We had this setting where $$ r = \dim H_{\mathcal{L}_ \emptyset^\perp}^1(G_F, (\mathrm{ad}^0 \bar{r})(1)). $$
Proposition 1. For every $N \gt 0$, there exists a set $Q_N$ of primes of $F$ such that
- $Q_N \cap (R \cup \lbrace v \mid \ell \rbrace) = \emptyset$,
- $\lvert Q_N \rvert = r$,
- for each $v \in Q_N$ we have $q_v = \lvert k(v) \rvert \equiv 1 \pmod{\ell^N}$,
- if $v \in Q_N$ then $\bar{r}(\mathrm{Frob}_ v)$ has distinct eigenvalues,
- $H_{\mathcal{L}_ {Q_N}^\perp}^1(G_F, (\mathrm{ad}^0 \bar{r})(1)) = 0$.
Last time we saw that it suffices to show that if $0 \neq \phi \in H_{\mathcal{L}_ \emptyset^\perp}^1(G_F, (\mathrm{ad}^0 \bar{r})(1))$ then there exists a $v \notin R \cup \lbrace v \mid \ell \rbrace$ such that
- $q_v \equiv 1 \pmod{\ell^N}$,
- $\bar{r}(\mathrm{Frob}_ v)$ has distinct eigenvalues,
- $\phi(\mathrm{Frob}_ v) \notin (\mathrm{Frob}_ v - 1) (\mathrm{ad}^0 \bar{r})(1)$.
By the Chebotarev density theorem, it suffices to prove that for all $0 \neq \emptyset \in H_{\mathcal{L}_ \emptyset^\perp}(G_F, (\mathrm{ad}^0 \bar{r})(1))$ there exists an element $\sigma_1 \in G_{F(\zeta_{\ell^N})}$ such that
- $\bar{r}(\sigma_1)$ has distinct eigenvalues,
- $\phi(\sigma_1) \notin (\sigma_1 - 1) \mathrm{ad}^0 \bar{r}(1)$.
Then we can choose $\mathrm{Frob}_ v = \sigma_1$. We now set $E_N = \bar{F}^{\ker \bar{r}}(\zeta_{\ell^N})$. Decomposing $\sigma_1 = \tau \tilde{\sigma}$, what we want now is an element
- $\sigma \in \operatorname{Gal}(E_N/F(\zeta_{\ell^N}))$ with lift $\tilde{\sigma} \in G_{F(\zeta_{\ell^N})}$,
- $\tau \in G_{E_N}$,
such that
- $\bar{r}(\sigma)$ has distinct eivengalues (i.e., $\mathrm{ad} \bar{r}(\sigma)$ has order prime to $\ell$),
- $\phi(\tau) + \phi(\tilde{\sigma}) \notin (\sigma-1) (\mathrm{ad}^0 \bar{r})(1)$.
To arrange the second condition, we just need $\sigma$ to have eigenvalue $1$ on the $\mathbb{F}$-subspace spanned by $\phi(G_{E_N})$. So it suffices to prove that
- $0 \neq \operatorname{res} \phi \in H^1(G_{E_N}, (\mathrm{ad}^0 \bar{r})(1)) = \Hom(G_{E_N}, (\mathrm{ad}^0 \bar{r})(1))$ (which would imply that $\phi(G_{E_N}) \neq 0$),
- for every $G_F$-invariant subspace $0 \neq W \subseteq (\mathrm{ad}^0 \bar{r})$, there exists a $\sigma \in \operatorname{Gal}(E_N/F(\zeta_{\ell^N}))$ such that $\sigma$ has eigenvalue $1$ on $W$, where $\mathrm{ad} \sigma$ is not $\ell$-power order.
This is now pure group theory, but let me try to give a self-contained account. For (2), we note that we have $$ \operatorname{Gal}(E_n/F(\zeta_{\ell^N})) \hookrightarrow \operatorname{Gal}(E_1/F(\zeta_\ell)) $$ with $\ell$-power order index. This reduces to the case when $N = 1$, because we can just take a $\ell$-power power.
Case 1: $\mathrm{ad}^0 \bar{r}$ is irreducible
Every $\sigma$ has eigenvalue $1$ on $\mathrm{ad}^0 \bar{r}$ and so we just need $\sigma \in G_{F(\zeta_\ell)}$ such that $\mathrm{ad} \bar{r}(\sigma)$ is not $\ell$-power order. We note that $$ (\mathrm{ad}^0 \bar{r})^{G_{F(\zeta_\ell)}} = 0 $$ because otherwise $\mathrm{ad}^0 \bar{r}$ would have a line invariant by $G_F$ as $\operatorname{Gal}(F(\zeta_\ell)/F)$ is abelian. This shows that $\mathrm{ad}^0 \bar{r}(G_{F(\zeta_\ell)})$ does not have $\ell$-power order (because otherwise it would be unipotent).
Case 2: $\mathrm{ad}^0 \bar{r}$ is reducible
Then there exists a character $\delta$ such that $\bar{r} \cong \bar{r} \otimes \delta$. By taking the determinant we see that $\delta^2 = 1$, but $\delta \neq 1$ because $\bar{r}$ is irreducible. Now taking $L = \bar{F}^{\ker \delta}$ we get $\bar{r} \cong \operatorname{Ind}_ {G_F}^{G_L} \theta$ and $\bar{r} \vert_{G_L} = \theta \oplus \theta^\prime$. Now we have $$ \mathrm{ad}^0 \bar{r} = (\operatorname{Ind}_ {G_F}^{G_L} \theta/\theta^\prime) \oplus \delta. $$
Case 2-1: $\operatorname{Ind}_ {G_F}^{G_L} \theta/\theta^\prime$ is irreducible
We could have $W = \delta$, in which case we need $\sigma$ such that $\delta(\sigma) = 1$, i.e., $\sigma \in G_L$ and $(\theta/\theta^\prime)(\sigma) \neq 1$. This can be done because $\theta \neq \theta^\prime$.
We also could have $W = \operatorname{Ind}_ {G_F}^{G_L} \theta/\theta^\prime$. We can look for $\sigma$ with $\delta(\sigma) = -1$. Then we have $$ (\theta/\theta^\prime)(\sigma^2) = \frac{\theta(\sigma^2)}{\theta(\sigma \sigma^2 \sigma^{-1})} = 1 $$ and so $$ \operatorname{Ind}_ {G_F}^{G_L}(\theta/\theta^\prime)(\sigma) \sim \begin{pmatrix} 0 & 1 \br (\theta/\theta^\prime)(\sigma^2) & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \br 1 & 0 \end{pmatrix}. $$ Because $\sigma \in G_{F(\zeta_\ell)}$, this can be done as $L \not\subseteq F(\zeta_\ell)$.
Case 2-2: $\mathrm{ad}^0 \bar{r} = \delta \oplus \chi \oplus \chi\delta$
In this case, because $\bar{r}$ is irreducible these are all distinct quadratic characters. In this case, we can just choose $\sigma$ such that $(\delta(\sigma), \chi(\sigma)) = (1, -1)$ and so on.
The other thing we needed was the vanishing of $$ H^1(\operatorname{Gal}(E_N/F), (\mathrm{ad}^0 \bar{r})(1)) = \ker(H^1(G_F, (\mathrm{ad}^0 \bar{r})(1)) \to H^1(G_{E_N}, (\mathrm{ad}^0 \bar{r})(1))). $$ We can use inflation-restriction to get $$ \begin{align} 0 &\to H^1(\operatorname{Gal}(E_1/F), (\mathrm{ad}^0 \bar{r})(1)) \to H^1(\operatorname{Gal}(E_N/F), (\mathrm{ad}^0 \bar{r})(1)) \br &\to H^1(\operatorname{Gal}(E_N/E_1), (\mathrm{ad}^0 \bar{r})(1))^{\operatorname{Gal}(E_1/F)}, \end{align} $$ but this last term is $$ \Hom(\operatorname{Gal}(E_N/E_1), (\mathrm{ad}^0 \bar{r})(1))^{\operatorname{Gal}(E_1/F)} = \Hom(\operatorname{Gal}(E_N/F_1), (\mathrm{ad}^0 \bar{r})(1)^{G_F}) = 0 $$ because $\bar{r}$ is irreducible.
So we are down to the case $N = 1$. Let $\Gamma = \im \bar{r} \subseteq \mathrm{GL}_ 2(\mathbb{F})$ and for simplicity let us suppose that $\mathbb{F} = \mathbb{F}_ \ell$ (there are references for other cases). What we need to show is that $$ H^1(\Gamma, \mathrm{ad}^0 \otimes \mathrm{det}^{-1}) = 0. $$ If $\lvert \Gamma \rvert$ is prime to $\ell$ we are done. If $\ell$ divides $\lvert \Gamma \rvert$ then up to conjugation $\Gamma$ contains the subgroup $$ S = \biggl\lbrace \begin{pmatrix} 1 & \ast \br 0 & 1 \end{pmatrix} \biggr\rbrace \subseteq \Gamma. $$ We then have the normalizer $N = N_\Gamma(S)$, which is just $\Gamma$ intersect with the upper-triangular matrices. Then $$ H^1(\Gamma, (\mathrm{ad}^0 \bar{r}) \otimes \mathrm{det}^{-1}) \hookrightarrow H^1(S, (\mathrm{ad}^0 \bar{r}) \otimes \mathrm{det}^{-1})^N. $$ The right hand side can be computed to be the line spanned by $(\begin{smallmatrix} 0 & 0 \br 1 & 0 \end{smallmatrix})$. So the left hand side is zero if there exists an element $$ \begin{pmatrix} \alpha & \ast \br 0 & \beta \end{pmatrix} \in \Gamma $$ with $\beta^2 \neq 1$, because this acts by $\beta^2$ on the line.
At this point we can appeal to some classification theorem of subgroups of $\mathrm{GL}_ 2(\mathbb{F})$, but this is easier. Because $\Gamma$ acts on $\mathbb{P}^1(\mathbb{F}_ \ell)$ with no fixed points but it has an $\ell$-cycle, the action is transitive. Using this we can check that $N$ must be all of the upper-triangular matrices.
Local deformation ring
Recall that for $v \in R$, there was this framed local deformation ring $R_{\chi_v}^\square$ that parametrized $$ r \colon G_{F_v} \to \mathrm{GL}_ 2(T) $$ where
- $r \bmod \mathfrak{m}_ T = 1$,
- $\det r = \epsilon_\ell^{-1}$,
- for $\sigma \in I_{F_v}$ we have $\operatorname{char}_ {r(\sigma)}(x) = (x - \chi_v(\sigma)) (x - \chi_v(\sigma)^{-1})$.
We had the local deformation ring $$ R_\chi^\mathrm{loc} = \bigotimes^\wedge_{v \in R} R_{\chi_v}^\square. $$ Here are the things we need to know.
- If $\operatorname{Spec} R_1^\mathrm{loc}$ has irreducible components $C_1, \dotsc, C_s$ (over $\mathcal{O}$) then their special fibers $c_i \times \operatorname{Spec} \mathbb{F}$ are irreducible and distinct and exhaust the irreducible components of $\operatorname{Spec}(R_1^\mathrm{loc} \otimes_\mathcal{O} \mathbb{F})$.
- If $\chi_v \neq 1$ for all $v$ then $\operatorname{Spec} R_\chi^\mathrm{loc}$ is irreducible and flat over $\mathcal{O}$.
Choose $\tau_v \in I_{F_v}$ lifting a generator of tame inertia and $\phi_v \in G_{F_v}$ lifting the Frobenius. These two together generate $G_{F_v}$ quotiented by wild inertia. Then $r(\tau_v) = \Sigma_v$ and $r(\phi_v) = \Phi_v$ determine $r$. More precisely, $R_{\chi_v}^\square$ parametrizes pairs $$ (\Phi_v, \Sigma_v) \in \ker(\mathrm{GL}_ 2(T) \to \mathrm{GL}_ 2(\mathbb{F})) $$ with $\det \Phi_v =q_v$ and $\Phi_v^{-1} \Sigma_v \Phi_v = \Sigma_v^{q_v}$ and $\operatorname{char}_{\Sigma_v}(x) = (x - \chi_v(\sigma_v)) (x - \chi_v(\sigma_v)^{-1})$. (We note that the conditions on $r$ forces it to be tamely ramified.)