The special orthogonal group is a manifold
Solution 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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