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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!

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    $\begingroup$ Consider $\gamma$ at $n$ random points in $[0,1]$, and calculate the determinant of those points: $\det\neq 0$ means that $\gamma$ definitely is not contained in an $n-1$-dimensional subspace; $\det =0$ means that all those points lie in some lower-dimensional subspace, and it may well contain the whole curve. $\endgroup$
    – Matt F.
    Dec 5, 2022 at 22:03
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    $\begingroup$ If the image of $\gamma$ is contained in a proper subspace then so is its derivative. You could use any linear algebra method to test if the tangent vectors are in a plane. $\endgroup$ Dec 6, 2022 at 11:13
  • $\begingroup$ Unless you tell us more about how your curve $\gamma:[0,1]\to\mathbb{R}^n$ is specified and what sorts of information about it can be easily computed, it's hard to give advice about an 'easy-to-check condition'. For example, it's a simple matter to construct examples of smooth $\gamma$ that don't lie in a proper subspace of $\mathbb{R}^n$ but for which uniform sampling from $[0,1]^m$ to test $m$ points $\gamma(t_1),\ldots,\gamma(t_m)$ for linear independence will have a 99.99999% probability of 'yes', but still $\gamma([0,1])$ does not lie in a proper subspace. $\endgroup$ Dec 9, 2022 at 10:37

2 Answers 2

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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.

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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.

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