one $P$. Since $\eta_0$ is $G$-invariant, if $P$ is calibrated then so is $\gamma\cdot P$ for any $\gamma\in G$, which usually means we have quite a few calibrated planes. We know by the *holonomy principle* (see, for example, [@Joycebook2 Proposition 2.5.2]) that we then get a parallel $k$-form $\eta$ on $M$ which is identified with $\eta_0$ at every point. Since $\nabla\eta=0$, we have $\d\eta=0$ and hence $\eta$ is a calibration. Moreover, we have a lot of calibrated tangent planes on $M$, so we can hope to find calibrated submanifolds.

Complex submanifolds
====================

We would now like to address the question: where does the calibration condition come from? The answer is from *complex geometry*. On $\R^{2n}=\C^n$ with coordinates $z_j=x_j+iy_j$, we have the complex structure $J$ and the distinguished Kähler 2-form $$\omega=\sum_{j=1}^n\d x_j\wedge \d y_j=\frac{i}{2}\sum_{j=1}^n\d z_j\wedge\d \overline{z}_j.$$ More generally we can work with a *Kähler manifold* $(M,J,\omega)$. Our first key result is the following.

\[cxcalib.thm\] On a Kähler manifold $(M,J,\omega)$, $\frac{\omega^k}{k!}$ is a calibration whose calibrated submanifolds are the complex $k$-dimensional submanifolds: i.e. submanifolds $N$ such that $J(T_pN)=T_pN$ for all $p\in N$.

Since $\d\omega^k=k\d\omega\wedge\omega^{k-1}=0$, Theorem \[cxcalib.thm\] follows immediately from the following result.

\[Wirtinger.thm\] For any unit vectors $e_1,\ldots,e_{2k}\in\C^n$, $$\frac{\omega^k}{k!}(e_1,\ldots,e_{2k})\leq 1$$ with equality 