On understanding $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$ in Taylor-Wiles method

Question

I'm currently learning Taylor-Wiles method and modularity lifting and comming up with following difficulties, which I think is based on understanding how (global and local) Galois groups act on $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$. Here:

1. $\overline{\rho}: G_{F,S} \rightarrow \mathrm{GL}_2(\mathbb{F})$ is the residue representation. Here $S$ is a finite set of finite places of $F$ containing $\{v | p\}$, where $F$ is a number field (a finite extension over $\mathbb{Q}$). Here $G_{F,S}$ is the Galois group of the maximum extension over $F$ which is unramified outside $S \sqcup \{ v | \infty \}$.
2. $\mathrm{ad}^0 \overline{\rho}$ consists of traceless matrices $M \in M_n(\mathbb{F})$ with $G_{F,S}$-conjugation action $\sigma \cdot X := \overline{\rho}(\sigma) X \overline{\rho}(\sigma)^{-1}$.
3. $\mathrm{ad}^0 \overline{\rho}(1)$ is its Tate twist $\mathrm{ad}^0 \overline{\rho} \otimes \mathbb{Z}_p(1)$.

Via the $G_{F,S}$-action on $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$, we can consider the $G_{F,S}$-cohomology with coefficients in $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$ accordingly.

My question 1: How to prove the following claim:

If $\overline{\rho}|_{G_{F(\zeta_p)}}$ is absolutely irreducible, then $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1)) = 0$. (Where $G_{F(\zeta_p)}$ is the absolute Galois group of $F(\zeta_p)$).

My attempt: The zeroth cohomology group $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1))$ consists of exactly the elements in $\mathrm{ad}^0 \overline{\rho}(1)$ fixed by $G_{F,S}$, so it is essential to figure out the action of $G_{F,S}$ on $\mathrm{ad}^0 \overline{\rho}(1)$. Then I have found someone claimed that

Elements in $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1))$ correspond to the intertwining operators $\overline{\rho} \rightarrow \overline{\rho}(1)$ between $G_{F(\zeta_p)}$-modules.

But how to prove this claim?. After proving this claim, Schur's lemma implies that $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1)) = 0$ and we are done.

My question 2: Let $v$ be a Taylor-Wiles prime of $F$. (i.e. $v$ is a finite place of $F$ satisfying the two conditions: the norm of $v$ (denoted by $q_v$) satisfies $q_v \equiv 1 \pmod{p}$; and $\overline{\rho}(\mathrm{Frob}_v) \in \mathrm{GL}_2(\mathbb{F})$ has distinct eigenvalues.) Let $F_v$ be the completion of $F$ wrt $v$ and $G_v$ be its (local) absolute Galois group. Via $$ G_v \hookrightarrow G_F \twoheadrightarrow G_{F,S} \xrightarrow{\overline{\rho}} \mathrm{GL}_2(\mathbb{F}), $$ we can consider the $G_v$ action on $\mathrm{ad}^0 \overline{\rho}(1)$. Then how to prove the following claim:

$H_0(G_{v}, \mathrm{ad}^0 \overline{\rho}(1)) = H_0(G_{v}, \mathrm{ad}^0 \overline{\rho}).$

So again the essential point is to understand how $G_v$ acts on $\mathrm{ad}^0 \overline{\rho}(1)$ and I have difficulty on this. I even guess

The actions of $G_{v}$ on $\mathrm{ad}^0 \overline{\rho}(1)$ and $\overline{\rho}$ are the same.

But I don't know how to prove this. (I feel like this is due to the condition that $q_v \equiv 1 \pmod{p}$.)

Answer 1

I think a lot of this is easier to see if we don't fix a basis. Consider $$\overline\rho\colon G_{F, S}\to \mathrm{GL}(V)$$ where $V$ is a two-dimensional $\mathbb F$-vector space (although the arguments here work in higher dimensions too). Then $\mathrm{ad}(\overline\rho)$ is the representation $$\mathrm{ad}(\overline\rho)\colon G_{F, S}\to \mathrm{GL(End}(V))$$ where $g\in G_{F, S}$ acts on $\phi\in \mathrm{End}(V)$ by sending it to the map $v\mapsto \overline\rho(g)\cdot\phi(\overline\rho(g)^{-1}\cdot v)$. As a representation, $\mathrm{End}(V)$ is canonically isomorphic to $V\otimes V^*$, which decomposes as $\mathbb 1 \oplus \mathrm{ad}^0(\overline\rho)$, where $\mathbb 1$ is the trivial representation.

In fact, in the two-dimensional case, $V^*\cong V\otimes \det(V)^{-1}$, so $$V\otimes V^* \cong (V\otimes V\otimes \det(V)^{-1}) \cong \mathrm{Sym}^2(\overline\rho)\otimes \det(\overline\rho)^{-1} \oplus\mathbb 1$$ and $\mathrm{ad}^0(\overline\rho) \cong \mathrm{Sym}^2(\overline\rho)\otimes \det(\overline\rho)^{-1}$.

But regardless, $\mathrm{ad}^0(\overline\rho)(1)$ is a subrepresentation of $\mathrm{ad}(\overline\rho)(1) \cong V\otimes\epsilon\otimes V^*\cong \mathrm{Hom}(V, V(1))$, where $\epsilon$ is the mod $p$ cyclotomic character. Hence, elements of $H^0(G_{F, S}, \mathrm{ad}^0(\overline\rho))\cong \mathrm{Hom}_{G_{F, S}}(V, V(1))$ naturally corresponds to intertwining operators between $\rho$ and $\rho(1)$ as $G_{F, S}$-modules.

If $V$ is irreducible as a $G_{F, S}$-module, then by Schur's lemma, $\mathrm{Hom}_{G_{F, S}}(V, V(1)) = 0$ unless $V(1)\cong V$ as $G_{F, S}$-modules, which (by Clifford theory) will happen if and only if $\overline{\rho}|_{G_{F(\zeta_p)}}$ is reducible. (Note that $G_{F(\zeta_p)}$ is the kernel of the mod $p$ cyclotomic character.)

For your second question, note that, if $q_v \equiv 1\pmod p$, then $\epsilon(\mathrm{Frob}_v) = q_v\equiv 1\pmod p$, so $\epsilon|_{G_v}$ is the trivial character.