Why does the torsion remain invariant under a change of orientation?

Question

There is this statement in the book of Differential Geometry of Curves and Surfaces by Do Carmo page 18 that I have a doubt in.

Notice that by changing orientation the binormal vector changes sign, since $b=t\wedge n$. It follows that $b\text{ }'(s)$, and, therefore, the torsion, remains invariant under a change of orientation.

My questions are:

1. Under a change of orientation $n(s)$ remain invariant for both magnitude and direction, am I correct?
2. since $b$ changes sign and $b\text{ }'(s)=\tau(s)n(s)$, and $n(s)$ remains invariant so $b\text{ }'$ should also change sign isn't it? And therefore the torsion $\tau(s)$ changes sign, am I correct?

But why does it say that the torsion remains invariant under a change of orientation? Does "remain invariant" mean it preserves the magnitude and direction of the torsion?

Thanks for the help and explanation!

Answer 1

Let's write $\gamma_1(s) = \gamma(-s)$, and look at $t_1, n_1, b_1$ the Frenet frame for this new curve, at $s = 0$.

Note that $$ t_1(s) = \gamma_1'(s) = \gamma'(-s)(-1) = -t(-s) $$ by the chain rule. Similarly, $n_1(s) = n(-s)$, and $b_1(s) = -b(-s)$. Hence $b_1'(s) = -b'(-s)(-1) = b'(-s)$.

It's the $-s$ inside the parentheses that you were missing.

Now \begin{align} b_1'(s) &= \tau_1(s) n_1(s) \\ b'(-s) &= \tau_1(s) n(-s)\\ b'(q) &= \tau_1(-q) n(q)\\ \end{align} But $b'(q) = \tau(q) n(q)$, so we conclude that $\tau(q) = \tau_1(-q)$, which is what the not-so-well expressed statement you quoted was trying to say.