# A Reductive Lie group whose commutator is not closed?

It seems to me every book has a different definition for reductive Lie groups. I've never seen anyone use the definition which seems most natural: a Lie group whose tangent algebra is reductive (i.e. a direct sum of an abelian and semisimple Lie algebras). Doesn't anyone discuss such groups?

More importantly and more concretely, is there a connected Lie group G whose tangent algebra is reductive but whose commutator is not closed?

• "Doesn't anyone discuss such groups?" Yes, see here, section Lie group case, and the references. – Dietrich Burde Aug 21 '16 at 19:19
• Interesting. Never seen reductive Lie groups defined this way. – Cronus Aug 21 '16 at 20:41

## 1 Answer

Let $S$ be a semisimple connected Lie group with infinite center (e.g., the universal covering of $\mathrm{SL}_2(\mathbf{R}$. Let $K$ be the circle group. Let $Z\subset S\times K$ be the graph of a homomorphism with dense image $Z(S)\to K$; this is a discrete central subgroup. Then $(S\times K)/Z$ is a reductive Lie group whose derived subgroup is dense and not closed.

On the other hand, it is not hard to prove that if a reductive Lie group has derived subgroup $S$ with finite center, then $S$ is closed.

• Ha! Of course! I knew this group as an example of a topologically perfect non-perfect group, but somehow failed to make the connected to this question... Let me see if I understand your last statement: if $G$ is a reductive Lie group whose commutator $S$ has finite centre, then we can denote the identity component of the centre of $G$ by $Z$ and obtain that $S\times Z\to G: (s,z)\mapsto s\cdot z$ is a finite covering map, hence a closed map. This means the image of $S\times{1}$ (which is a closed subset), which is $S$, is closed. Is that correct? – Cronus Aug 23 '16 at 0:19
• Yes this is the argument. – YCor Aug 23 '16 at 8:18
• Thank you very much, I appreciate it. – Cronus Aug 23 '16 at 11:11