We had $F/\mathbb{Q}$ a totally real field with even degree and $D/F$ a quaternion algebra split at the infinity places. We also had a finite set $R$ of finite places, for each $v \in R$ and $k(v)^\times \supseteq \Delta_v \xrightarrow{\chi_v} \mathcal{O}^\times$ where $L/\mathbb{Q}_ \ell$ has $\mathcal{O}/\lambda = \mathbb{F}$. Then we had for every $\mathcal{O}$-algebra $A$ the space $$ S(U_\Delta(R), A)_ \chi = \lbrace \varphi : \varphi(\delta g z u) = \chi(u) \varphi(g) \text{ for } \delta \in D^\times, z \in \mathbb{A}_ F^{\infty\times}, u \in U_\Delta(R) \rbrace. $$ We had the torsion-free finite over $\mathcal{O}$ Hecke algebra $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi = \prod_\mathfrak{m} \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}. $$
For each $\mathfrak{m}$ we were able to construct a torsion Galois representation $$ \bar{r}_ \mathfrak{m} \colon G_F \to \GL_2(k(\mathfrak{m})), $$ unramified away from $R \cup \lbrace v \mid \ell \rbrace$ satisfying $\tr \bar{r}_ \mathfrak{m}(\mathrm{Frob}_ v) = T_v$ and $\det \bar{r}_ \mathfrak{m} = \epsilon_\ell^{-1}$. We said $\mathfrak{m}$ is non-Eisenstein if $\bar{r}_ \mathfrak{m}$ is absolutely irreducible. In this case, we were able to lift it to $$ r_\mathfrak{m} \colon G_F \to \GL_2(\mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}) $$ satisfying the properties
- $\det r_\mathfrak{m} = \epsilon_\ell^{-1}$,
- unramified away from $R \cup \lbrace v \mid \ell \rbrace$,
- Fontaine–Laffaille with Hodge–Tate weights $\lbrace 0, 1 \rbrace$ at $v \mid \ell$,
- for $v \in R$, tamely ramified at $v$ with the property that if $\sigma \in I_{F_v}$ maps to $a \in \Delta_v$ then its characteristic polynomial is $\operatorname{char}_ {r_\mathfrak{m}(\sigma)}(T) = (T-\chi_v^\mathrm{gal}(\sigma)) (T-\chi_v^\mathrm{gal}(\sigma)^{-1})$.
We also saw that $\mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}$ is generated by $T_v$ for $v \notin R \cup \lbrace v \mid \ell \rbrace \cup S$ for any finite set $S$ of finite places.
For simplicity, we extended $L$ so that $k(\mathfrak{m}) = \mathbb{F}$, and for all $\sigma \in G_F$ the eigenvalues of $\bar{r}_ \mathfrak{m}(\sigma)$ are also contained in $\mathbb{F}$.
For $v \in R$, suppose there exist $a \in F_v^\times \cap \mathcal{O}_ {F,v}$ and $\sigma \in G_{F_v}$ such that $\sigma \mapsto \mathrm{Art}(\sigma) \in G_{F_v}^\mathrm{ab}$ and the polynomial $$ X^2 - \tr \bar{r}_ \mathfrak{m}(\sigma) X + \det \bar{r}_ \mathfrak{m}(\sigma) $$ has roots $\alpha_v, \beta_v \in \mathbb{F}$ with $\alpha_v / \beta_v \notin \lbrace 1, \lvert a \rvert_ v^{\pm 1} \rbrace$. For every $\pi \subseteq \mathcal{A}(D^\times \backslash (D \otimes \mathbb{A}_ \mathbb{Q}^\infty)^\times)_ 0$ such that $$ (0) \neq \pi_\mathfrak{m}^{U_\Delta(R),\chi} \subseteq S(U_\Delta(R), \bar{\mathbb{Q}}_ \ell)_ {\chi,\mathfrak{m}}, $$ recall there were three possibilities for $\pi$ and $\pi_v^{U_\Delta(R)_ v, \chi_v}$. Because of this eigenvalue condition, we see that
- $\pi$ is infinite-dimensional,
- $\pi_v^{U_\Delta(R)_ v, \chi_v}$ is $2$-dimensional,
and consequently $r_\pi \vert_{G_{F_v}}$ factors through $G_{F_v}^\mathrm{ab}$. On the other hand, we have $$ r_\mathfrak{m} \colon G_F \to \GL_2(\mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}) \subseteq \GL_2(\tilde{\mathbb{T}}) \subseteq \GL_2\biggl( \prod_{\pi_\mathfrak{m}^{U_\Delta(R),\chi} \neq 0} \bar{\mathbb{Q}}_ \ell\biggr) $$ and hence $r_\mathfrak{m} \vert_{G_{F_v}}$ factors through $G_{F_v}^\mathrm{ab}$. By Hensel’s lemma, it follows that $$ X^2 - \tr r_\mathfrak{m}(\sigma) X + \det r_\mathfrak{m}(\sigma) = (X - A_v) (X - B_v) $$ for elements $A_v, B_v \in \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}$ lifting $\alpha_v, \beta_v$.
Consider $$ e_{\alpha_v} = \frac{r_\mathfrak{m}(\sigma) - B_v}{A_v - B_v}, \quad e_{\beta_v} = \frac{r_\mathfrak{m}(\sigma) - A_v}{B_v - A_v}. $$ These satisfy $e_{\alpha_v} + e_{\beta_v} = 1$ and $e_{\alpha_v} e_{\beta_v} = 0$. Hence we can decompose $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}^{\oplus 2} = e_{\alpha_v} \mathbb{T}^2 \oplus e_{\beta_v} \mathbb{T}^2, $$ where the fact that $r_\mathfrak{m} \vert_{G_{F_v}}$ factors through $G_{F_v}^\mathrm{ab}$ implies that both components are preserved by all $\tau \in G_{F_v}$. This shows that there are two characters $$ \chi_{\alpha_v}, \chi_{\beta_v} \colon G_{F_v} \to \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}^\times $$ such that $$ r_\mathfrak{m} \vert_{G_{F_v}} \sim \begin{pmatrix} \chi_{\alpha_v} & 0 \br 0 & \chi_{\beta_v} \end{pmatrix}. $$
We can also do a similar thing on the automorphic forms side. Recall for $b \in F_v^\times \cap \mathcal{O}_ {F,v}$ there is an operator $U_b$ acting on $S(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}$. This satisfies the polynomial $$ (X - \chi_{\alpha_v}(\mathrm{Art}(b))) (X - \chi_{\beta_v}(\mathrm{Art}(b))). $$ This is because we can reduce to $\bar{\mathbb{Q}}_ \ell$-coefficients, and those where $\pi_\mathfrak{m}^{U_\Delta(R), \chi} \neq 0$, and then $\pi_v^{\mathrm{Iw}_ v^1}$ is $2$-dimensional where the $U_b$ action as the same characteristic polynomial as $r_\pi(\mathrm{Art}(b))$. We can then construct the idempotents $$ \tilde{e}_ {\alpha_v} = \frac{U_a - B_v}{A_v - B_v}, \quad \tilde{e}_ {\beta_v} = \frac{U_a - A_v}{B_v - A_v}. $$ Then we can define $$ S(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}} = \tilde{e}_ {\alpha_v} S(U_\Delta(R), \mathcal{O}) \oplus \tilde{e}_ {\beta_v} S(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}} $$ where both summands are preserved by $U_b$ for all $b \in F_v^\times \cap \mathcal{O}_ {F,v}$ and $U_b$ acts on each component by $\chi_{\alpha_v}(\mathrm{Art}(b))$ and $\chi_{\beta_v}(\mathrm{Art}(b))$.
Choosing Taylor–Wiles primes
Let $R$ be a finite set of finite places, where for all $v \in R$ we have $q_v = \lvert k(v) \rvert \equiv 1 \pmod{\ell}$. This is the places where $r$ can be ramified. We choose an auxiliary set of primes $Q$, disjoint from $R$, such that $q_v \equiv 1 \pmod{\ell}$ for all $v \in Q$.
- For $v \in R$, we will set $\Delta_v = k(v)^\times$ for $v \in R$, because I can ensure that the ramifications are unipotent. To make the argument work, we will allow $\chi_v$ to be any $\ell$-power order.
- For $v \in Q$, we will set $\Delta_v = \ker(k(v)^\times \twoheadrightarrow k(v)^\times_\ell)$, where $k(v)^\times_\ell$ is the maximal $\ell$-power order quotient. Here we will $\chi_v = 1$.