Recall we had $S = \lbrace 2, \ell \rbrace$ and $\bar{r} \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(\mathbb{F})$ and lifted it to $$ r^\mathrm{univ} \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(R^\mathrm{univ}) $$ satisfying three properties, $\det r^\mathrm{univ} = \epsilon_\ell^{-1}$ and $r^\mathrm{univ} \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} 1 & \ast \br 0 & \epsilon_\ell^{-1} \end{pmatrix} \otimes \delta$ with $\delta^2 = 1$ unramified, and $r^\mathrm{univ} \vert_{G_{\mathbb{Q}_ \ell}}$ Fontaine–Laffaille. We showed that we have a presentation $$ 0 \to I \to \mathcal{O}[[x_1, \dotsc, x_d]] \twoheadrightarrow R^\mathrm{univ} $$ where $d = \dim_\mathbb{F} H_\mathscr{L}^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r})$ is minimal and $I/\mathfrak{m}I \hookrightarrow H_\mathscr{l}^2(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r})$. It followed that $$ \dim R^\mathrm{univ} \ge 1 + \dim H_\mathscr{L}^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) - \dim H_\mathscr{L}^2(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}). $$
How were these Selmer groups defined? We had two sets $\mathscr{L} = \lbrace L_2, L_\ell \rbrace$ where $L_v \subseteq H^1(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r})$ corresponding to $\tilde{L}_ v \subseteq Z^1(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r})$. Then we had this chain complex $$ \begin{align} C^0(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) &\to C^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \br &\to C^2(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \oplus \bigoplus_{v \in S} C^1(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r}) / \tilde{L}_ v \br &\to C^3(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r}) \oplus \bigoplus_{v \in S} C^2(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r}) \to \dotsb, \end{align} $$ where the maps are $(\phi, (\psi_v)) \mapsto (\partial \phi, (\phi \vert_{G_{\mathbb{Q}_ v}})$. We can now define $$ H_\mathscr{L}^i(G_{\mathbb{Q}, S}, \mathrm{ad}^0 \bar{r}) $$ to be the cohomology of this complex.
If we look at the short exact sequence of chain complexes given by the local to mixed to global, we get a long exact sequence $$ \begin{align} 0 &\to H_\mathscr{L}^0(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \to H^0(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \to 0 \br &\to H_\mathscr{L}^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \to H^1(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \to \bigoplus_{v \in S} H^1(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r}) / L_v \br &\to H_\mathscr{L}^2(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \to H^2(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \to \bigoplus_{v \in S} H^2(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r}) \br &\to H_\mathscr{L}^3(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) \to 0, \end{align} $$ where we are using the fact that $H^3(G_{\mathbb{Q},S}, -) = 0$ as long as the coefficient is not $2$. (This should be understood as number field being “open.”) We also have $H_\mathscr{L}^i(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) = 0$ for $i \gt 3$.
By Poitou–Tate duality, we have $$ H_\mathscr{L}^0(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) = H^0(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}), \quad H_\mathscr{L}^3(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}) = H^0(G_{\mathbb{Q},S}, (\mathrm{ad}^0 \bar{r})^\vee(1))^\vee. $$ Here we have $(\mathrm{ad}^0 \bar{r})^\vee = \mathrm{ad}^0 \bar{r}$ because of the trace pairing, and there is a perfect pairing $$ H^1(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r}) \times H^1(G_{\mathbb{Q}_ v}, (\mathrm{ad}^0 \bar{r})^\vee(1)) \to H^2(G_{\mathbb{Q}_ v}, \mathbb{F}(1)) \cong \mathbb{F}. $$ We now let $L_v^\perp$ be the annihilator of $L_v$. We can then identify $$ H_\mathscr{L}^i(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}(1)) = H_{\mathscr{L}^\perp}^{3-i}(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r})^\vee. $$ So if we write $$ h_i = \dim_\mathbb{F} H^i(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}), \quad h_\mathscr{L}^i = \dim_\mathbb{F} H_\mathscr{L}^i(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}), \quad h_v^i = \dim_\mathbb{F} H^i(G_{\mathbb{Q}_ v}, \mathrm{ad}^0 \bar{r}) $$ and also $l_v = \dim L_v$, then we have $$ 1 + h_\mathscr{L}^1 - h_\mathscr{L}^2 = 1 + h^0 - h_\mathscr{L}^3 + (h^1 - h^2 - h_0) - \sum_{v \in S} (h_v^1 - h_v^2 - h_v^0) + \sum_{v \in S} (l_v - h_v^0). $$ We know that $h^1 - h^2 + h^0 = \dim \mathrm{ad}^0 \bar{r} - \dim (\mathrm{ad}^0 \bar{r})^{c=1}$ and $h_v^1 - h_v^2 + h_v^0 = \delta_{v,\ell} \mathrm{dim} \mathrm{ad}^0 \bar{r}$. Then this simplifies to $$ 1 + h^0 - h_\mathscr{L}^3 + \sum_{v \in S - \lbrace \ell \rbrace} (l_v - h_v^0) + (l_v - h_l^0) - \dim (\mathrm{ad}^0 \bar{r})^{c=1}. $$
- Because $\bar{r}$ is absolutely irreducible we have $h^0 = 0$.
- Because $h_\mathscr{L}^3 = \dim H^0(G_{\mathbb{Q},S}, \mathrm{ad}^0 \bar{r}(1))$ and this is zero because $\bar{r} \ncong \bar{r}(1)$ as $\bar{\epsilon}_ \ell^2 \neq 1$. (Here we are using $\ell \gt 3$.)
- We have $l_v - h_v^0 = 0$ for $v \neq \ell$.
- We have $l_\ell - h_\ell^0 = \dim (\mathrm{ad}^0 \bar{M} / \mathrm{Fil}^0 \mathrm{ad}^0 \bar{M}) = 1$.
- We have $\dim (\mathrm{ad}^0 \bar{r})^{c=1} = 1$.
So at the end, we get $$ \dim R^\mathrm{univ} \ge 1. $$
Remark 1. The last quantity $(l_\ell - h_\ell^0) - \dim(\mathrm{ad}^0 \bar{r})^{c = 1}$ is always $\le 0$. In our case, it equals zero, but this happens only when we are looking at a Shimura variety and the Hodge–Tate weights are distinct. When this number becomes negative, this won’t give a very useful bound.
Now suppose that we know that $\mathcal{O} \to R^\mathrm{univ}$ is finite. The kernel must be zero, because otherwise $R^\mathrm{univ}$ will be finite over some $\mathcal{O}/\lambda^n$. By going up, there exists a prime ideal $\mathfrak{p} \subseteq R^\mathrm{univ}$ such that $\mathfrak{p} \cap \mathcal{O} = (0)$. Then $\mathcal{O} \hookrightarrow R^\mathrm{univ} / \mathfrak{p}$ with field of fractions $L \hookrightarrow L^\prime$ a finite extension. Now $r^\mathrm{univ}$ pushed forward to $L^\prime$ defines a lift of $\bar{r}$ satisfying (A), (B), (C).
Here is the rough strategy. Suppose that $F/\mathbb{Q}$ is a finite extension such that $F/\mathbb{Q}$ is unramified at $\ell$ and $F \cap \bar{\mathbb{Q}}^{\ker \bar{r}} = \mathbb{Q}$. Then $\bar{r}(G_{\mathbb{Q}}) = \bar{r}(G_F)$ and so we can look at a similar deformation problem for $\bar{r} \vert_{G_F}$. Here we are going to impose the conditions
- (A) $\det r = \epsilon_\ell^{-1}$,
- (C) the Fontaine–Laffaille condition at all places above $\ell$.
(We drop condition (B) because we needed something like $\mathrm{Frob}_ 2$ having distinct eigenvalues, and it’s easier to drop the condition altogether.) Then there is a universal deformation ring $R_F^\mathrm{univ}$ and a map $R_F^\mathrm{univ} \to R^\mathrm{univ}$. We will then show that this map is finite.