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?


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.

  • $\begingroup$ I think I understand this argument except the last line, "...as commutator of elements of $g'$ generate the Lie algebra of $K$, what we want follows". Why is this good enough? What if $g'$ only has these paths with tangent vectors giving $\mathfrak{g}'$ but doesn't contain a neighborhood of $K$ near the identity? $\endgroup$ – freeRmodule Aug 3 '16 at 6:08
  • $\begingroup$ There is a neighborhood the identity in $K$ is entirely composed of elements which are exponentials of elements in its Lie algebra. $\endgroup$ – Mariano Suárez-Álvarez Aug 3 '16 at 15:16
  • $\begingroup$ Yes, but how do we know the paths in $g'$ include these exponential paths? What if, for instance, none of the paths tangent to $X\in\mathfrak{g}$ in $g'$ intersect the exponential $exp(tX)$ except at $e$? $\endgroup$ – freeRmodule Aug 3 '16 at 16:08
  • $\begingroup$ I do not understand what you wrote. $\endgroup$ – Mariano Suárez-Álvarez Aug 3 '16 at 16:11
  • $\begingroup$ I don't understand why the fact about exponentials you wrote helps us unless we know, for instance, that all the exponentials are contained in $g'$ (at least in a some neighborhood of $e$). $\endgroup$ – freeRmodule Aug 3 '16 at 16:45

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.

  • 2
    $\begingroup$ What you wrote does not prove anything, as far as I can tell. $\endgroup$ – Mariano Suárez-Álvarez Aug 3 '16 at 4:37
  • $\begingroup$ I'm confused by this answer as well. I'm not sure why it's imposed by the regularity of multiplication and inversion. Perhaps you could elaborate? $\endgroup$ – freeRmodule Aug 3 '16 at 4:40
  • $\begingroup$ Apparently you need that $G$ is simply connected to ensure that it is closed (and hence Cartan's theorem, aka Closed subgroup theorem is applicable; see en.wikipedia.org/wiki/Closed_subgroup_theorem) Sourse: 2nd answer here: mathoverflow.net/questions/90216/commutator-of-closed-subgroups $\endgroup$ – Justin Benfield Aug 3 '16 at 4:57
  • $\begingroup$ Thanks. I think I get this case. I'm interested in what can be said in general, though. $\endgroup$ – freeRmodule Aug 3 '16 at 5:00
  • $\begingroup$ That the commutator subgroup is closed is precisely the non-trivial fact that has to be proved here, and in fact depends on simply-connectedness. $\endgroup$ – Mariano Suárez-Álvarez Aug 3 '16 at 5:08

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