if and only if $\operatorname{Span}\{e_1,\ldots,e_{2k}\}$ is a complex $k$-plane in $\C^n$.

Before proving this we make the following observation.

\[starcab.lem\] If $\eta$ is a calibration and $*\eta$ is closed then $*\eta$ is a calibration. Moreover an oriented tangent plane $P$ is calibrated by $\eta$ if and only if there is an orientation on the orthogonal complement $P^{\perp}$ so that it is calibrated by $*\eta$.

Suppose that $\eta$ is a calibration $k$-form on $(M,g)$ with $\d\!*\!\eta=0$. Let $p\in M$. Take any $n-k$ orthonormal tangent vectors $e_{k+1},\ldots,e_n$ at $p$. Then there exist $e_1,\ldots,e_k\in T_pM$ so that $\{e_1,\ldots,e_n\}$ is an oriented orthonormal basis for $T_pM$. Since $\{e_1,\ldots,e_n\}$ is an oriented orthonormal basis, we can use the definition of the Hodge star to calculate $$*\eta(e_{k+1},\ldots,e_n)=\eta(e_1,\ldots,e_k)\leq 1.$$ Hence $*\eta$ is a calibration by Definition \[calibration.dfn\]. Moreover, the oriented plane $P=\operatorname{Span}\{e_{k+1},\ldots,e_n\}$ is calibrated by $*\eta$ if and only if there is an orientation on $\operatorname{Span}\{e_1,\ldots,e_k\}=P^{\perp}$ so that it is calibrated by $\eta$, since $\eta(e_1,\ldots,e_k)=\pm *\eta(e_{k+1},\ldots,e_n)=\pm 1$.

We can now prove Wirtinger’s inequality.

We see that $|\frac{\omega^k}{k!}|^2=\frac{n!}{k!(n-k)!}$ and $\operatorname{vol}_{\C^n}=\frac{\omega^n}{n!}$ so $*\frac{\omega^k}{k!}=\frac{\omega^{n-k}}{(n-k)!}$. Hence, by Lemma \[starcab.lem\], it is enough to study the case where $k\leq \frac{n}{2}$.

Let $P$ be any $2k$-plane in $\C^n$ with $2k\leq n$. We shall 