The Cayley parametrization of $O(n),$ as in my answer to this question makes one wonder: which algebraic groups are actually rational? I am sure this is very well understood, just not by me...
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asked Mar 3, 2012 at 19:58
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5$\begingroup$ As the commentw on Ryan's answers indicate, you need to say more precisely what you mean by "algebraic group" and by "rational". There are results going back to Rosenlicht (in older algebraic geometry language) about which varieties are "rational" in the sense that the function field is a purely transcendental field. The linguistic issues tend to get in the way of clarity. $\endgroup$– Jim HumphreysCommented Mar 3, 2012 at 23:15
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5$\begingroup$ The answer to this MO query mathoverflow.net/questions/76882/… cites a paper showing there are three-dimensional commutative algebraic groups over ${\bf Q}$ that are rational over $\overline{\bf Q}$ but not over ${\bf Q}$! $\endgroup$– Noam D. ElkiesCommented Mar 4, 2012 at 5:11
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$\begingroup$ @Noam: thanks, that's very interesting! $\endgroup$– Igor RivinCommented Mar 5, 2012 at 1:17
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$\begingroup$ Note that any two-dimensional linear algebraic group $G$ over a field $k$ of characteristic 0 is $k$-rational. Indeed, since ${\rm char}(k)=0$, we may assume that $G$ is reductive. Since there are no semisimple groups of dimension $\le2$, our group $G$ is a torus. Finally, in Section 4.9 of Voskresenskii's book it is proved that any $k$-torus of dimension one or two is $k$-rational. $\endgroup$– Mikhail BorovoiCommented Jul 3, 2017 at 1:27
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3 Answers
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A reductive group $G$ over a ground field $k$ need not be a rational variety over $k$ (although the group $G_{/\ell}$ obtained by scalar extension is a rational $\ell$ variety, where $\ell$ is an algebraic closure of $k$). Maybe a good place to look concerning these matters is Merkurjev's 1996 Publ. Math. IHES paper "R-equivalence and rationality problem for semisimple adjoint classical algebraic groups".
In the intro to that paper, Merkurjev makes some interesting observations:
- Chevalley showed in the 50s that there are algebraic tori over local fields which are not rational varieties.
- algebraic tori over dimension <= 2 are always rational varieties; thus, any reductive group of rank <=2 is rational. (Chevalley proved that the variety of maximal tori in a connected group is a rational variety, so the rationality of $G$ boils down to the rationality of some maximal torus of $G$).
- any semisimple group over a number field which is a counterexample to weak approximation is an example of a non-rational group.
In Merkurjev's paper, you will find explicit examples of groups $G$ which are not stably rational -- a $k$-variety $X$ is stably rational if $X \times \operatorname{Aff}^d$ is a rational $k$-variety for some $d \ge 0$. One way to show that $G$ is not stably rational (and hence not rational) is to show that the "group of $R$-equivalence classes" for $G$ is non-trivial; this group of $R$-equivalence classes was introduced by Manin. Merkurjev's paper is devoted to the computation of the group of $R$-equivalence classes for semisimple, adjoint, classical groups.
answered Mar 5, 2012 at 13:20
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The following answer applies to affine algebraic groups over algebraically closed fields, or more generally for quasi-split groups, as Scott Carnahan suggests. I don't really know about these rationality (in the number-theoretic sense) questions.
Under these conditions, every algebraic group is an extension of a reductive group by a unipotent group (since a reductive group is one whose unipotent radical, i.e. maximal normal unipotent connected subgroup, is trivial). Every unipotent group is rational, since it is isomorphic as a variety to $\mathbb{A}^n$. Every reductive group is rational by the Bruhat decomposition. Every Zariski-locally trivial bundle over a rational variety whose fibers are rational, is rational.
answered Mar 3, 2012 at 21:32
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$\begingroup$ By "rationality" shouldn't we also mean that the group multiplication, inverse, and identity are defined over $\mathbb{Q}$? In that case, an elliptic curve whose $j$-invariant isn't rational can't be rational, right? $\endgroup$ Commented Mar 3, 2012 at 21:56
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$\begingroup$ The Bruhat decomposition is usually stated for affine reductive groups, unlike this elliptic curve. $\endgroup$ Commented Mar 3, 2012 at 22:06
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$\begingroup$ Sorry, I keep forgetting that "algebraic group" means something different to some people. I have in mind only affine algebraic groups. $\endgroup$ Commented Mar 3, 2012 at 22:10
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$\begingroup$ (I consider the larger class to be just "group schemes") $\endgroup$ Commented Mar 3, 2012 at 22:11
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$\begingroup$ I think this only peoves geometric rationality - you may need the group to be quasi-split to invoke Bruhat over the ground field. $\endgroup$– S. Carnahan ♦Commented Mar 3, 2012 at 22:20
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Suppose G is a connected affine algebraic group over a field k. If G is reductive or k is perfect, then G is unirational. A reference is Borel, Linear Algebraic Groups, Theorem 18.2. This doesn't quite give rationality, but is the best result in general that I know of, and is sufficient for some applications. (Ryan's answer of course gives more precise information under stronger hypotheses).
An affine counterexample over an imperfect field of characteristic p is given by y^p=x+tx^p where t is not a p-th power in k. (This is a subgroup of Ga2).
Now if G is connected and not affine, then it surjects onto an abelian variety. So if G were rational, then we could get a nonconstant morphism from P^1 to an abelian variety. Now let me just quote https://math.stackexchange.com/a/114611 so no non-affine conntected G can be rational, or even unirational.
answered Mar 3, 2012 at 23:09