Hermitian form, fundamental $2$-form of Kahler structure on $\mathbb{C}^n$

Question

I've come across the following (it is an excerpt of Stolzenberg's lecture notes 19):

Wirtinger's Inequality.

Let $L$ be a complex linear space and let $M$ be a real even-dimensional subspace. Let $H$ be a positive definite Hermitian form on $L$ . Then $H = S + iA$ where $S$ is symmetric and $A$ is alternating . Let $\{ m_1,...,m_{2k} \} $ be a basis of $M$ which is orthonormal with respect to $S$ . Then $$ | A ^k ( m_1 , . . . , m_{2 > k} ) | \le k !$$ with equality holding precisely when $M$ is a complex $k$-dimensional subspace of $L$ . (Here $A^k$ is the $k$-th exterior power of $A$ .)

By a suitable shuffling of the basis it can be arranged that $A ^k ( > m_1 , . . . , m_{2 k} ) \ge 0$

Let $z_1, . . . , z_n$ be coordinates on complex $n$ - space $\mathbb{C}^n$ and set

$$\omega = \frac{i}{2} \sum_{j=1}^n dz_j \wedge d \overline{z}_j$$

This is the fundamental $2$-form of the standard Kahler structure on $\mathbb{C}^n$ and at each point $p \in \mathbb{C}^n$ $\omega_p$ is the alternating part of the positive definite Hermetian form:

$$\sum_{j=1}^n dz_{j(p)} \cdot d \overline{z}_{j(p)}$$

on the tangent space to $\mathbb{C}^n$ at $p$.

Therefore , if $\mathcal{M}$ is any smoothn $2k$-dimensional manifold immersed in $\mathbb{C}^ n$ , Wirtinger's Inequality implies immediately that:

$$\int_{\mathcal{M}} \frac{1}{k!} \omega ^k \le \int_{\mathcal{M}} 1 \ d > \mathcal{M} = \text{Volume} _{2k}(\mathcal{M})$$

with equality is and only if $ \mathcal{M}$ is a complex $k$-dimensional manifold.

Also each $\frac{1}{k!} \omega ^k $ is an exact $2k$ form.

My questions are:

Could you explain to me how we use the fact that $\omega$ is the fundamental $2$-form of the standard Kahler structure on $\mathbb{C}^n$ to apply the Wirtinger inequality here?

Is this somehow connected to this Second fundamental form ?

Could you explain to me what the fundamental $2$-form of the standard Kahler structure is? Or recommend a good source in which I could read about it?

In Werner Ballmann's Lectures on Kahler Manifolds the author defines the associated Kahler form (which I presume could be the same as the fundamental $2$-form) in this way:

Let $M$ be a complex manifold with complex structure $J$ and compatible Riemannian metric $g = < \cdot, \cdot > $ (so $<JX, JY>= < X, Y > $). The alternating $2$-form $\omega(X, Y ) := g(JX, Y )$ is called the associated Kahler form. We say that $g$ is a Kahler metric and if $\omega$ is closed, we say that $(M, g)$ is a Kahler manifold.

On page 48, the author states that we can

view $TM$ together with $J$ as a complex vector bundle over $M$, and let $h$ be a Hermitian metric on $TM$. Then $g = Re h$ is a compatible metric on $M$ and $Imh$ is the associated Kahler form: $$g(JX, Y ) = \Re h(JX, Y ) = \Re h(iX, Y ) = \Re(−ih(X, Y )) = \Im h(X, Y )$$ If $g$ is a compatible Riemannian metric on $M$ and $\omega$ is the associated Kahler form, then $h = g + i\omega$ is a Hermitian metric on $TM$.

Also, a Riemannian metric is a general notion. Looking at what I've pasted into the frame above, do we need to consider Riemannian metrics compatible with the complex structure in general or not necessarily?

I would be very grateful for all your insight.

Answer 1

This is all "basic stuff" (this question should probably be asked on Mathematics Stack Exchange).

Let $M$ be a manifold. A Riemannian metric $g$ and an almost complex structure $I$ are compatible (i.e. $I$ is $g$-orthogonal) if and only if $g$ is the real part of a (necessarily unique) Hermitian metric $h$. When that is the case, the imaginary part of $h$ is a nondegenerate $(1,1)$-form on $(M,I)$, sometimes called the fundamental $2$-form (completely unrelated to second fundamental forms for submanifolds of Riemannian manifolds). I personally prefer the convention with a minus sign: $$ h = g - i\omega$$

For instance on $M = \mathbb{C}$, $h = dz \otimes d\bar{z}$, $g = dz\,d\bar{z} = dx^2 + dy^2$, and $\omega = \frac{i}{2}dz \wedge d\bar{z} = dx \wedge dy$.

When $I$ is integrable (so $(M,I)$ is a complex manifold) and $\omega$ is closed (so it's a symplectic form), the structure $(I,g)$ (or $(I,h)$) is called Kähler, and $\omega$ is called the Kähler form. But that does not matter here.

It is not hard to show that the top exterior power of $\omega$ divided by $n!$ where $n$ is the complex dimension of $M$ is precisely the volume form of the Riemannian metric $g$: $$ \frac{\omega^{\wedge n}}{n!} = \operatorname{vol}_g$$ So your Wirtinger's inequality is actually an equality when $2k = n$. In fact is is an equality if and only if $(m_1, \dots, m_{2k})$ span a complex linear subspace.

All of this is just linear algebra, it does not matter that you are working with manifolds. By restricting your attention to one tangent space, you can assume that $M$ is a vector space and everything is linear.