Proof that geodesics have zero curvature with Cartan's moving frames method I tried to use Cartan's moving frames method to prove that any minimizing-length curve in $\mathbb{R}^2$ has zero curvature. Here below is my idea of proof. I am asking for a proof-verification and for some specific doubts.

The set of all orthonormal frames could be identified with the euclidean plane group $E(2)$. We have the principal bundle $\pi: E(2) \to \mathbb{R}^2$.
We fix two points $A, B \in \mathbb{R}^2$ and suppose that it exists a 1-manifold $C \subseteq \mathbb{R}^2$ such that $\partial C = \{A, B\}$ and $C$ has minimal length.
It exists $f: C \to E(2)$ such that $\pi f = \text{id}_C$ and $f^*\omega^2 = 0$. This is the classic Frenet-Serret-Cartan frame associated to a curve and we have two possible choice for the orientation. We define $F = f(C) \cong C$. We have that $\int_C f^*\omega^1$ is minimum because $C$ has minimal length.
We consider a variation with fixed end-points in $\mathbb{R}^2$, i.e. a vector field $X \in \mathfrak{X}(\mathbb{R}^2)$ such that $X_A = X_B = 0$. Let be $\phi_t: \mathbb{R}^2 \to \mathbb{R}^2$ the flow of $X$ for $t \in (-\epsilon, \epsilon)$. We define the variations of $C$ as $C_t = \phi_t(C) \cong C$. For each $C_t$ it exists the unique frame $f_t: C_t \to E(2)$ such that $\pi f = \text{id}_{C_t}$, $f_t^*\omega^2 = 0$ and $f_t$ has the same orientiation of $f$. Obviously we have that $C_0 = C$ and $f_0 = f$.
Because $C$ has minimal length we have that $\frac{\text{d}}{\text{d}t}\int_{C_t}f_t^*\omega^1|_{t=0} = 0$ and so that $\frac{\text{d}}{\text{d}t}\int_{F}(f_t\phi_t\pi)^*\omega^1|_{t=0} = 0$ after a change of variable. If we define $Y: F \to TE(2)$ such that $Y(p) = \frac{\partial}{\partial t}(f_t\phi_t\pi)(p)\big|_{t=0}$ we have that
0 = $\frac{\text{d}}{\text{d}t}\int_{F}(f_t\phi_t\pi)^*\omega^1|_{t=0} = \int_{F}\frac{\partial}{\partial t}(f_t\phi_t\pi)^*\omega^1|_{t=0} = \int_{F}\mathcal{L}_Y\omega^1 = \int_{\partial F}i_Y\omega^1 + \int_F i_Y\text{d}\omega^1$
I think that the first integral vanishes. In fact I guess that $T\pi \circ Y = X \circ \pi$ so $T\pi \circ Y(\partial F) = X \circ \pi (\partial F) = X (\partial C) = 0$. Hence $0 = \omega^1(Y) = i_Y\omega^1$ on $\partial F$.
For the second integral we have that $\text{d}\omega^1 = \omega^1_2\wedge\omega^2$ and so $i_Y\text{d}\omega^1 = Y^1_2\omega^2 - Y^2\omega^1_2$ where $Y^1, Y^2, Y^1_2$ are the local components of $Y$. Because $\omega^2$ on $F$ vanishes we have that $0 = - \int_FY^2\omega^1_2 = - \int_F Y^2 k \omega^1$ where $k: F \to \mathbb{R}$ is the curvature of $C$. Because $Y^2$ depends only of $X^2$ and $X^2$ could be choosen randomly and also because $\omega^1 \neq 0$ always we can conclude that $k = 0$.

My really first question is: is this proof correct?
My doubts regarding the proof are:

*

*Does $f_t$ always exist? More important: is it $\mathcal{C}^\infty$ in $t$?

*Does Lie derivative definition work for $Y$ even if $Y$ is defined only on $F$ and not on all $E(2)$?

*Are $X$ and $Y$ $\pi$-related? Is true that $i_Y\omega^1$ vanishes on $\partial F$?

*Is true and $Y^2$ could be everything and $Y^2(p) = X^2(\pi(p))$? Can we conclude that it is necessary that $k=0$ if the integral vanishes for all $Y^2$?

Other questions regard possible generalizations:

*

*Can we modify this proof for $C$ not being a $1$-manifold but a parametrized curve?

*Does this proof  work for a 2 Riemannian manifold with principal bundle of orthonormal frames $\pi: O(M) \to M$ and the Levi-Civita connection $\omega^1_2$?

*Can we avoid using local components of $Y$ in the proof? Because maybe this could be a problem if the topology of $O(M)$ is more complicated of those of $E(2) \cong \mathbb{R}^2 \times \mathbb{S}^1$.

 A: $\newcommand{\R}{\mathbb{R}}$
Here's another way to write out the calculation (CORRECTED):
Let $f_0: [0,1]\rightarrow E(2)$ be the Frenet frame of a curve. In particular, if $s$ denotes the coordinate on $[0,1]$, then
\begin{align*}
f_0^*\omega^1 &= ds\\
f_0^*\omega^2 &= 0\\
f_0^*\omega^1_2 &= k\omega^1 = k\,ds,
\end{align*}
where $k$ is the curvature.
Define a variation to be a map $f: [0,1] \times (-\delta,\delta) \rightarrow E(2)$ such that $f(\cdot,0) = f_0$. For each $t$, let $f_t = f(\cdot, t)$.
Let $(s,t)$ denote the coordinates on $D = [0,1]\times (-\delta,\delta)$, $S = \partial_s$, and $T = \partial_t$.  We also assume the frame along $f_t$ has $e_1$ tangent to the curve, i.e.,
$$
\langle f_t^*\omega^1, S\rangle = 0.
$$
Also, note that if $\theta$ is a differential form on $E(2)$, then
$$
\langle S, f^*_t\theta\rangle = \langle S, f^*\theta\rangle,
$$
and, if $\eta$ is a $1$-form on $D$,
$$
\partial_t\eta = \mathcal{L}_T\eta = d\langle T,\eta\rangle + \langle T, d\eta\rangle
$$
Therefore, the derivative of the length of the curve parameterized by $f_t$ is
\begin{align*}
\frac{d}{dt}\int_{s=0}^{s=1} \langle S,f_t^*\omega^1\rangle\,ds
&= \frac{d}{dt}\int_{s=0}^{s=1} \langle S, f^*\omega^1\rangle\,ds\\
  &= \int_{s=0}^{s=1} \langle S, \partial_t(f^*\omega^1)\rangle\,ds\\
  &= \int_{s=0}^{s=1} \langle S, \mathcal{L}_T(f^*\omega^1)\rangle\,ds\\
  &= \int_{s=0}^{s=1} \langle S, d\langle T,f^*\omega^1\rangle\rangle + \langle T\otimes S, d(f^*\omega^1)\rangle\,ds\\
  &= \int_{s=0}^{s=1} \partial_s(\langle T,f^*\omega^1\rangle) + \langle T\otimes S, f^*(d\omega^1)\rangle\,ds\\
  &= \left.\langle f_*T,\omega^1\rangle\right|_{s=0}^{s=1} + \int_{s=0}^{s=1} -\langle T\otimes S,f^*(\omega^1_2\wedge\omega^2)\rangle\,ds\\
\end{align*}
If we assume that $f$ fixes the endpoints of the curve, then $f_*T(0,0)= \partial_t f(0,0)= 0$ and $f_*T(1,0)= \partial_t f(1,0)= 0$.
Therefore, 
\begin{align*}
\left.\frac{d}{dt}\right|_{t=0}\int_{s=0}^{s=1} \langle S, f_t^*\omega^1\rangle\,ds
  &= \int_{s=0}^{s=1} -\langle T,f^*\omega^1_2\rangle\langle S,f^*\omega^2\rangle
    + \langle T,f^*\omega^2\rangle \langle S, f^*(\omega^1_2)\rangle\, ds\\
  &= \int_{s=0}^{s=1} -\langle T,f^*\omega^1_2\rangle\langle S,f_0^*\omega^2\rangle
    + \langle T,f^*\omega^2\rangle \langle S, f_0^*(\omega^1_2)\rangle\, ds\\
  &= \int_{s=0}^{s=1} \langle T,f^*\omega^2\rangle \langle S, f_0^*(\omega^1_2)\rangle\, ds\\
  &= \int_{s=0}^{s=1} \langle T,f^*\omega^2\rangle \langle S, k\,ds\rangle\, ds\\
  &= \int_{s=0}^{s=1} \langle f_*T,\omega^2\rangle k\, ds\\
  &= \int_{s=0}^{s=1} \langle X,\omega^2\rangle k\, ds,
\end{align*}
where $X = \partial_tf(\cdot,0)$. If this vanishes for any variation $f$ that fixes the endpoints, then $k$ must vanish on the curve.
