Checking that the image of a curve is not contained in a hyperplane Let $\gamma : [0,1] \to \mathbb R^n$ be a smooth curve, $n \geq 2$. I would like to find an easy to check condition such that the image of $\gamma$ is not contained in an $n-1$ dimensional linear subspace of $\mathbb R^n$, i.e. for every such subspace $H \subset \mathbb R^n$ it holds that
$$
\gamma([0,1]) \cap (\mathbb R^n \setminus H) \neq \emptyset.
$$
Is there any theory in this direction?
Any help would be highly appreciated!
 A: I would check existence of one nondegenerate point; it means that $\alpha'(t_0),\dots,\alpha^{(n-1)}(t_0)$ are lineraly independent.
If such point exists, then find a vector $w$ perpendicular to all $\alpha'(t_0),\dots,\alpha^{(n-1)}(t_0)$, and then check that $\langle w,\alpha'\rangle =0$ --- if yes, then yes; if no, then no.
If your curve has only degenerate points, then I would try to use elementary geometry.
A: A curve $\alpha$ in $\mathbb{R}^3$ is called non-degenerate if $\alpha'$ and $\alpha''$ are linearly independent at every point.
A curve parametrized by arc-length is a Frenet curve if $\alpha''\neq 0$ everywhere (i.e. if it has non-vanishing curvature).
One can prove that a curve $\alpha$ is non-degenerate iff its arc-length parametrization is a Frenet curve.
The non-degenerae curve $\alpha$ is contained in a plane if and only if its torsion equals zero. That is $\alpha$ is contained in a plane if and only if
$$
\det (\alpha',\alpha'',\alpha''')=0
$$
at all points.
You can find this result is almost any textbook on differential geometry of curves and surfaces. I am pretty sure the result has a generalization to curves in higher dimensions, but I do not remember details.
