# curvature$=0$ implies straight line?

The fundamental theory of differential geometry states that:

If there is a given curvature $\bar{\kappa}(s)>0$ and torsion $\bar{\tau}(s)$ which both of them are differentiable and continuous in $(a,b)$, then:

1.there exists an arc length parametrized curve $C:r=r(s)$ such that its curvature $\kappa(s)=\bar{\kappa}(s)$ and its torsion $\tau(s)=\bar{\tau}(s)$

2.the above curve $C$ is the only curve in Euclidean space (up to rigid transformation)

The question I want to ask is that why it emphasizes that $\bar{\kappa}(s)>0$ in $(a,b)$ ? Isn't that $\bar{\kappa}(s)=0$ means the curve is a straight line? If it is a straight line, the curve must be the only curve in the space.

• The negation of $\kappa(s) > 0$ is not that it is identically $0$. The assumption ensures, for example, that $\kappa$ does not change sign, presumably because in that case the uniqueness statement of the theorem cannot be guaranteed anymore. – Thomas May 4 '14 at 11:13
• @Thomas: Isn't that curvature can only be non-negative? By the way, can the statement become '$\kappa \ge 0$'? And isn't that if $\kappa =0$ at a point implies $\kappa =0$ at the neighborhood of a point? – Y.H. Chan May 4 '14 at 11:18
• In two dimensions, curvature can have a sign. In three dimensions, it can become $0$, and of course you can start out with a prescribed curvature function with isolated zeros. The assumption just states that that case is not covered by the theorem. I doubt that a point with $\kappa =0$ implies that it vanishes in a neighbourhood. – Thomas May 4 '14 at 11:27
• If $\bar{\kappa} = 0$ over a neighborhood within $(a,b)$ then I think the torsion is undefined within that neighborhood. – David K May 4 '14 at 11:50