Commutator of closed subgroups

Question

Suppose we have a simply-connected Lie group $G$. Let $G_1$ and $G_2$ be two closed and connected subgroups of $G$. Is it true that the commutator $[G_1,G_2]$ is a closed subgroup of $G$?

Answer 1

[EDIT: This is utterly wrong. See the comments.]

No. Let's make an example in which both $G_1$ and $G_2$ are one-dimensional.

Start by choosing $x$ and $y$ in $\frak{sl}_2(\mathbb R)$ such that the subgroup determined by $[x,y]$ is a circle (the square of this matrix has negative trace) but the subgroups generated by $x$ and by $y$ are isomorphic to $\mathbb R$ (their squares have positive trace).

Now in $\frak{sl}_2(\mathbb R)\times \frak{sl}_2(\mathbb R)$ consider the elements $(x,x)$ and $(y,cy)$, where $c$ is some irrational number. These give closed noncompact one-dimensional subgroups $G_1$ and $G_2$, but their commutator gives a dense line in a torus.

$SL_2(\mathbb R)\times SL_2(\mathbb R)$ is not simply connected, so embed it in $SL_4(\mathbb R)$ and take that, or rather its double cover, to be $G$.

Answer 2

Maybe I can frame the question further, though I'm not a Lie group specialist. From the viewpoint of topological groups, just requiring one of the two subgroups to be connected will force the commutator group here to be connected. But closure is a more delicate issue. Requiring a simply connected Lie group $G$ at least avoids a standard type of counterexample: the quotient of a simply connected nilpotent Lie group by a discrete subgroup can yield a connected Lie group $G$ for which $[G,G]$ itself fails to be closed.

For arbitrary linear algebraic groups (initially over an algebraically closed field) the situation is more elementary: here the connectedness of one of the two closed subgroups is enough to make the commutator group both closed and connected. (Note that one is not dealing with topological groups because the topology is not Hausdorff. But in the analytic topology, the groups here are actually complex Lie groups if one works over $\mathbb{C}$.)

There are some (but not many) books that treat the structure of Lie groups in general, including older books like The Structure of Lie Groups by Hochschild and Chapter 3 of Bourbaki's treatiste Groupes et algebres de Lie. Though such books always have some discussion of commutation, I'm unaware of any precise results on simply connected groups that would answer the question here one way or the other. But I've never delved into the extensive "exercises" for Section 9 of Bourbaki.

Most (but not all) real semisimple Lie groups do turn out eventually to be algebraic groups, where it's more likely that the answer to the question is yes after adapting from $\mathbb{C}$ to $\mathbb{R}$.

Answer 3

It is true.

1st case: $G$ is linear, i.e., isomorphic to a closed subgroup of matrices. Here I don't need simple connectedness — I even need not to assume it, and I only assume that $G_1,G_2$ are immersed connected subgroups, possibly not closed.

Note that the case when $G_1,G_2$ are Zariski-closed is standard, as has already be mentioned. But I'm not assuming this.

First a simple group-theoretic fact: $G_1, G_2$ both normalize $C=[G_1,G_2]$. Indeed, write the commutator as $[g,h]=ghg^{-1}h^{-1}$ (the choice of convention does not affect the definition of $[G_1,G_2]$). Then $[g,st]=[g,s]s[g,t]s^{-1}$. Hence, writing $C=[G_1,G_2]$, taking $g\in G_1$, $s,t\in G_2$, we see that $s[g,t]s^{-1}\in C$. Viewing this for each $s$, and all $g,t$, we see that $s$ normalizes $C$. Since this is true for each $s$, we see that $G_2$ normalizes $C$. Similarly $G_1$ normalizes (e.g., because $[G_1,G_2]=[G_2,G_1]$).

Next, it is a known fact that the commutator subgroup of every Lie subalgebra is algebraic. Hence, after modding out, we can suppose that $C$ is abelian.

At this point, we can enlarge $G$ and suppose $G=\mathrm{GL}_n(\mathbf{C})$.

Let $U$ be the (complex) Zariski-closure of $C$; this is a Zariski-closed connected abelian group. decompose the $C$-action into weight blocks $V_\chi$ ($\chi\in\mathrm{Hom}(U,\mathbf{C}^*)$). In the block $V_\chi$, $U$ acts by the scalar $V_\chi$ times some unipotent element (said otherwise, in $V_\chi$, we can triangulate the $U$-action so that $u$ acts as an upper triangular matrix with constant diagonal $\chi(u)$).

Since $G_i$ normalizes $C$, it normalizes $U$, and hence preserves the weight decomposition, i.e., preserves each $V_\chi$. Hence $[G_1,G_2]$ has determinant one in each $V_\chi$. But the determinant of $u\in U$ on $V_\chi$ is $\chi(u)^{\dim(V_\chi)}$. Since $C$ is Zariski-dense in $U$, this has to be $=1$. Since $U$ is connected, this is constant only if $\chi=1$. Thus $\mathbf{C}^n=V_1$. In other words, $U$ is upper unipotent. Then every real Lie subalgebra of $\mathfrak{u}$ defines a (real Zariski-closed) subgroup. Hence $[G_1,G_2]=C$ is real Zariski closed (and unipotent).

In general (i.e. going back before the reduction to $C$ abelian), we reach the conclusion that $[G_1,G_2]$ is real Zariski-closed and has a unipotent abelianization (i.e., its reductive Levi factor is semisimple).

General case: every simply connected Lie group $G$ has a quotient $G/Z$ by a discrete central subgroup that is linear. Then $[G_1,G_2]$ is a lift of $[G_1Z/Z,G_2Z/Z]$, which is closed (using the 1st case, where I need not to assume that the subgroups are closed!), and hence is closed as well.

Answer 4

$\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$To complement YCor's answer, here is a counterexample using a $G$ which is neither linear nor simply connected. (YCor showed that, if $G$ is either linear or simply connected, the conclusion is true.) It is closely inspired by Tom Goodwillie's answer that didn't work.

Define $$H = \begin{bmatrix} 1&\RR&\RR/\ZZ \\ 0&1&\RR \\ 0&0&1 \end{bmatrix}$$ $$H_1 = \begin{bmatrix} 1&\RR&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \qquad H_2 = \begin{bmatrix} 1&0&0 \\ 0&1&\RR \\ 0&0&1 \end{bmatrix} \qquad Z_H = \begin{bmatrix} 1&0&\RR/\ZZ \\ 0&1&0 \\ 0&0&1 \end{bmatrix}.$$ Then it is easy to check that $(H_1, H_2) = Z_H$, more precisely, $$\left( \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}, \begin{bmatrix} 1&0&0 \\ 0&1&y \\ 0&0&1 \end{bmatrix}\right) = \begin{bmatrix} 1&0&xy \\ 0&1&0 \\ 0&0&1 \end{bmatrix}.$$

Now, let $G = H \times H$, and put $$G_1 = \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \subset H_1 \times H_1 \subset G$$ $$G_2 = \begin{bmatrix} 1&0&0 \\ 0&1&y \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix} 1&0&0 \\ 0&1&\sqrt{2} y \\ 0&0&1 \end{bmatrix} \subset H_2 \times H_2 \subset G.$$ Then $G_1$ and $G_2$ are both closed in $G$, but $(G_1, G_2)$ is the immersed subgroup $(z, \sqrt{2} z)$ in $Z_H \times Z_H \cong (\RR/\ZZ)^2$, which is not closed.