Product of two algebraic subgroups of a (solvable) group = another algebraic subgroup?

Question

Let $G$ be a linear algebraic group over a field $K$. (Say $K=\mathbb{F}_q$ or $K=\mathbb{C}$; do not assume $K$ is algebraically closed or of characteristic $0$.) Let $H_1$, $H_2$ be algebraic subgroups of $G$. Consider the multiplication map

$\phi:H_1\times H_2\to G$.

The image of $\phi$ is a constructible set, i.e., a variety $H$ with perhaps a few varieties of lower dimension deleted from it. (This is a special case of a result of Chevalley's.)

Question: when is $H_1(K) H_2(K)$ equal to $H(K)$?

There are two issues here: closure (i.e., really getting a variety rather than a constructible set as the image) and rationality.

Getting more specific, since the question above may be too hairy in general:

(a) Assume that $G$ is solvable. Does that help? Can we then answer the question in the affirmative?

(b) Say, furthermore, that both $H_1$ and $H_2$ are in the same unipotent subgroup of $G$, or that $H_1$ is unipotent and $H_2$ is a subgroup of a corresponding maximal torus. Does that help?

Answer 1

The following is an answer to a previous version of the question, which asked whether there exists an algebraic subgroup $H$ of $G$ such that $H(K)=H_1(K)H_2(K)$:

There are two necessary conditions on $H_1, H_2$:

First, the set $\Gamma:=H_1(K)H_2(K)$ has to be a subgroup of $G(K)$.

Also, since any algebraic subgroup of an algebraic group over a field is closed, the set $\Gamma$ has to closed in $G$.

These conditions are also sufficient. This follows from from the following fact: Let $K$ be a field and $\Gamma$ a subgroup of $\operatorname{GL}_n(K)$ which is closed (for the Zariski toplogy on $\operatorname{GL}_n(K)$). Then there exists an algebraic subgroup $G$ of $\operatorname{GL}_n$ such that $G(K)=\Gamma$. This is (part of) Theorem 4.8 of these notes of Milne: http://www.jmilne.org/math/CourseNotes/aag.html

Answer 2

This is not always true. Example: Let $G = \mathbb{G}_m^2$ with $H_1$ the subtorus $\{ (x,x) \in \mathbb{G}_m^2 \}$ and $H_2$ the subtorus $\{ (y,y^{-1}) \in \mathbb{G}_m^2 \}$. Note that $H_1 H_2 = G$. However, given $(u,v) \in G(K)$, we can write $(u,v)$ as $(xy, x y^{-1})$ if and only if $uv$ is square. So, if there are nonsquare elements of $K$, then $H_1(K) H_2(K) \neq G(K)$.

Here is a general sort of approach to these questions.

Let $F = H_1 \cap H_2$. Let $Y = H_1 H_2$. (Note that these are intersections and products of varieties.) Let $m$ be the map $H_1 \times H_2 \to Y$. For $x \in Y(K)$, the fiber $m^{-1}(x)$ is a torsor for $F$; namely, let $F$ act on $m^{-1}(x)$ by $(h_1, h_2) \mapsto (h_1 f^{-1}, f h_2)$.

So, if $H^1(\mathrm{Gal}(K), F)$ vanishes, then all torsors for $F$ over $K$ are trivial, and $m^{-1}(x)$ has a point, which is what you wanted.

For example, if the ambient group $G$ is unipotent, and we are in characteristic zero, then $F$ will automatically be unipotent. In particular, $F$ will have a filtration by $\mathbb{G}_a$'s and, as $H^1(\mathrm{Gal}(K), \mathbb{G}_a)=0$, we see that the statement is true in this case.

Answer 3

Maybe it's worth pointing out that the question contained in the header has an obvious negative answer (and is not the main question being asked). The easiest counterexample would be the product of the two one-dimensional unipotent subgroups in the $3 \times 3$ upper triangular unipotent group which correspond to simple roots for the special linear group. This is a closed set (of dimension 2) but not a subgroup. On the other hand, the product of all three positive root groups in this situation is the entire upper triangular unipotent group. But here as in many natural solvable groups you are building a group step-by-step as product of a normal subgroup and another group.

Answer 4

The following is a partial answer to the question whether $H_1 H_2$ is closed in $G$:

This is not true in general. For example, if $G$ is reductive, $T$ a maximal torus of $G$ and $B$ a Borel subgroup containing $T$, in the Bruhat decomposition of $G$ into double cosets $BwB$ for $w$ in the Weyl group with respect to $T$, for all but one $w$ the cell $BwB$ is not closed in $G$. For such a $w$ the product of the subgroups $B$ and $wBw^{-1}$ is not closed.

However, if $H_1$ and $H_2$ are unipotent, then $H_1 H_2$ is closed. This follows from the fact that any orbit under the action of a unipotent group on an affine variety is closed. This includes the first part of your (b).

$H_1 H_2$ is also closed in the situation of the second part of (b): Let $H_1$ be unipotent and $H_2$ a subgroup of a maximal torus $T$ which normalizes $H_1$ (I assume that's what you mean by corresponding maximal torus.) Since $H_1 H_2$ is a disjoint union of translates of $H_1^0 H_2$ we may assume that $H_1$ is connected. Then there exists a Borel subgroup $B$ of $G$ containing $T$ and $H_1$. Let $U$ be the unipotent radical of $B$. The product morphism $U\times T\to B$ is an isomorphism of varieties and under this isomorphism, $H_1 H_2$ corresponds to $H_1\times H_2$. Thus $H_1H_2$ is closed in $B$ and hence in $G$.

Answer 5

If your subgroups are $\{exp(At)\}_t$ and $\{exp(Bt)\}_t$ and $[A,B]$, $A$, $B$ are linearly independent then your statement is false.