# Lecture 9 :::{.remark} Recall that we've seen the following reps of $\GL_2(K)$ for $K\in \Field_{\padic}$: - $I(\chi_1, \chi_2) \subseteq \Grp(\GL_2(K), \CC_\disc)$ with $\chi_1/\chi_2 \neq \norm{\wait}^{\pm 1}$, - $S(\chi, \chi\times \norm{\wait})$, a subset/quotient of $I$, - $\mathrm{BC}_L^K(\psi) \in \Grp(L\units, \CC\units)$ admissible with $\psi \neq \psi \circ \sigma$, - $\chi \circ \det$, a subset/quotient of $I$ which is 1-dimensional. For $p>2$, these are all of the smooth admissible irreducible reps. ::: :::{.exercise title="?"} Prove that $I$ is smooth and admissible. ::: ## Conductors :::{.remark} Let $\pi$ be a smooth admissible irreducible representation of $\GL_2(K)$ 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 $\GL_2$, $\pi$ is generic iff $\dim \pi = \infty$. ::: :::{.theorem title="Casselman"} For $n\geq 0$, define \[ G \da U_1(\mfp_K^n) = \ts{M\da \matt a b c d \in \GL_2(\OO_K) \st M \equiv \matt * * 0 1 \mod \mfp^n} \leq \GL_2(K) ,\] which is a local analog of $\Gamma_1(p^n)$ for modular forms. These are all compact and open, and therefore by admissibility \[ d(\pi, n) \da \dim \pi^G < \infty ,\] where the RHS denotes the $G\dash$invariants. There is a *conductor* $\mff(\pi)\in \ZZ_{\geq 0}$ such that \[ d(\pi, n) = \max(0, 1+n-\mff(\pi)) \qquad n\geq 0 ,\] i.e. the sequence is zeros for a fixed number of $n$ depending on $\pi$ and increases linearly beyond that. ::: :::{.warnings} Conductors are delicate! ::: :::{.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 $\GL_2(\OO_K)$), but the groups get smaller as $n$ increases so $d(\pi, n)$ generally increases. ::: :::{.remark} A glimpse at the proof: for interesting reps of $\GL_2(K)$, one can use *Whittaker models* that realize $\pi$ as a set of functions on $K$. ::: :::{.exercise title="?"} Compute the conductors for $I$ and $S$: - $\mff(I(\chi_1, \chi_2)) = \mff(\chi_1) + \mff(\chi_2)$ - $\mff(S(\chi, \chi \norm{})) = 1$ if $\mff(\chi) = 0$ (the unramified case) and is $2\mff(\chi)$ if $\mff(\chi) > 0$ (the ramified case). ::: ## Central Characters :::{.exercise title="?"} Show that Schur's lemma holds: if $\pi$ is an irreducible admissible representation of $\GL_2(K)$ then there is an admissible character $\chi_\pi: K\units \to \CC\units$ such that for all $\lambda \in K\units = Z(\GL_n(K))$, using the diagonal embedding \[ K\units &\to \GL_n(K) \\ \lambda &\mapsto \diag(\lambda, \lambda,\cdots, \lambda) ,\] the action is given by \[ \lambda . \pi = \chi(\pi)\lambda \] where $\chi(\pi)$ is a scalar called the **central character**. ::: :::{.exercise title="?"} Show that - $\chi_{I(\chi_1, \chi_2)} = \chi_1 \chi_2$. - $\chi_{S(\chi, \chi \norm{})} = \chi_1 \chi_2$. - $\chi_{\phi \circ \det} = \phi^2$. ::: ## Local Langlands for $\GL_2$ :::{.remark} We'll assume LLC for $\GL_1$. Some notation: for $\chi_i \in \Grp(K\units,\CC\units)$, attach $\rho_i\in \Grp(W_K, \CC\units)$. Note that $I(\chi_1,\cdots, \chi_n)$ makes sense for $\GL_n(K)$, and the fudge factor here is $\rho$, the half-sum of positive weights. There would be a corresponding semisimple Galois representation given by $\rho_1\oplus\cdots\rho_n$. ::: :::{.remark} We'll match up infinite dimensional reps of $\GL_2(K)$ with 2-dimensional reps of $G_K$: | Weil-Deligne Representations | Galois Representations | |------------------------------------------------------------------------------------------------------------------- |----------------------------------------------------------------------------------------------------------------------- | | $\pi$ | $\rho$ | | $I(\chi_1, \chi_2)$ | $(\rho_0 = \rho_1 \oplus \rho_2, N=0)$ | | $S(\chi_1, \chi_1\cdot \norm{\wait})$ | $\qty{ \rho_0 = \matt{\norm{\wait}\rho_2}{0}{0}{\rho_1}, N = \matt 0 1 0 0}$ | | $\chi_1 \circ \det$ | $\qty{ \rho_0 = \matt{\rho_1 \cdot \norm{\wait}^{1\over 2}}{0}{0}{\rho_1 \cdot \norm{\wait}^{-{ 1\over 2} } }, N=0 }$ | | $\mathrm{BC}_L^K(\psi), \psi \in \Grp(L\units, \CC\units) \leadsto_{\text{CFT }} \sigma \in \Grp(W_L, \CC\units)$ | $\qty{\rho_0 = \Ind_{W_L}^{W_K} \sigma, N=0}$ | | Conductors $\mff(\pi)$ | $\mff(\rho_0, N)$ | | Central characters $\chi_\pi$ | $\det \rho_0$ | | | | | | | 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 **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 $\QQ_2$. This can't happen for $\QQpadic$ for $p>2$: - Wild inertia must be trivial, since it is pro-$p$, - Tame inertia must be cyclic - The unramified extension must also be cyclic So this would force $S_4$ to have a cyclic subgroup with a cyclic quotient, which doesn't exist. Thus no extension of a \(p\dash \)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{tikzcd} &&&& {\GL_2(\CC)} \\ \\ {W_{\QQ_2}} && {S_4} && {\PGL_2(\CC)} \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} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJXX3tcXFFRXzJ9Il0sWzIsMiwiU180Il0sWzQsMiwiXFxQR0xfMihcXENDKSJdLFs0LDAsIlxcR0xfMihcXENDKSJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMywiXFxleGlzdHMiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMywyLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XV0=) The indicated lift exists by a theorem of Tate, and is measured by an element of $H^2(W_{\QQ_2}; \CC\units) = 1$. Using this, one gets a 2-dimensional representation $W_{\QQ_2} \to \GL_2(\CC)$ 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. ::: :::{.remark} A good way of getting irreducible reps is inducing up. ::: :::{.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. ::: :::{.definition title="Conductors and determinants of a Weil-Deligne representation"} The **conductor** of a Weil-Deligne representation $(\rho_0, N)$ is defined as \[ \mff(\rho_0, N) = \mff(\rho_0) + \dim\qty{V^{S} \over (\ker N)^S},\, S \da I_{\bar K/K} ,\] where the second term is $0$ when $N=0$. The **determinant** is the 1-dimensional representation \[ \det(\rho_0) \in \Grp(W_K, \CC\units) .\] ::: :::{.exercise title="A computation"} Check that if the LLC for $\GL_2$ hold, there are matchings - $\pi \mapstofrom (\rho_0, N)$ - $\mff(\pi) \mapstofrom f(\rho_0, N)$ - $\chi_\pi \mapstofrom \det \rho$ by LCFT ::: :::{.exercise title="?"} Check that if $p > 2$, all of the $F\dash$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 $k$. ::: :::{.remark} Conductors measure to what extent something is ramified. Here $\mff(\pi)$ controls a dimension. ::: :::{.remark "Title on the unramified case"} Consider the $\mff(\pi) = 0$ unramified case. This forces - $\pi = I(\chi_1, \chi_2)$ (these are called **principal series representations**), and the characters factor: $\chi_i: K\units\to K\units/\OO_K\units \to \CC\units$. The $\chi_i$ are 1-dimensional and unramified. - $\pi = \det \circ \chi$ where $\chi$ factors as above. On the $W_K$ side, this forces $\rho = \rho_1 \oplus \rho_2$ and $\rho_i: W_K\to W_K/I_{\bar K/K}$ and $N=0$. ::: :::{.remark} Supposing $\dim \pi = \infty$ and $\mff(\pi) = 0$, the invariants $\pi^G$ for $G=\GL_2(\OO_K)$ 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 $G \in \Alg\Grp\slice K$ be connected reductive and unramified, e.g. $G = \GL_n$, and let $\pi$ be a smooth admissible representation of $G(K)$. We say $\pi$ **is unramified** if there exists a hyperspecial maximal compact subgroup $H\leq G(K)$ (e.g. $H = \GL_n(\OO_K)$) with $\pi^H\neq 0$. Note that $G$ being unramified means there is a model over $\OO_K$, so a group scheme over $\spec \OO_K$, and $G(\OO_K)$ is a hyperspecial maximal compact subgroup. ::: :::{.remark} The invariants $\pi^H$ here will be the finite piece we pick out of an infinite dimensional representation -- unfortunately it is not $\GL_2(K)$ invariant The trick: use Hecke operators. Let $G = \GL_2(K)$, 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. $U = U_1(\mfp_K^n)$ or $\GL_2(\OO_K)$, and if $g\in G$ then there exists a Hecke operator \[ [UgV] \in \mods{\CC}(\pi^V &\to \pi^U)\\ x&\mapsto \sum_{1\leq i\leq r} g_i x ,\] where we take a coset decomposition $UgV = \disjoint_{1\leq i\leq r} g_i V$, which must be finite using compactness and that $V$ is open. This is a type of averaging process or a trace. ::: :::{.exercise title="?"} Show $UgV \in \pi^U$ and is independent of the choices of $g_i$. ::: :::{.remark} Now just consider $\GL_2(K)$ with $\mff(\pi) = 0$ and $U=V=\GL_2(\OO_K)$. Define a map \[ T \da [UgV]\in \mods{\CC}(\pi^U \to \pi^U), \qquad g \da \matt{\pi_K} 0 0 1 S \da [UhV]\in \mods{\CC}(\pi^U \to \pi^U), \qquad h \da \matt{\pi_K} 0 0 {\pi_K} .\] Since $\dim \pi^U = 1$, $T$ and $S$ act by scalars $t,s\in \CC$. ::: :::{.exercise title="Forces unraveling definitions"} If $\pi = I( \lambda_1, \lambda_2)$, $\chi_1/\chi_2 = \norm{\wait}^{\pm 1}$, and $\mff(\pi) = 0$, find the values of $t$ and $s$. Possible solutions: - $t = \sqrt{q_K} \cdot (\alpha+ \beta)$ where $q_K$ is the size of the residue field - $s = \chi_\pi(\pi_K) = \alpha \beta$ where \( \alpha = \chi_1(\pi_K), \beta = \chi_2(\pi_K) \). Also do the 1-dimensional unramified case. As a consequence, show that if $\pi$ is an admissible irrep of $\GL_2(K)$ and $\mff(\pi) = 0$, 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 $N=0$, $\rho_0: W_K\to W_K/I_{\bar K/K} \cong \ZZ$, which contains $\Frob$, and we can further embed $\ZZ\embeds \GL_2(\CC)$. Then $\rho_0(\Frob)$ has characteristic polynomial \[ x^2 - - {t\over \sqrt{q_K}}x + s .\] ::: :::{.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. ::: :::{.exercise title="?"} More generally for $G = \GL_n(K)$, one can take $T_i = U \diag(\pi_K, \cdots, \pi_K, 1, \cdots, 1) U$ with $i$ copies of $\pi_K$ where $U \da \GL_n(\OO_K)$. Then the $T_i$ eigenvalue is $t_i$, what is $\charpoly_{\rho_0(\Frob)}(x)$? ::: :::{.remark} If $G = G(K)$ with $K$ unramified with $\pi$ an unramified representation of $G$, the Langlands reinterpretation of the **Satake correspondence** associates to $\pi$ a semisimple conjugacy class $C$ in ${}^L G(\CC)$. In this case, we'll have $\pi \to \rho_0 \cdot \rho_0(\Frob) = C$. 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. :::