# Is there a minimal graph in $\mathbb{R}^3$ which is not area-minimizing?

Let $\Omega\subset\mathbb{R}^2$ be an open subset such that $\partial\Omega$ is a closed, simple curve.

I'm trying to find an example of an $u:\overline{\Omega}\to\mathbb{R}$ such that $\Sigma:=\text{graph}(u)$ is a minimal surface and, yet, there exists another minimal surface $\Sigma'$ with $\partial\Sigma'=\partial\Sigma$ and $\text{Area}(\Sigma')<\text{Area}(\Sigma)$.

Does such an example exist?

This is an excellent question. The answer is "no." The way to see this is via a calibration argument.

## Background on Calibrations

Def: Let $$(M^n, g)$$ be a Riemannian manifold. A calibration on $$M$$ is a $$p$$-form $$\varphi \in \Omega^p(M)$$ satisfying:

1. $$d\varphi = 0$$, and
2. $$|\varphi(v_1, \ldots, v_p)| \leq 1$$ for every orthonormal set $$\{v_1, \ldots, v_p\}$$ in $$T_xM$$.

An oriented $$p$$-dim subspace $$V \subset T_xM$$ is calibrated by $$\varphi$$ iff $$\varphi(v_1, \ldots, v_p) = 1$$ for some oriented orthonormal basis $$\{v_1, \ldots, v_p\}$$ of $$V$$.

An oriented $$p$$-dim submanifold $$N^p \subset M^n$$ is calibrated by $$\varphi$$ iff each tangent space $$T_xN \subset T_xM$$ is calibrated by $$\varphi$$.

The following theorem is due to F. Reese Harvey and H. Blaine Lawson (1982):

Fundamental Theorem on Calibrations: Let $$(M^n, g)$$ be a Riemannian manifold and $$\varphi$$ a calibration. Let $$N, N' \subset M$$ be two compact, oriented, $$p$$-dim submanifolds with $$\partial N = \partial N'$$ and $$N$$ homologous to $$N'$$. If $$N$$ is calibrated by $$\varphi$$, then $$\text{Area}(N) \leq \text{Area}(N')$$.

Proof: Using that $$N$$ is calibrated first, then Stokes' Theorem second, then the definition of calibration third, we have $$\text{Area}(N) = \int_{N} \varphi = \int_{N'} \varphi \leq \text{Area}(N'). \ \ \ \lozenge$$

## Application: Graphical Minimal Surfaces in $$\mathbb{R}^3$$

Let $$u \colon \overline{\Omega} \to \mathbb{R}$$ be such that $$\Sigma := \text{Graph}(u)$$ is a minimal surface in $$\mathbb{R}^3$$. Regarding $$\Sigma$$ as the level set $$\{v(x,y) = 0\}$$, where $$v(x,y) := z - u(x,y)$$, we see that a unit normal vector field to $$\Sigma$$ is $$N_u = \frac{\nabla v}{\Vert \nabla v \Vert} = \frac{(-u_x, -u_y, 1)}{\sqrt{1 + (u_x)^2 + (u_y)^2}}.$$ Define the $$2$$-form $$\varphi_u \in \Omega^2(\mathbb{R}^3)$$ via $$\varphi_u(X,Y) = \det(X,Y, N_u) = (X \times Y) \cdot N_u.$$ The following exercise, together with the Fundamental Theorem above, gives the result:

Exercise: (a) The $$2$$-form $$\varphi_u$$ is a calibration on $$\mathbb{R}^3$$. That is:

1. $$d\varphi_u = 0$$, and
2. $$|\varphi_u(X,Y)| \leq 1$$ for every orthonormal set $$\{X,Y\}$$ in $$T_x\mathbb{R}^3$$.

(b) The surface $$\Sigma = \text{Graph}(u)$$ is calibrated by $$\varphi_u$$. That is: If $$\{X,Y\}$$ is an oriented orthonormal set having $$X,Y \in T_x\Sigma$$, then $$\varphi_u(X,Y) = 1$$.

• The "usual" form of Stokes' theorem won't work because the region that $N$ and $N'$ bound could be something wild (and it isn't hard to come up with examples where there's no clear "in between" region). What is the correct form that one should apply? Oct 7 '17 at 3:27
• I suspect the answer might contain "see Federer" or "see Giaquinta-Modica-Soucek"... Oct 7 '17 at 3:40
• @JesseMadnick, I don't understand why $\partial N=\partial N'$ implies $\int_N\varphi=\int_{N'}\varphi$. I would understand it if $N-N'=\partial R$ for some $R$, which is not the same as $\partial N=\partial N'$, right? Jun 16 '19 at 23:24
• @rmdmc89: Yes, $N - N' = \partial R$ for some $R$. Now Stokes' Theorem gives $$\int_N \varphi = \int_{N'} \varphi + \int_{\partial R} \varphi = \int_{N'} \varphi + \int_{R} d\varphi = \int_{N'} \varphi$$ since $d\varphi = 0$. Jun 17 '19 at 3:40
• Oh, I see. In my original post, I somehow forgot to mention that the Fundamental Theorem requires $N$ and $N'$ to be homologous. That was careless. OK, it's fixed now. Nov 16 '20 at 11:54

The calibration argument works perfectly if $$\bar\Omega$$ is convex; if not, then there are counterexamples.

If $$\bar\Omega$$ is not convex, then the minimizer can escape from $$\bar\Omega\times \mathbb R$$ (where the form $$\varphi_u$$ is defined). In this case we have no control on its area.

For example, a minimal graph $$z=u(x,y)$$ might not minimize area if $$u$$ is defined on domain $$\bar\Omega\subset \mathbb R^2$$ as on the picture: Indeed, imagine that boundary values are zero except the horizontal lines where they depend only on the $$x$$-coordinate. In this case (for suitable $$u$$) it is cheaper to take a disc in the $$(x,y)$$-plane with two thin belts approaching the center from both sides. (Projection of an area minimizer to $$(x,y)$$ plane will enter in the gaps between half-discs.)