Defining a differentiable curve's torsion

47 Views Asked by At

In exercise 10, page 26 of Manfredo's Differentiable Geometry of Curves and Surfaces, we are asked to look at the following differentiable curve: $$\alpha(t)=\left\{\begin{array}{ll} (t,0,e^{-\frac{1}{t^2}}) &, \mbox{if}\, t>0 \\ (t,e^{-\frac{1}{t^2}},0)&, \mbox{if}\, t<0\\ (0,0,0) &, \mbox{if}\, t=0 \end{array} \right.$$ The d part of this question asks

Show that τ can be defined so that τ ≡ 0, even though α is not a plane curve.

I have calculated the binormal vector to be $b(t)=(0,1,0)$ for $t>0$ and $(0,0,1)$ for $t<0$, and in both these cases the torsion $\tau=1$. What might he mean when saying that it can defined to be zero?