Let $\gamma(s)$ be a unit-speed curve in $\mathbb{R}^3$. Let $t = \dot{\gamma}(s)$, $n = \frac{\dot{t}}{\left \| \dot{t} \right \|}$ and $b= t \times n$. The vectors $(t,n,b)$ form what is called a Frenet frame in a point $s$. Define the curvature as $k = \dot{t} \cdot n$ and the torsion as $\tau = - \dot{b} \cdot n$. We can change the sign of $t$ and $n$ arbitrarily, obtaining (for example) the new frame $(t^\ast=-t,n^\ast=-n,b^\ast=b)$. My teacher said that curvature's sign depends on the choice of $t$ but torsion's sign remains unchanged after arbitrary sign switching on $t$ and $n$... but that seems not to be true. In fact we have $$\tau= - \dot{b} \cdot n$$ realted to the first frame and $$\tau^\ast = - \dot{b^\ast} \cdot n^\ast = -\dot{b} \cdot (-n) = -\tau \quad$$ for the second I wrote. So, where is my mistake?


1 Answer 1


First of all, curvature's sign is always positive (unless one is dealing with a plane curve, in which case the signed definition of curvature is different from the general one you've given here).

I'm slightly confused about your reference to a "choice of $t$"; you just defined $t$ to be the tangent vector to $\gamma,$ which involves no choice. Neither is there a choice involved in your definitions of $n$ or $b,$ so one cannot willy-nilly change the signs of $t,$ $n,$ or $b.$

However, if we change the direction of $\gamma$ to get a curve $\Gamma$ that covers the same range as $\gamma,$ but backwards (so $\Gamma(s) = \gamma(c-s)$ for some constant $c$), then we have $\dot{\Gamma}(s) = -\dot{\gamma}(c-s).$ That is, at any given point of the curves' common image, the tangent vectors of $\gamma$ and $\Gamma$ point in opposite directions. Then the Frenet frame of $\Gamma$ would be, as you have written, $(t^\ast(s),n^\ast(s),b^\ast(s)) = (-t(s),-n(s),b(s)).$ So the Frenet equations for $\gamma$ are $$\dot{t}(s) = k(s) n(s)$$ $$\dot{n}(s) = -k(s) t(s) + \tau(s) b(s)$$ $$\dot{b}(s) = -\tau(s) n(s).$$ The Frenet equations for $\Gamma$ are, of course, $$\dot{t^\ast}(s) = k^\ast(s) n^\ast(s),$$ $$\dot{n^\ast}(s) = -k^\ast(s) t^\ast(s) + \tau^\ast(s) b^\ast(s),$$ $$\dot{b^\ast}(s) = -\tau^\ast(s) b^\ast(s).$$

In what follows, for convenience I will write $S = c-s,$ and all derivatives are taken with respect to $s.$ In particular, $\dot{S} = -1.$ Now, $\dot{t^\ast}(s) = -\dot{t}(S),$ so $k^\ast(s) = k(S)$ and $n^\ast(s) = -n(S).$ On the other hand, $b^\ast(s) = t^\ast(s) \times n^\ast(s) = -t(S) \times -n(S) = b(S).$ Furthermore, $\tau^\ast(s) = -\dot{b^\ast}(s) = -b^\ast(s) \cdot n^\ast(s) = b(S) \cdot n(S) = -\tau(S).$ So $$\dot{t^\ast}(s) = - k(S) n(S),$$ $$\dot{n^\ast}(s) = k(S) t(S) - \tau(S) b(S),$$ $$\dot{b^\ast}(s) = \tau(S) n(S).$$

I hope this sufficiently answers your question.


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