Recall our setting where $F$ a totally real field of even degree, $\ell$ unramified in $F$, a quaternion algebra $D/F$ ramified precisely at the infinite places, $R$ a finite set of finite places of $F$, the group $U_\Delta(R) \subseteq \GL_2(\hat{\mathcal{O}}_ F)$ for $\Delta = \prod_{v \mid R} \Delta_v$ where $\Delta_v \subseteq k(v)^\times$ and $\chi = \prod_{v \mid R} \chi_v$ where $\chi_v \colon \Delta_v \to \mathcal{O}^\times$. Here $\mathcal{O} = \mathcal{O}_ L$ is the coefficient where $L / \mathbb{Q}_ \ell$ is a finite extension with $\mathcal{O} / \lambda = \mathbb{F}$.
For $A$ a Noetherian $\mathcal{O}$-algebra, we had the space of automorphic forms $$ S(U_\Delta(R), A)_ \chi = \lbrace \varphi \colon D^\times \backslash (D \otimes \mathbb{A}^\infty)^\times / (\mathbb{A}_ F^\infty)^\times \to A, \varphi(gu) = \chi(u) \varphi(g) \text{ for } u \in U_\Delta(R) \rbrace. $$ This is finite free over $A$, and has an action of $T_v$ for $V \notin R \cup \lbrace v \mid \ell \rbrace$. Then we had $$ \mathbb{T}(U_\Delta(R), A)_ \chi \subseteq \End_A(S(U_\Delta(R), A)_ \chi) $$ generated by the $T_v$. This is still finite torsion free over $A$, and we had $$ S(U_\Delta(R), A)_ \chi = S(U_\Delta(R), \mathcal{O})_ \chi \otimes_\mathcal{O} A, \quad \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi \otimes_\mathcal{O} A \twoheadrightarrow \mathbb{T}(U_\Delta(R), A)_ \chi. $$ Because $\mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi$ is a finite algebra over $\mathcal{O}$, it is a product of local rings and hence we can write $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi = \prod_\mathfrak{m} \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}, \quad S(U_\Delta(R), \mathcal{O})_ \chi = \bigoplus_\mathfrak{m} S(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}. $$
For such a maximal ideal $\mathfrak{m} \subseteq \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi$ we have by faithfulness $$ 0 \neq S(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}} \otimes_\mathcal{O} \bar{\mathbb{Q}}_ \ell = \bigoplus_{\pi \subseteq \mathcal{A}(D^\times \backslash (D \otimes \mathbb{A}^\infty)^\times / \mathbb{A}_ F^{\infty\times})} \pi_\mathfrak{m}^{U_\Delta(R),\chi}. $$ This in particular shows that this is nonzero for one nonzero $\pi$. The corresponding Galois representation $$ r(\pi) \colon G_F \to \GL_2(\mathcal{O}_ {\bar{\mathbb{Q}}_ \ell}) \subseteq \GL_2(\bar{\mathbb{Q}}_ \ell) $$ has the property that $\bar{r}(\pi) \colon G_F \to \GL_2(\bar{\mathbb{F}}_ \ell)$ is unramified outside $R \cup \lbrace v \mid \ell \rbrace$. Here, for $v \notin R \cup \lbrace v \mid \ell \rbrace$ we have that $\tr \bar{r}(\pi)(\mathrm{Frob}_ v)$ is the eigenvalue of $T_v$ on $\pi^{U_\Delta(R),\chi}$, which is the eigenvalue of $T_v$ on $\pi_v^{\GL_2(\mathcal{O}_ {F,v})}$.
From this we see that $\mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi$ acts on $\pi_\mathfrak{m}^{U_\Delta(R),\chi}$ by scalars. So from $\pi$ we obtain $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}} \to \mathcal{O}_ {\bar{\mathbb{Q}}_ \ell} \twoheadrightarrow \bar{\mathbb{F}}_ \ell, \quad T_v \mapsto \tr \bar{r}(\pi)(\mathrm{Frob}_ v). $$ This factors through the residue field $k(\mathfrak{m})$ because of Chebotarev.
Fact 1. If $r \colon \Gamma \to \GL_2(\bar{k})$ is semisimple and $\tr r$ is valued in $k$, and if $\operatorname{Br}(k^\prime) = 0$ for all $k^\prime / k$ finite, then $r$ is conjugate to a representation $\Gamma \to \GL_2(k)$.
Applying this to our situation, we obtain $$ \bar{r}_ \mathfrak{m} \colon G_F \to \GL_2(k(\mathfrak{m})). $$ This is unramified outside $R \cup \lbrace v \mid \ell \rbrace$ and satisfies $\det \bar{r}_ \mathfrak{m} = \epsilon_\ell^{-1}$ and $\tr \bar{r}_ \mathfrak{m}(\mathrm{Frob}_ v) = T_v \bmod{\mathfrak{m}}$ for $v \notin R \cup \lbrace v \mid \ell \rbrace$. This characterizes $\bar{r}_ \mathfrak{m}$ completely, and so it is also independent of $\pi$.
Definition 2. The maximal ideal $\mathfrak{m}$ is called non-Eisenstein if $\bar{r}_ \mathfrak{m}$ is absolutely irreducible.
Now we have an embedding $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}} \hookrightarrow \prod_{\pi_\mathfrak{m}^{U_\Delta(R),\chi} \neq 0} \bar{\mathbb{Q}}_ \ell. $$ Because the left hand side is finite, this factors through some product $\prod_\pi \mathcal{O}_ {L_\pi}$ for some $L_\pi / L$ finite. But because the $\mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}$ is a local ring, this actually lands in $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}} \hookrightarrow \tilde{\mathbb{T}} = \prod_\pi^{k(\mathfrak{m})} \mathcal{O}_ {L_\pi} = \Bigl\lbrace (a_\pi) \in \prod_\pi \mathcal{O}_ {L_\pi} : a_\pi \bmod{k(\pi)} \text{ is independent of } \pi \Bigr\rbrace. $$
Upon enlarging $L_\pi$, we can make $r(\pi)$ factor through $$ r(\pi) \colon G_F \to \GL_2(\mathcal{O}_ {L_\pi}). $$ Then $r(\pi) \bmod{\lambda_{L_\pi}}$ has the same traces of $\bar{r}_ \mathfrak{m}$, and hence we may conjugate them so that $r(\pi) \bmod{\lambda_{L_\pi}} = \bar{r}_ \mathfrak{m}$.
Now taking the product of them we obtain $$ \prod r(\pi) \colon G_F \to \GL_2(\tilde{\mathbb{T}}). $$ We have also seen Carayol’s lemma, which says that in the non-Eisenstein case if the traces live in a subring $\mathbb{T}$ then it can be conjugated to a representation valued in $\mathbb{T}$. This shows that we actually have a representation $$ r_\mathfrak{m} \colon G_F \to \GL_2(\mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}). $$ This has the following properties:
- $r_\mathfrak{m} \bmod{\mathfrak{m}} = \bar{r}_ \mathfrak{m}$,
- $\det r_\mathfrak{m} = \epsilon_\ell^{-1}$,
- $r_\mathfrak{m}$ is unramified outside $R \cup \lbrace v \mid \ell \rbrace$, and for $v \notin R \cup \lbrace v \mid \ell \rbrace$ we have $\tr r_\mathfrak{m}(\mathrm{Frob}_ v) = T_v$,
- $r_\mathfrak{m}$ is Fontaine–Laffaille for all $v \mid \ell$ with Hodge–Tate numbers $\lbrace 0, 1 \rbrace$,
- if $v \in R$ and $\sigma \in I_{F_v}$ maps to $\Delta_v$ under $I_{F_v} \twoheadrightarrow I_{F_v^\mathrm{ab}/F_v} \cong \mathcal{O}_ {F_v}^\times \twoheadrightarrow k(v)^\times$ then $r_\mathfrak{m}(\sigma)$ has characteristic polynomial $(x - \chi_v^\mathrm{gal}(\sigma)) (x - \chi_v^\mathrm{gal}(\sigma)^{-1})$, where $\chi_v^\mathrm{gal}(\sigma)$ is just $\chi$ applied to the image of $\sigma$ in $\Delta_v$.
Remark 3. If $S$ is a finite set of place of $F$ and $\mathfrak{m}$ is non-Eisenstein, then $\mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}$ is generated over $\mathcal{O}$ by $T_v$ for $v \notin S \cup R \cup \lbrace v \mid \ell \rbrace$. This is because the subalgebra generated by these $T_v$ contains $\tr r_\mathfrak{m}(\mathrm{Frob}_ v)$ for $v \notin S \cup R \cup \lbrace v \mid \ell \rbrace$, and hence by Chebotarev contains $\tr r_\mathfrak{m}(\sigma)$ for all $\sigma \in G_F$. So it also contains for $v \in S - (R \cup \lbrace v \mid \ell \rbrace)$ the element $T_v = \tr r_\mathfrak{m}(\mathrm{Frob}_ v)$.