$N$ is calibrated by $\eta$ and suppose for simplicity that $N$ is compact. We will show that $N$ is homologically volume-minimizing.

Suppose that $N'$ is homologous to $N$. Then there exists a compact manifold $K$ with boundary $-N\cup N'$ and, since $\d\eta=0$, we have by Stokes’ Theorem that $$0=\int_K\d\eta=\int_{N'}\eta-\int_N\eta.$$ We deduce that $$\begin{aligned}
\operatorname{Vol}(N)&=\int_N\eta=\int_{N'}\eta\leq\operatorname{Vol}(N').\end{aligned}$$ We then have the result by the definition of minimal submanifold.

We conclude this introduction with the following elementary result.

There are no compact calibrated submanifolds in $\R^n$.

Suppose that $\eta$ is a calibration and $N$ is compact and calibrated by $\eta$. Then $\d\eta=0$ so by the Poincaré Lemma $\eta=\d\zeta$, and hence $$\operatorname{Vol}(N)=\int_N\eta=\int_N\d\zeta=0$$ by Stokes’ Theorem.

Although there are many calibrations, having calibrated submanifolds greatly restricts the calibrations you want to consider. The calibrations which have calibrated submanifolds have special significance and there is a particular connection with special holonomy, due to the following observations.

Let $G$ be the holonomy group of a Riemannian metric $g$ on an $n$-manifold $M$. Then $G$ acts on the $k$-forms on $\R^n$, so suppose that $\eta_0$ is a $G$-invariant $k$-form. We can always rescale $\eta_0$ so that $\eta_0|_P\leq \operatorname{vol}_P$ for all oriented $k$-planes $P$ and equality holds for at least 