on the condition "$G$ is defined over $\mathbb{Q}$"

Question

This might be a stupid question, but I cannot understand the "technical condition" when studying some basics of arithmetic groups, that is an algebraic group is defined over $\mathbb{Q}$. Specifically, Borel-Harish-Chandra's theorem says that if $G$ is a semisimple group with no compact factor, then $G$ is defined over $\mathbb{Q}$. Can someone give me an explanation that why should we pose such a condition first when talking about algebraic group? What about $\mathbb{Q}$ being replaced by other number fields? Thanks!

Answer 1

See here; what's below is essentially taken directly from there. An algebraic group $G$ over a number field $K$ is a subgroup $G ⊂ GL_n(K)$, which is defined as the set of common zeroes of a set of polynomials in the matrix coefficients where in addition these polynomials have coefficients in $K$.

Answer 2

The statement you state is confusing, since you don't specify the ground field, and also there is a confusion between "defined over $\mathbf{Q}$" and "definable over $\mathbf{Q}$". The second means that the group admits a $\mathbf{Q}$-structure (for instance, it means that its Hopf algebra can be obtained from a Hopf algebra over $\mathbf{Q}$ by extending scalars to the ground field; the first means you fix such a structure (this is what's needed when you want to construct arithmetic lattices).

Any semisimple group over $K=\mathbf{R}$ or $\mathbf{C}$ is definable over $\mathbf{Q}$ (in several non $\mathbf{Q}$-isomorphic ways). This is not true in general. For instance in $K=\mathbf{Q}(t)$, the semisimple group $SO(q)$ where $q$ is the quadratic form $X^2+Y^2+tZ^2$ is not definable over $\mathbf{Q}$.

Answer 3

Here is a Galois-cohomological proof of the Borel-HC theorem, valid in the general connected reductive (not just semisimple) case and not requiring any avoidance of compact factors. (There is some hard input below for which references are given in the book of Planatov and Rapinchuk on arithmetic of algebraic groups.)

Possibly the methods below will be too high-level to be useful for you (you don't indicate your background in algebraic geometry or algebraic groups), in which case I apologize in advance, but it provides a systematic viewpoint for analyzing questions of "descent of the ground field" as in this theorem (so there is some merit in being aware of the existence of such methods). I have no idea what method of proof was used by Borel and Harish-Chandra (I never heard of this result before seeing your question), so I don't know why they imposed an avoidance of compact factors; maybe the Galois cohomological formalism (or theory of algebraic groups over fields) was not sufficiently developed in those days?

Let $G$ be a connected reductive $\mathbf{R}$-group, and let $G_0$ be the corresponding split $\mathbf{R}$-group (e.g., if $G = {\rm{SO}}(q)$ for a non-degenerate quadratic space $(V,q)$ over $\mathbf{R}$ of dimension $n \ge 3$ then $G_0 = {\rm{SO}}_n$ is the special orthogonal group of the orthogonal sum of $n/2$ hyperbolic planes when $n$ is even and of $(n-1)/2$ such planes and the space $(\mathbf{R}, x^2)$ when $n$ is odd). There is a split connected reductive group $H_0$ over $\mathbf{Q}$ which is a $\mathbf{Q}$-descent of $G_0$ (nothing special about $\mathbf{R}$ here; could be any field of characteristic 0). The "automorphism variety" $A = {\rm{Aut}}_{H_0/\mathbf{Q}}$ is a $\mathbf{Q}$-group locally of finite type that fits into an exact sequence of $\mathbf{Q}$-groups $$1 \rightarrow H_0^{\rm{ad}} \rightarrow A \rightarrow \Gamma \rightarrow 1$$ where $\Gamma$ is the automorphism group of the based root datum attached to $H_0$; this also splits as a semi-direct product via a choice of pinning of $H_0$, so concretely $A$ is a disjoint union of copies of $H_0^{\rm{ad}}$ indexed by $\Gamma$ in a suitable manner. (If $G$ is semisimple then $\Gamma$ is finite.)

The isomorphism class of $G$ corresponds to an element $[G]$ in the Galois cohomology set ${\rm{H}}^1(\mathbf{R}, A)$, and the meaning of the Borel-HC theorem (without the "compact factor" hypothesis) is that the image of the natural map $${\rm{H}}^1(\mathbf{Q},A) \rightarrow {\rm{H}}^1(\mathbf{R},A)$$ hits $[G]$. The image of $[G]$ in ${\rm{H}}^1(\mathbf{R},\Gamma) = {\rm{Hom}}({\rm{Gal}}(\mathbf{C}/\mathbf{R}),\Gamma)$ certainly lifts to ${\rm{H}}^1(\mathbf{Q},\Gamma)$ (e.g., using an imaginary quadratic field), and we can lift the latter back into ${\rm{H}}^1(\mathbf{Q},A)$ via a choice of semi-direct product splitting of $A \twoheadrightarrow \Gamma$ using a pinning. Now applying the "twisting method" in non-abelian Galois cohomology (as discussed in section 5 of Chapter I of Serre's book on Galois cohomology) allows us to reduce to the case that $[G]$ has trivial image in ${\rm{H}}^1(\mathbf{R},\Gamma)$ at the cost of replacing $H_0$ with a (quasi-split) $\mathbf{Q}$-form $H'_0$ and replacing $A$ with $A' = {\rm{Aut}}_{H'_0/\mathbf{Q}}$ and replacing $\Gamma$ with $\Gamma' = A'/(H'_0)^{\rm{ad}}$.

Letting $G'_0 = (H'_0)_{\mathbf{R}}$, now $G$ is classified by an element of ${\rm{H}}^1(\mathbf{R},A')$ that comes from ${\rm{H}}^1(\mathbf{R},(H'_0)^{\rm{ad}})$. Hence, it suffices to show that the natural map $${\rm{H}}^1(\mathbf{Q}, (H'_0)^{\rm{ad}}) \rightarrow {\rm{H}}^1(\mathbf{R}, (H'_0)^{\rm{ad}})$$ is surjective. But it is a general (deep) fact that for any connected linear algebraic group $H$ over a number field $F$, the natural map ${\rm{H}}^1(F,H) \rightarrow \prod_{v|\infty} {\rm{H}}^1(F_v,H)$ is surjective. (For a proof, see Theorem 6.17 in the book on algebraic groups by Platanov and Rapinchuk; it involves using work of Borovoi on ${\rm{H}}^1(\mathbf{R},\cdot)$ of connected semisimple groups to reduce to the case when $H$ is an $F$-torus via an argument based on rationality of the "variety of maximal tori" over the ground field $F$; see the proof of Corollary 3 at the end of section 7.1.) Taking $F = \mathbf{Q}$ and $H = (H'_0)^{\rm{ad}}$ does the job.