Let $\gamma : (-1,1) \longrightarrow \Bbb M_n(\Bbb R)$ be a $C^1$ map such that $\gamma (t) \in O(n),\ \forall t.$ Then show that for all $t,$ $\dot {\gamma} (t)$ is a skew-symmetric matrix.
My attempt $:$ Let us consider two maps $f : (-1,1) \longrightarrow O(n) \times O(n)$ defined by $t \mapsto (\gamma (t),\gamma(t)),\ t \in (-1,1)$ and another map $g : \Bbb M_n (\Bbb R) \times \Bbb M_n (\Bbb R) \longrightarrow \Bbb M_n(\Bbb R)$ defined by $(A,B) \mapsto AB^{\top},\ A,B \in \Bbb M_n (\Bbb R).$ Then $(g \circ f) (t) = I,\ \forall t \in \Bbb (-1,1),$ where $g$ is a bounded bilinear map (Since $\Bbb M_n(\Bbb R)$ is finite dimensional). Hence $(g \circ f)'(t) = 0,\ \forall t \in (-1,1).$ On the other hand by chain rule for differentiation we have for all $t \in (-1,1)$ $$\begin{align*} (g \circ f)' (t) & = D(g \circ f) (t) (1) \\ & = Dg(f(t)) (D(f(t)) (1)) \\ & = Dg(\gamma(t),\gamma(t)) (f'(t)) \\ & = Dg(\gamma(t),\gamma(t)) (\dot {\gamma} (t), \dot {\gamma} (t)) \\ & = g (\gamma(t),\dot {\gamma} (t)) + g (\dot {\gamma} (t), \gamma (t)) \\ & = \gamma (t) {\dot {\gamma} (t)}^{\top} + \dot {\gamma} (t) {\gamma (t)}^{\top} . \end{align*}$$
So we have for all $t \in (-1,1)$ $$\gamma (t) {\dot {\gamma} (t)}^{\top} + \dot {\gamma} (t) {\gamma (t)}^{\top} = 0.$$
So we get $\dot {\gamma} (t) {\gamma (t)}^{\top} $ is a skew-symmetric matrix for all $t \in (-1,1).$ From here how do I show that $\dot {\gamma} (t)$ is skew-symmetric for all $t \in (-1,1)\ $?
I think the problem is wrongly posed. If I consider the map $\gamma : t \mapsto \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}$ then we find that for all $t$ $$\dot {\gamma} (t) = \begin{pmatrix} -\sin t & -\cos t \\ \cos t & -\sin t \end{pmatrix}$$ which is clearly not skew-symmetric as it's diagonal entries are not necessarily zero. Am I missing something? Any suggestion regarding this will be highly appreciated.
Thanks for your time.