Recall we had $S = \lbrace 2, \ell \rbrace$ and $\bar{r} \colon G_{\mathbb{Q},S} \to \GL_2(\bar{\mathbb{F}}_ \ell)$ satisfying the usual conditions. We found a finite Galois totally real $F/\mathbb{Q}$ disjoint from $\bar{\mathbb{Q}}^{\ker \bar{r}}(\zeta_\ell)$ such that $\bar{r}$ is automorphic over $F$. Then using the automorphy lifting theorem again, we found a lift $r \colon G_{\mathbb{Q},S} \to \GL_2(\mathcal{O})$ for some $\mathcal{O} \subseteq L / \mathbb{Q}_ \ell$ finite, satisfying the usual conditions with the additional property that $r \vert_{G_F}$ is automorphic.
Because of base change, we know that if $F \supseteq E \supseteq \mathbb{Q}$ has $F/E$ soluble, then $r \vert_{G_E}$ is automorphic. Then after choosing an isomorphism $i : \bar{\mathbb{Q}}_ \ell \cong \bar{\mathbb{Q}}_ 3$, we had a corresponding $3$-adic representation $$ r_E^\prime \colon G_E \to \GL_2(\bar{\mathbb{Q}}_ 3) $$ that is unramified outside $6$, satisfies $\det r_E^\prime = \epsilon_3^{-1}$ and the usual conditions, is irreducible. This had the property that $r_{\sigma E}^\prime = r_E^\prime \circ \mathrm{conj}_ \sigma$ and if $E \supseteq E^\prime$ then $r_{E^\prime}^\prime \vert_{G_E} = r_E^\prime$. If $v \nmid 6\ell$, we also had the eigenvalues of $r_E^\prime(\mathrm{Frob}_ v)$ matching with the eigenvalues of $r \vert_{G_E}(\mathrm{Frob}_ v)$.
To descend this, we looked at $\mathsf{Rep}_ {K,p}$ the Grothendieck ring of continuous semisimple $\bar{\mathbb{Q}}_ p$-representations of $G_K$. This had all the properties such as Frobenius reciprocity and Mackey’s formula and so on. Brauer now tells us that we can look write the trivial representation of $\Gal(F/\mathbb{Q})$ as $$ [\mathrm{triv}_ {\Gal(F/\mathbb{Q})}] = \sum_{\substack{\mathbb{Q} \subseteq E \subseteq F \br F/E \text{ soluble}}} n_E [\operatorname{ind}_ {E/\mathbb{Q}} \psi_E], \quad n_E \in \mathbb{Z}, \quad \psi_E \colon \Gal(F/E) \to \bar{\mathbb{Q}}_ \ell^\times. $$ This implies that we have $$ [r] = \sum_E n_E \operatorname{Ind}_ {E/\mathbb{Q}}([\psi_E] \operatorname{Res}_ {F/E}[r]). $$ We know what these restrictions look like $3$-adically, so we may now define $$ A_3 = \sum_E n_E \operatorname{Ind}_ {E/\mathbb{Q}}([i\psi_E] [r_E^\prime]). $$ Now if we can check that $(A_3, A_3) = 1$ and $\tr_1 A_3 = 2$, then $A_3 = [r^\prime]$ for some $r^\prime \colon G_\mathbb{Q} \to \GL_2(\bar{\mathbb{Q}}_ 3)$, so we are happy.
The point is that all these computations will be mirrored on the other side. We can calculate $(A_3, A_3)$ as $$ \begin{align} (A_3, A_3) &= \sum_{E,E^\prime} n_E n_{E^\prime} (\operatorname{ind}_ {E/\mathbb{Q}}([i\psi_E] [r_E^\prime]), \operatorname{ind}_ {E^\prime/\mathbb{Q}} ([i\psi_{E^\prime}], [r_{E^\prime}^\prime]))_ {\mathbb{Q},3} \br &= \sum_{E,E^\prime} n_E n_{E^\prime} ([i\psi_E] [r_E^\prime], \operatorname{res}_ {E/\mathbb{Q}} \operatorname{ind}_ {E^\prime/\mathbb{Q}} ([i\psi_{E^\prime}], [r_{E^\prime}^\prime]))_ {\mathbb{Q},3} \br &= \sum_{E,E^\prime} n_E n_{E^\prime} \sum_{\sigma \in G_{E^\prime} \backslash G_\mathbb{Q} / G_E} ([i\psi_E] [r_E^\prime], \operatorname{ind}_ {(\sigma^{-1} E^\prime) E / E} \mathrm{conj}_ \sigma^{-1} \mathrm{res}_ {E^\prime (\sigma E)/E^\prime} [i \psi_{E^\prime}] [r_{E^\prime}^\prime])_ {E,3} \br &= \sum_{E,E^\prime} n_E n_{E^\prime} \sum_{\sigma \in G_{E^\prime} \backslash G_\mathbb{Q} / G_E} ((\mathrm{res}_ {(\sigma^{-1} E^\prime) E / E} [i\psi_E]) [r_{(\sigma^{-1} E^\prime) E}^\prime], (\mathrm{conj}_ \sigma^{-1} \mathrm{res}_ {E^\prime (\sigma E)/E^\prime} [i \psi_{E^\prime}]) [r_{(\sigma^{-1} E^\prime) E}^\prime])_ {(\sigma^{-1} E^\prime) E,3} \end{align} $$ using Frobenius reciprocity, Mackey formula, and the well-behavior of $r_E^\prime$ under restriction. But now, each term is pairing between two irreducible representations, and so it is $1$ if they are isomorphic and $0$ if they are not isomorphic. Whether they are isomorphic can be tested by looking at the traces over Frobenii lifts $\mathrm{Frob}_ v$ for $v \nmid 6\ell$ places of $(\sigma^{-1} E^\prime) / E$. This shows that we can replace $r^\prime$ with $r \vert_{G_E}$ everywhere and this becomes $([r], [r])_ {\mathbb{Q}, \ell} = 1$. We also calculate $$ \tr_1 A_3 = \sum_E n_E \tr_1 \operatorname{Ind}_ {E/\mathbb{Q}}([i\psi_E] [r_e^\prime]) = \sum_E 2 n_E [E:\mathbb{Q}] = \tr_1 [r] = 2. $$ Using similar methods, we can also check that $$ \begin{align} \operatorname{res}_ {E/\mathbb{Q}} A_3 &= \sum_{E^\prime} n_{E^\prime} \operatorname{res}_ {E/\mathbb{Q}} \operatorname{ind}_ {E^\prime/\mathbb{Q}} ([i\psi_{E^\prime}] [r_{E^\prime}^\prime]) \br &= \sum_{E^\prime} \sum_{\sigma \in G_{E^\prime} \backslash G_\mathbb{Q} / G_E} n_{E^\prime} \operatorname{ind}_ {(\sigma^{-1} E^\prime) E/E} ((\mathrm{conj}_ \sigma^{-1} \mathrm{res}_ {E^\prime(\sigma E)/E^\prime} [i\psi_{E^\prime}]) [r_{(\sigma^{-1} E^\prime) E}^\prime]) \br &= \sum_{E^\prime} \sum_{\sigma \in G_{E^\prime} \backslash G_\mathbb{Q} / G_E} n_{E^\prime} (\operatorname{ind}_ {(\sigma^{-1} E^\prime) E/E} (\mathrm{conj}_ \sigma^{-1} \mathrm{res}_ {E^\prime(\sigma E)/E^\prime} [i\psi_{E^\prime}])) [r_{E}^\prime] \br &= (\operatorname{res}_ {E/\mathbb{Q}}[\mathrm{triv}]) [r_E^\prime] = [r_E^\prime] \end{align} $$ because $[r_{(\sigma^{-1} E^\prime)/E}] = \mathrm{res}_ {(\sigma^{-1} E^\prime) E/E} [r_E^\prime]$.
So we have found a representation $r^\prime \colon G_\mathbb{Q} \to \GL_2(\bar{\mathbb{Q}}_ 3)$ such that $r^\prime \vert_{G_E} = r_E^\prime$ for all $F \supseteq E$ soluble. Given any place $\bar{w}$ of $\bar{\mathbb{Q}}$ living over $w$ of $F$ and $v$ of $\mathbb{Q}$, we have $$ F \supseteq F^{\Gal(F/\mathbb{Q})_ w} \supseteq \mathbb{Q}. $$ Then $$ r^\prime \vert_{G_{\mathbb{Q},\tilde{w}}} = r_{F^{\Gal(F/\mathbb{Q})_ w}}^\prime \vert_{G_{\mathbb{Q},\tilde{w}}}. $$ This shows that $r^\prime$ is unramified outside $6$, and that for $v \nmid 6\ell$ the $r$ and $r^\prime$ have the same eigenvalues for $\mathrm{Frob}_ v$. For $v = 2$ we see that $r^\prime \vert_{G_{\mathbb{Q}_ 2}}$ has the desired form, and $v = 3$ we have that $r^\prime \vert_{G_{\mathbb{Q}_ 3}}$ is Fontaine–Laffaille with Hodge–Tate weights $0, 1$. It also satisfies $\det r^\prime = \epsilon_3^{-1}$, and so $r^\prime$ is reducible by the theorem. Then $r^\prime$ is reducible. Roughly what we are doing in this argument is that we are using that Galois groups of local fields are soluble, so that the we can “dig a soluble hole” to make the place inert.
The remaining goal is to prove the automorphy lifting theorem, which is the main topic of this course.