Let $L/\mathcal{Q}_ \ell$ be a finite extension with ring of integers $\mathcal{O}$ and residue $\mathcal{O}/\lambda = \mathbb{F}$. Recall $\mathcal{C}_ \mathcal{O}$ is the category of complete local Noetherian $\mathcal{O}$-algebras $R$ such that $\mathcal{O}/\lambda \cong R/\mathfrak{m}$.
Let $S = \lbrack 2, \ell \rbrace$ and let $\mathbb{Q}_ S / \mathbb{Q}$ be the maximal extension unramified outside $S$, and write $G_{\mathbb{Q},S} = \Gal(\mathbb{Q}_ S/\mathbb{Q})$. We were looking at the functor $$ D \colon \mathcal{C}_ \mathcal{O} \to \mathsf{Sets} $$ where the $R$-points are liftings $r \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(R)$ of $\bar{r}$ such that
- (A) $\det r = \epsilon_\ell^{-1}$,
- (B) $r \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} 1 & \ast \br 0 & \epsilon_\ell^{-1} \end{pmatrix}$,
- (C) for all $J \subseteq R$ open $r \bmod{J}$ is in the image of the Fontaine–Laffaille functor $\mathbb{G}$,
up to the equivalence relation $r \sim A r A^{-1}$ for $A \in \ker(\mathrm{GL}_ 2(R) \to \mathrm{GL}_ 2(R/\mathfrak{m}_ R))$.
Proposition 1. The functor $D$ is represented by some $(R^\mathrm{univ}, [r^\mathrm{univ}])$.
The idea is that if we forget these three conditions (A), (B), (C), this functor $D_0$ is pro-representable when $\bar{r}$ is irreducible. To show it is representable, we need some bound on the tangent space. This happens when for every open subgroup $H \subseteq G_{\mathbb{Q},S}$ the quotient $$ H / \langle h_1 h_2 h_1^{-1} h_2^{-1}, h_1^\ell : h_1, h_2 \in H \rangle $$ is finite. This is indeed true in our case because $H$ corresponds to some finite extension $K/\mathbb{Q}$ and then the quotient corresponds to some class group by class field theory.
In general, let’s say that we have a condition $\mathcal{P}$ we want to impose. We want to check that
- (i) $\bar{r}$ has the three properties (A), (B), (C), and these properties are preserved under conjugation,
- (ii) if $r$ is a lift of $\bar{r}$ to $R$ with the property and if $f \colon R \to S$ is a morphism of $\mathcal{O}$-algebras then $f(r)$ as the property,
- (iii) if $r$ is a lift of $\bar{r}$ to $R$, and there are $I_i \subseteq R$ nested ideals with $\bigcap I_i = (0)$ such that $r \bmod{I_i}$ all have the property then so does $r$,
- (iv) if $R_1, R_2$ and $r_1, r_2$ are two liftings with the property, if $I_i \subseteq R_i$ are ideals with isomorphisms $R_1 / I_1 \cong R_2 / I_2$ and $r_1 \bmod{I_1} \sim r_2 \bmod{I_2}$ under this isomorphism, then $r = r_1 \times r_2 = G_\mathbb{Q} \to \mathrm{GL}_ 2(R_1 \times_{R_1/I_1} R_2)$ has the property,
- (v) if $R \hookrightarrow S$ and $r$ is a lifting of $\bar{r}$ to $R$, if $f(r)$ has the property then so does $r$.
Now the claim is that if these are satisfied, then we can look at $$ \mathfrak{X} = \lbrace I \subseteq R_0^\mathrm{univ} : r_0^\mathrm{univ} \bmod{I} \text{ has property} \rbrace $$ and (iii) plus Zorn’s lemma tells us that $\mathfrak{X}$ has a minimal element in $I^\mathrm{min}$. Now (iv) implies that for all $I, J \in \mathfrak{X}$ then $I \cap J \in \mathfrak{X}$ and so we can use $I^\mathrm{min} \subseteq I$ for all $I \in \mathfrak{X}$. Now we can define $$ R^\mathrm{univ} = R_0^\mathrm{univ} \vert_{I^\mathrm{min}}. $$ Now using (v) we can check that this is indeed the universal one.
Proposition 2. The conditions (A), (B), (C) satisfy these five conditions (i)–(v).
The determinant condition (A) should be easy. But already for (B) this is not trivial.
Example 3. Let $\sigma \in G_\mathbb{Q}$ such that $\bar{r}(\sigma) = I_2$. Let us consider the condition (D) that states that $$ r(\sigma) \sim \begin{pmatrix} 1 & \ast \br 0 & 1 \end{pmatrix}. $$ We claim that this does not satisfy (iv). Let us lift it to $R_1 = \mathbb{F}[\epsilon]$ and $R_2 = \mathbb{F}[\delta]$ with $\epsilon^2 = \delta^2 = 0$ where $$ r_1(\sigma) = \begin{pmatrix} 1 & \epsilon \br 0 & 1 \end{pmatrix}, \quad r_2(\sigma) = \begin{pmatrix} 1 & 0 \br \delta & 1 \end{pmatrix}. $$ These are both conjugate to an upper-triangular matrix. But the glued thing $$ r(\sigma) = \begin{pmatrix} 1 & \epsilon \br \delta & 1 \end{pmatrix} $$ is not conjugate to an upper-triangular matrix.
So for (B), if $r$ lifts $\bar{r}$ and $\phi \in G_{\mathbb{Q}_ 2}$ lifts $\mathrm{Frob}_ 2$ then there exists a basis $e_1, e_2$ of $R^2$ such that $r(\phi) e_1 = \alpha e_1$ and $r(\phi) e_2 = \beta e_2$ where $\alpha \equiv 1 \pmod{\mathfrak{m}}$ and $\beta \equiv 2 \pmod{\mathfrak{m}}$ because $(R, \mathfrak{m})$ is henselian. This $e_1, e_2$ are unique up to multiplication by $R^\times$, and so we can rigidity the situation. More precisely, the point is that if $$ r \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} 1 & \ast \br 0 & \epsilon_\ell^{-1} \end{pmatrix} $$ then it also has the same form with respect to the basis $e_1, e_2$, because the change of basis matrix has to be upper-triangular. Now all the properties can be verified using these basis.
The condition (C) was on all Artinian quotients, so we can assume all rings are Artinian. For instance for (iv) we use that $R_1 \times_{R_1/I_1} R_2 \subseteq R_1 \times R_2$ and that the image of $\mathbb{G}$ is closed under subquotients.