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

Knowing that $\mathrm{O}(n,\mathbb{R})$ is a closed submanifold (of the general linear group) and that $\mathrm{SO}(n,\mathbb{R})$ is one of its subgroups with the same dimension, is there a quick way (possibly using only basic arguments, i.e. without any reference to Lie theory) to see that $\mathrm{SO}(n,\mathbb{R})$ is a manifold?

share|cite|improve this question
Certainly a connected component of a manifold is a manifold... – Qiaochu Yuan Sep 21 '11 at 19:11
up vote 11 down vote accepted

The map $\det:O(n,\mathbb{R}) \rightarrow \{-1,+1\}$ is a continuous map. $\{+1\}$ is open in $\{-1,+1\}$.

So $SO(n,\mathbb{R})=\det^{-1}(\{+1\})$ is an open set in $O(n,\mathbb{R})$. An open set in a manifold is a manifold

share|cite|improve this answer

Your Answer


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