Let, $\alpha:I\rightarrow\mathbb{R}^{3}$ be a curve. There are three possibilities for the reflection. Say the $\tau_{\alpha}$ and the $\tau_{\beta_i}$ be the torsions of the $\alpha$ and $\beta_{i}$.
$(1)$ $\beta_1$ is the reflection of the $\alpha$ for the point, $\mathbf{p}$
$(2)$ $\beta_2$ is the reflection of the $\alpha$ for the line, $\ell$
$(3)$ $\beta_3$ is the reflection of the $\alpha$ for the Plane, $P$
What is the relationship between the $\tau_{\alpha}$ and the each $\tau_{\beta_{i}}$ respectively? Could you suggest each answer and reason with the geometric explanation?
P.S.: In my guess, $\tau_{\alpha} = -\tau_{\beta_{i}}$ for all $i \in \{1,2,3\}$ only depending on my geometric idea.
Best Answer
I think you are not exactly right.
Since reflection is an isometry, the torsion should only differ by a sign.
In a calculative approach, we could use the torsion derived from derivatives of $r(t)$. wiki $$ \tau = \frac{\det([r',r'',r'''])}{\|r'\times r''\|^2} $$ Consider the matrix $[r',r'',r''']$, in 3d coordinate form $$ T=\begin{bmatrix} x'&y'&z'\\ x''&y''&z''\\ x'''&y'''&z'''\\ \end{bmatrix} $$ With a proper choice of coordinates system / axis, the effect of reflection could be represented by changing the sign of $x,y,z$, which change signs of the columns of $T$.
So $\tau_{\alpha} = -\tau_{\beta_{1}}=-\tau_{\beta_{3}}=\tau_{\beta_{2}}$