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We have that $G'$ is an almost-direct product of $\eta(G)$ and the almost-direct product of the duals of the simple factors of $G'$ not contained in $\eta(G)$, which allows us to reduce (1) and (3) to the case where $\eta$ is surjective.

I think you probably want to assume that $\eta$ is a quotient by a smooth, connected, normal subgroup followed by a central isogeny; this is automatic if $\eta$ is separable. If you do not want to reduce to this case, then you have to handle something like the exceptional isogeny $\operatorname{SO}_{2n + 1, k} \to \operatorname{Sp}_{2n, k}$ when $\operatorname{char} k$ is $2$, whose dual homomorphism I guess should be a map $\operatorname{SO}_{2n + 1, \mathbb C} \to \operatorname{Sp}_{2n, \mathbb C}$; but I think that there are no non-trivial such homomorphisms.

When $\eta$ is a quotient by a smooth, connected, normal subgroup $N$ of $G$, we have a similar decomposition of $G$ as the almost-direct product of $N$ with a canonical complement. Thus, if it is OK to make the reduction above, then we need only handle the case where $\eta$ is an isogeny.

In this case, if $(B, T, \mathcal X)$ is a pinning of $G_{\overline k}$, then $(B', T', \mathcal X') \mathrel{:=} (\eta_{\overline k}(B), \eta_{\overline k}(T), \eta_{\overline k}(\mathcal X))$ is a pinning of $G'_{\overline k}$. We compute $\Psi_0(G_{\overline k})$ and $\Psi_0(G'_{\overline k})$ in terms of these pinnings, and then let $\Psi_0(\eta)$ be the obvious map. If I understand correctly how $\Gamma_{\overline k/k}$ acts, then this is $\Gamma_{\overline k/k}$-equivariant: if we choose $\sigma \in \Gamma_{\overline k/k}$ and let $g \in G(\overline k)$ be such that $\sigma(B, T, \mathcal X)$ equals $\DeclareMathOperator\Ad{Ad}(g B g^{-1}, g T g^{-1}, \Ad(g)\mathcal X)$, then $\sigma(B', T', \mathcal X') = (\eta_{\overline k}(\sigma B), \eta_{\overline k}(\sigma T), \eta_{\overline k}(\sigma\mathcal X))$ equals $(g'B'g^{\prime\,{-1}}, g'T'g^{\prime\,{-1}}, \Ad(g')\mathcal X')$, where $g'$ equals $\eta_{\overline k}(g)$. Then $\sigma$, viewed as a morphism $X_*(T) \to X_*(T)$, respectively $X_*(T') \to X_*(T')$, sends $\lambda$, respectively $\lambda'$, to $\DeclareMathOperator\Int{Int}\Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$, respectively $\Int(g')^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$; and then $\Psi_0(\eta)(\sigma\lambda) = \eta \circ \Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$ equals $\Int(g')^{-1} \circ \sigma \circ \eta \circ \lambda \circ \sigma^{-1} = \sigma(\Psi_0(\eta)(\lambda))$.

Still in the case where $\eta$ is a central isogeny, we have that $\widehat{\Psi_0(\eta)}$ is also a $\Gamma_{\overline k/k}$-equivariant, separable isogeny of root data, so that it is dual to a (necessarily central) $\Gamma_{\overline k/k}$-equivariant isogeny $\hat\eta : \hat{G'} \to \hat G$, as you say uniquely determined by choices of $\Gamma_{\overline k/k}$-fixed pinnings. Since two different pinnings of $\hat{G'}$ are uniquely $\hat{G'}/\operatorname Z(\hat{G'})$-conjugate, the corresponding $\Gamma_{\overline k/k}$-equivariant homomorphisms are $(\hat{G'}/\operatorname Z(\hat{G'}))^{\Gamma_{\overline k/k}}$-conjugate; but it is not clear to me why the conjugacy can be lifted to $\smash{\hat{G'}}^{\Gamma_{\overline k/k}}$. If $\operatorname Z(\hat{G'})$ is connected (equivalently, if the derived group of $G'$ is simply connected), then one can probably show that $H^1(\Gamma_{\overline k/k}, \operatorname Z(\hat{G'}))$ is trivial and deduce a suitable lifting; but I don't know about the case of disconnected dual centre.

We have that $G'$ is an almost-direct product of $\eta(G)$ and the almost-direct product of the duals of the simple factors of $G'$ not contained in $\eta(G)$, which allows us to reduce (1) and (3) to the case where $\eta$ is surjective.

I think you probably want to assume that $\eta$ is a quotient by a smooth, connected, normal subgroup followed by a central isogeny; this is automatic if $\eta$ is separable. If you do not want to reduce to this case, then you have to handle something like the exceptional isogeny $\operatorname{SO}_{2n + 1, k} \to \operatorname{Sp}_{2n, k}$ when $\operatorname{char} k$ is $2$, whose dual homomorphism I guess should be a map $\operatorname{SO}_{2n + 1, \mathbb C} \to \operatorname{Sp}_{2n, \mathbb C}$; but I think that there are no non-trivial such homomorphisms.

When $\eta$ is a quotient by a smooth, connected, normal subgroup $N$ of $G$, we have a similar decomposition of $G$ as the almost-direct product of $N$ with a canonical complement. Thus, if it is OK to make the reduction above, then we need only handle the case where $\eta$ is an isogeny.

In this case, if $(B, T, \mathcal X)$ is a pinning of $G_{\overline k}$, then $(B', T', \mathcal X') \mathrel{:=} (\eta_{\overline k}(B), \eta_{\overline k}(T), \eta_{\overline k}(\mathcal X))$ is a pinning of $G'_{\overline k}$. We compute $\Psi_0(G_{\overline k})$ and $\Psi_0(G'_{\overline k})$ in terms of these pinnings, and then let $\Psi_0(\eta)$ be the obvious map. If I understand correctly how $\Gamma_{\overline k/k}$ acts, then this is $\Gamma_{\overline k/k}$-equivariant: if we choose $\sigma \in \Gamma_{\overline k/k}$ and let $g \in G(\overline k)$ be such that $\sigma(B, T, \mathcal X)$ equals $\DeclareMathOperator\Ad{Ad}(g B g^{-1}, g T g^{-1}, \Ad(g)\mathcal X)$, then $\sigma(B', T', \mathcal X') = (\eta_{\overline k}(\sigma B), \eta_{\overline k}(\sigma T), \eta_{\overline k}(\sigma\mathcal X))$ equals $(g'B'g^{\prime\,{-1}}, g'T'g^{\prime\,{-1}}, \Ad(g')\mathcal X')$, where $g'$ equals $\eta_{\overline k}(g)$. Then $\sigma$, viewed as a morphism $X_*(T) \to X_*(T)$, respectively $X_*(T') \to X_*(T')$, sends $\lambda$, respectively $\lambda'$, to $\DeclareMathOperator\Int{Int}\Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$, respectively $\Int(g')^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$; and then $\Psi_0(\eta)(\sigma\lambda) = \eta \circ \Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$ equals $\Int(g')^{-1} \circ \sigma \circ \eta \circ \lambda \circ \sigma^{-1} = \sigma(\Psi_0(\eta)(\lambda))$.

Still in the case where $\eta$ is a central isogeny, we have that $\widehat{\Psi_0(\eta)}$ is also a $\Gamma_{\overline k/k}$-equivariant, separable isogeny of root data, so that it is dual to a (necessarily central) $\Gamma_{\overline k/k}$-equivariant isogeny $\hat\eta : \hat{G'} \to \hat G$, as you say uniquely determined by choices of $\Gamma_{\overline k/k}$-fixed pinnings. Since two different pinnings of $\hat{G'}$ are uniquely $\hat{G'}/\operatorname Z(\hat{G'})$-conjugate, the corresponding $\Gamma_{\overline k/k}$-equivariant homomorphisms are $(\hat{G'}/\operatorname Z(\hat{G'}))^{\Gamma_{\overline k/k}}$-conjugate. Then, as you point out in a comment, Lemma 1.6 of Kottwitz - Stable trace formula: cuspidal tempered terms shows that they are $\smash{\hat{G'}}^{\Gamma_{\overline k/k}}\vphantom{\hat{G'}}$-conjugate.

We have that $G'$ is an almost-direct product of $\eta(G)$ and the almost-direct product of the duals of the simple factors of $G'$ not contained in $\eta(G)$, which allows us to reduce (1) and (3) to the case where $\eta$ is surjective.

I think you probably want to assume that $\eta$ is a quotient by a smooth, connected, normal subgroup followed by a central isogeny; this is automatic if $\eta$ is separable. If you do not want to reduce to this case, then you have to handle something like the exceptional isogeny $\operatorname{SO}_{2n + 1, k} \to \operatorname{Sp}_{2n, k}$ when $\operatorname{char} k$ is $2$, whose dual homomorphism I guess should be a map $\operatorname{SO}_{2n + 1, \mathbb C} \to \operatorname{Sp}_{2n, \mathbb C}$; but I think that there are no non-trivial such homomorphisms.

When $\eta$ is a quotient by a smooth, connected, normal subgroup $N$ of $G$, we have a similar decomposition of $G$ as the almost-direct product of $N$ with a canonical complement. Thus, if it is OK to make the reduction above, then we need only handle the case where $\eta$ is an isogeny.

In this case, if $(B, T, \mathcal X)$ is a pinning of $G_{\overline k}$, then $(B', T', \mathcal X') \mathrel{:=} (\eta_{\overline k}(B), \eta_{\overline k}(T), \eta_{\overline k}(\mathcal X))$ is a pinning of $G'_{\overline k}$. We compute $\Psi_0(G_{\overline k})$ and $\Psi_0(G'_{\overline k})$ in terms of these pinnings, and then let $\Psi_0(\eta)$ be the obvious map. If I understand correctly how $\Gamma_{\overline k/k}$ acts, then this is $\Gamma_{\overline k/k}$-equivariant: if we choose $\sigma \in \Gamma_{\overline k/k}$ and let $g \in G(\overline k)$ be such that $\sigma(B, T, \mathcal X)$ equals $\DeclareMathOperator\Ad{Ad}(g B g^{-1}, g T g^{-1}, \Ad(g)\mathcal X)$, then $\sigma(B', T', \mathcal X') = (\eta_{\overline k}(\sigma B), \eta_{\overline k}(\sigma T), \eta_{\overline k}(\sigma\mathcal X))$ equals $(g'B'g^{\prime\,{-1}}, g'T'g^{\prime\,{-1}}, \Ad(g')\mathcal X')$, where $g'$ equals $\eta_{\overline k}(g)$. Then $\sigma$, viewed as a morphism $X_*(T) \to X_*(T)$, respectively $X_*(T') \to X_*(T')$, sends $\lambda$, respectively $\lambda'$, to $\DeclareMathOperator\Int{Int}\Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$, respectively $\Int(g')^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$; and then $\Psi_0(\eta)(\sigma\lambda) = \eta \circ \Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$ equals $\Int(g')^{-1} \circ \sigma \circ \eta \circ \lambda \circ \sigma^{-1} = \sigma(\Psi_0(\eta)(\lambda))$.

Still in the case where $\eta$ is a central isogeny, we have that $\widehat{\Psi_0(\eta)}$ is also a $\Gamma_{\overline k/k}$-equivariant, separable isogeny of root data, so that it is dual to a (necessarily central) $\Gamma_{\overline k/k}$-equivariant isogeny $\hat\eta : \hat{G'} \to \hat G$, as you say uniquely determined by choices of $\Gamma_{\overline k/k}$-fixed pinnings. Since two different pinnings of $\hat{G'}$ are uniquely $\hat{G'}/\operatorname Z(\hat{G'})$-conjugate, the corresponding $\Gamma_{\overline k/k}$-equivariant homomorphisms are $(\hat{G'}/\operatorname Z(\hat{G'}))^{\Gamma_{\overline k/k}}$-conjugate; but it is not clear to me why the conjugacy can be lifted to $\smash{\hat{G'}}^{\Gamma_{\overline k/k}}$.

We have that $G'$ is an almost-direct product of $\eta(G)$ and the almost-direct product of the duals of the simple factors of $G'$ not contained in $\eta(G)$, which allows us to reduce (1) and (3) to the case where $\eta$ is surjective.

I think you probably want to assume that $\eta$ is a quotient by a smooth, connected, normal subgroup followed by a central isogeny; this is automatic if $\eta$ is separable. If you do not want to reduce to this case, then you have to handle something like the exceptional isogeny $\operatorname{SO}_{2n + 1, k} \to \operatorname{Sp}_{2n, k}$ when $\operatorname{char} k$ is $2$, whose dual homomorphism I guess should be a map $\operatorname{SO}_{2n + 1, \mathbb C} \to \operatorname{Sp}_{2n, \mathbb C}$; but I think that there are no non-trivial such homomorphisms.

When $\eta$ is a quotient by a smooth, connected, normal subgroup $N$ of $G$, we have a similar decomposition of $G$ as the almost-direct product of $N$ with a canonical complement. Thus, if it is OK to make the reduction above, then we need only handle the case where $\eta$ is an isogeny.

In this case, if $(B, T, \mathcal X)$ is a pinning of $G_{\overline k}$, then $(B', T', \mathcal X') \mathrel{:=} (\eta_{\overline k}(B), \eta_{\overline k}(T), \eta_{\overline k}(\mathcal X))$ is a pinning of $G'_{\overline k}$. We compute $\Psi_0(G_{\overline k})$ and $\Psi_0(G'_{\overline k})$ in terms of these pinnings, and then let $\Psi_0(\eta)$ be the obvious map. If I understand correctly how $\Gamma_{\overline k/k}$ acts, then this is $\Gamma_{\overline k/k}$-equivariant: if we choose $\sigma \in \Gamma_{\overline k/k}$ and let $g \in G(\overline k)$ be such that $\sigma(B, T, \mathcal X)$ equals $\DeclareMathOperator\Ad{Ad}(g B g^{-1}, g T g^{-1}, \Ad(g)\mathcal X)$, then $\sigma(B', T', \mathcal X') = (\eta_{\overline k}(\sigma B), \eta_{\overline k}(\sigma T), \eta_{\overline k}(\sigma\mathcal X))$ equals $(g'B'g^{\prime\,{-1}}, g'T'g^{\prime\,{-1}}, \Ad(g')\mathcal X')$, where $g'$ equals $\eta_{\overline k}(g)$. Then $\sigma$, viewed as a morphism $X_*(T) \to X_*(T)$, respectively $X_*(T') \to X_*(T')$, sends $\lambda$, respectively $\lambda'$, to $\DeclareMathOperator\Int{Int}\Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$, respectively $\Int(g')^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$; and then $\Psi_0(\eta)(\sigma\lambda) = \eta \circ \Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$ equals $\Int(g')^{-1} \circ \sigma \circ \eta \circ \lambda \circ \sigma^{-1} = \sigma(\Psi_0(\eta)(\lambda))$.

Still in the case where $\eta$ is a central isogeny, we have that $\widehat{\Psi_0(\eta)}$ is also a $\Gamma_{\overline k/k}$-equivariant, separable isogeny of root data, so that it is dual to a (necessarily central) $\Gamma_{\overline k/k}$-equivariant isogeny $\hat\eta : \hat{G'} \to \hat G$, as you say uniquely determined by choices of $\Gamma_{\overline k/k}$-fixed pinnings. Since two different pinnings of $\hat{G'}$ are uniquely $\hat{G'}/\operatorname Z(\hat{G'})$-conjugate, the corresponding $\Gamma_{\overline k/k}$-equivariant homomorphisms are $(\hat{G'}/\operatorname Z(\hat{G'}))^{\Gamma_{\overline k/k}}$-conjugate; but it is not clear to me why the conjugacy can be lifted to $\smash{\hat{G'}}^{\Gamma_{\overline k/k}}$. If $\operatorname Z(\hat{G'})$ is connected (equivalently, if the derived group of $G'$ is simply connected), then one can probably show that $H^1(\Gamma_{\overline k/k}, \operatorname Z(\hat{G'}))$ is trivial and deduce a suitable lifting; but I don't know about the case of disconnected dual centre.

We have that $G'$ is an almost-direct product of $\eta(G)$ and the almost-direct product of the duals of the simple factors of $G'$ not contained in $\eta(G)$, which allows us to reduce (1) and (3) to the case where $\eta$ is surjective.

I think you probably want to assume that $\eta$ is a quotient by a smooth, connected, normal subgroup followed by a central isogeny; this is automatic if $\eta$ is separable. If you do not want to reduce to this case, then you have to handle something like the exceptional isogeny $\operatorname{SO}_{2n + 1, k} \to \operatorname{Sp}_{2n, k}$ when $\operatorname{char} k$ is $2$, whose dual homomorphism I guess should be a map $\operatorname{SO}_{2n + 1, \mathbb C} \to \operatorname{Sp}_{2n, \mathbb C}$; but I think that there are no non-trivial such homomorphisms.

When $\eta$ is a quotient by a smooth, connected, normal subgroup $N$ of $G$, we have a similar decomposition of $G$ as the almost-direct product of $N$ with a canonical complement. Thus, if it is OK to make the reduction above, then we need only handle the case where $\eta$ is an isogeny.

In this case, if $(B, T, \mathcal X)$ is a pinning of $G_{\overline k}$, then $(B', T', \mathcal X') \mathrel{:=} (\eta_{\overline k}(B), \eta_{\overline k}(T), \eta_{\overline k}(\mathcal X))$ is a pinning of $G'_{\overline k}$. We compute $\Psi_0(G_{\overline k})$ and $\Psi_0(G'_{\overline k})$ in terms of these pinnings, and then let $\Psi_0(\eta)$ be the obvious map. If I understand correctly how $\Gamma_{\overline k/k}$ acts, then this is $\Gamma_{\overline k/k}$-equivariant: if we choose $\sigma \in \Gamma_{\overline k/k}$ and let $g \in G(\overline k)$ be such that $\sigma(B, T, \mathcal X)$ equals $\DeclareMathOperator\Ad{Ad}(g B g^{-1}, g T g^{-1}, \Ad(g)\mathcal X)$, then $(\eta_{\overline k}(\sigma B), \eta_{\overline k}(\sigma T), \eta_{\overline k}(\sigma\mathcal X))$ equals $\sigma(g'B'g^{\prime\,{-1}}, g'T'g^{\prime\,{-1}}, \Ad(g')\mathcal X')$, where $g'$ equals $\eta_{\overline k}(g)$. Then $\sigma$, viewed as a morphism $X_*(T) \to X_*(T)$, respectively $X_*(T') \to X_*(T')$, sends $\lambda$, respectively $\lambda'$, to $\DeclareMathOperator\Int{Int}\Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$, respectively $\Int(g')^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$; and then $\Psi_0(\eta)(\sigma\lambda) = \eta \circ \Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$ equals $\Int(g')^{-1} \circ \sigma \circ \eta \circ \lambda \circ \sigma^{-1} = \sigma(\Psi_0(\eta)(\lambda))$.

Still in the case where $\eta$ is a central isogeny, we have that $\widehat{\Psi_0(\eta)}$ is also a $\Gamma_{\overline k/k}$-equivariant, separable isogeny of root data, so that it is dual to a (necessarily central) $\Gamma_{\overline k/k}$-equivariant isogeny $\hat\eta : \hat{G'} \to \hat G$, as you say uniquely determined by choices of $\Gamma_{\overline k/k}$-fixed pinnings. Since two different pinnings of $\hat{G'}$ are uniquely $\hat{G'}/\operatorname Z(\hat{G'})$-conjugate, the corresponding $\Gamma_{\overline k/k}$-equivariant homomorphisms are $(\hat{G'}/\operatorname Z(\hat{G'}))^{\Gamma_{\overline k/k}}$-conjugate; but it is not clear to me why the conjugacy can be lifted to $\smash{\hat{G'}}^{\Gamma_{\overline k/k}}$.

We have that $G'$ is an almost-direct product of $\eta(G)$ and the almost-direct product of the duals of the simple factors of $G'$ not contained in $\eta(G)$, which allows us to reduce (1) and (3) to the case where $\eta$ is surjective.

I think you probably want to assume that $\eta$ is a quotient by a smooth, connected, normal subgroup followed by a central isogeny; this is automatic if $\eta$ is separable. If you do not want to reduce to this case, then you have to handle something like the exceptional isogeny $\operatorname{SO}_{2n + 1, k} \to \operatorname{Sp}_{2n, k}$ when $\operatorname{char} k$ is $2$, whose dual homomorphism I guess should be a map $\operatorname{SO}_{2n + 1, \mathbb C} \to \operatorname{Sp}_{2n, \mathbb C}$; but I think that there are no non-trivial such homomorphisms.

When $\eta$ is a quotient by a smooth, connected, normal subgroup $N$ of $G$, we have a similar decomposition of $G$ as the almost-direct product of $N$ with a canonical complement. Thus, if it is OK to make the reduction above, then we need only handle the case where $\eta$ is an isogeny.

In this case, if $(B, T, \mathcal X)$ is a pinning of $G_{\overline k}$, then $(B', T', \mathcal X') \mathrel{:=} (\eta_{\overline k}(B), \eta_{\overline k}(T), \eta_{\overline k}(\mathcal X))$ is a pinning of $G'_{\overline k}$. We compute $\Psi_0(G_{\overline k})$ and $\Psi_0(G'_{\overline k})$ in terms of these pinnings, and then let $\Psi_0(\eta)$ be the obvious map. If I understand correctly how $\Gamma_{\overline k/k}$ acts, then this is $\Gamma_{\overline k/k}$-equivariant: if we choose $\sigma \in \Gamma_{\overline k/k}$ and let $g \in G(\overline k)$ be such that $\sigma(B, T, \mathcal X)$ equals $\DeclareMathOperator\Ad{Ad}(g B g^{-1}, g T g^{-1}, \Ad(g)\mathcal X)$, then $\sigma(B', T', \mathcal X') = (\eta_{\overline k}(\sigma B), \eta_{\overline k}(\sigma T), \eta_{\overline k}(\sigma\mathcal X))$ equals $(g'B'g^{\prime\,{-1}}, g'T'g^{\prime\,{-1}}, \Ad(g')\mathcal X')$, where $g'$ equals $\eta_{\overline k}(g)$. Then $\sigma$, viewed as a morphism $X_*(T) \to X_*(T)$, respectively $X_*(T') \to X_*(T')$, sends $\lambda$, respectively $\lambda'$, to $\DeclareMathOperator\Int{Int}\Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$, respectively $\Int(g')^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$; and then $\Psi_0(\eta)(\sigma\lambda) = \eta \circ \Int(g)^{-1} \circ \sigma \circ \lambda \circ \sigma^{-1}$ equals $\Int(g')^{-1} \circ \sigma \circ \eta \circ \lambda \circ \sigma^{-1} = \sigma(\Psi_0(\eta)(\lambda))$.

Still in the case where $\eta$ is a central isogeny, we have that $\widehat{\Psi_0(\eta)}$ is also a $\Gamma_{\overline k/k}$-equivariant, separable isogeny of root data, so that it is dual to a (necessarily central) $\Gamma_{\overline k/k}$-equivariant isogeny $\hat\eta : \hat{G'} \to \hat G$, as you say uniquely determined by choices of $\Gamma_{\overline k/k}$-fixed pinnings. Since two different pinnings of $\hat{G'}$ are uniquely $\hat{G'}/\operatorname Z(\hat{G'})$-conjugate, the corresponding $\Gamma_{\overline k/k}$-equivariant homomorphisms are $(\hat{G'}/\operatorname Z(\hat{G'}))^{\Gamma_{\overline k/k}}$-conjugate; but it is not clear to me why the conjugacy can be lifted to $\smash{\hat{G'}}^{\Gamma_{\overline k/k}}$.