# Calculus of variation and isoperimetric problem with differential forms and moving frames

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 $$\omega^3=0$$.)

• 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 – Ted Shifrin Mar 24 at 23:00
• @TedShifrin I edited my original post explaining my reasoning! Thanks for your time, I always hope in your help – Marco All-in Nervo Mar 25 at 14:10