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?
Curvature and torsion changes related to Frenet frame choice
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differential-geometry