# Special orthogonal group as a manifold

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?

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Certainly a connected component of a manifold is a manifold... –  Qiaochu Yuan Sep 21 '11 at 19:11

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