## The curvature and torsion: how to distinguish the shape of a curve

### The Fundamental Theorem of Curves

In the planar case there is one more concept: the curvature (with sign). In this case, a plane curve is determined by its curvature with sign, up to some euclidean motion. This fact suggests a similar fact for space curves: a space curve is determined by its curvature and torsion, up to some euclidean motion. This is indeed true but requires an additional condition to ensure uniqueness.

FUNDAMENTAL THEOREM OF CURVES:

Let $$P_{0}$$ in $$\mathbb{R}^{3}$$ and let $$TF_{0}$$ be an orthonormal frame with positive orientation. Further, let $$k,\tau:\, I\rightarrow\mathbb{R}^{3}$$ be continuous functions, with $$k>0$$ in $$I$$.

Then, there is one, and only one, $$C^{2}$$ curve, $$c:\, I\rightarrow\mathbb{R}^{3}$$, parametrized by arc length, whose curvature and torsion are precisely $$k$$ and $$\tau$$, whose starting point is $$P_{0}$$ and whose initial Frenet-Serret frame is $$TF_{0}$$.

(See, for example: M. Spivak, A comprehensive introduction to differential geometry (vol. 2); 1999)

Observations: The assumption that curvature is strictly positive is essential to guarantee the uniqueness of the curve. In fact:

(1) Consider the following functions in $$\mathbb{R}^{3}$$:

$f(t)=\left(t^{3},t,0\right);$ $g(t)=\begin{cases} \left(t^{3},t,0\right), & \mbox{if }t<0\\ \left(0,t,t^{3}\right), & \mbox{if }t\geq0 \end{cases}.$

The two functions have the same curvature, which is zero in $$t=0$$. The torsion in both cases is zero at every point where is defined (that is, all points except $$t=0$$). The initial point and the initial Frenet-Serret frame also coincide in the two curves (note that the curves coincide until instant $$t=0$$). But there is no Euclidean motion that takes one curve to the other, as is shown in the following picture:

(2) Another example similar to the previous one:

$f(t)=\begin{cases} \left(0,0,0\right), & \mbox{if }t=0\\ \left(t,0,5e^{-t^{-2}}\right), & \mbox{if }t\neq0 \end{cases};$ $g(t)=\begin{cases} \left(t,5e^{-t^{-2}},0\right), & \mbox{if }t<0\\ \left(0,0,0\right), & \mbox{if }t=0\\ \left(t,0,5e^{-t^{-2}}\right), & \mbox{if }t>0 \end{cases}.$

(A. Gray, Modern Geometry of Curves and Surfaces, pp. 142-145)

Such kind of examples justifies the option for not considering that the torsion is zero when the curvature is zero. With this option we still have that a curve has null torsion in an interval if and only if the curve is planar in that interval, otherwise not (recall, for example, functions $$g$$ presented above).