# What is the torsion and curvature in modern language?

I think it can not be the torsion and curvature in the connection context, for these two are anti-symmetric in two variables, so that they must vanish on a one-dimensional space (tangent space of a curve).

So what should these two notions be in a "modern" context? How can we arrive at them?

Also, I am confused about the method of moving frames. Is the Frenet-Serret frame amoung realizations of the method? It seems the F-S frame is used to study the embedding of a curve into $\mathbb{R}^3$. Does general method of moving frames only serve to study embedding of curves into some ambient manifolds?

• Are you asking what torsion and curvature mean with regards to curves? – Eric O. Korman Jun 14 '13 at 20:45
• @Eric, exactly. – Anonymous Coward Jun 28 '16 at 1:04
• A moving frame only makes sense In an ambient space, so no wonder... – Mariano Suárez-Álvarez Jun 28 '16 at 1:08
• Isn't sectional curvature good enough? I mean if you accept Gaussian curvature. – user40276 Jun 30 '16 at 16:00
• I'm not sure exactly what you're asking. The curvature of curves in Riemannian manifolds is defined very similarly to the classical case - see e.g. en.wikipedia.org/wiki/Geodesic_curvature. In three dimensions torsion should work the same, while in higher dimension it'll look a bit different but I guess there should still be some invariant measuring deviation from planarity. – Anthony Carapetis Jul 1 '16 at 3:37

What is modern? Here is an answer which is probably not modern, but, might be of interest.

Suppose $\alpha: I \subseteq \mathbb{R} \rightarrow \mathbb{R}^n$ is a regular arclength parametrized curve. We say $\alpha$ is a Frenet curve if $\alpha', \alpha'', \dots , \alpha^{(n-1)}$ are linearly independent. Furthermore, a set of $n$-vector fields $E_1, E_2, \dots , E_n$ is a Frenet frame of $\alpha$ if the following three conditions are met:

1.) $E_1, E_2, \dots, E_n$ are orthonormal and positively oriented,

2.) for each $k=1,2, \dots , n-1$ we have $\text{span} \{ E_1, \dots, E_k \} = \text{span} \{ \alpha', \dots, \alpha^{(k)} \}$

3.) $\alpha^{(k)} \cdot E_k > 0$ for $k=1,\dots , n-1$.

The condition of linear independence in $n=3$ simply amounts to the assumption $\alpha'' \neq 0$. In other words, a regular non-linear arclength-parametrized curve is a Frenet curve and you can verify that $T,N,B$ frame is a Frenet frame. The Frenet Serret equations also have a generalization to Frenet curves in $\mathbb{R}^n$. In particular:

Let $\alpha$ be a Frenet curve and $E_1,E_2, \dots , E_n$ a Frenet frame then there exist non-negative curvature functions $\kappa_i = E_i' \cdot E_{i+1}$ for $i=1,2, \dots , n-2$ and torsion $\kappa_{n-1} = E_{n-1}' \cdot E_{n}$ for which the following differential equations hold true: $$\left[ \begin{array}{l} E_1' \\ E_2' \\ \vdots \\ E_{n-1}' \\ E_n' \end{array} \right] = \left[ \begin{array}{cccccc} 0 & \kappa_1 &0 & \cdots & 0 & 0 \\ -\kappa_1 & 0 &\kappa_2 & \ddots & \ddots & \vdots \\ \vdots & -\kappa_2 &0 & \ddots & \ddots & \vdots \\ \vdots & \ddots &\ddots & \ddots & 0 &\kappa_{n-1} \\ 0 & 0 &\cdots & \cdots & -\kappa_{n-1} & 0 \end{array} \right] \left[ \begin{array}{l} E_1 \\ E_2 \\ \vdots \\ E_{n-1} \\ E_n \end{array} \right]$$ What is the meaning of these curvature functions?

In short, they simply describe how the frame evolves along the curve in terms of the frame itself. Curvature and torsion and higher curvatures are frame dependent concepts.

I'm not sure the history of all this, I personally found these things in Kuhnel's Differential Geometry: Curves - Surfaces - Manifolds

Notice, curvature, torsion, and the higher curvatures in the equations above all describe departures from planar motion at the infinitesimal level.

Intuitively, if two curves share the same curvature functions then they have the same shape. There is some isometry which moves one curve to another which is the "same". In principle, we can define curves in terms of these functions.