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$u: \Omega \subset \mathbb R^2 \rightarrow \mathbb R$ is a $C^2$ function. Graph of $u$ is $$ G_u=\{(x,y,u(x,y)) : (x,y)\in \Omega\} $$ And the upward pointing unit normal is $N$. $\omega$ is the two-form on $\Omega\times \mathbb R$ given by that for $X,Y\in \mathbb R^3$ $$ \omega(X,Y) = \det (X,Y,N) $$ Then we have $$ \omega(\partial_x, \partial_y) = \frac{1}{\sqrt{1+|\nabla u|^2}} \\ \omega(\partial_y, \partial_z) = \frac{-u_x}{\sqrt{1+|\nabla u|^2}} \\ \omega(\partial_x, \partial_z) = \frac{u_y}{\sqrt{1+|\nabla u|^2}} $$ $\partial_x, \partial_y, \partial_z$ are the standard basis of $\mathbb R^3$. Then, how to show given any orthogonal unit vectors $X,Y$ at a point $(x,y,z)$ we have $$ |\omega(X,Y)|\le 1 $$ And equality holds iff $$ X,Y\subset T_{(x,y,u(x,y))} G_u $$

This question is from 1.1 of T.H. Colding and W.P. Minicozzi's Minimal surfaces.

Martin Sleziak
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Enhao Lan
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This supposed to be obvious, because $\det(X,Y,N)$ is the volume of the parallelepiped formed by vectors $X$, $Y$, and $N$. Vectors $X$ are $Y$ orthogonal, and if they lie in the tangent plane, the parallelepiped is a unit cube. For the converse you need to use a formula for the triple product, or the fact that the cube has the maximal volume among all parallelepipeds with equal sides.

Yuri Vyatkin
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