Lue’s strategy is to, instead of looking at crystalline deformations, look at all deformations which is often connected.
Fact 1 (Paskunas). Let $\ell \gt 3$ and $\bar{r} \colon G_{\mathbb{Q}_ \ell} \to \mathrm{GL}_ 2(\mathbb{F})$.
- If $\bar{r}$ is irreducible when $R_{\bar{r}, \epsilon_\ell^{-1}}^\square$ is formally smooth (as $H^2(G_{\mathbb{Q}_ \ell}, \mathrm{ad}^0 \bar{r}) = H^0(G_{\mathbb{Q}_ \ell}, (\mathrm{ad}^0 \bar{r})(1))^\vee$ vanishes) and hence is isomorphic to $\mathcal{O}[[x_1, \dotsc, x_6]]$.
- If we have $\bar{r} = \begin{pmatrix} \chi & \ast \br 0 & \chi^{-1} \bar{\epsilon}_ \ell^{-1} \end{pmatrix}$ non-split, then as long as $\chi^2 \neq \bar{\epsilon}_ \ell^{-2}$ it is again formally smooth for the same reason and we get $\mathcal{O}[[x_1, \dotsc, x_6]]$.
- In the case when $\bar{r} = \begin{pmatrix} \chi \bar{\epsilon}_ \ell^{-1} & \ast \br 0 & \chi^{-1} \end{pmatrix}$ is non-split and $\chi^2 = 1$, we get the simplest possible singularity $$ \mathcal{O}[[A_1, A_2, A_3, x, y, z, w]] / (xy - zw). $$
- If $\bar{r} = \chi \oplus \chi^{-1} \bar{\epsilon}_ \ell^{-1}$, then as long as $\chi^2 \neq \bar{\epsilon}_ \ell^{-2}$ and $\chi^2 \neq 1$ it is again $\mathcal{O}[[x_1, \dotsc, x_6]]$.
- If $\bar{r} = \chi \oplus \chi^{-1} \bar{\epsilon}_ \ell^{-1}$ with $\chi^2 = 1$ we have the same $\mathcal{O}[[\cdots]] / (xy-zw)$.
Let us start describing the situation we want to treat. Let $F/\mathbb{Q}$ be totally real of even degree. We also want $\ell$ to split completely at $F$ (because we only have a theory for $\mathrm{GL}_ 2(\mathbb{Q}_ \ell)^n$). As always, $L/\mathbb{Q}_ \ell$ is finite and $\mathcal{O}/\lambda = \mathbb{F}$. For simplicity we assume $L$ contains all images of $F(\zeta_\ell) \hookrightarrow \bar{L}$. Let $$ \bar{r} \colon G_F \to \mathrm{GL}_ 2(\mathbb{F}) $$ and assume that for all $\sigma \in G_F$ the eigenvalues of $\bar{r}(\sigma)$ lie in $\mathbb{F}$. We also assume that
- $\det \bar{r} = \bar{\epsilon}_ \ell^{-1}$,
- $\bar{r}$ is unramified away from $\ell$ (which we can usually arrange using soluble base change),
- there is $R$ a finite set of primes of $F$ such that if $v \in R$ then $q_v \equiv 1 \pmod{\ell}$ and $\bar{r} \vert_{G_{F_v}} = 1$.
Given a character $\chi = \prod_{v \in R} \chi_v$ where $\chi_v \colon k(v)^\times \to \mathcal{O}^\times$ of $\ell$-power order, we can consider for each $v \in R$ the universal lifting ring $R_{\chi_v}^\square$ of $\bar{r} \vert_{G_v}$ for lifts $r$ with $\det r = \epsilon_\ell^{-1}$ and for $\sigma \in I_{F_v}$ having $\tr r(\sigma) = \chi_v(\sigma) + \chi_v(\sigma)^{-1}$. For $v \mid \ell$, we will look at the universal deformation ring $R_v^\square$ with only the determinant condition.
Now we can put them together to $$ R_\chi^\mathrm{loc} = \bigotimes_{v \mid \ell}^\wedge R_v^\square \otimes^\wedge \bigotimes_{v \in R}^\wedge R_{\chi_v}^\square. $$ This has dimension $6\lvert \lbrace v \mid \ell \rbrace \rvert + 3 \lvert R \rvert + 1$. We also consider $Q$ a finite set of primes of $F$ such that
- $q_v \equiv 1 \pmod{\ell}$,
- $\bar{r}(\mathrm{Frob}_ v)$ has distinct eigenvalues $\alpha_v \neq \beta_v \in \mathbb{F}$ (and hence $Q \cap (R \cup \lbrace v \mid \ell \rbrace) = \emptyset$).
As before, we will consider the global universal deformation ring $R_{Q,\chi}^\mathrm{univ}$ for lifts $r$ of $\bar{r}$ such that
- $\det r = \epsilon_\ell^{-1}$,
- $r$ is unramified away from $R \cup Q \cup \lbrace v \mid \ell \rbrace$,
- if $v \in R$ and $\sigma \in I_{F_v}$ then $\tr r(\sigma) = \chi_v(\sigma) + \chi_v(\sigma)^{-1}$.
There is also a version that was framed at $R \cup \lbrace v \mid \ell \rbrace$ and then $R_{Q,\chi}^\square$ was a power series ring over $R_{Q,\chi}^\mathrm{univ}$ over $4 \lvert R \cup \lbrace v \mid \ell \rbrace \rvert - 1$ variables.
Because of how we chose $Q$, we see that for $v \in Q$ we have $$ r^\mathrm{univ} \vert_{G_{F_v}} \sim \psi_{\alpha_v} \oplus \psi_{\beta_v} $$ where $(\psi_{\alpha_v} \bmod{\mathfrak{m}})(\mathrm{Frob}_ v) = \alpha_v$. Then on the inertia this gives $$ \psi_{\alpha_v} \vert_{I_{F_v}} \colon H_v \to R_{Q,\chi}^\mathrm{univ}, $$ where $H_v$ is the maximal $\ell$-power quotient of $k(v)^\times$ and $H_Q = \prod_{v \in Q} H_v$. Then we had $$ \tilde{\mathfrak{a}}_ Q \subseteq \Lambda_Q = \mathcal{O}[H_Q][[A_{v,i,j} : v \in R \cup \lbrace v \mid \ell \rbrace, 1 \le i, j \le 2]] / (A_{v_0,1,1}) \to R_{Q,\chi}^\square $$ where we have non-canonically $$ R_{Q,\chi}^\square = R_{Q,\chi}^\mathrm{univ} \otimes_{\mathcal{O}[H_Q]} \Lambda, \quad R_{Q,\chi}^\square / \tilde{\mathfrak{a}}_ Q \cong R_{\emptyset,\chi}^\mathrm{univ}. $$
We now have $$ R_\chi^\mathrm{loc}[[x_1, \dotsc, x_s]] \twoheadrightarrow R_{Q,\chi}^\square $$ where the computation yields $$ s = \lvert R \rvert + \lvert Q \rvert - 1 + \dim H^1_{\mathcal{L}_ Q^\perp}(G_{F, \lbrace v \mid \ell \rbrace \cup R}, (\mathrm{ad}^0 \bar{r})(1)). $$ So its Krull dimension is $$ 6 \lvert \lbrace v \mid \ell \rbrace \rvert + 4 \lvert R \rvert + \lvert Q \rvert + \dim H^1_{\mathcal{L}_ Q^\perp}. $$ On the other hand, $\Lambda_Q$ has dimension $$ 4 \lvert \lbrace v \mid \ell \rbrace \rvert + 4 \lvert R \rvert, $$ where when we do patching we will have $\lvert Q \rvert$ more variables and $\dim H^1$ will become zero. But even so, there is a difference between $4$ and $6$ of the coefficient of $\lvert \lbrace v \mid \ell \rbrace \rvert$, and we need to understand what compensates for that.
Completed space of modular forms
Let $D/F$ be a quaternion algebra ramified exactly at the infinite places, so that $D \otimes_\mathbb{Q} \mathbb{A}^\infty \cong M_{2 \times 2}(\mathbb{A}_ F^\infty)$. Away from infinity, we will consider the level $$ \prod_{v \nmid \ell} U_{Q,v} = U_Q^\ell \subseteq \mathrm{GL}_ 2(\mathbb{A}_ F^{\infty,\ell}), \quad U_{Q,v} = \begin{cases} \mathrm{GL}_ 2(\mathcal{O}_ {F,v}) & v \notin Q \cup R \cup \lbrace v \mid \ell \rbrace, \br \lbrace (\begin{smallmatrix} a & b \br c & d \end{smallmatrix}) : v \mid c \rbrace & v \in R, \br \lbrace (\begin{smallmatrix} a & b \br c & d \end{smallmatrix}) : v \mid c, a/d \in \ker(k(v)^\times \twoheadrightarrow H_v) \rbrace & v \in Q. \end{cases} $$ We then have a character $$ \chi \colon U_Q^\ell \to \mathcal{O}^\times; \quad \begin{pmatrix} a_v & b_v \br c_v & d_v \end{pmatrix}_ v \mapsto \prod_{v \in R} \chi_v(a_v/d_v \bmod{v}). $$ At $\ell$, we will take any open compact subgroup $U_\ell \subseteq \mathrm{GL}_ 2(\mathcal{O}_ {F,\ell})$.
For $A$ an $\mathcal{O}$-module, we have the space of modular forms $$ S(U_Q^\ell U_\ell, A)_ \chi = \lbrace \varphi \colon D^\times \backslash \mathrm{GL}_ 2(\mathbb{A}_ F^\infty) / (\mathbb{A}_ F^\infty)^\times \to A : \varphi(gu) = \chi(u) \varphi(g) \rbrace. $$ As before, this decomposes as some invariants, but the action is actually trivial because $$ (g^{-1} D^\times g (\mathbb{A}_ F^\infty)^\times \cap U_Q^\ell U_\ell) / ((\mathbb{A}_ F^\infty)^\times \cap U_Q^\ell U_\ell) $$ is a finite group of order prime to $\ell$ (which follows from that $\ell \gt 3$ and $\ell$ is unramified in $F$). So we actually have $$ S(U_Q^\ell U_\ell, A)_ \chi = \bigoplus_{g \in D^\times \backslash \mathrm{GL}_ 2(\mathbb{A}_ F^\times) / (\mathbb{A}_ F^\times) U_Q^\ell U_\ell} A. $$
We then pass to infinite level at $\ell$ and consider $$ S^\mathrm{sm}(U_Q^\ell, A)_ \chi = \lbrace \varphi \colon D^\times \backslash \mathrm{GL}_ 2(\mathbb{A}_ F^\infty) / (\mathbb{A}_ F^\infty)^\times : \varphi \text{ locally constant}, \varphi(gu) = \chi(u) \varphi(g) \rbrace $$ and similarly $$ S^\mathrm{cts}(U_Q^\ell,\mathcal{O} \text{ or } L)_ \chi = \lbrace \cdots : \varphi \text{ continuous}, \cdots \rbrace. $$ This is something Emerton would call completed cohomology. We easily check that $$ S^\mathrm{cts}(U_Q^\ell, L)_ \chi = S^\mathrm{cts}(U_Q^\ell, \mathcal{O})_ \chi [1/\ell], \quad S^\mathrm{cts}(U_Q^\ell, \mathcal{O})_ \chi \cong \Hom_ \mathcal{O}(L/\mathcal{O}, S^\mathrm{sm}(U_Q^\ell, L/\mathcal{O})_ \chi). $$ The first equality is because $D^\times \backslash \mathrm{GL}_ 2(\mathbb{A}_ F^\infty) / (\mathbb{A}_ F^\times)^\times$ is compact. For the second, we can decompose with respect to the cosets in $D^\times \backslash \mathrm{GL}_ 2(\mathbb{A}_ F^\infty) / (\mathbb{A}_ F^\infty)^\times U_Q^\ell \mathrm{GL}_ 2(\mathcal{O}_ {F,\ell})$ and check that $$ C^\mathrm{cts}(\mathrm{PGL}_ 2(\mathcal{O}_ {F,\ell}), \mathcal{O}) \cong \Hom_\mathcal{O} (L/\mathcal{O}, C^\mathrm{sm}(\mathrm{PGL}_ 2(\mathcal{O}_ {F,\ell}), L/\mathcal{O})). $$ This is a general fact that holds for profinite sets.
There is also its dual $$ M(U_Q^\ell, \mathcal{O})_ \chi = \Hom(S^\mathrm{sm}(U_Q^\ell, L/\mathcal{O})_ \chi, L/\mathcal{O}), $$ sometimes called completed homology. This is a direct sum of stuff that looks like $\mathcal{O}[[\mathrm{PGL}_ 2(\mathcal{O}_ {F,\ell})]]$. Note that there is a $\mathrm{GL}_ 2(F_\ell)$-action on all of these spaces. We then check that $$ S^\mathrm{sm}(U_Q^\ell, L/\mathcal{O})_ \chi \in \mathsf{Mod}^\mathrm{sm}_ {\mathrm{GL}_ 2(F_\ell),1}(\mathcal{O}), \quad M(U_Q^\ell, \mathcal{O}) \in \mathsf{Mod}^\mathrm{augpro}_ {\mathrm{GL}_ 2(F_\ell),1}(\mathcal{O}). $$ These are also related to each other by $$ M(U_Q^\ell, \mathcal{O})_ \chi \cong \Hom_\mathcal{O}(S^\mathrm{cts}(U_Q^\ell, \mathcal{O})_ \chi, \mathcal{O}), \quad S^\mathrm{cts}(U_Q^\ell, \mathcal{O})_ \chi \cong \Hom_\mathcal{O}^\mathrm{cts}(M(U_Q^\ell, \mathcal{O})_ \chi, \mathcal{O}). $$