom \correspond{ \text{Certain $n\dash$dimensional reps of a group} \\ A \text{ related to }\Gal(\bar K /K) } .\] 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 $\GL_n$ 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. ::: :::{.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 $\Gal(L/K) \in \Fin\Grp$ (which is a theorem) are the auts of $L$ fixing $K$ pointwise, and there is a correspondence \[ \Gal(L/\wait):\, \correspond{ \text{Subfields } K \leq M \leq L } &\mapstofrom \correspond{ \text{Subgroups } 0 \leq \Gal(L/M) \leq \Gal(L/K) } \\ M &\mapsto \ts{\sigma\in \Gal(L/M) \st \ro{\sigma}{M} = \id_M} .\] ::: :::{.definition title="Galois Extensions"} For the infinite case, let $K\in \Field$ and let $L/K$ be algebraic, possibly with $[L:K] = \infty$. We say $L$ is again Galois iff normal and separable. ::: :::{.warnings} Note that the cardinalities need not match up here: $L$ can be a countably infinite extension with an uncountable Galois group. ::: :::{.remark} For a fixed element $\lambda \in L$ and $\sigma\in \Gal(L/K)$, 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 $\Res_{L/M}: \Gal(L/K)\to \Gal(M/K)$ and $\sigma(\lambda)$ is determined by $\tilde\sigma(\lambda)$ for $\tilde \sigma = \Res_{L/M}(\sigma)$. 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 \[ \Gal(L/K) \embeds \prod_{\substack{ K \subseteq M \subseteq L \\ M/K \text{ finite Galois }}} \Gal(M/K) \implies \Gal(L/K) \cong \cocolim_{\substack{K \subseteq M \subseteq L \\ M/K \text{ finite Galois}}} M \in \Grp ,\] and it turns out that $\Gal(L/K)$ is a closed subspace of $\prod \Gal(M/K)$ in the subspace topology. ::: :::{.warnings} Each $\Gal(M/K)$ has the discrete topology, but $\Gal(L/K)$ 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 $U_i\neq X_i$. ::: :::{.proposition title="FTGT"} If $L/K$ is Galois, i.e. algebraic normal and separable, with $\Gal(L/K)$ given the projective limit topology, there is a bijection \[ \Gal(L/\wait):\, \correspond{ \text{Subfields } K \leq M \leq L } &\mapstofrom \correspond{ \text{Closed subgroups of }\Gal(L/K) } \] 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! ::: :::{.example title="?"} Take $K=\QQ$ and $L = \Union_{n\geq 1} L_n$ where $L_n\da \QQ(\zeta_{p^n}) \subseteq \CC$. There is a canonical isomorphism $\Gal(L_n/\QQ) = C_{p^n}\units$, so $G_L \embeds \prod_{n\geq 1} C_{p^n}\units$. Note that this is not a surjection in general: note that the $L_i$ are filtered, and we can restrict $\Gal(L/K)\to \Gal(L_n/K)$ by just restricting automorphisms. So writing $\phi_n \da \ro{\phi}{L_n}$ for any $\phi \in \Gal(L/K)$, knowing $\phi_n$ implies knowing $\phi_{\leq n}$ (by restriction). Under the canonical identification, $\phi_n \in C_{p^n}\units$, and for $m\leq n$ we have $\phi_m = \phi_n \mod p^m$. In this case, there is a homeomorphism \[ \Gal(L/\QQ) = \cocolim C_{p^n}\units \da \ZZpadic\units \injects \ZZpadic ,\] noting that $\ZZpadic\units \injects \ZZpadic$ is a topological group in a topological ring, and the subspace topology works here since inversion is continuous. ::: :::{.example title="?"} Let $K$ be finite, say $\size K = q$, and $L = \bar K$. Noting that $L_n \da \FF_{q^n} \subseteq L_m \da \FF_{q^m} \iff n\divides m$, we can form the filtered colimit \[ L = \colim_n L_n = \Union_{n\geq 1} L_n .\] Recall that $\Gal(L_n/\FF_q) = C_n = \gens{\Frob_q}$ where $\Frob_q(x) \da x^q$ has order $n$. So $\Gal(L/K) \embeds \prod C_n$; which subset is it? If $g_n$ is in the image, then $g_n \mod m = g_m$ for all $m\divides n$, so \[ \Gal(L/K) = \cocolim_n C_n = \hat\ZZ ,\] the profinite integers. ::: :::{.warnings} \[ \FF_q \subseteq \FF_{q^2} \not\subseteq \FF_{q^3} \] for dimension reasons, so one can't form the usual directed system here. ::: ## Local Fields :::{.remark} For us, the examples of local fields will be $K/\QQpadic$ finite extensions. To be happy with $\QQpadic$, see exercises in Cassels' book on elliptic curves. Choose an algebraic closure $\bar K/K$, which is unique but not up to unique isomorphism. Slight issue with isomorphisms here: picking two algebraic closures $\bar K, \bar K'$ and two random isomorphisms $i: \bar K \to \bar K'$ and $j: \bar K'\to \bar k$, it may not be that $ij = \id$. On $\Gal(\bar K/K)$, 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). ::: :::{.remark} Recall that there is a (normalized) valuation $v: \QQpadic \to \ZZ\union\ts{\infty}$ where $v(p^n u) = n$ for $u\in \ZZpadic\units$. Similarly $K \contains \OO_K \contains \mfp_K$ its unique maximal/prime ideal, which is principal and generated by a uniformizer $\mfp_K = \gens{\pi_K}$. Thus for every $k\in K$ we can write $k = \pi_K^n u$, and we have a valuation \[ v: K\units &\to \ZZ \\ \pi_K &\mapsto 1 \\ \pi_K^n u &\mapsto n .\] ::: :::{.definition title="Unramified extensions"} Let $L/K$ be algebraic, possibly infinite. The valuation $v_K$ extends to $L\units \to \QQ$, and $L \contains \OO_L = \ts{0} \union \ts{\lambda \in L \st v(\lambda) \geq 0}$. Note $\OO_L \contains \mfp_L$ its unique maximal ideal, which need not be principal here. Write $\kappa_L = \OO_L/\mfp_L$, which is an algebraic extension of $\kappa_K \da \OO_K/\mfp_K$, which here is a finite field. If $L/K$ is Galois, there is a surjective map \[ \Gal(L/K)\surjects \Gal(\kappa_L/\kappa_K) .\] This need not be injective in general, so we say $L/K$ is **unramified** iff this is a bijection. ::: :::{.proposition title="?"} Let $K/\QQpadic$ be a finite extension and $\mfp_K \in \spec \OO_K$ with $\mfp_K = \gens{\pi_K}$ generated by a uniformizer. TFAE: - $\mfp_L = \pi_K \OO_L$, so the same element generates the maximal ideal of $L$, - $v_K: L\units \to \QQ$ has image $\ZZ$. - $L/K$ is unramified. ::: :::{.remark} The compositum of two unramified extensions of $K$ is again unramified, and if $L/K$ there is a unique unramified subextension $M$ with $L \contains M \contains K$. Moreover $\Gal(M/K) = \Gal(\kappa_M/\kappa_L)$, since all unramified extensions are Galois, and this will be a procyclic group. :::