The special orthogonal group is a manifold

Question

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 $\mathbb R^{n^2}$.

Answer 1

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.

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