I want to show that $O(m) = \{A \in M_m(\Bbb{R}) \simeq \Bbb{R}^{m^2}: AA^t = I\}$ is a submanifold. According to my definition of submanifold, I must find a submersion $f: U \to \Bbb{R}^n$ where $U$ is a neighborhood of $X \in O(m)$ such that $O(m) \cap U = f^{-1}(\{0\})$.
I defined $f: M_m(\Bbb{R}) \to Sym_m(\Bbb{R})$ by $f(A) = AA^t - I$. I showed that $f$ is differenciable. If $A \in O(m)$, then I can show that $Df(A)$ is surjective, but no if $A \not\in O(m)$. I know is enough to use the Regular Value Theorem, but I want to use only that definition. Also, I google some proofs, but they all use the Regular Value Theorem.
Can someone help me?
I don't know whether you view this as "applying the regular value theorem", but if $Df(A)$ is surjective, then continuaity of $Df$ implies that there is an open neighborhood $U_A$ of $A$ in $M_m(\mathbb R)$ such that $Df(B)$ is surjective for all $B\in U_A$. (Viewing $Df(A)$ as a matrix, there is an ivertible square submatrix of maximal size and you can take $U_A$ to be the set on which the determinant of that submatrix is non-zero.) Then put $U:=\cup_AU_A$ and restrict $f$ to $U$.