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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:

  1. Does $f_t$ always exist? More important: is it $\mathcal{C}^\infty$ in $t$?
  2. Does Lie derivative definition work for $Y$ even if $Y$ is defined only on $F$ and not on all $E(2)$?
  3. Are $X$ and $Y$ $\pi$-related? Is true that $i_Y\omega^1$ vanishes on $\partial F$?
  4. 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:

  1. Can we modify this proof for $C$ not being a $1$-manifold but a parametrized curve?
  2. 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 $\omega^1_2$?
  3. 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 $E(2) \cong \mathbb{R}^2 \times \mathbb{S}^1$.
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    $\begingroup$ Use compactly supported perturbations, defined by compactly supported vectors fields $X$ supported away from the two end points. Then the $f_t$ always exist. $\endgroup$
    – Ben McKay
    Aug 2, 2019 at 14:12
  • $\begingroup$ One trick, when doing the variation is to note that the variation is really a map $[0,1]\times (-delta,\delta) \rightarrow \mathbb{R}^2$ and you can pull all of your calculations back to $[0,1]\times (-delta,\delta)$. This shows that your calculations using the Lie derivative are indeed valid. $\endgroup$
    – Deane Yang
    Aug 2, 2019 at 17:45
  • $\begingroup$ @BenMcKay and is it $\mathcal{C}^\infty$ in $t$? $\endgroup$
    – CNS709
    Aug 2, 2019 at 21:51
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    $\begingroup$ Smoothness results from Picard's existence and uniqueness theorem, which ensures smoothness of the solutions of smooth determined systems of ordinary differential equations. $\endgroup$
    – Ben McKay
    Aug 3, 2019 at 6:14

1 Answer 1

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$\newcommand{\R}{\mathbb{R}}$ 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.

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  • $\begingroup$ I think that you mean $f: [0, 1] \times (-\delta, \delta) \to E(2)$ and not $\to \mathbb{R}^2$. In this case I think you have to impose that $f^*\omega^2 = 0$ and maybe also that $f^*\omega^1 \neq 0$ (for every $t$ $f$ can't move just the frame letting fixed the point). But in this case the vector field $X$ is not "any vector field", or maybe it's not explicit (it's the same of my $Y^2$). Finally, I prefer consider variation of curve and then construct the frame, because the minimum is achieved varying the curve. What do you think? $\endgroup$
    – CNS709
    Aug 2, 2019 at 21:49
  • $\begingroup$ 1) Yes, it should be to $E(2)$. Thanks. 2) I don't see where in my calculation I need $f^*w^2 = 0$. Could you point that out? 3) The interesting point is that $f^*w^1 \ne 0$ isn't needed. It doesn't matter if $f$ is degenerate. In any case, it holds automatically for $t$ sufficiently close to $0$. 4) Given any vector field $X$ along the curve, you can find an $f$ such that $\partial_tf(s,0) = X(f_0(s))$. 5) I start with a curve, lift it to its Frenet frame, and then vary the framed curve. I don't assume that the frame is the Frenet frame along the curve $f(\cdot,t)$ when $t \ne 0$. $\endgroup$
    – Deane Yang
    Aug 2, 2019 at 22:05
  • $\begingroup$ In fact, it's important that $f^*\omega^2$ not necessarily be zero. That's needed for the last step. $\endgroup$
    – Deane Yang
    Aug 2, 2019 at 22:13
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    $\begingroup$ You don't need it for the calculation, you need it because if $f^*\omega^2 \neq 0$ you have that $f^*\omega^1$ is no more the arc-length form; so it does not hold that $\frac{d}{dt}\int f^*\omega^1 = 0$. In your way you can consider variations that just rotate the frame and does not change the curve. You can end with a frame 90-degree rotated that has $f^*\omega^1 = 0$ everywhere and so the entire calculation does not hold. $\endgroup$
    – CNS709
    Aug 2, 2019 at 22:26
  • $\begingroup$ You're right about the arclength issue. I need to fix that. $\endgroup$
    – Deane Yang
    Aug 2, 2019 at 22:56

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