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!

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    $\begingroup$ The connected reductive group is defined by some equations with coefficients in your local field. Take a global field dense in your local field and approximate them arbitrarily closely by elements of your global field. I think there will be a Krasner's Lemma phenomenon where if you get close enough the reductive group will be unchanged. $\endgroup$
    – Will Sawin
    Mar 4, 2015 at 15:18
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    $\begingroup$ @WillSawin: I think you will have a hard time proving that your approximation is a group to begin with. Would you care to provide a complete argument? $\endgroup$ Mar 4, 2015 at 16:46
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    $\begingroup$ Yes, by inner twisting one can reduce to the problem of algebraizing torsors for any smooth affine group over any henselian valued field whatsoever. This requires no input from number theory or serious algebraic geometry at all. I'll post a complete proof later, when I find time. I do not believe that the method suggested by Will can be turned into a proof; it seems too soft. $\endgroup$
    – user74230
    Mar 4, 2015 at 19:57
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    $\begingroup$ @DaveWitteMorris: the result you mention proves a much stronger conclusion (in particular that adjoint/simply connected semisimple groups are $\mathbf{Q}$-definable). But this stronger conclusion is not true for arbitrary semisimple connected groups over $\mathbf{Q}_p$. For the question I would tend to believe that the material in Borel-Serre's 1964 paper "Theoremes de finitude en Cohomologie Galoisienne" answer the question (at least in char. 0). $\endgroup$
    – YCor
    Mar 4, 2015 at 21:12
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    $\begingroup$ @QuestionMark Just take the variety of all affine algebraic groups embedded in some affine space by equations of bounded degree. This is defined over $\mathbb Z$, and has a local point. The problem is that I don't know an appropriate continuity statement. As user74230 points out, if you avoid this enormous space of groups and just focus on the space of groups in $GL_n$ that are conjugate, over an algebraically closed field, to $H$, you get continuity using the implicit function theorem. $\endgroup$
    – Will Sawin
    Mar 4, 2015 at 23:02

2 Answers 2


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).



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.


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