Is every connected reductive group over a local field already defined over a global field?

Question

Let $K$ be a local field, e.g. $\mathbb{Q}_p$ or $\mathbb{F}_p((t))$. Let $G$ be a connected reductive group over $K$. Is it true that $G$ is already defined over a global field? More precisely, does there exist a global field $F$, a place $v$ in $F$ with $F_v\simeq K$ and a connected reductive group $\tilde{G}$ over $F$ such that as groups $$\tilde{G}\otimes_F F_v\simeq G?$$ I am particularly interested in the case where $K$ is of equal characteristic. Thanks in advance!

Answer 1

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

Answer 2

I can't answer your question completely, but this extended comment may be helpful. Probably the answer will be "yes", judging at least from the old classification by Tits summarized in the proceedings of the 1965 AMS Summer Institute at Boulder (published by AMS in 1966 as vol. 9 in their series Proc. of Symposia in Pure Math., which was for a while freely available at the AMS website). Tits was interested in sorting out the possible forms of a simple algebraic group over various kinds of fields, and in his summary tables one sees case-by-case that existence of a group defined over what he calls a "$\mathfrak{p}$-adic" field implies existence of such a group over some number field (but not vice versa). He also provides some general theorems in arbitrary characteristic, with proofs later filled in and somewhat corrected in thesis work by his student Martin Selbach at Bonn (1976). But it may be impossible to find a complete treatment anywhere in the literature. (Will's comment is worth following up.)

P.S. It's worth adding that a typical (connected) reductive group is an almost-direct product of an algebraic torus with one or more simple algebraic groups. In particular, you'd want to look separately at the case of a torus.