Let $(F,g)$ be a closed Surface, $(M,h)$ a Riemannian 3-Manifold and $f: F \to M$ a smooth immersion. Denote by $f^*(h)$ the pullback metric on $TF$ induced by $f$ and let $dV_g$ and $dV_{f^*(h)}$ be the respective volume forms on $F$.The Area of $F$ inside $M$ is defined as the Area of $(F,f^*(h))$, i.e $Area(F,f):= \int_F dV_{f^*(h)}$. Further, the Dirichlet energy of $f$ with respect to $g$ is defined as $E(f,g) := \frac{1}{2} \int_F |df|^2 dV_g$. Now it seems to be a general known fact that $Area(F,f) \leq E(f,g)$ for any (riemannian) metric $g$ on $F$ as I have read this very claim in several papers about least area surfaces. In the special case that $M = \mathbb R^3$ and $(F,g)$ is a flat surface (which is very restrictive, of course), this inequality boils down to a simple computation on the norm of a particular cross product, which is quite simple to do. However, I have no idea how to do the general case, as there seem to be way too many variables to deal with. Does anybody have an idea ?
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1$\begingroup$ This is just the integration of a pointwise inequality, and it only involves the first derivative of the map $f$ and the values of the metrics $g$ and $h$ at a fixed point in $F$. Thus, checking it in the 'flat' case that you have already done proves it in the general case. By the same calculation, you also get that equality holds if and only if $f$ is a conformal immersion. $\endgroup$– Robert BryantCommented Aug 26, 2015 at 15:50
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$\begingroup$ I've tried this, but i can't quite deal with the arbitrariness of the metric $g$ on $F$. In appropriate charts, the area integrand should be $\sqrt{|\partial f/ \partial x_1|^2_h|\partial f/ \partial x_2|^2_h - <\partial f/\partial x_1,\partial f,\partial x_2>_h^2}$, while the energy integrand is $\sum_{i,j=1}^2 g^{ij}<\partial f/\partial x_i,\partial f/ \partial x_j>_h$. But since I have no control about how the $g^{ij}$ might look, i am stuck at a certain point in a chain of inequailites. $\endgroup$– H1ghfiv3Commented Aug 26, 2015 at 16:51
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$\begingroup$ Calculate at a point: Let $e_1,e_2$ be a $g$-orthonormal basis of $T_xF$ and let $f_i = df(e_i)\in T_{f(x)}M$. Then you just need to show that $$\tfrac12\bigl(|f_1|^2+|f_2|^2\bigr) \ge \sqrt{\ |f_1|^2|f_2|^2-(f_1\cdot f_2)^2\ }$$ where the inner products come from $h$ on $T_{f(x)}M$. But this is obvious (and doesn't depend on $M$ being $3$-dimensional, by the way). Note, too, that you get equality if and only if $f_1\cdot f_2 = |f_1|^2-|f_2|^2 = 0$. $\endgroup$– Robert BryantCommented Aug 26, 2015 at 17:01
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$\begingroup$ Of course, I forgot about geodesic normal coordinates. This solves the problem. Thank you, Sir. $\endgroup$– H1ghfiv3Commented Aug 26, 2015 at 17:59
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$\begingroup$ You're welcome, but I should point out that it's not actually a use of geodesic normal coordinates. The derivatives of the metrics $g$ and $h$ play no role in the formulae for the integrands and hence are irrelevant for the inequality. It really is just linear algebra at a point. $\endgroup$– Robert BryantCommented Aug 26, 2015 at 19:02
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