We had $L/\mathbb{Q}_ \ell$ finite with $\mathcal{O} = \mathcal{O}_ L$ and $\mathcal{O} / \lambda = \mathbb{F}$, and $F$ was a totally real field with a finite set of finite primes $R$ such that for all $v \in R$ we have $q_v \equiv 1 \pmod{\ell}$ and $\chi_v \colon k(v)^\times \to \mathcal{O}^\times$ of $\ell$-power order. We were looking at the universal lifting ring $R_{\chi_v}^\square$ for $\mathrm{triv} \colon G_{F_v} \to \mathrm{GL}_ 2(\mathbb{F})$ such that
- for $\sigma \in I_{F_v}$ we have $\operatorname{char}_ {r(\sigma)}(x) = (x - \chi_v(\sigma)) (x - \chi_v(\sigma)^{-1})$,
- $\det r = \epsilon_\ell^{-1}$.
Then we were trying to study $$ R_\chi^\mathrm{loc} = \bigotimes_{v \in R}^\wedge R_{\chi_v}^\square. $$
Proposition 1. 1. If $\chi_v \neq 1$ for all $v \in R$ then $\operatorname{Spec} R_\chi^\mathrm{loc}$ is irreducible of dimension $1 + 3\lvert R \rvert$ with generic point of characteristic $0$.
- If $\chi_v = 1$ for all $v \in R$ and $C_1, \dotsc, C_r$ are the irreducible components of $\operatorname{Spec} R_1^\mathrm{loc}$ then $C_i$ have dimension $1 + 3 \lvert R \rvert$, generic point of characteristic zero, and irreducible special fibers that are distinct.
We still needed to prove in the case $\chi_v \neq 1$ for all $v$ that $R_\chi^\mathrm{loc}$ has connected smooth rigid generic fiber.
The smoothness is a fairly formal argument. What we want is that if $\mathfrak{p}$ is a maximal ideal of $R_\chi^\mathrm{loc}[1/\ell]$ then $R_\chi^\mathrm{loc}[1/\ell]_ \mathfrak{p}^\wedge$ is a power series ring over $k(\mathfrak{p})$. We can do this in an elementary way. We have this isomorphism $$ R_ \chi^\mathrm{loc}[1/\ell] / \mathfrak{p} \cong k(\mathfrak{p}). $$ First using deformation theory we check $$ R_\chi^\mathrm{loc}[1/\ell] / \mathfrak{p}^2 \cong k(\mathfrak{p})[[x_1, \dotsc, x_r]] / (x_1, \dotsc, x_r)^2. $$ Then we want to successively build $$ f_s \colon R_\chi^\mathrm{loc}[1/\ell] / \mathfrak{p}^s \to k(\mathfrak{p})[[x_1, \dotsc, x_r]] / (x_1, \dotsc, x_r)^s. $$ Then $f_s$ is automatically surjective and then by counting dimension we see that $f_s$ is an isomorphism.
As before, we set $\tau_v \in I_{F_v}$ a generator of tame inertia, $\phi_v \in G_{F_v}$ lift of Frobenius, and $\zeta_v = \chi_v(\tau_v)$. What we want is to lift $A/\mathfrak{p}^s \cong B/I$ to a map $A/\mathfrak{p}^{s+1} \to B$. Over $R_\chi^\mathrm{loc}[1/\ell]$, we can find eigenvectors $$ r_v(\tau_v) e_{1,v} = \zeta_v e_{1,v}, \quad r_v(\tau_v) e_{2,v} = \zeta_v^{-1} e_{2,v}. $$ Then we will have $$ r_v(\phi_v) e_{1,v} = \alpha_{1,v} e_{1,v}, \quad r_v(\phi_v) e_{2,v} = \alpha_{2,v} e_{2,v} $$ for $\alpha_{i,v} \in B/I$ that reduces to $\bar{\alpha}_ {i,v} = \alpha_{i,v} \bmod{\mathfrak{p}}$ in $k(\mathfrak{p})$. Now choose $\tilde{\alpha}_ {i,v} \in B$ lifting $\alpha_{i,v}$ modulo $\mathfrak{p}^s$. Writing $\mathcal{O}_ {k(\mathfrak{p})}^1$ for the set of $\beta \in \mathcal{O}_ {k(\mathfrak{p})}$ whose image in its residue field is in $\mathbb{F}$, we consider $$ B^0 = \mathcal{O}_ {k(\mathfrak{p})}^1[\tilde{\alpha}_ {i,v} - \bar{\alpha}_ {i,v}] \subseteq B $$ where we note that $\tilde{\alpha}_ {i,v} - \bar{\alpha}_ {i,v}$ is nilpotent. Then $B^0$ is a complete local Noetherian $\mathcal{O}$-algebra with residue field $\mathbb{F}$. Now we can lift $r_v$ to $$ r_v^1 \colon G_{F_v} \to \mathrm{GL}_ 2(B^0) $$ given by
- $r_v^1(\tau_v) e_{1,v} = \zeta_v e_{1,v}$ and $r_v^1(\tau_v) e_{2,v} = \zeta_v^{-1} e_{2,v}$ and
- $r_v^1(\phi_v) e_{1,v} = \tilde{\alpha}_ {1,v} e_{1,v}$ and $r_v^1(\phi_v) e_{2,v} = \tilde{\alpha}_ {2,v} e_{2,v}$.
We now need connectedness of $R_\chi^\mathrm{loc}[1/\ell]$. To do this, we make a chain that connects the representations.
Claim 2. Suppose $\mathfrak{p}$ is a maximal ideal of $R_\chi^\mathrm{loc}[1/\ell]$. Then there exists a maximal ideal $\mathfrak{p}^\prime$ in the same irreducible component such that for all $v \in R$ the representation $r_v \bmod{\mathfrak{p}^\prime}$ is upper triangular and $$ \tau_v \mapsto \begin{pmatrix} \zeta_v & \ast \br 0 & \zeta_v^{-1} \end{pmatrix}. $$
Fix a primitive $0 \neq e_{1,v}$ of $\mathcal{O}_ {k(\mathfrak{p})}^2$ such that $$ r_v(\tau_v) e_{1,v} = \zeta_v e_{1,v}, \quad r_v(\phi_v) e_{1,v} = \alpha_{1,v} e_{1,v}. $$ (Note that we can’t do the same thing for the $\zeta_v^{-1}$-eigenvector if we want both things to be integral.) Extend this to a basis $e_{1,v}, e_{2,v}$ of $\mathcal{O}_ {k(\mathfrak{p})}^2$ and write $A_v = (e_{1,v} ; e_{2,v})$. Then we have $$ A_v^{-1} r_v(\tau_v) A_v = \begin{pmatrix} \zeta_v & \ast \br 0 & \zeta_v^{-1} \end{pmatrix}, \quad A_v^{-1} r_v(\sigma_v) A_v = \begin{pmatrix} \alpha_{1,v} & \ast \br 0 & \ast \end{pmatrix}. $$
Once we have this, we can look at the ring $$ S = \mathcal{O}_ {k(\mathfrak{p})}\langle x_{v,i,j}, y_v \rangle / (\det x_{v,i,j} y_v - 1). $$ Then there is a map $$ \operatorname{Spec} S \to \mathcal{M}_ v; \quad ((x_{v,i,j})^{-1} r_v(\tau_v) (x_{v,i,j}), (x_{v,i,j})^{-1} r_v(\sigma_v) (x_{v,i,j})). $$ Modulo $\ell$, we are not changing the matrix because both $r_v(\tau_v)$ and $r_v(\sigma_v)$ are trivial. That is, this factors as a map $$ \operatorname{Spec} S \to \operatorname{Spec} R_\chi^\mathrm{loc}. $$ Now we are connecting the two points $\mathfrak{p}$ and $\mathfrak{p}^\prime$ corresponding to conjugation by $A_v$ through a connected rigid analytic variety.
Claim 3. With $\mathfrak{p}$ as above, there exists a $\mathfrak{p}^\prime$ in the same connected component of $\operatorname{Spec} R_\chi^\mathrm{loc}[1/\ell]$ with each $r_v \bmod{\mathfrak{p}^\prime}$ being diagonal and $r_v(\tau_v) \bmod{\mathfrak{p}^\prime} = \operatorname{diag}(\zeta_v, \zeta_v^{-1})$.
We can carry the exact same argument with $$ S = \operatorname{Spec} \mathcal{O}_ {k(\mathfrak{p})}\langle y_v \rangle \to \operatorname{Spec} R_\chi^\mathrm{loc}; \quad (\operatorname{diag}(1, y_v^{-1}) r(\phi_v) \operatorname{diag}(1, y_v), \operatorname{diag}(1, y_v^{-1}) r(\tau_v) \operatorname{diag}(1, y_v)), $$ noting that once $r_v$ is already upper triangular we actually don’t need $y_v$ to be invertible. Now degenerating from $y_v = 1$ to $y_v = 0$ gives the claim.
Claim 4. For $\mathfrak{p}$ as above, it is in the same connected component as $\mathfrak{p}_ 0$ which has $$ r_{\mathfrak{p}_ 0}(\tau_v) = \operatorname{diag}(\zeta_v, \zeta_v^{-1}), \quad r_{\mathfrak{p}_ 0}(\phi_v) = \operatorname{diag}(1, q_v). $$
At this point, we have $$ (r_v \bmod{\mathfrak{p}})(\tau_v) = \operatorname{diag}(\zeta_v, \zeta_v^{-1}), \quad (r_v \bmod{\mathfrak{p}})(\phi_v) = \operatorname{diag}(\alpha_v, q_v/\alpha_v). $$ Here, we have $\alpha_v - 1$ is nilpoent, so the space of such things is $\mathcal{O}[[\alpha_v - 1]]$, which is already connected after inverting $\ell$.