As always, we are starting with $\ell \gt 3$ and a representation $\bar{r} \colon G_{\mathbb{Q},S} \to \GL_2(\mathbb{F}_ \ell)$ where $S = \lbrace 2, \ell \rbrace$, which we are assuming satisfies $\det \bar{r} = \bar{\epsilon}_ \ell^{-1}$ and $r \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell^{-1} \end{pmatrix} \otimes \delta$ and $r \vert_{G_{\mathbb{Q}_ \ell}}$ is Fontaine–Laffaille with Hodge–Tate numbers $0, 1$, and is irreducible. We deduced from automorphy lifting that there is a finite extension $L/\mathbb{Q}_ \ell$ and a lift $r \colon G_{\mathbb{Q},S} \to \GL_2(\mathcal{O}_ L)$ satisfying the same conditions. There was also a totally real finite Galois extension $F/\mathbb{Q}$ such that for all $F \supseteq E \supseteq \mathbb{Q}$ with $F/E$ soluble, the restriction $r \vert_{G_E}$ is automorphic.
Let’s say this comes from the automorphic representation $\pi_E$. Fixing isomorphisms $i \colon \bar{\mathbb{Q}}_ \ell \cong \bar{\mathbb{Q}}_ 3$ and $j \colon \bar{\mathbb{Q}}_ \ell \cong \mathbb{C}$, we have $$ r_{\ell,j^{-1}}(\pi_E) \cong r \vert_{G_E}. $$ Let us also write $$ r_{3,ij^{-1}}(\pi_E) = r_E^\prime \colon G_{E,\lbrace 2,3 \rbrace} \to \GL_2(\bar{\mathbb{Q}}_ 3) $$ so that
- $\det r_E^\prime = \epsilon_3^{-1}$,
- at $v \mid 2$ we have $r_E^\prime \vert_{G_{E_v}} \sim \begin{pmatrix} 1 & \ast \br 0 & \epsilon_3^{-1} \end{pmatrix} \otimes \delta$ for $\delta$ unramified,
- at $v \mid 3$ we have $r_E^\prime \vert_{G_{E_v}}$ is Fontaine–Laffaille with Hodge–Tate weights $0, 1$,
- at $v \nmid 6\ell$ we have $\tr r_E^\prime(\mathrm{Frob}_ v) = i \tr r \vert_{G_E}(\mathrm{Frob}_ v)$.
Claim 1. The representation $r_E^\prime$ is irreducible.
If not, we would have $r_E^\prime \cong \chi_1 \oplus \chi_2$ where each $\chi_i$ is Fontaine–Laffaille over $3$.
Fact 2. If $E$ is a totally real field and $\chi \colon G_E \to \bar{\mathbb{Q}}_ p^\times$ is a continuous Fontaine–Laffaille (or de Rham) character then $\chi = \epsilon_p^n \psi$ for some $\psi$ of finite order.
So we can write $\chi_1 = \epsilon_3^{n_1} \psi_1$ and $\chi_2 = \epsilon_3^{n_2} \psi_2$. Then we have for all but finitely many Frobenius elements $$ \tr r \vert_{G_E}^\mathrm{ss} = i^{-1} \tr r_E^\prime(\mathrm{Frob}_ v) = i^{-1} (\lvert k(v) \rvert^{-n_1} \psi_1(\mathrm{Frob}_ v) + \lvert k(v) \rvert^{-n_2} \psi_2(\mathrm{Frob}_ v)). $$ This is a contradiction.
Generalities on ℓ-adic representations
To descend this, we do the following. For any number field $F$ and $\ell$ a prime, we consider the category $$ \mathsf{GR}_ {\ell,F} $$ of continuous semisimple linear representations of $G_F$ on finite dimensional $\bar{\mathbb{Q}}_ \ell$-vector spaces. This also has a tensor product, because in characteristic zero, the tensor product of semisimple representations is semisimple. (This can be shown by taking the Zariski closure and reducing to a statement about algebraic groups.) If $I$ is the index set for the isomorphism classes of irreducible representations, then there is a unique decomposition $V \cong \bigoplus_{i \in I} V_i^{\oplus n_i}$. If $E/F$ is a finite extension, there are restriction and induction functors $$ \mathrm{res}_ {E/F} \colon \mathsf{GR}_ {\ell,F} \to \mathsf{GR}_ {\ell,E}, \quad \mathrm{ind}_ {E/F} \colon \mathsf{GR}_ {\ell,E} \to \mathsf{GR}_ {\ell,F}, $$ and this is another abstract representation theory fact.
We can look at the Grothendieck group $\mathrm{Rep}_ {F,\ell}$, which is the free abelian group with basis $[V_i]$ for $i \in I$. There is a natural ring homomorphism $$ \tr \colon \mathrm{Rep}_ {F,\ell} \hookrightarrow \prod_{\sigma \in G_F} \bar{\mathbb{Q}}_ \ell. $$ We can also define a symmetric bilinear pairing $$ (-, -) \colon \mathrm{Rep}_ {F,\ell} \times \mathrm{Rep}_ {F,\ell} \to \mathbb{Z} $$ by declaring that $[V_i]$ form an orthonormal basis.
For any $\sigma \in G_\mathbb{Q}$, there is a map $$ \mathrm{conj}_ \sigma \colon \mathrm{Rep}_ {F,\ell} \to \mathrm{Rep}_ {\sigma F,\ell} $$ coming from composition with $\mathrm{conj}_ \sigma^{-1} \colon G_{\sigma F} \to G_F$. There is also for finite $F^\prime/F$ the maps $$ \mathrm{res}_ {F^\prime/F} \colon \mathrm{Rep}_ {F,\ell} \to \mathrm{Rep}_ {F^\prime,\ell}, \quad \mathrm{ind}_ {F^\prime/F} \colon \mathrm{Rep}_ {F^\prime,\ell} \to \mathrm{Rep}_ {F,\ell}. $$ These satisfy the obvious relations
- $\tr_\tau \circ \mathrm{conj}_ \sigma = \tr_{\sigma^{-1} \tau \sigma}$,
- $(\mathrm{conj}_ \sigma A, \mathrm{conj}_ \sigma B)_ {\sigma F, \ell} = (A, B)_ {F,\ell}$,
- $\tr_\tau \circ \mathrm{res}_ {F^\prime/F} = \tr_\tau$,
- $\mathrm{res}_ {\sigma F^\prime / \sigma F} \circ \mathrm{conj}_ \sigma = \mathrm{conj}_ \sigma \circ \mathrm{res}_ {F^\prime/F}$,
- $\tr_\sigma \mathrm{ind}_ {F^\prime/F} = \sum_{\tau \in G_F/G_{F^\prime}, \tau \sigma \tau^{-1} \in G_{F^\prime}} \tr_{\tau \sigma \tau^{-1}}$,
- $\mathrm{ind}_ {F^\prime/F} (A \mathrm{res}_ {F^\prime/F} B) = (\mathrm{ind}_ {F^\prime/F} A) B$,
- $(\mathrm{ind}_ {F^\prime/F} A, B)_ {F,\ell} = (A, \mathrm{res}_ {F^\prime/F} B)_ {F^\prime,\ell}$,
- for $F^\prime \supseteq F \subseteq F^{\prime\prime}$ the Mackey formula $$ \mathrm{res}_ {F^{\prime\prime}/F} \mathrm{ind}_ {F^\prime/F} A = \sum_{\sigma \in G_{F^\prime} \backslash G_F / G_{F^{\prime\prime}}} \mathrm{ind}_ {(\sigma^{-1} F^\prime) F^{\prime\prime} / F^{\prime\prime}} \circ \mathrm{conj}_ \sigma^{-1} \circ \mathrm{res}_ {F^\prime (\sigma F^{\prime\prime}) / F^\prime} A. $$
Let us go back to our case of $F, E, \ell, 3$. Brauer tells us that we can write $$ [\mathrm{triv}_ {G_\mathbb{Q}}] = \sum_{F/E \text{ soluble}} n_E (\operatorname{ind}_ {E/\mathbb{Q}} \psi_E), \quad \psi_E \colon G_E \twoheadrightarrow \Gal(F/E) \to \bar{\mathbb{Q}}_ \ell^\times. $$ This tells us that we have $$ [r] = \sum_E n_E \operatorname{ind}_ {E/\mathbb{Q}} [\psi_E \otimes r \vert_{G_E}] \in \mathrm{Rep}_ {\mathbb{Q},\ell}. $$ We can now define $$ A_3 = \sum_E n_E \operatorname{ind}_ {E/\mathbb{Q}} ([i\psi_E] [r_E^\prime]) \in \mathrm{Rep}_ {\mathbb{Q},3}. $$ Then we can check that $(A_3, A_3) = 1$ and this will give us $A_3 = [r^\prime]$.