We were in the process of proving the following theorem.
Theorem 1. Let $\ell \gt 3$ be a prime, and let $\bar{r} \colon G_{\mathbb{Q},S} \to \GL_2(\bar{\mathbb{F}}_ \ell)$ with $S = \lbrace 2, \ell \rbrace$ satisfy
- $\det \bar{r} = \epsilon_\ell^{-1}$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ \ell}}$ is Fontaine–Laffaille with Hodge–Tate weights $0, 1$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell^{-1} \end{pmatrix} \otimes \delta$ for $\delta^2 = 1$ unramified.
Then there exists a totally real finite Galois extension $F/\mathbb{Q}$ satisfying $F \cap \bar{\mathbb{Q}}^{\ker \bar{r}}(\zeta_\ell) = \mathbb{Q}$ such that $\bar{r}$ is automorphic over $F$ with weight $0$ and level $U_0(S_1)$ for some $S_1$ not containing $\ell$.
Last time we considered a finite extension $M / \mathbb{Q}$ an imaginary quadratic field such that $2$, $\ell$ are unramified in $M$ and some prime $p \gt 3$ is ramified, and then constructed a character $$ \theta \colon \mathbb{A}_ M^\times \to N^\times $$ where $N \supseteq M$ is finite. For $\ell^\prime$ that splits completely in $N$ and $\lambda^\prime \mid \ell^\prime$, we constructed an $\ell$-adic character $$ \theta_{\lambda^\prime} \colon \mathbb{A}_ M^\times / M^\times M_\infty^\times \to \mathbb{Z}_ {\ell^\prime}^\times. $$ This satisfied the property that
- $\det \operatorname{Ind}_ {G_M}^{G_\mathbb{Q}} \theta_{\lambda^\prime} = \epsilon_\ell^{-1}$,
- $\operatorname{Ind}_ {G_M}^{G_\mathbb{Q}} \theta_{\lambda^\prime}$ is unramified at $2$ and $\ell$ (in fact only ramified at $\ell^\prime$, $p$, and primes ramified in $M/\mathbb{Q}$),
- $\operatorname{Ind}_ {G_M}^{G_\mathbb{Q}} \bar{\theta}_ {\lambda^\prime} \vert_{G_{\mathbb{Q}(\zeta_{\ell^\prime})}}$ is absolutely irreducible because for $p = v \bar{v}$ we have $\bar{\theta}_ {\lambda^\prime} \vert_{I_v}$ of order exactly $p-1$,
- $\operatorname{Ind}_ {G_M}^{G_\mathbb{Q}} \theta_{\lambda^\prime} \vert_{G_{\mathbb{Q}_ {\ell^\prime}}}$ is Fontaine–Laffaille with Hodge–Tate weights $0, 1$.
Now let $T/\mathbb{Q}$ be the moduli space of elliptic curve $E$ with isomorphisms $E[\ell]^\vee \cong \bar{r}$ and $E[\ell^\prime]^\vee \cong \operatorname{Ind}_ {G_M}^{G_\mathbb{Q}} \bar{\theta}_ {\lambda^\prime}$ compatible with the Weil pairing and the determinant pairing (which we choose). Over $\mathbb{C}$, this is just $$ T(\mathbb{C}) \cong \Gamma(\ell \ell^\prime) \backslash \mathcal{H}, $$ so in particular it is geometrically connected. We have $T(\mathbb{R}) \neq \emptyset$, because $$ \bar{r}(c) \sim \begin{pmatrix} 1 & 0 \br 0 & -1 \end{pmatrix}, \quad (\operatorname{Ind}_ {G_M}^{G_\mathbb{Q}} \bar{\theta}_ {\lambda^\prime})(c) \sim \begin{pmatrix} 1 & 0 \br 0 & -1 \end{pmatrix}. $$
Claim 2. There exist unramified extensions $L/\mathbb{Q}_ \ell$ and $L^\prime/\mathbb{Q}_ {\ell^\prime}$ with $T(L) \neq \emptyset$ and $T(L^\prime) \neq \emptyset$.
This follows from the following lemma.
Lemma 3. Let $\ell$ be an odd prime, and let $\bar{r} \colon G_{\mathbb{Q}_ \ell} \to \GL_2(\bar{\mathbb{F}}_ \ell)$ be Fontaine–Laffaille with Hodge–Tate weights $0, 1$ and $\det \bar{r} = \bar{\epsilon}_ \ell^{-1}$. Then there exists $L/\mathbb{Q}_ \ell$ unramified and $E/L$ an elliptic curve with good reduction together with an isomorphism $E[\ell]^\vee \cong \bar{r}$ under which the Weil pairing corresponds to the determinant pairing.
In the case when $\bar{r}$ is irreducible, we have $\bar{r} \cong \operatorname{Ind}_ {G_{\mathbb{Q}_ {\ell^2}}}^{G_{\mathbb{Q}_ \ell}} \phi$ for some $\phi$ satisfying $$ \phi \vert_{I_{\mathbb{Q}_ {\ell^2}}} \colon I_{\mathbb{Q}_ {\ell^2}} \to I_{\mathbb{Q}_ {\ell^2}^\mathrm{ab} / \mathbb{Q}_ {\ell^2}} \cong \mathbb{Z}_ {\ell^2}^\times \twoheadrightarrow \mathbb{F}_ {\ell^2}^\times $$ and $\phi(\ell) = -1$. Now if we take $E/\mathbb{Q}_ {\ell^2}$ any elliptic curve with good supersingular reduction, we have $E[\ell] \cong \bar{r}$. Now to match the pairing, we alter by an element of $\mathbb{F}_ {\ell^2}^\times$ that commutes with the Galois action, using the fact that $N_{\mathbb{F}_ {\ell^2}/\mathbb{F}_ \ell} \colon \mathbb{F}_ {\ell^2}^\times \to \mathbb{F}_ \ell^\times$ is surjective.
In the second case, for $\bar{r}$ reducible, we know that we have $$ \bar{r} \sim \begin{pmatrix} \bar{\chi} & \ast \br 0 & \bar{\chi}^{-1} \bar{\epsilon}_ \ell^{-1} \end{pmatrix}, $$ where $\bar{\chi}$ is unramified, $\ast$ is peu ramifié. Here, if we choose the alternating paring $\langle (x, y), (u, v) \rangle = xv - yu$ then $\ast \in H^1(G_{\mathbb{Q}_ \ell}, \mathbb{F}_ \ell(\bar{\chi}^2 \bar{\epsilon}_ \ell))$ is well-defined by the pairing.
Choose an $n$ such that $\bar{\chi} \vert_{G_{\mathbb{F}_ {\ell^n}}} = 1$ and an ordinary elliptic curve $\bar{E} / \mathbb{F}_ {\ell^n}$ with a point of order $\ell$. Writing $T_\ell \bar{E} \cong \mathbb{Z}_ \ell(\psi)$ for some $\psi \colon G_{\mathbb{F}_ {\ell^n}} \to (1 + \ell \mathbb{Z}_ \ell)$, Serre–Tate theory tell us that lifts of $\bar{E}$ to $\mathbb{Z}_ {\ell^n}$ correspond to elements $$ x \in H^1(G_{\mathbb{Q}_ {\ell^n}}, \mathbb{Z}_ \ell(\psi^2 \epsilon_\ell)). $$ The elliptic curve $E_x$ corresponding to $x$ has Tate module $$ T_\ell E_x \sim \begin{pmatrix} \psi^{-1} \epsilon_\ell & \ast \br 0 & \psi \end{pmatrix} $$ such that the Weil pairing is $\langle (x, y), (u, v) \rangle = xv - yu$.
Now what we need is that the image of $$ H^1(G_{\mathbb{Q}_ {\ell^n}}, \mathbb{Z}_ \ell(\psi^2 \epsilon_\ell)) \to H^1(G_{\mathbb{Q}_ {\ell^n}}, \mathbb{F}_ \ell(\epsilon_\ell)) \cong \mathbb{Q}_ {\ell^n}^\times / (\mathbb{Q}_ {\ell^n}^\times)^\ell $$ contains $\mathbb{Z}_ {\ell^n}^\times / (\mathbb{Z}_ {\ell^n}^\times)^\ell$. Using the long exact sequence on cohomology, we see that it suffices to prove that the composition $$ \mathbb{Z}_ {\ell^n}^\times / (\mathbb{Z}_ {\ell^n}^\times)^\ell \to H^1(G_{\mathbb{Q}_ {\ell^n}}, \mathbb{F}_ \ell(\epsilon_\ell)) \to H^2(G_{\mathbb{Q}_ {\ell^n}}, \mathbb{Z}_ \ell(\psi^2 \epsilon_\ell))[\ell] $$ is zero.
The easiest way to see this is to look at its Tate dual. Then what we need is that the composition $$ H^0(G_{\mathbb{Q}_ {\ell^n}}, (\mathbb{Q}_ \ell/\mathbb{Z}_ \ell)(\psi^{-2})) / \ell \to \Hom(G_{\mathbb{Q}_ {\ell^n}}, \mathbb{F}_ \ell) \to \Hom(I_{\mathbb{Q}_ {\ell^n}}, \mathbb{F}_ \ell) $$ is zero. Here, the boundary map is given by sending $x$ to $\sigma \mapsto (\psi^{-2}(\sigma) - 1)(x/\ell)$. But if $\sigma \in I_{\mathbb{Q}_ {\ell^n}}$ then $\psi^{-2}(\sigma) - 1 = 0$ because $\psi$ is unramified.
Proposition 4. If $K$ is a number field, and $K^\mathrm{avoid}/K$ is finite Galois. Let $S$ be a finite set of places of $K$ and $T/K$ is a smooth geometrically connected curve (or variety). For $v \in S$, suppose $L_v^\prime / K_v$ is a finite Galois extension and let $\Omega_v \subseteq T(L_v^\prime)$ be a nonempty open (in the $v$-adic topology) that is $\Gal(L_v^\prime/K_v)$-invariant. Then there exists $L/K$ a finite Galois extension with $L \cap K^\mathrm{avoid} = K$ and $P \in T(L)$ such that for $w \mid v \in S$ we have $L_v^\prime \cong L_w$ and $P \in T(L_w) \cong T(L_v^\prime)$ is in $\Omega_v$.