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This question follows this one. I want to apply the calculus of variation with differential forms to three classical problem: 1. arc-length minimizing curve (geodesics) 2. area-minimizing surfaces (minimal surface) 3. isoperimetric problem.

  1. (Geodesics) I consider as space the space of all orthogonal frames (Cartan's moving frames) over $\mathbb{R^2}$ (or a surface). In this space (SO(2)) there are three 1-form $\omega_1, \omega_2, \omega_{12}$. The idea is minimizing $\omega_1$ with constraint $\omega_2 = 0$. This gives the correct result ($\omega_{12} = 0$, i.e geodesic curvature vanishes) but... is the idea correct?

  2. (Minimal surfaces) In the same way I consider the space of all orthogonal frames over $\mathbb{R}^3$. The idea is minimize $\omega_1 \wedge \omega_2$ with constraint $\omega_3 = 0$. But then when I write the Euler-Lagrange equation $i_Xd(\omega_1 \wedge \omega_2 + \lambda\omega_3) = 0$ I note that I am mixing 2-forms and 1-forms. I again got the correct result (i.e $H = 0$), but this time the approach seems very wrong to me.

  3. (Isoperimetric problem) I want to fix area (integral of $\omega_1 \wedge \omega_2$) inside a curve and minimize the length of the curve (integral of $\omega_1$). This time I have no idea. I suppose I have to express the area inside the curve as the integral of a 1-form (such as $dxdy \to (xdy-ydx)/2$, but I have no idea how to do that with moving frames.

EDIT

@TedShifrin All these variational problems are intended to be with the fixed-boundary constraint (in fact, these are the problems I have considered in the cited previous post: the Euler-Lagrange equation $i_Xd(\Lambda + \lambda_i\phi^i) = 0$ works only for these). Now I'll try to explain my idea better. My idea was the same you said: I want to find a curve $C$ that minimizes $\int_C\omega^1$. But then I noticed a thing: $\omega^1$ is not a 1-form in $\mathbb{R}^2$! I mean: for every variation of the curve $C_t$ we have a different $\omega^1$: let's say $\omega_t^1$ (instead, in the previous post $pdq - Hdt$ is a fixed 1-form in $T^*Q\times \mathbb{R}$). Then I asked my self: what is this? So $\omega_1$ is a well-defined fixed 1-form in $SO(2)$ and what I want to minimize is really the pullback of it by means of the Frenet frame. I mean: for every variation of the curve $C_t$, I have a (canonical) Frenet frame $F_t: C_t \to SO(2)$ and so what I want is $\frac{d}{dt}\int_{C_t}F_t^*\omega^1 = 0$, i.e I want $\frac{d}{dt}\int_{F_t}\omega^1 = 0$. So now the problem is well-defined, with $M = SO(2)$, $\Lambda = \omega^1$ and what I'm looking for is a "curve of adapted frames". But, if before I could vary the curve as I want, now that I am minimizing in the space of frames, I could vary the "curve of frame" only between the adapted ones, i.e those with $\omega^2=0$. So, in my opinion, in this variational problem I have to consider the ideal $I = (\omega^2)$, instead of $I = (0)$. (Then I made the same reasoning with the second case, but as now I could not remember if the Frenet frame for surfaces has any conditions other than $\omega^3=0$.)

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  • $\begingroup$ I disagree with your approach. You want to minimize $\int_C \omega_1$ working with adapted frames, the constraint being fixed boundary. Similarly, for minimal surfaces you want to minimize $\int_S \omega_1\wedge\omega_2$ with the constraint being fixed boundary, again working with adapted frames. You certainly can't add a $1$-form and a $2$-form, let alone differentiate their sum. :P $\endgroup$ – Ted Shifrin Mar 24 at 23:00
  • $\begingroup$ @TedShifrin I edited my original post explaining my reasoning! Thanks for your time, I always hope in your help $\endgroup$ – Marco All-in Nervo Mar 25 at 14:10

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