I'm having trouble differentiating $(M \circ \alpha)(t)$ where $\alpha:(a,b)\to \mathbb{R}^n$ is a regular curve and $M:\mathbb{R}^n \to \mathbb{R}^n$ is a rigid motion.

For context, I'm reading Do Carmo's Differential Geometry, and in the section on the local theory of curves, he states the following:

Fundamental Theorem of the Local Theory of Curves
If $k(s)>0$ and $\tau(s)$ are differentiable functions on some interval $I=(a,b)\owns s$, then there exists a regular curve $\alpha:I\to\mathbb{R}^3$ parametrized by arc length such that $k(s)=\lvert \alpha''(s)\rvert$ is the curvature and $\tau(s)$ is the torsion of $\alpha$. Moreover, any other such curve $\bar\alpha:I\to\mathbb{R}^3$ differs from $\alpha$ by a rigid motion; that is, there exists some $\rho\in\mathrm{O}(3)$ with $\det\rho>0$ and some $c\in\mathbb{R}^3$ such that $\alpha=\rho\circ \bar{\alpha}+c$.

He forgoes the proof of the existence of $\alpha$, but he does offer a sketch of a proof of the part on uniqueness. I say "sketch" because his proof starts off by claiming

$$\int_{t_0}^{t} \left\lvert \dfrac{d\alpha(t')}{dt'} \right\rvert dt' = \int_{t_0}^{t} \left\lvert \dfrac{d(M\circ\alpha)(t')}{dt'} \right\rvert dt'$$

for $t_0<t$ in $I$. He doesn't actually verify this, however; instead waving his hands about the symmetries of the dot and cross products.

How would I verify this second result?


By writing out the coordinates of the matrix-vector product $M \circ \alpha$, we see that $$ \frac{d(M \circ \alpha)(t')}{dt'} = M \circ \frac{d\alpha(t')}{dt'}. $$ In fact, this is true for any Matrix $M$, not just for a rigid motion.

Now, if $M$ is a rigid motion, we have $M^TM = E$, where the superscript $T$ denotes transposition, and $E$ is the identity matrix. From this, we get $$ \left| \frac{d(M \circ \alpha)(t')}{dt'} \right| = \left(\left(\frac{d(M \circ \alpha)(t')}{dt'}\right)^T\left(\frac{d(M \circ \alpha)(t')}{dt'}\right)\right)^{1/2} = \left(\left(M \circ \frac{d\alpha(t')}{dt'}\right)^T\left(M \circ \frac{d\alpha(t')}{dt'}\right)\right)^{1/2} = \left(\left(\frac{d\alpha(t')}{dt'}\right)^TM^TM\left(\frac{d\alpha(t')}{dt'}\right)\right)^{1/2} = \left(\left(\frac{d\alpha(t')}{dt'}\right)^T\left(\frac{d\alpha(t')}{dt'}\right)\right)^{1/2} = \left| \frac{d\alpha(t')}{dt'} \right|. $$ Integrating this equality proves the second yellow box in your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.