Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Yesterday i heard that a connected Lie group $G$ is simply connected (i.e. $\pi_1(G)=0$) if and only if $G$ is simple as a group, i.e. $G$ has no nontrivial normal subgroup. That sounds too good to be wrong, it explains why one would call this property "simply connectedness". Could someone briefly explain why this is true or give me some reference where i could find it?

Greetings, atajh

share|cite|improve this question
up vote 9 down vote accepted

Unfortunately this is not true. For instance $\mathbb{R}^n$ is simply connected and, being an infinite abelian group, is very far from being simple. When $n > 1$ it is not even simple as a Lie group, i.e., it has nontrivial connected, closed normal subgroups.

Also the other direction is false: $\operatorname{PSL}_2(\mathbb{R})$ is simple as an abstract group, but not simply connected, since $\operatorname{SL}_2(\mathbb{R}) \rightarrow \operatorname{PSL}_2(\mathbb{R})$ is a degree $2$ connected covering map. (If am not mistaken, then at least for linear groups what is true is that an abstractly simple group is of adjoint type, i.e., the dual condition to being simply connected.)

Anyway, you did the right thing: if you have a healthy level of interactions with other members of the mathematical community (students, faculty, etc.) then you will "hear things" which are quite new to you. Not everything you hear is actually true -- or especially, what you heard / remembered is not actually true -- so it's important to try to verify / disprove these things that you hear, or ask someone else about them.

share|cite|improve this answer
Thanks a lot, to both of you. – atajh Dec 2 '11 at 10:39

It's not true. $SU(2)$ is simply connected but is not simple as a group -- the center $\{\pm I\}$ is nontrivial. (There seems to be a concept called a simple Lie group which is not just Lie groups that happen to be simple, in that it asks for connected normal subgroups, so SU(2) does qualify there).

On the other hand $SO(3)$ is not simply connected, but is a simple group.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.