We had $D/F$ and a finite set of places $R \amalg Q \amalg \lbrace v \mid \ell \rbrace$. For $v \in R$, we had $q_v \equiv 1 \pmod{\ell}$ and $\Delta_v = k(v)^\times$ and $\chi_v \colon k(v)^\times \to \mathcal{O}^\times$ of $\ell$-power order. Then we had a Hecke algebra $\mathbb{T}(U_\Delta(R), \mathbb{F})_ \chi$ that is actually independent of $\chi$. Associated to a maximal ideal $\mathfrak{m}$, there is $\bar{r}_ \mathfrak{m}$ an irreducible representation, unramified outside $\ell$ and $\det \bar{r}_ \mathfrak{m} = \bar{\epsilon}_ \ell^{-1}$ and $\bar{r}_ \mathfrak{m} \vert_{G_{F_v}} = 1$ for all $v \in R$.
For $v \in Q$, we also have $q_v \equiv 1 \pmod{\ell}$ and $\bar{r}_ \mathfrak{m}(\mathrm{Frob}_ v$ has distinct eigenvalues $\alpha_v, \beta_v$. Here, we choose $\Delta_v$ to be the maximal prime-to-$\ell$ subgroup of $k(v)^\times$ and $\chi_v = 1$. Then we further have $$ \mathbb{T}(\chi, Q; \mathcal{O}) = \mathbb{T}(U_{\Delta_Q}(R \cup Q), \mathcal{O})_ {\chi,\mathfrak{m}} \twoheadrightarrow \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}. $$ We also defined $$ S(\chi, Q; \mathcal{O}) = \Bigl( \prod_{v \in Q} \tilde{e}_ {\alpha_v} \Bigr) S(U_{\Delta_Q}(R \cup Q), \mathcal{O})_ {\chi,\mathfrak{m}}, $$ which $\mathbb{T}(\chi, Q; \mathcal{O})$ acted on. Then there was a Galois representation $$ G_F \to \GL_2(\mathbb{T}(\chi, Q; \mathcal{O})) $$ where for each $v \in Q$ we had for $v \in Q$ we had $$ r_Q^\mathrm{mod} \vert_{G_{F_v}} \sim \begin{pmatrix} \chi_{\alpha_v} & 0 \br 0 & \chi_{\beta_v} \end{pmatrix}. $$ Here, $H_Q = \prod_{v \in Q} k(v)^\times / \Delta_v$ is the maximal $\ell$-power quotient of $\prod_{v \in Q} k(v)^\times$ and it acts on $S(\chi, Q; \mathcal{O})$, where the map to $\mathbb{T}(\chi, Q; \mathcal{O})$ is via $\prod_{v \in Q} \chi_{\alpha_v}$.
We showed that $S(\chi, Q; \mathcal{O})$ is a finite free $\mathcal{O}[H_Q]$-module. If we define $\Delta_v^0 = k(v)^\times$, then we have an isomorphism $$ \tr_{H_Q} \colon S(\chi, Q; \mathcal{O}) / \mathfrak{a}_ Q \cong S(\chi, Q; \mathcal{O})^{H_Q} = \Bigl( \prod_{v \in Q} \tilde{e}_ {\alpha_v} \Bigr) S(U_{\Delta_Q^0}(R \cup Q), \mathcal{O})_ {\chi,\mathfrak{m}}. $$
Claim 1. We have an isomorphism $\Bigl( \prod_{v \in Q} \tilde{e}_ {\alpha_v} \Bigr) S(U_{\Delta_Q^0}(R \cup Q), \mathcal{O})_ {\chi,\mathfrak{m}} \cong S(\chi, \emptyset; \mathcal{O})$.
To show this, we defined a trace map $t$ from the left to the right and also the map $e = \prod_{v \in Q} \tilde{e}_ {\alpha_v}$ from the right to the left. Then we checked that $t \circ e$ is an automorphism.
Claim 2. The two sides have the same rank over $\mathcal{O}$.
To check this, we can tensor with $\bar{\mathbb{Q}}_ \ell$. On the left hand side, the dimension is $$ \sum_\pi \dim_{\bar{\mathbb{Q}}_ \ell} \pi_\mathfrak{m}^{U_\Delta(R),\chi} = \sum_\pi \dim_{\bar{\mathbb{Q}}_ \ell} \pi_\mathfrak{m}^{Q,U_\Delta(R),\chi} $$ because $\pi_v^{\GL_2(\mathcal{O}_ v)}$ are $1$-dimensional. On the right hand side, we have $$ \sum_\pi \dim_{\bar{\mathbb{Q}}_ \ell} \Bigl( \pi_\mathfrak{m}^{Q,U_\Delta(R),\chi} \otimes \bigotimes_{v \in Q} \tilde{e}_ {\alpha_v} \pi_v^{\mathrm{Iw}_ v} \Bigr) = \sum_\pi \pi_\mathfrak{m}^{Q,U_\Delta(R),\chi}, $$ which is what we want. Here, the reason is that if $\pi_v^{\mathrm{Iw}_ v}$ is nonzero but $\pi_v^{\GL_2(\mathcal{O}_ {F,v})}$ is zero, but $\pi_v$ has to be Steinberg, but we have excluded this possibility because $\alpha_v \neq \beta_v$ and $q_v \equiv 1 \pmod{\ell}$.
The universal Galois deformation
We now want to introduce the corresponding universal deformation rings. This is the deformation ring for lifts $r$ of $\bar{r}_ \mathfrak{m}$, such that
- $r$ is unramified outside $R \cup Q \cup \lbrace v \mid \ell \rbrace$,
- $\det r = \epsilon_\ell^{-1}$,
- if $v \mid \ell$ then $r \vert_{G_{F_v}}$ is Fontaine–Laffaille,
- if $v \in R$ and $\sigma \in I_{F_v}$ then $r(\sigma)$ has characteristic polynomial $(X - \chi_v^\mathrm{gal}(\sigma)) (X - \chi_v^\mathrm{gal}(\sigma)^{-1})$.
This has no condition at $Q$ aside from the fact that $\det r = \epsilon_\ell^{-1}$. Then we have a map $$ R_{\chi,Q}^\mathrm{univ} \twoheadrightarrow \mathbb{T}(\chi, Q; \mathcal{O}), $$ where surjectivity comes from the fact that $\mathbb{T}$ is generated by $\tr r^\mathrm{univ}(\mathrm{Frob}_ v)$.
If $v \in Q$ and $\varphi \in G_{F_v}$ lifts $\mathrm{Frob}_ v$, we have that $\bar{r}_ \mathfrak{m}(\varphi)$ has distinct eigenvalues $\alpha_v, \beta_v \in \mathbb{F}$. So by Hensel’s lemma $r^\mathrm{univ}(\varphi)$ also have distinct eigenvalues $A_v, B_v \in R_{\chi,Q}^\mathrm{univ}$. So we can choose a basis such that this looks like $$ r^\mathrm{univ}(\varphi) \sim \begin{pmatrix} A_v & 0 \br 0 & B_v \end{pmatrix}. $$
Because $r^\mathrm{univ}(I_{F_v})$ is a pro $\ell$-group, the representation $r^\mathrm{univ} \vert_{I_{F_v}}$ factors through $I_{F_v} \twoheadrightarrow \mathbb{Z}_ \ell(1)$. Choosing a tame generator $\sigma$, let us write $$ r^\mathrm{univ}(\sigma) = \begin{pmatrix} x & y \br z & w \end{pmatrix}, \quad x-1, w-1, y, z \in \mathfrak{m}. $$ Let $I = (y, z) \subseteq R_{\chi,Q}^\mathrm{univ}$ be the ideal generated by $y, z$. Then we have the identity $$ r^\mathrm{univ}(\varphi)^{-1} r^\mathrm{univ}(\sigma) r^\mathrm{univ}(\varphi) = r^\mathrm{univ}(\sigma)^{q_v} $$ that becomes $$ \begin{pmatrix} x & B_v / A_v y \br A_v / B_v z & w \end{pmatrix} \equiv \begin{pmatrix} x^{q_v} & y \br z & w^{q_v} \end{pmatrix} \pmod{\mathfrak{m}I}. $$ Because $B_v/A_v - 1$ and $A_v/B_v - 1$ are both invertible, we see that $y, z \in \mathfrak{m}I$. Now by Nakayama we conclude that $I = 0$.
Conclusion 3. For every $v \in Q$, we have $$ r^\mathrm{univ} \vert_{G_{F_v}} = \begin{pmatrix} \chi_{\alpha_v} & 0 \br 0 & \chi_{\beta_v} \end{pmatrix}, $$ where $(\chi_{\alpha_v} \bmod{\mathfrak{m}})(\mathrm{Frob}_ v) = \alpha_v$ and similarly for $\beta$.
If we give an $H_Q$-action on $R_{\chi,Q}^\mathrm{univ}$ via $\prod_{v \in Q} \chi_{\alpha_v} \circ \mathrm{Art}$, then the map $$ R_{\chi,Q}^\mathrm{univ} \twoheadrightarrow \mathbb{T}(\chi, Q; \mathcal{O}) $$ is $\mathcal{O}[H_Q]$-linear. We also have a natural map $$ R_{\chi,Q}^\mathrm{univ} / \mathfrak{a}_ Q \xrightarrow{\cong} R_{\chi,\emptyset}^\mathrm{univ} $$ that is an isomorphism because they have the same points because of the previous analysis.