Proposition about curves in $S^2$ Let $\gamma_1,\gamma_2:(a,b)\to S^2$ be unit speed curves in $S^2=\{\vec{v}\in\mathbb{R^3}:\vec{v}\cdot\vec{v}=1\}$. Then the following two statements are equivalent:
(1) There is a $3\times 3$ orthogonal matrix $M$ such that $\gamma_2\equiv M\gamma_1.$
(2) $\beta_j:(a,b)\to \mathbb{R}, \beta_j=\det(\gamma_j,\gamma_j',\gamma_j'')$ satisfy  $\beta_1\equiv \beta_2$ or $\beta_1\equiv -\beta_2.$
I saw this proposition last night, but I still don't know how to prove it. Could you help me with it? Thanks in advance.

1=>2 is easy, just compute (and use $|\det(M)|=1$)
$$
\det(M\gamma_1,M\gamma_1',M\gamma_1'')=\det(M)\det(\gamma_1,\gamma_1',\gamma_1'')
$$
2=>1 is not so easy
Fix $s_0\in (a,b)$. 
Since $\gamma_1(s_0)$ and $\gamma_2(s_0)$ have unit length, there is a $M\in O(3)$ such that
$\gamma_2(s_0)=M\gamma_1(s_0)$. Now compare $\gamma_2$ and $M\gamma_1$
and use http://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas
Edit: better link: http://en.wikipedia.org/wiki/Fundamental_theorem_of_curves
Edit (sorry, this is not a full solution!): Since $\gamma(t)$ is parametrized by arclenth, the tangent is
$$
T(s)=\gamma'(s)
$$
and the curvature $\kappa$ and the normal $N$ are given by
$$
\kappa(s)=|T'(s)|=|\gamma''(s)| \qquad T'(s)=\kappa(s) N(s)
$$
The Binormal is
$$
B(s)=T(s)\times N(s) \quad\text{and the Torsion $\tau$ is given by } B'(s)=\tau (s) N(s)
$$
Now $\det(\gamma,\gamma',\gamma'')=\det(\gamma,T,\kappa N)=\kappa\det(\gamma,T,N)$ fixes $\kappa$
Another hint: Differentiate $\gamma(s)^2=1$ several times to fix $\tau$.
