Last time we defined what it means for an $\ell$-adic representation $r \colon G_F \to \GL_n(\bar{\mathbb{Q}}_ \ell)$ or a mod-$\ell$ representation $\bar{r} \colon G_F \to \GL_n(\bar{\mathbb{F}}_ \ell)$ to be automorphic, and we stated that this can be detected after base change.
Theorem 1 (automorphy lifting). Let $F$ be a totally real number field, and let $\ell \gt 3$ be a prime unramified in $F$ (which implies $[F(\zeta_\ell) : F] \gt 2$). Write $S = S_1 \amalg S_\ell$ be a finite set of finite places of $F$, where $S_\ell = \lbrace v \mid \ell \rbrace$ and $S_1$ is some other set. Let $r \colon G_{F,S} \to \GL_2(\bar{\mathbb{Q}}_ \ell)$ be a continuous representation, which is conjugate to some $G_{F,S} \to \GL_2(\mathcal{O}_ {\bar{\mathbb{Q}}_ \ell})$ and consider its semisimplified reduction $\bar{r} \colon G_{F,S} \to \GL_2(\bar{\mathbb{F}}_ \ell)$. Assume that
- $\bar{r} \vert_{G_{F(\zeta_\ell)}}$ absolutely irreducible and $\bar{r}$ is
- automorphic (meaning $\bar{r}$ is $\bar{r}_ {\ell,i}(\pi)$ for some regular algebraic cuspidal automorphic representation $\pi$),
- of weight $0$ (meaning its Harish-Chandra weights are $\mathrm{HC}_ \tau = \lbrace 0, -1 \rbrace$ for all $\tau$),
- and level $U_0(S_1)$ (meaning that $\pi_v$ is unramified if $v \notin S_1$ and $\pi_v^{\mathrm{Iw}_ v} \neq (0)$ if $v \in S_1$),
- if $v \in S_\ell$ then $r \vert_{G_{F_v}}$ is Fontaine–Laffaille with $\mathrm{HT}_ \tau(\bar{r}) = \lbrace 0, 1 \rbrace$ for all $\tau$,
- if $v \in S_1$ then $v \vert_{I_{F_v}}$ is unipotent.
Then $r$ is also automorphic of weight $0$ and level $U_0(S_1)$.
Here is the consequence we want to draw.
Theorem 2. Suppose $\ell \gt 3$ and let $\bar{r} \colon G_{\mathbb{Q},S} \to \GL_2(\mathbb{F}_ \ell)$ with $S = \lbrace 2, \ell \rbrace$ such that
- $\det \bar{r} = \bar{\epsilon}_ \ell^{-1}$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ \ell}}$ is Fontaine–Laffaille with Hodge–Tate numbers $\lbrace 0, 1 \rbrace$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ 2}} = \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell \end{pmatrix} \otimes \delta$ with $\delta^2 = 1$ unramified.
Then there exists $F/\mathbb{Q}$ a finite totally real Galois extension in which $\ell$ is unramified, with $F \cap \bar{\mathbb{Q}}^{\ker \bar{r}}(\zeta_\ell) = \mathbb{Q}$, such that $\bar{r} \vert_{G_F}$ is automorphic of weight $0$ and level $U_0(S_1)$ for some finite set of places $S_1$ containing the primes above $2$.
Choose $M/\mathbb{Q}$ an imaginary quadratic field in which $2$ and $\ell$ are unramified but ramified at some prime bigger than $3$. Choose a prime $p \nmid 6\ell$ which splits in $M$.
Lemma 3. There exists a finite extension $N/M$ and a continuous character $\theta \colon \mathbb{A}_ M^\times \to N^\times$ such that
- $\theta \vert_{M^\times} = \id$,
- $\theta \vert_{\mathbb{A}_ \mathbb{Q}^\times}(x) = \lVert x \rVert^{-1} x_\infty \delta_{M/\mathbb{Q}}(x)$, where $\delta_{M/\mathbb{Q}} \colon \mathbb{A}_ \mathbb{Q}^\times / \mathbb{Q}^\times N_{M/\mathbb{Q}} \mathbb{A}_ M^\times \to \lbrace \pm 1 \rbrace$ is the quadratic character,
- $\theta$ is unramified except at $p$ and the primes that ramify in $M/\mathbb{Q}$,
- if $v \mid p$ then $\lvert \theta(\mathcal{O}_ {M,v}^\times) \rvert = p-1$.
Proof.
We can first map $$ \theta \colon U = \mathbb{C}^\times \times \prod_{v \text{ unramified}} \mathcal{O}_ {M,v}^\times \times \prod_{v \text{ ramified}} (1 + \pi_v \mathcal{O}_ {M,v}) \to M(\zeta_{p-1})^\times $$ satisfying (3) and (4), by letting it be trivial everywhere except at $v \mid p$ and for $v \mid p$ setting $\theta \vert_{\mathcal{O}_ {M,v}^\times} = \chi$ of order $p-1$ and $\theta \vert_{\mathcal{O}_ {M,\bar{v}}^\times} = \chi^{-1}$. Now we extend $\theta$ to $U \mathbb{A}_ \mathbb{Q}^\times$ satisfying (2), (3), (4). To do this, we need to check that they agree on $$ U \cap \mathbb{A}_ \mathbb{Q}^\times = \mathbb{R}^\times \times \prod_{v \text{ unramified}} \mathbb{Z}_ v^\times \times \prod_{v \text{ ramified}} (1 + v \mathbb{Z}_ v). $$ But $\theta$ is trivial on this because we have made $\chi$ and $\chi^{-1}$ to cancel out. On the other hand, the formula in (2) is also trivial because the only possible thing is the contribution at infinity, but the signs of $x_\infty$ and $\delta_{M/\mathbb{Q}}(x)$ cancel out. Finally, for (1) we note that if $\alpha \in M^\times \cap U\mathbb{A}_ \mathbb{Q}^\times$ then $\alpha / c \alpha \in M^\times \cap U = \lbrace \pm 1 \rbrace \cap U = \lbrace 1 \rbrace$ tells us that $\alpha \in \mathbb{Q}^\times$. This shows that we finally have a map $$ \theta \colon M^\times U \mathbb{A}_ \mathbb{Q}^\times \to M(\zeta_n)^\times $$ satisfying (1)–(4). Now we can extend this character because $M^\times U \mathbb{A}_ \mathbb{Q}^\times \subseteq \mathbb{A}_ M^\times$ has finite index.
We choose $\ell^\prime$ such that $\ell^\prime \nmid 2 \ell p(p-1)$ at which $M$ is unramified and $\ell^\prime$ spits completely in $N$. Choose $\lambda^\prime \mid \ell^\prime$ a prime of $N$. We can then construct the $\ell$-adic character $$ \theta_{\lambda^\prime} \colon G_M^\mathrm{ab} \cong \mathbb{A}_ M^\times / M^\times M_\infty^\times \to N_{\lambda^\prime}^\times \cong \mathbb{Z}_ \ell^\times; \quad x \mapsto \theta(x) x_{\lambda^\prime}^{-1}. $$
In general, if $\Delta \subseteq \Gamma$ is index $2$ and $\chi$ is a character of $\Delta$ then we check that $$ \det \operatorname{Ind}_ \Delta^\Gamma \chi = (\chi \circ \mathrm{tr}) \delta_{\Gamma/\Delta}, $$ where the transfer map $\mathrm{tr} \colon \Gamma^\mathrm{ab} \to \Delta^\mathrm{ab}$ sends $\epsilon \mapsto \epsilon \gamma \epsilon \gamma^{-1}$ for some $\gamma \in \Gamma - \Delta$ and $\delta_{\Gamma/\Delta} \colon \Gamma/\Delta \cong \lbrace \pm 1 \rbrace$ is the quadratic character.
In our case, we will see that $$ \det \operatorname{Ind}_ {G_M}^{G_\mathbb{Q}} \theta_{\lambda^\prime} = \theta_{\lambda^\prime} \vert_{\mathbb{A}_ \mathbb{Q}^\times} \delta_{M/\mathbb{Q}} = \epsilon_{\ell^{-1}}^{-1}. $$