# Showing 0 curvature

I'm trying to show that a curve, $\gamma$, of general type in $\mathbb{R}^n$ that is contained in an $(n-1)$-dimensional affine subspace has curvature $\kappa_{n-1}=0$. My thought process was to consider the general curvature function $\chi_{n-1}(t)=\dfrac{\langle \textbf{e}_{n-1}'(t),\textbf{e}_n(t)\rangle}{|| \ \gamma'(t) \ ||}$ where $\textbf{e}_i$ are the Frenet vectors, but I haven't been able to come up with any convincing argument so far. It seems intuitive that the last curvature must be zero since the curve is contained in the affine plain. Any help would be great. Thanks

Let's think about how this works for a curve in $\Bbb R^3$. We show that a curve $\gamma$ has torsion identically $0$ if and only if $\gamma$ lies in an affine plane. Namely: If there is a constant unit vector $\mathbf C$ so that $\mathbf C\cdot(\gamma-\gamma_0) = 0$, then we differentiate a few times to find that $\mathbf e_1\cdot\mathbf C = \mathbf e_2\cdot\mathbf C = 0$, and so $\mathbf C = \pm\mathbf e_3$; thus, $\mathbf e_3' = \mathbf 0$, whence torsion is $0$. Conversely, if torsion is $0$, $\mathbf e_3$ is constant, say $\mathbf e_3=\mathbf C$, and we infer that $\alpha\cdot\mathbf C$ is constant as well. (Throughout all this you want to assume $\gamma$ is arclength-parametrized for convenience.)
In $n$ dimensions, the same argument will work. Assuming $\gamma$ is generic (so that all the Frenet frame vectors are defined), if there is a unit constant vector $\mathbf C$ so that $\mathbf C\cdot(\gamma-\gamma_0) = 0$, differentiating $n-1$ times will give $\mathbf e_i\cdot\mathbf C = 0$ for $i=1,\dots,n-1$, and so, once again $\mathbf C = \pm\mathbf e_n$. Conversely, if $\kappa_{n-1}=0$, then $\mathbf e_n$ will be constant, and we deduce that $\gamma$ lies in an affine plane with $\mathbf e_n$ as its normal vector.