 Solution

Let ${\mathrm{Sym}}_n$ be the vector space of real symmetric matrices and considerAt $R\in O(n)$,This derivative is surjective: for $S\in {\mathrm{Sym}}_n$, choosing $H=RS/2$ gives $D_R\Phi (H)=S$. Thus $I$ is a regular value and the orthogonal group $O(n)={\Phi }^{-1}(I)$ is a submanifold of codimensionThe determinant takes only the values $\pm 1$ on $O(n)$, so the special orthogonal group $SO(n)=\{R\in O(n):\det R=1\}$ is an open-and-closed submanifold of $O(n)$. Therefore

The tangent space is the kernel of $D_R\Phi$:where the Special orthogonal Lie algebra $\mathfrak{so}(n)$ is the vector space of skew-symmetric matrices. This proves the special orthogonal group as a submanifold description.