TITLE: Geometric meaning of parallelepiped formed by consecutive derivatives
QUESTION [0 upvotes]: Do parallelepipeds formed by consecutive derivatives of a function from $\mathbb{R} \to \mathbb{R}^n$ have a nice geometirc interpretation? I encountered one in do Carmo's Differential Geometry of Curves and Surfaces if you slightly change some of the equations and was wondering whether it had a geometric meaning.

Intuitively, I think of the fundamental theorem of the local theory of curves as meaning that  a curve is uniquely determined up to rigid motion by the radius of curvature (how big of a circle I'm locally going around) and the torsion (how quickly the plane inside which I'm locally turning around a circle is changing. The arc-length parameterization of the space curve must be infinitely differentiable (differentiable in do Carmo) and the radius of curvature must be nonzero. Restricting the radius of curvature to be nonzero makes intuitive sense because straight lines would be allowed without this restriction, and those things do not seem to have a well defined osculating plane geometrically.
I was messing around with different ways of restating the theorem and tried to get the right hand sides to be simple expressions in terms of $\alpha$. I came up with the following:
$I$ is a closed interval in $\mathbb{R}$ with lower bound $a$ and upper bound $b$.
$$ I \stackrel{\text{def}}{=} [a,b] \tag{101} $$
$ \alpha $ is infinitely differentiable.
$$ \alpha \in I \to \mathbb{R}^3 \;\;\;\text{and}\;\;\; \alpha \in C^{\infty} \tag{102} $$
We pick the curve to start at the origin and initially move up instead of saying that the curve is uniquely determined up to a rigid motion.
$$ \alpha(a) = \vec{0} \tag{103} $$
$$ \alpha'(a) \cdot \hat{z} = 1 \tag{104} $$ 
$\alpha$ is parameterized by arc length.
$$ \left|\alpha'(\xi)\right| = 1 \;\;\;\text{forall $\xi$ in $[a,b)$} \tag{105} $$
Let the square of the curvature be defined as the square of the length of the acceleration of $\alpha$ . This makes intuitive sense geometrically, you would expect the radius of convergence to halve if the acceleration doubled.
$$ \kappa^2(s) \stackrel{\text{def}}{=} \alpha''(s) \cdot \alpha''(s) \tag{106} $$
And then we get to this crazy thing. This expression is taken from an exercise in do Carmo, asking the reader to prove that the torsion is given by the expression $ - \frac{\alpha'(s) \wedge \alpha''(s) \cdot \alpha'''(s)}{\lvert k(s) \rvert^2} $. I'm writting $\kappa$ instead of $k$ and defining torsion to be the negation of what do Carmo calls the torsion, which he says many authors do on page 18.
Using a determinant instead of the equivalent triple product, we get a nice formula for the product of the curvature squared and the torsion.
$$ \kappa^2 \tau \stackrel{\text{def}}{=} \det \begin{bmatrix} \vert & \vert & \vert \\ \alpha' & \alpha'' & \alpha''' \\ \vert & \vert & \vert \end{bmatrix} \tag{107} $$
Do parallelepipeds formed by consecutive derivatives like in the right hand side of (107) have a nice geometric meaning?
I'm stuck trying to figure out what it means for the right hand side of (107) to be zero or nonzero.
I think $\alpha'$ and $\alpha''$ cannot be linearly dependent by the hypotheses of the theorem. But I don't have a good sense of why $\alpha'''$ being a linear combination of $\alpha'$ and $\alpha''$ has anything to do with the osculating plane not changing at this moment in time.

REPLY [1 votes]: Here is what I think: if $\alpha'''$ is contained in the osculating plane (that is what your condition says), then $\alpha''$ is changing (at the given instant) in the osculating plane (I mean, it is not departing from that plane transversally) and the osculating plane itself is stationary at that moment of time. Because of that, the torsion is zero.
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