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I'm trying to prove that a curve $\gamma$ in $\mathbb{R}^n$ is contained in a hyperplane if and only if $\kappa_{n-1} = 0$, where $$\kappa_{i} = \frac{\langle e_{i}',e_{i+1}\rangle}{\| \gamma' \|}$$ is the $i^{th}$ curvature function of $\gamma$ (here $e_{i}$ is the $i$th vector in the Frenet Frame). My strategy is to show that $\kappa_{n-1} = 0$ if and only if the unit normal vector of the osculating hyperplane and the distance of the osculating hyperplane from the origin are constant. However, I'm not sure how to proceed. I can see why this result is true in 3 dimensions but am struggling to reframe it in a way that is easy to deal with in $n$ dimensions.

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  • $\begingroup$ what is the definition of the i-th curvature function? $\endgroup$ – Xipan Xiao Oct 3 '13 at 20:07
  • $\begingroup$ If $e_{i}$ is the $i$th vector in the Frenet Frame, then $\kappa_{i} = \frac{\langle e_{i}', e_{i+1}\rangle}{\| \gamma' \|}. $\endgroup$ – Ben Perez Oct 3 '13 at 20:11
  • $\begingroup$ What curvature vanishes for all curves in 3 dimensional Frenet frame? $\endgroup$ – Narasimham Sep 24 '16 at 11:04
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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\eps}{\varepsilon}\newcommand{\Neg}{\phantom{-}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$Let $\eps > 0$, and $\gamma:(-\eps, \eps) \to \Reals^{n}$ a curve of class $C^{n}$ in $\Reals^{n}$, parametrized by arc length (for simplicity, and without loss of generality), whose first $(n - 1)$ derivatives are linearly independent. Define the Frenet-Serret frame $(\Basis_{i}(s))_{i=1}^{n-1}$ by applying Gram-Schmidt to the ordered basis $$ \bigl(\gamma'(s), \gamma''(s), \dots, \gamma^{(n-1)}(s)\bigr), $$ let $\Basis_{n}(s)$ denote the unique unit field orthogonal to $\Basis_{i}(s)$ for each $i = 1, \dots, n-1$ and forming a positive orthonormal frame. (Orientation isn't important below, it merely fixes one of two choices.)

For $1 \leq i \leq n - 1$, define $$ \kappa_{i}(s) = \Brak{\Basis_{i}'(s), \Basis_{i+1}(s)}, $$ so that $$ \left[\begin{array}{@{}c@{}} \Basis_{1}' \\ \Basis_{2}' \\ \Basis_{3}' \\ \vdots \\ \Basis_{n}' \\ \end{array}\right] = \left[\begin{array}{@{}ccccc@{}} 0 & \Neg \kappa_{1}(s) & & & \\ -\kappa_{1}(s) & 0 & \Neg \kappa_{2}(s) & & \\ 0 & -\kappa_{2}(s) & 0 & \ddots & \\ & & \ddots & 0 & \Neg \kappa_{n-1}(s) \\ & & & -\kappa_{n-1}(s) & 0 \\ \end{array}\right] \left[\begin{array}{@{}c@{}} \Basis_{1} \\ \Basis_{2} \\ \Basis_{3} \\ \vdots \\ \Basis_{n} \\ \end{array}\right]. \tag{1} $$ In words, $\Basis_{i}'(s)$ lies in the plane spanned by $\Basis_{i-1}(s)$ and $\Basis_{i+1}(s)$.

If $\kappa_{n-1}(s) \equiv 0$, then $\Basis_{n}'(s) \equiv 0$, i.e., $\Basis_{n}(s) = \Basis_{n}$ is constant, and $$ \Brak{\gamma(s) - \gamma(0), \Basis_{n}} \equiv 0. $$ (The left-hand side vanishes at $0$, and its derivative vanishes identically by (1).) This means $\gamma$ lies in the hyperplane $$ \Brak{\mathbf{x} - \gamma(0), \Basis_{n}} = 0. $$

Conversely, if $\gamma$ lies in some hyperplane $$ \Brak{\mathbf{x} - \gamma(0), \Basis} = 0, $$ then every derivative $\gamma^{(i)}(s)$ is orthogonal to $\Basis$. Particularly, $\kappa_{n-1}(s) = \Brak{\Basis_{n-1}'(s), \Basis} \equiv 0$ by (1).

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    $\begingroup$ @Sandy.Davidson: If $\kappa_{n}(s) \equiv 0$, the last equation in (1) implies $\Basis_{n}(s)$ is constant. Since $\Brak{\gamma'(s), \Basis_{n}(s)} = \Brak{\Basis_{1}(s), \Basis_{n}(s)} \equiv 0$ in general, the fundamental theorem of calculus gives the equation you asked about, namely$$\Brak{\gamma(s) - \gamma(0), \Basis_{n}} = \int_{0}^{s} \Brak{\gamma'(t), \Basis_{n}}\, dt = 0.$$ $\endgroup$ – Andrew D. Hwang Sep 25 '16 at 21:32
  • $\begingroup$ @Sandy.Davidson: The function$$\Brak{\gamma'(t), \Basis_{n}} = \frac{d}{dt} \Brak{\gamma(t), \Basis_{n}}$$depends only on one variable, $t$. It vanishes identically by (1), so its integral vanishes; the integral has the stated value by the ordinary one-variable fundamental theorem of calculus. The unit vector $\Basis_{n}$ is normal to the hyperplane containing the image of $\gamma$. (Answering here instead of under your question so the comments don't get separated from their context. :) $\endgroup$ – Andrew D. Hwang Sep 28 '16 at 21:30

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