We had a prime $\ell \gt 3$, the finite set $S = \lbrace 2, \ell \rbrace$ of primes, and $$ \bar{r} \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(\mathbb{F}_ \ell) $$ satisfying
- $\det \bar{r} = \bar{\epsilon}_ \ell^{-1}$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} 1 & \ast \br 0 & \epsilon_2^{-1} \end{pmatrix} \otimes \delta$ for $\delta^2 = 1$ unramified,
- $\bar{r} \vert_{G_{\mathbb{Q}_ \ell}} = \mathbb{G}(\bar{M})$ where $\bar{M} \in \mathcal{MF}$ with $\dim \mathrm{gr}^i \bar{M} = 1$ only when $i = 0, 1$.
We were looking for $L / \mathbb{Q}_ \ell$ with $\mathcal{O} / \lambda = \mathbb{F}$ and a lift $r \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(\mathcal{O})$ with the same properties. Our strategy was to fix one such $L$ for now and looking at the category $\mathcal{C}_ \mathcal{O}$ of complete Noetherian local $\mathcal{O}$-algebras $R$ such that $\mathcal{O}/\lambda \cong R/\mathfrak{m}$, and considering the functor of lifts $r \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(R)$ satisfying
- (A) $\det r = \epsilon_\ell^{-1}$,
- (B) $r \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} 1 & \ast \br 0 & \epsilon_\ell^{-1} \end{pmatrix} \otimes \delta$ with $\delta^2 = 1$ unramified,
- (C) for all open $J \subseteq R$ we have $r \bmod{J} \in \im(\mathbb{G})$,
up to conjugation by $\ker(R \to \mathbb{F})$. We showed last time that this was represented by some $$ r^\mathrm{univ} \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(R^\mathrm{univ}). $$ What we to show is that there is a ring homomorphism $R^\mathrm{univ} \to \mathcal{O}_ {\bar{L}}$.
Let $\mathfrak{m} \subseteq R^\mathrm{univ}$ be the maximal ideal. Because $R^\mathrm{univ}$ is Noetherian, the $\mathbb{F}$-vector space $\mathfrak{m} / \langle \lambda, \mathfrak{m}^2\rangle$ is finite-dimensional, and so has a basis $\bar{y}_ 1, \dotsc, \bar{y}_ d$ for some $d$. If we lift them to $y_i \in \mathfrak{m}$, then we have ring homomorphism $$ \mathcal{O}[[x_1, \dotsc, x_d]] \twoheadrightarrow R^\mathrm{univ} $$ sending $x_i$ to $y_i$.
Lemma 1. If $A, B \in \mathcal{C}_ \mathcal{O}$ and $\phi \colon A \to B$ satisfies $\mathfrak{m}_ A / (\lambda, \mathfrak{m}_ A^2) \twoheadrightarrow \mathfrak{m}_ B / (\lambda, \mathfrak{m}_ B^2)$ then $A \twoheadrightarrow B$.
Proof.
We do some kind of successive approximation.
Now this has some kernel $I \subseteq \mathcal{O}[[x_1, \dotsc, x_d]]$. What we will now show is that $$ \Hom_\mathbb{F}(\mathfrak{m} / \langle \lambda, \mathfrak{m}^2 \rangle, \mathbb{F}) \xrightarrow{\cong} H_\mathscr{L}^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) $$ and also that $$ \Hom_\mathbb{F}(I / \mathfrak{m}I, \mathbb{F}) \hookrightarrow H_\mathscr{L}^2(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}). $$ From this we can conclude that we have a bound $$ \dim R^\mathrm{univ} \ge 1 + \dim H_\mathscr{L}^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) - \dim H_\mathscr{L}^2(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}). $$ The right hand side is going to be $1$. We will also later show that $R^\mathrm{univ}$ is finite over $\mathcal{O}$, and then there exists a map $R^\mathrm{univ} \to \bar{L}$.
Let us now start constructing $\mathscr{L} = \lbrace L_2, L_\ell \rbrace$ and defining the Selmer groups $H_\mathscr{L}^1$ and $H_\mathscr{L}^2$.
The tangent space of the universal deformation ring
For the first fact, we note that $$ \Hom_\mathbb{F}(\mathfrak{m} / \langle \lambda, \mathfrak{m}^2 \rangle, \mathbb{F}) \cong \Hom_{\mathcal{O}\mathsf{-alg}}(R^\mathrm{univ}, \mathbb{F}[\epsilon]), $$ where $\phi \colon R^\mathrm{univ} \to \mathbb{F}[\epsilon]$ correspond to $(\phi \vert_\mathfrak{m}) / \epsilon$. Now this can be interpreted as the set of lifts $$ r \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(\mathbb{F}[\epsilon]), \quad r \bmod{\epsilon} = \bar{r} $$ satisfying (A), (B), (C), up to conjugation by $I_2 + \epsilon M_{2 \times 2}(\mathbb{F})$. Writing $r = (1 + \phi \epsilon) \bar{r}$ the condition that $r$ is a representation correspond to that $$ \phi \colon G_{\mathbb{G},S} \to M_{2 \times 2}(\mathbb{F}) $$ is a cocycle, i.e., $$ \phi(\sigma \tau) = \phi(\sigma) + \mathrm{ad}(\bar{r}(\sigma))(\phi(\tau)). $$ Condition (A) says that $\tr \phi(\sigma) = 0$, and so let us write $\mathrm{ad}^0 \bar{r}$ for the set of traceless elements of $M_{2 \times 2}(\mathbb{F})$ with the action of $G_{\mathbb{Q},S}$ via $\sigma(A) = \bar{r}(\sigma) A \bar{r}(\sigma)^{-1}$. Now condition (B) will be some condition on the local classes $$ \mathrm{res}_ 2[\phi] \in L_2 \subseteq H^1(G_{\mathbb{Q}_ 2}, \mathrm{ad}^0 \bar{r}), \quad \mathrm{res}_ \ell[\phi] \in L_\ell \subseteq H^1(G_{\mathbb{Q}_ \ell}, \mathrm{ad}^0 \bar{r}). $$ Finally, when we conjugate $r$ by some $1 + \epsilon A$, we change $$ \phi \to \phi + A - \mathrm{ad} \bar{r} A. $$ Here, scalars don’t do anything so we may as well assume that $A$ is traceless.
Definition 2. For $S$ a finite set of primes, $M$ a $G_{\mathbb{Q},S}$-module, and $L_v \subseteq H^1(G_{\mathbb{Q}_ v}, M)$ for each $v \in S$, we define $$ H_ {\lbrace L_v \rbrace}^1(G_{\mathbb{Q},S}, M) = \lbrace x \in H^1(G_{\mathbb{Q},S}, M) : \mathrm{res}_ v x \in L_v \text{ for all } v \in S \rbrace. $$ These are called Selmer groups.
So at the end, we will have $$ \Hom_\mathbb{F}(\mathfrak{m} / \langle \lambda, \mathfrak{m}^2 \rangle, \mathbb{F}) \cong H_{\lbrace L_2, L_\ell \rbrace}^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}). $$
The local conditions
So let us look at the condition at $2$. Choose a Frobenius lift $\varphi$ and basis $\bar{e}_ 1$ and $\bar{e}_ 2$ such that $\bar{r}(\varphi)(\bar{e}_ 1) = \bar{e}_ 1$ and $\bar{r}(\varphi)(\bar{e}_ 2) = 2 \bar{e}_ 2$, so we have $$ \bar{r} \vert_{G_{\mathbb{Q}_ 2}} = \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell^{-1} \end{pmatrix}. $$ We need to be able to find a basis $$ e_1 = \begin{pmatrix} 1 + a\epsilon \br c\epsilon \end{pmatrix}, \quad e_2 = \begin{pmatrix} b\epsilon \br 1 + d\epsilon \end{pmatrix} $$ on $\mathbb{F}[\epsilon]^{\oplus 2}$ such that $G_{\mathbb{Q}_ 2}$ acts via upper triangular matrices with $1, \epsilon_\ell^{-1}$ on the diagonals. If we write out the conditions, this is $$ (1 + \phi(\sigma) \epsilon) \bar{r}(\sigma) \biggl( 1 + \epsilon \begin{pmatrix} a & b \br c & d \end{pmatrix} \biggr) = \biggl( 1 + \epsilon \begin{pmatrix} a & b \br c & d \end{pmatrix} \biggr) \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell(\sigma)^{-1} \end{pmatrix}. $$ Taking the coefficient of $\epsilon$, we get $$ \phi(\sigma) + \mathrm{ad} \bar{r}(\sigma) \begin{pmatrix} a & b \br c & d \end{pmatrix} - \begin{pmatrix} a & b \br c & d \end{pmatrix} = \begin{pmatrix} 0 & \ast \br 0 & 0 \end{pmatrix}. $$ Here, again we may as well assume that $a + d = 0$. So this just means that $\phi$ has the same cohomology class as the right hand side.
So we consider $$ \mathrm{ad}^0 \bar{r} \supseteq \begin{pmatrix} 0 & \mathbb{F} \br 0 & 0 \end{pmatrix} = \Hom(\bar{\epsilon}_ \ell^{-1}, 1) \cong \mathbb{F}(\bar{\epsilon}_ \ell) $$ as representations of $G_{\mathbb{Q}_ 2}$. If we quotient this out, we get $$ \mathrm{ad}^0 \bar{r} / \Hom(\bar{\epsilon}_ \ell^{-1}, 1) \supseteq \biggl\lbrace \begin{pmatrix} a & 0 \br 0 & -a \end{pmatrix} \biggr\rbrace \cong \mathbb{F}, $$ and that quotient is $$ \mathrm{ad}^0 \bar{r} / (\Hom(\bar{\epsilon}_ \ell^{-1}, 1), \mathbb{F}) \cong \mathbb{F}(\bar{\epsilon}_ \ell^{-1}). $$ So this $\mathrm{ad}^0 \bar{r}$ has a three-step filtration with known graded pieces. Now we are saying that the class of $\phi$ lies in $$ L_2 = \im(H^1(G_{\mathbb{Q}_ 2}, \Hom(\bar{\epsilon}_ \ell^{-1}, 1)) \to H^1(G_{\mathbb{Q}_ 2}, \mathrm{ad}^0 \bar{r})). $$
While we are here, let us also figure out the dimension. We have a short exact sequence $$ 0 \to \Hom(\bar{\epsilon}_ \ell^{-1}, 1) \to \mathrm{ad}^0 \bar{r} \to Q \to 0, $$ and hence we get $$ \begin{align} 0 &= H^0(G_{\mathbb{Q}_ 2}, \mathbb{F}(\bar{\epsilon}_ \ell)) \to H^0(G_{\mathbb{Q}_ 2}, \mathrm{ad}^0 \bar{r}) \to H^0(G_{\mathbb{Q}_ 2}, Q) \cong \mathbb{F} \br &\to H^1(G_{\mathbb{Q}_ 2}, \mathbb{F}(\bar{\epsilon}_ \ell)) \cong \mathbb{Q}_ 2^\times \otimes_\mathbb{Z} \mathbb{F} \cong \mathbb{F} \to L_2 \to 0. \end{align} $$