Area-minimizing current

Remarks

A local version is used for currents in an open set $U$. There one compares $T$ with $S$ only when $T - S$ is the boundary of an integral current with compact support in $U$. For a compact oriented $C^1$ submanifold, the mass of the associated current is its area. So an area-minimizing submanifold has least area among all oriented competitors with the same boundary, smooth or not.

Example: let $D$ be the closed unit disc in the plane $x_3 = 0$ of $\R^3$, oriented by $e_1 \wedge e_2$, and let $\omega = dx_1 \wedge dx_2$, a closed differential form of comass $1$. For every integral $2$-current $S$ with $\partial S = \partial [\![D]\!]$, one has $S - [\![D]\!] = \partial R$ for an integral $3$-current $R$, since every integral cycle in $\R^3$ bounds. Then $S(\omega) - [\![D]\!](\omega) = R(d\omega) = 0$ by the definition of the boundary, as in Stokes' theorem. So $\mathbf{M}(S) \ge S(\omega) = [\![D]\!](\omega) = \pi = \mathbf{M}([\![D]\!])$, and $[\![D]\!]$ is area-minimizing.

Non-example: the upper unit hemisphere, oriented so that its boundary is $\partial [\![D]\!]$, has mass $2\pi > \pi$, so it is not area-minimizing.