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Let $k$ be a finite field of characteristic $p\neq 2$ (in fact, one only needs to consider the case $p\in\{3,5\}$), let $\Sigma$ be a finite set of primes containing $\infty$ and $p$, and

$$\rho_{0}:{\rm Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})\rightarrow {\rm GL}_{2}(k)$$

an absolutely irreducible representation, meaning $\rho_{0}\otimes\overline{k}$ cannot be written as the direct sum of two one-dimensional subrepresentations, where $\mathbb{Q}_{\Sigma}$ is the largest Galois extension of $\mathbb{Q}$ unramified outside the primes in $\Sigma$. Assume that if $\tau\in\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})$ is complex conjugation, then $\det(\rho_{0}(\tau))=-1$. Then $\rho$ induces a projective representation

$$\tilde{\rho}_{0}:\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})\longrightarrow\operatorname{PGL}_{2}(k).$$

Assume this projective representation has dihedral image, meaning $$\operatorname{image}(\tilde{\rho}_{0})\cong\left<s,r\mid s^{2}=r^{m}=(sr)^{2}=1\right>$$ for some $m\in\mathbb{N}$, and assume further that $\rho_{0}|_{\mathbb{Q}(\sqrt{(-1)^{\frac{p-1}{2}}p})}$ is absolutely irreducible.

The action of $\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})$ on $k^{2}$ induces an action on $V_{\lambda}=\operatorname{Hom}(k^{2},k^{2})$, namely by conjugation. (What that $\lambda$ stands for is of no significance here.) Since $p\neq 2$, one has a direct sum of $k[\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})]$-modules

$$V_{\lambda}=W_{\lambda}\oplus k\text{,}$$

where $W_{\lambda}$ denotes the space of $\operatorname{trace}$-$0$ matrices and $k$ is the space of scalar multiplications.

Let $K_{1}$ be the splitting field of $\rho_{0}$ (i.e. $\operatorname{Gal}(\mathbb{Q}_{\Sigma}/K_{1})=\operatorname{ker}(\rho_{0})$), and $$G:=\operatorname{Gal}(K_{1}/\mathbb{Q})=\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})/\operatorname{ker}(\rho_{0})=\operatorname{image}(\rho_{0}).$$

Since $\overline{\rho}_{0}$ has dihedral image, $\rho_{0}\otimes\overline{k}=\operatorname{Ind}_{H}^{G}(\chi)$ for some character $\chi$.

Question: Wiles now makes the following claims.

(1.) Under the above conditions, $W_{\lambda}\otimes\overline{k}=\delta\otimes\operatorname{Ind}_{H}^{G}(\chi/\chi')$ where $\chi'$ is the quadratic twist of $\chi$ by any element of $G\setminus H$ and $\delta$ is the quadratic character $G\longrightarrow G/H$ (what does that even mean - $W_{\lambda}\otimes\overline{k}$ decomposes as a quadratic character + something else?)

(2.) since $M(\zeta_{p^{n}})$ is Abelian over $\mathbb{Q}$, where $\mathbb{Q}\subseteq M\subseteq K_{1}$ such that $G/H=\operatorname{Gal}(M/\mathbb{Q})$, one always finds for any $n\in\mathbb{N}$ an $x\in\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ which fixes $\mathbb{Q}(\zeta_{p^{n}})$ and $\tilde{\rho}_{0}(x)\neq 1$ as long as $m\neq 2$ (i.e. $\operatorname{image}(\rho_{0})\neq\mathbb{Z}/2\times\mathbb{Z}/2$).

Why is that the case? Maybe I'm not seeing the wood for all the trees here, but who knows. EDIT: Also, should it not be $\operatorname{Gal}(\mathbb{Q}_{\Sigma}/\mathbb{Q})$ instead of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$?

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    $\begingroup$ (1): Note that as $\rho_0$ is induced from a character on a subgroup $H \leq G$, and has dimension $2$, $H$ is an index two subgroup of $G$, and thus the quadratic character $G \rightarrow G/H$ has a meaning. I agree that there's a dimension problem though. Also, your $G$ is probably not the kernel of $\rho$ -- instead, $Gal(\mathbb{Q}_{\Sigma}/K_1)$ is the kernel of $\rho_0$, I assume? $\endgroup$
    – Aphelli
    Dec 28, 2021 at 21:28
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    $\begingroup$ Anything about Wiles' proof is probably more suited for MathOverflow $\endgroup$ Dec 28, 2021 at 21:54
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    $\begingroup$ @MathematicsStudent1122: maybe not though. It's a fairly elementary question overall. For (1), I see two separate cases: if the complex conjugation is in the distinguished subgroup or not. In the first case, it commutes to everything in the projective image, and thus $W_{\lambda}$ can be decomposed as the sum of the conjugation-invariant part (another one of dimension $1$), and the one where conjugation acts as $-1$. If the conjugation is represented by $diag(1,-1)$ (and thus the distinguished subgroup is made with diagonal matrices), the latter is the space of anti-diagonal matrices. $\endgroup$
    – Aphelli
    Dec 28, 2021 at 22:00
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    $\begingroup$ @MathematicsStudent1122 I know where you are coming from, but I think this should actually be pretty straightforward linear algebra with a little extra flavor of groups acting on the vector space. $\endgroup$ Dec 28, 2021 at 22:00
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    $\begingroup$ @Mindlack: Can you elaborate on that a little? $\endgroup$ Dec 28, 2021 at 22:02

1 Answer 1

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(1): First, let's define some notations. Let's call $G$ the Galois group, $D$ its dihedral image, $C$ its distinguished cyclic subgroup, $H$ its inverse image in $G$. We fix some $s \in G$ whose image in $D$ is the $s$ of the dihedral presentation.

The quotient of $H$ by its scalar matrices (a central subgroup) is $C$ cyclic, so that $H$ is abelian. Then, we can note that $\rho_0$ is given on $H$ by direct sum of two characters $\chi$ and $\chi'$.

We change the basis for $\rho_0$ so that $k(1,0)$, $k(0,1)$ are the lines stable under $H$ (with characters $\chi,\chi'$) and $\rho_0(s)=\begin{bmatrix}0 & 1\\1&0\end{bmatrix}$.

Then we can easily see that $V = kI_2 \oplus k\cdot \mathrm{diag}(1,-1) \oplus A$, where $A$ is the space of antidiagonal matrices. Clearly, $G$ acts trivially on $kI_2$, $H$ acts trivially on the second summand, but $s$ acts by $-1$, so it is the quadratic character $G \rightarrow G/H$.

As for the last summand, it has a canonical basis ($\begin{bmatrix}0 & 1\\0&0\end{bmatrix}$ and its transpose) that $s$ permutes, and $\mathrm{diag}(a,b)$ acts by $\mathrm{diag}(a/b,b/a)$ on this basis, and we recognize the induced representation from $H$ to $G$ by $\chi/\chi'$ (provided that the twist of $\chi/\chi'$ is $\chi'/\chi$, which mostly follows from the dihedral presentation).

(2): I'm interpreting the question as follows: assume $m \neq 2$ and let $n\geq 1$. Why is there an $x \in \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that $x$ fixes $\mathbb{Q}(\zeta_{p^n})$ and $\overline{\rho_0(x)} \neq 1$?

Answer: assume that there is no such $x$: then $G_{\mathbb{Q}(\zeta_{p^n})} \subset G_{K_2}$, where $K_2 \subset K_1$ is the subextension such that $Gal(K_1/K_2)$ is mapped by $\rho_0$ to the scalar matrices. In other words, $K_2 \subset \mathbb{Q}(\zeta_{p^n})$. In particular, $K_2/\mathbb{Q}$ is abelian. But the Galois group of $K_2/\mathbb{Q}$ is isomorphic to $D$ (ie is dihedral), so isn't abelian except if $m \leq 2$. We get a contradiction and we're done.

For the edit question, I don't know Wiles' paper very well, but as far as I understand, the goal is to prove that some Galois representations are modular using deformation theory. But usually, one needs some finiteness properties to use deformation theory (see Mazur's 1989 paper in "Galois group over Q" iirc), which are not satisfied by $G_Q=Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, but are satisfied by Galois groups with "restricted ramification" ie $Gal(\mathbb{Q}_{\Sigma}/\mathbb{Q})$.

(exercise/example: use cyclotomic extensions to show that $H^1(G_Q,\mathbb{F}_p):= \mathrm{Hom}(G_Q,\mathbb{F}_p)$ is infinite).

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  • $\begingroup$ How donI know $H$ acts trivially on the secind summand? $\endgroup$ Dec 28, 2021 at 23:46
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    $\begingroup$ Because by construction (see the beginning) $\rho_0(H)$ contains only diagonal matrices. $\endgroup$
    – Aphelli
    Dec 28, 2021 at 23:50
  • $\begingroup$ aaaaaaaaaaaaah. $\endgroup$ Dec 28, 2021 at 23:51
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    $\begingroup$ Well, I suppose that this part of Wiles’ paper isn’t the most detailed one. $\endgroup$
    – Aphelli
    Dec 29, 2021 at 0:16
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    $\begingroup$ That can be said about many parts of his paper. From what I know he was in a hurry when he wrote that, for he feared someone else might find the missing step to close the gap that was still open in his older proof... $\endgroup$ Dec 29, 2021 at 0:18

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