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 $q_v \equiv 1 \pmod{p}$.)


1 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.

  • $\begingroup$ Thank you so much for your great answer. Such considerations without fixed basis are great! Yet it is not quite clear for me on the paragraph "In fact, in the two-dimensional cases ...". I'm wondering what is the isomorphism $V^{\ast} \cong V \otimes \det(V)^{-1}$ and I wanna make this isomorphism clear in order to see why this is specific for the 2-dimensional case. Moreover, I haven't seen how $\mathrm{Sym}^2(\overline{\rho})$ appeared there. Thank you again for your great answer and so sorry if these questions are kind of elementary. $\endgroup$
    – Hetong Xu
    Commented Mar 10, 2021 at 3:21
  • $\begingroup$ You can actually see it by just manipulating $2\times 2$ matrices (the inverse transpose of a $2\times 2$ matrix is conjugate to its adjugate matrix). But abstractly, there is always a pairing $\wedge^a V\otimes \wedge^b V\mapsto \wedge^{a+b}V$. If $V$ is $n$-dimensional, then $\wedge^n V$ is a one-dimensional vector space on which $\mathrm{GL}(V)$ acts by the determinant. In the $2$-dimensional case, we take $a = b = 1$ to obtain an intertwining map $V\otimes V\to \det(V)$ and thus an isomorphism $V\cong V^*\otimes \det(V)$. In general, $\wedge^a V \cong \wedge^{n-a}V^*\otimes \det(V)$ $\endgroup$
    – Mathmo123
    Commented Mar 10, 2021 at 8:23
  • $\begingroup$ We then get $V\otimes V^*\cong (V\otimes V)\otimes \det(V)^{-1}$. But $V\otimes V \cong \mathrm{Sym}^2 V \oplus \wedge^2 V \cong \mathrm{Sym}^2 V \oplus \det(V)$. $\endgroup$
    – Mathmo123
    Commented Mar 10, 2021 at 8:25

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