Cartan's method of moving frames and the classification of curves in $\mathbb{R}^n$ up to isometry

I am writing this small post to illustrate a general feature of Lie groups which some people, like me a few weeks ago, may not be aware of and has remarkable consequences at the level of uniqueness of certain geometric objects. Hope you enjoy this as much as I did.

 In this post, the Lie groups involved are matrix Lie groups, and, in particular, we will be interested in $G=\mathrm{O}(n)$. 

Suppose we have a smooth map $f: M\to G$, where $M$ is a connected smooth manifold. Then we have the following result.

Theorem 1: Suppose $f,f': M\to G$ are smooth maps with the property $$f^*\omega= f^{'*}\omega,$$ where $\omega=g^{-1}dg$ is the Maurer-Cartan form. Then, there exists $h\in G$ such that $f(x)=hf'(x)$.

The proof is actually very simple.

Proof:  Note that $f(x)f'^{-1}(x)=h(x)$ is well defined, and now observe that

$$dh= df f{'-1}+fdf^{'-1}= f\left(f^{-1}df\right)f{'-1} -f\left( f^{'-1}df ^{'}\right)f^{'-1}=f\left(f^{*}\omega -f^{'*}\omega\right)f^{'-1}=0,$$  

from which the result trivially follows. $\square$

As an application of this, suppose we have a regular curve $\gamma: I\to \mathbb{R}^n$, where $I$ is an interval. By regular, one means that $|\gamma'|\neq 0$ everywhere (an immersion). Also assume that $\gamma(I)$ is not contained in a proper linear subspace $\mathbb{R}^{k}\subset \mathbb{R}^{n}$. Then we can define a map $f: I\to \mathrm{O}(n)$ by taking the orthogonal matrix  $f=[e_1,\dots ,e_n]$ whose columns $e_i$, $i=1,\dots,n$, form the orthogonal basis---an orthogonal frame---obtained from $E=[\gamma',\gamma'',\dots, \gamma^{(n)}]$ by the Gram-Schmidt orthogonalization procedure. Now the matrix $A$ relating $f$ and $E$ is, by construction, upper triangular $e_{i}=\sum_{j=1}^n \gamma^{(j)}A_{ji}=\sum_{j=1}^{i}\gamma^{(j)}A_{ji}$. The inverse of an upper triangular matrix is upper triangular and hence, we may write

 $de_i=\sum_{j=1}^{i}\gamma^{(j+1)}A_{ji}dt +\sum_{j=1}^{i}\gamma^{(j)} dA_{ji}= \sum_{j=1}^{i}\sum_{l=1}^{j+1}e_{l}A^{-1}_{l, j+1}A_{j,i}dt+\sum_{j=1}^{i}\sum_{l=1}^{i} e_{l}A^{-1}_{lj}dA_{ji}=:\sum_{j=1}^{n}e_j\theta_{ji}.$

It is then clear that the matrix of $1$-forms $f^*\omega=f^{-1}df=[\theta_{ij}]_{1\leq i,j\leq n}$ satisfies $\theta_{ji}=0$ for $j>i+1$. But now since $f^tf=I_n$ it follows that $\theta_{ij}=-\theta_{ji}$. We conclude that 

$$de_{i}= e_{i-1}\theta_{i-1,i}+e_{i+1}\theta_{i+1,i}.$$

Without loss of generality, we can assume the curve is parameterized by the arclength, meaning that the tangent vector $\gamma'(t)$ has unit length.  Introducing the curvature functions by the formula $\kappa_{i} dt:=\theta_{i+1,i}$, $i=1,\dots,n-1$, we may write

$$\frac{de_{i}}{dt}= -e_{i-1} \kappa_{i-1} + e_{i+1} \kappa_i.$$

From Theorem 1, it follows that if two curves $\gamma$ and $\widetilde{\gamma}$ share the same curvature functions then they are related at most by a rotation and translation, both constant. To see this,  let $f$ and $\widetilde{f}$ be the orthogonal frame fields associated with $\gamma$ and $\widetilde{\gamma}$. Because the curvature functions are the same, it follows that $f$ and $\widetilde{f}$ different by a constant rotation. The first column of $f$ and $\widetilde{f}$ are the unit speed tangent vectors for $\gamma$ and $\widetilde{\gamma}$, respectively. Upon integration this means that $\gamma$ and $\widetilde{\gamma}$ differ, at most, by a rotation and a translation, both constant. In fact, it is also not hard, using the existence and uniqueness of solutions of ODEs, to show existence of a curve with specified curvature functions.

Out of curiosity, let us take the case $n=2$. Then 

$$e_1=\gamma', \text{ and } e_{2}=\frac{\gamma''}{||\gamma''||},$$

so that $\kappa:=\kappa_1=||\gamma''||$ is the standard curvature function of the curve. Here $e_2$ is usually known as the unit normal to the curve. The theorem then tells us that two regular curves sharing the same curvature function differ at most by isometries of $\mathbb{R}^2$.

Finally, take $n=3$, then

$$e_1=\gamma',\;  e_{2}=\frac{\gamma''}{||\gamma''||},\; e_3= \pm e_1\times e_2,$$

where here we used the fact that since $e_3$ is a unit vector orthogonal to both $e_1$ and $e_2$ it must be proportional to the cross product (which is also a unit vector and hence the proportionality constant must be of this form). The third vector is known as the binormal vector the curve. The new curvature function $\tau:=\kappa_2$ is (up to a sign convention) the so-called torsion of the curve and the theorem says that  two regular curves in $\mathbb{R}^3$ sharing the same curvature and torsion differ at most by an isometry of $\mathbb{R}^3$.