If $G$ is a Lie group, is $[G,G]$ always Lie subgroup

Question

Let $G$ be a connected Lie group, $[G,G]$ its derived subgroup (i.e. subgroup generated by commutators of elements in $G$). Here is my question:

Is $[G,G]$ always a Lie subgroup, meaning a does it have the structure of a Lie group such that the inclusion map $[G,G]\rightarrow G$ is an immersion? (i.e. I do not require $[G,G]$ to be a submanifold with respect to the induced topology) If so, why?

Answer 1

One shows this as follows, for a connected simply connected Lie group $G$.

• Let $g$ be the Lie algebra of $G$ and $g'$ it derived Lie algebra, which is an ideal in $g$. Then $g/g'$ is an abelian Lie algebra, corresponding to an abelian, connected simply connected Lie group $H$. The map $g\to g/g'$ corresponds to a morphism of groups $f:G\to H$. Since $H$ is abelian, the derived subgroup $G'$ of $G$ is contained in the kernel $K$ of $f$. The Lie algebra of $K$ is $g'$: this follows from it construction. Moeover, the group $K$ is connected:

• Next, one shows that a path-connected neigborhood of $1$ in $K$ is contained in $G'$. if $x$ and $y$ are in $g$, then then the curve $t\mapsto [\exp(\sqrt t x),\exp(\sqrt ty)]$ takes values in $G'$ and its tangent vector at $0$ is $[x,y]$ (one has to extend the curve to a neighborhood of zero, this formula only works for positive $t$); as commutator of elements of $g$ generate the Lie algebra of $K$, what we want follows. Using this we see that in act $G'$ and $K$ coincide.

In all, this is not a trivial at all.

Answer 2

If $G$ is simply connected, then $[G,G]$ does have the structure of a Lie group. The fact that it is a subgroup of the given Lie group $G$ is trivial. The reason that it is, in fact, a Lie subgroup is a consequence of being topologically closed via the Closed subgroup theorem.

Answer 3

Yes, it is always a Lie group in the sense that it has the structure of a Lie group for which the inclusion map is an immersion.

More generally:

Theorem (Greenberg) When $H, K \subset G$ are Lie subgroups, their commutator subgroup $[H, K]$ is a Lie subgroup. Its Lie algebra is the smallest subalgebra of $\mathfrak g$ containing $[\mathfrak h, \mathfrak k]$ that is invariant under $\operatorname{Ad}_H$ and $\operatorname{Ad}_K$.

• L. Greenberg, Commutator groups and algebras. Journal of Research of the National Bureau of Standards — B. Mathematical Sciences. Vol. 73B, No.3, July–September 1969, pp. 247–249. https://nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn3p247_A1b.pdf

In particular, $[G, G]$ is a Lie subgroup with Lie algebra $[\mathfrak g, \mathfrak g]$.