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user74230
user74230
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Pick a global field $E$ and finite place $w$ with $E_w=K$. The fraction field $k$ over $E$ of the henselization of the "algebraic" local ring at $w$ is the direct limit of finite separable sub extensions $F/E$ for which the place $v$ on $F$ from the valuation on $k$ satisfies $F_v=K$. Thus, it suffices to "algebraize" $G$ over $k$. We therefore forget about number field and focus on a henselian valued field $k$ with completion denoted $K$ and aim to descend a connected reductive $K$-group $G$ to a connected reductive $k$-group.

The Galois groups of $k$ and $K$ are naturally isomorphic, so if $R$ denotes the root datum of $G_{K_s}$ then ${\rm{H}}^1(k, {\rm{Aut}}(R)) \rightarrow {\rm{H}}^1(K, {\rm{Aut}}(R))$ is bijective. This says that every quasi-split connected reductive $K$-group descends to a quasi-split connected reductive $k$-group which is moreover unique up to isomorphism (since these H$^1$'s classify quasi-split forms with a given geometric root datum). Every connected reductive $K$-group $G$ has a (unique) quasi-split inner form $G_0$, so $G$ is obtained from $G_0$ via twisting against a class in ${\rm{H}}^1(K, G_0^{\rm{ad}})$ for the adjoint semisimple $G_0^{\rm{ad}} := G_0/Z_{G_0}$. Thus, it suffices to show that ${\rm{H}}^1(k, H) \rightarrow {\rm{H}}^1(K,H_K)$ is surjective for every smooth connected affine $k$-group $H$ (such as $H$ being the quasi-split $k$-descent of $G_0^{\rm{ad}}$). Even better:

Theorem: For any smooth affine $k$-group $H$, the natural map $${\rm{H}}^1(k,H) \rightarrow {\rm{H}}^1(K,H)$$ is bijective.

Proof: By Galois-twisting, injectivity reduces to triviality of the kernel. In other words, if $E$ is an $H$-torsor over $k$ which has a $K$-point then it has a $k$-point. More generally, if $X$ is a smooth $k$-scheme then $X(k)$ is dense in $X(K)$ for the valuation topoloy. This is Zariski-local on $X$, so we can assume there is an etale map $f:X \rightarrow \mathbf{A}^n_k$. The open image $V=f(X)$ is dense open in $\mathbf{A}^n_k$, so $V(k)$ is dense in $V(K)$ due to density of $k$ in $K$. By the Zariski-local structure theorem for etale morphisms and the $K$-analytic inverse function theorem, for each $x \in X(K)$ and $v=f(x)\in V(K)$, every $v'$ sufficiently near $v$ admits $x'\in f^{-1}(v')$ near $x$ in $X(K)$. By openness of $X(K) \rightarrow V(K)$, for any $x \in X(K)$ and open $\Omega \subset X(K)$ around $x$ we can find an open $U \subset V(K)$ around $v=f(x)$ such that every $u \in U$ is the image of a $K$-point in $\Omega$. Consider such $u \in V(k) \cap U$ (as exists by density of $V(k)$ in $V(K)$). The fiber scheme $f^{-1}(u)$ is finite etale over $k$, so the equivalence of Galois theories of $k$ and $K$ shows that every $K$-point in $f^{-1}(u)$ comes from a unique $k$-point of $f^{-1}(u)$. Hence, we can find a $k$-point in $\Omega$. This completes the proof of injectivity.

For surjectivity, choose a closed $k$-subgroup inclusion $j:H \hookrightarrow {\rm{GL}}_n:=G$ and let $X=G/H$ (a smooth $k$-scheme). Thus, there is a natural surjection $$G(k)\backslash X(k) \rightarrow {\rm{H}}^1(k,H)$$ and likewise for $K$ (since ${\rm{H}}^1(k, {\rm{GL}}_n)=1$ and likewise for $K$). It therefore suffices to show that the natural map $$X(k) \rightarrow G(K) \backslash X(K)$$ is surjective. Since $X(k)$ is dense in $X(K)$ by the above, it suffices to show that all $G(K)$-orbits in $X(K)$ are open. But each orbit map $G \rightarrow X$$G_K \rightarrow X_K$ through a $K$-point is a smooth map since $H$ is smooth, so the induced map on $K$-points is open (hence has open image) by the $K$-analytic implicit function theorem (using the Zariski-local structure of smooth morphisms).

QED

Pick a global field $E$ and finite place $w$ with $E_w=K$. The fraction field $k$ over $E$ of the henselization of the "algebraic" local ring at $w$ is the direct limit of finite separable sub extensions $F/E$ for which the place $v$ on $F$ from the valuation on $k$ satisfies $F_v=K$. Thus, it suffices to "algebraize" $G$ over $k$. We therefore forget about number field and focus on a henselian valued field $k$ with completion denoted $K$ and aim to descend a connected reductive $K$-group $G$ to a connected reductive $k$-group.

The Galois groups of $k$ and $K$ are naturally isomorphic, so if $R$ denotes the root datum of $G_{K_s}$ then ${\rm{H}}^1(k, {\rm{Aut}}(R)) \rightarrow {\rm{H}}^1(K, {\rm{Aut}}(R))$ is bijective. This says that every quasi-split connected reductive $K$-group descends to a quasi-split connected reductive $k$-group which is moreover unique up to isomorphism (since these H$^1$'s classify quasi-split forms with a given geometric root datum). Every connected reductive $K$-group $G$ has a (unique) quasi-split inner form $G_0$, so $G$ is obtained from $G_0$ via twisting against a class in ${\rm{H}}^1(K, G_0^{\rm{ad}})$ for the adjoint semisimple $G_0^{\rm{ad}} := G_0/Z_{G_0}$. Thus, it suffices to show that ${\rm{H}}^1(k, H) \rightarrow {\rm{H}}^1(K,H_K)$ is surjective for every smooth connected affine $k$-group $H$ (such as $H$ being the quasi-split $k$-descent of $G_0^{\rm{ad}}$). Even better:

Theorem: For any smooth affine $k$-group $H$, the natural map $${\rm{H}}^1(k,H) \rightarrow {\rm{H}}^1(K,H)$$ is bijective.

Proof: By Galois-twisting, injectivity reduces to triviality of the kernel. In other words, if $E$ is an $H$-torsor over $k$ which has a $K$-point then it has a $k$-point. More generally, if $X$ is a smooth $k$-scheme then $X(k)$ is dense in $X(K)$ for the valuation topoloy. This is Zariski-local on $X$, so we can assume there is an etale map $f:X \rightarrow \mathbf{A}^n_k$. The open image $V=f(X)$ is dense open in $\mathbf{A}^n_k$, so $V(k)$ is dense in $V(K)$ due to density of $k$ in $K$. By the Zariski-local structure theorem for etale morphisms and the $K$-analytic inverse function theorem, for each $x \in X(K)$ and $v=f(x)\in V(K)$, every $v'$ sufficiently near $v$ admits $x'\in f^{-1}(v')$ near $x$ in $X(K)$. By openness of $X(K) \rightarrow V(K)$, for any $x \in X(K)$ and open $\Omega \subset X(K)$ around $x$ we can find an open $U \subset V(K)$ around $v=f(x)$ such that every $u \in U$ is the image of a $K$-point in $\Omega$. Consider such $u \in V(k) \cap U$ (as exists by density of $V(k)$ in $V(K)$). The fiber scheme $f^{-1}(u)$ is finite etale over $k$, so the equivalence of Galois theories of $k$ and $K$ shows that every $K$-point in $f^{-1}(u)$ comes from a unique $k$-point of $f^{-1}(u)$. Hence, we can find a $k$-point in $\Omega$. This completes the proof of injectivity.

For surjectivity, choose a closed $k$-subgroup inclusion $j:H \hookrightarrow {\rm{GL}}_n:=G$ and let $X=G/H$ (a smooth $k$-scheme). Thus, there is a natural surjection $$G(k)\backslash X(k) \rightarrow {\rm{H}}^1(k,H)$$ and likewise for $K$ (since ${\rm{H}}^1(k, {\rm{GL}}_n)=1$ and likewise for $K$). It therefore suffices to show that the natural map $$X(k) \rightarrow G(K) \backslash X(K)$$ is surjective. Since $X(k)$ is dense in $X(K)$ by the above, it suffices to show that all $G(K)$-orbits in $X(K)$ are open. But each orbit map $G \rightarrow X$ is a smooth map since $H$ is smooth, so the induced map on $K$-points is open (hence has open image) by the $K$-analytic implicit function theorem (using the Zariski-local structure of smooth morphisms).

QED

Pick a global field $E$ and finite place $w$ with $E_w=K$. The fraction field $k$ over $E$ of the henselization of the "algebraic" local ring at $w$ is the direct limit of finite separable sub extensions $F/E$ for which the place $v$ on $F$ from the valuation on $k$ satisfies $F_v=K$. Thus, it suffices to "algebraize" $G$ over $k$. We therefore forget about number field and focus on a henselian valued field $k$ with completion denoted $K$ and aim to descend a connected reductive $K$-group $G$ to a connected reductive $k$-group.

The Galois groups of $k$ and $K$ are naturally isomorphic, so if $R$ denotes the root datum of $G_{K_s}$ then ${\rm{H}}^1(k, {\rm{Aut}}(R)) \rightarrow {\rm{H}}^1(K, {\rm{Aut}}(R))$ is bijective. This says that every quasi-split connected reductive $K$-group descends to a quasi-split connected reductive $k$-group which is moreover unique up to isomorphism (since these H$^1$'s classify quasi-split forms with a given geometric root datum). Every connected reductive $K$-group $G$ has a (unique) quasi-split inner form $G_0$, so $G$ is obtained from $G_0$ via twisting against a class in ${\rm{H}}^1(K, G_0^{\rm{ad}})$ for the adjoint semisimple $G_0^{\rm{ad}} := G_0/Z_{G_0}$. Thus, it suffices to show that ${\rm{H}}^1(k, H) \rightarrow {\rm{H}}^1(K,H_K)$ is surjective for every smooth connected affine $k$-group $H$ (such as $H$ being the quasi-split $k$-descent of $G_0^{\rm{ad}}$). Even better:

Theorem: For any smooth affine $k$-group $H$, the natural map $${\rm{H}}^1(k,H) \rightarrow {\rm{H}}^1(K,H)$$ is bijective.

Proof: By Galois-twisting, injectivity reduces to triviality of the kernel. In other words, if $E$ is an $H$-torsor over $k$ which has a $K$-point then it has a $k$-point. More generally, if $X$ is a smooth $k$-scheme then $X(k)$ is dense in $X(K)$ for the valuation topoloy. This is Zariski-local on $X$, so we can assume there is an etale map $f:X \rightarrow \mathbf{A}^n_k$. The open image $V=f(X)$ is dense open in $\mathbf{A}^n_k$, so $V(k)$ is dense in $V(K)$ due to density of $k$ in $K$. By the Zariski-local structure theorem for etale morphisms and the $K$-analytic inverse function theorem, for each $x \in X(K)$ and $v=f(x)\in V(K)$, every $v'$ sufficiently near $v$ admits $x'\in f^{-1}(v')$ near $x$ in $X(K)$. By openness of $X(K) \rightarrow V(K)$, for any $x \in X(K)$ and open $\Omega \subset X(K)$ around $x$ we can find an open $U \subset V(K)$ around $v=f(x)$ such that every $u \in U$ is the image of a $K$-point in $\Omega$. Consider such $u \in V(k) \cap U$ (as exists by density of $V(k)$ in $V(K)$). The fiber scheme $f^{-1}(u)$ is finite etale over $k$, so the equivalence of Galois theories of $k$ and $K$ shows that every $K$-point in $f^{-1}(u)$ comes from a unique $k$-point of $f^{-1}(u)$. Hence, we can find a $k$-point in $\Omega$. This completes the proof of injectivity.

For surjectivity, choose a closed $k$-subgroup inclusion $j:H \hookrightarrow {\rm{GL}}_n:=G$ and let $X=G/H$ (a smooth $k$-scheme). Thus, there is a natural surjection $$G(k)\backslash X(k) \rightarrow {\rm{H}}^1(k,H)$$ and likewise for $K$ (since ${\rm{H}}^1(k, {\rm{GL}}_n)=1$ and likewise for $K$). It therefore suffices to show that the natural map $$X(k) \rightarrow G(K) \backslash X(K)$$ is surjective. Since $X(k)$ is dense in $X(K)$ by the above, it suffices to show that all $G(K)$-orbits in $X(K)$ are open. But each orbit map $G_K \rightarrow X_K$ through a $K$-point is a smooth map since $H$ is smooth, so the induced map on $K$-points is open (hence has open image) by the $K$-analytic implicit function theorem (using the Zariski-local structure of smooth morphisms).

QED

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user74230
user74230
  • 1.4k
  • 8
  • 10

Pick a global field $E$ and finite place $w$ with $E_w=K$. The fraction field $k$ over $E$ of the henselization of the "algebraic" local ring at $w$ is the direct limit of finite separable sub extensions $F/E$ for which the place $v$ on $F$ from the valuation on $k$ satisfies $F_v=K$. Thus, it suffices to "algebraize" $G$ over $k$. We therefore forget about number field and focus on a henselian valued field $k$ with completion denoted $K$ and aim to descend a connected reductive $K$-group $G$ to a connected reductive $k$-group.

The Galois groups of $k$ and $K$ are naturally isomorphic, so if $R$ denotes the root datum of $G_{K_s}$ then ${\rm{H}}^1(k, {\rm{Aut}}(R)) \rightarrow {\rm{H}}^1(K, {\rm{Aut}}(R))$ is bijective. This says that every quasi-split connected reductive $K$-group descends to a quasi-split connected reductive $k$-group which is moreover unique up to isomorphism (since these H$^1$'s classify quasi-split forms with a given geometric root datum). Every connected reductive $K$-group $G$ has a (unique) quasi-split inner form $G_0$, so $G$ is obtained from $G_0$ via twisting against a class in ${\rm{H}}^1(K, G_0^{\rm{ad}})$ for the adjoint semisimple $G_0^{\rm{ad}} := G_0/Z_{G_0}$. Thus, it suffices to show that ${\rm{H}}^1(k, H) \rightarrow {\rm{H}}^1(K,H_K)$ is surjective for every smooth connected affine $k$-group $H$ (such as $H$ being the quasi-split $k$-descent of $G_0^{\rm{ad}}$). Even better:

Theorem: For any smooth affine $k$-group $H$, the natural map $${\rm{H}}^1(k,H) \rightarrow {\rm{H}}^1(K,H)$$ is bijective.

Proof: By Galois-twisting, injectivity reduces to triviality of the kernel. In other words, if $E$ is an $H$-torsor over $k$ which has a $K$-point then it has a $k$-point. More generally, if $X$ is a smooth $k$-scheme then $X(k)$ is dense in $X(K)$ for the valuation topoloy. This is Zariski-local on $X$, so we can assume there is an etale map $f:X \rightarrow \mathbf{A}^n_k$. The open image $V=f(X)$ is dense open in $\mathbf{A}^n_k$, so $V(k)$ is dense in $V(K)$ due to density of $k$ in $K$. By the Zariski-local structure theorem for etale morphisms and the $K$-analytic inverse function theorem, for each $x \in X(K)$ and $v=f(x)\in V(K)$, every $v'$ sufficiently near $v$ admits $x'\in f^{-1}(v')$ near $x$ in $X(K)$. By openness of $X(K) \rightarrow V(K)$, for any $x \in X(K)$ and open $\Omega \subset X(K)$ around $x$ we can find an open $U \subset V(K)$ around $v=f(x)$ such that every $u \in U$ is the image of a $K$-point in $\Omega$. Consider such $u \in V(k) \cap U$ (as exists by density of $V(k)$ in $V(K)$). The fiber scheme $f^{-1}(u)$ is finite etale over $k$, so the equivalence of Galois theories of $k$ and $K$ shows that every $K$-point in $f^{-1}(u)$ comes from a unique $k$-point of $f^{-1}(u)$. Hence, we can find a $k$-point in $\Omega$. This completes the proof of injectivity.

For surjectivity, choose a closed $k$-subgroup inclusion $j:H \hookrightarrow {\rm{GL}}_n:=G$ and let $X=G/H$ (a smooth $k$-scheme). Thus, there is a natural surjection $$G(k)\backslash X(k) \rightarrow {\rm{H}}^1(k,H)$$ and likewise for $K$ (since ${\rm{H}}^1(k, {\rm{GL}}_n)=1$ and likewise for $K$). It therefore suffices to show that the natural map $$X(k) \rightarrow G(K) \backslash X(K)$$ is surjective. Since $X(k)$ is dense in $X(K)$ by the above, it suffices to show that all $G(K)$-orbits in $X(K)$ are open. But each orbit map $G \rightarrow X$ is a smooth map since $H$ is smooth, so the induced map on $K$-points is open (hence has open image) by the $K$-analytic implicit function theorem (using the Zariski-local structure of smooth morphisms).

QED