How can we show that $SO(n)$ is an $n^2$-manifold. It would be tempting to say that $SO(n)$ is an open set of $\mathbb R^{n^2}$ but this is not the case since $SO(n)$ is given as the intersection of preimages of singletons. But singletons are closed in $\mathbb R$ hence $SO(n)$ is closed in $\mathbb R^{n^2}$.
The special orthogonal group is a manifold
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2palio, shouldn't you correct the $n^2$ as pointed out to you in the comments? – 2014-06-28
1 Answers
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Let $f : M_n(\mathbb{R}) \longrightarrow S_n(\mathbb{R})$ defined by $f(A) = ~^tAA$
$O_n(\mathbb{R}) = f^{-1}(\{I_n\})$. Then check that $I_n$ is a regular value of $f$.
So $O_n(\mathbb{R})$ is a submanifold of $M_n(\mathbb{R})$, it's dimension is $\dim(M_n(\mathbb{R}) - \dim(S_n(\mathbb{R})) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$ . Since $SO_n(\mathbb{R})$ is a connected component of $O_n(\mathbb{R})$ it is a submanifold to.
Ask me if you want more details
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0Yes but in this case, since $SO_n(\mathbb{R}) = \det^{-1}(\mathbb{R}_+^*)$ it is open – 2012-03-25