Simons cone

Example

Write points of $\R^8$ as $(x, y)$ with $x, y \in \R^4$. The Simons cone is the hypersurface $$C = \set{(x, y) \in \R^4 \times \R^4 : \abs{x} = \abs{y}}.$$ It is a cone: if $(x, y) \in C$ and $t > 0$, then $(tx, ty) \in C$. It is smooth away from the origin, and the origin is its only singular point.

Check that $C \setminus \set{0}$ is a minimal surface, that is, has mean curvature zero. The cone is the zero set of $u(x, y) = \abs{x}^2 - \abs{y}^2$, and $\nabla u = 2(x, -y)$ vanishes only at the origin. The unit normal is $\nu = (x, -y)/r$ with $r = \sqrt{\abs{x}^2 + \abs{y}^2}$, and the mean curvature is, up to a constant factor, $\operatorname{div} \nu$. Since $\operatorname{div}(x, -y) = 4 - 4 = 0$ and $(x, -y) \cdot \nabla (1/r) = -(\abs{x}^2 - \abs{y}^2)/r^3$, one gets $\operatorname{div} \nu = -(\abs{x}^2 - \abs{y}^2)/r^3$, which is $0$ on $C$.