Suppose you are given an antisymmetric matrix $X$. Then $A(t)=e^{Xt}$ is a curve of orthonormal matrices, with $A(0)=Id$. Is it possible to construct a curve whose Frenet frame vectors $T,N,B$ are the columns of $A$? $dA/dt(0)=X$ is forced to be tridiagonal, according to Frenet's formulas, and not just a general antisymmetric matrix. This seems to imply that such curve doesn't always exist. This seems counterintuitive to me. Can someone help clarify the situation? Are the Frenet formulas qualitatively different when using a general parameter (not the arc length)? I believe they are still tridiagonal.
This seems to indicate that curves are not "flexible" enough to accommodate the continuous evolution of an orthonormal frame as the evolution of their Frenet frame.
I realized that the answer is actually simple: the knowledge of $r'(s)$ (the unit tangent vector) completely determines the curve up to translations. In particular it uniquely defines the normal and binormal vector. But there are infinitely many other orthogonal evolutions with the given $T(s)=r'(s)$. In other words, the restriction of the orthogonal transformations to the orthogonal complement of $T$ can be chosen arbitrarily. In 2D, however, the evolution of the tangent vector uniquely determines the whole curve in $SO_2$.