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

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.

• There are multiple questions in your post; I hope I've either: addressed them; given a sufficiently complete sketch that you can fill in missing details; or cited suitable references. – Andrew D. Hwang Apr 1 '15 at 22:48

$\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$Good sources for the material below include:

• R. O. Wells, Differential Analysis on Complex Manifolds.

• P. Griffiths and J. Harris, Principles of Algebraic Geometry.

The literature contains numerous conventions about factorials in wedge products, signs, and factors of two. I've tried to be internally consistent, but there always remains the possibility of errors.

Let $\Cpx^{n}$ have complex coordinates $z_{j} = x_{j} + iy_{j}$ for $1 \leq j \leq n$, let $\bar{z}_{j} = x_{j} - iy_{j}$, and note that for each $j$ we have \begin{align*} dz_{j} \otimes d\bar{z}_{j} &= (dx_{j} + i\, dy_{j}) \otimes (dx_{j} - i\, dy_{j}) \\ &= (dx_{j} \otimes dx_{j} + dy_{j} \otimes dy_{j}) - i (dx_{j} \otimes dy_{j} - dy_{j} \otimes dx_{j}) \\ &= (dx_{j} \otimes dx_{j} + dy_{j} \otimes dy_{j}) - 2i dx_{j} \wedge dy_{j}. \end{align*} That is, if $$H = \sum_{j=1}^{n} dz_{j} \otimes d\bar{z}_{j} = g - 2i\omega,$$ then $g$ is the Euclidean inner product on $\Cpx^{n} \simeq \Reals^{2n}$ and $\omega$ is the standard symplectic form: $$g = \sum_{j=1}^{n} (dx_{j} \otimes dx_{j} + dy_{j} \otimes dy_{j}),\qquad \omega = \sum_{j=1}^{n} dx_{j} \wedge dy_{j}. \tag{1}$$ (The pairing $H$ is Hermitian, i.e., is complex linear in the first argument, and satisfies $H(Y, X) = \overline{H(X, Y)}$ for all $X$ and $Y$ in $\Cpx^{n}$. The latter encodes symmetry of the real part of $H$ and skew-symmetry of the imaginary part of $H$.)

Note also that $$d\mu = \frac{\omega^{k}}{k!} = dx_{1} \wedge dy_{1} \wedge \dots \wedge dx_{n} \wedge dy_{n}$$ is the Euclidean volume form of $\Cpx^{n}$. That is, the answer to the first question is that the top exterior power of the imaginary part of the standard Hermitian structure on $\Cpx^{n}$ is the Euclidean volume form on $\Cpx^{n} \simeq \Reals^{2n}$. (Caution: Some authors define the volume form to be $dx_{1} \wedge \dots dx_{n} \wedge dy_{1} \wedge \dots \wedge dy_{n} = (-1)^{n(n-1)/2}\, d\mu$.)

Let $M$ is a smooth, immersed, oriented $2k$-manifold in $\Cpx^{n}$. The integral $$\int_{M} \frac{\omega^{k}}{k!}$$ can be evaluated using a partition of unity $\{\rho_{\alpha}\}$ subordinate to a cover $\{U_{\alpha}\}$ by coordinate neighborhoods. In each open set $U_{\alpha}$ of the cover, let $\{\dd_{1}, \dots, \dd_{2k}\}$ denote the Euclidean coordinate frame (in the ordering compatible with the volume form convention above, and suppressing the dependence on $\alpha$). If $dV$ denotes the Euclidean volume form on $\Cpx^{k}$ and $\phi_{\alpha}:U_{\alpha} \to M$ is the inverse of the chart mapping to $U_{\alpha}$, then Wirtinger's inequality applied to the $2$-form $A = \phi_{\alpha}^{*} \omega$ on $U_{\alpha}$ gives \begin{align*} \int_{M} \frac{\omega^{k}}{k!} &= \sum_{\alpha} \int_{M} \rho_{\alpha}\, \frac{\omega^{k}}{k!} \\ &= \sum_{\alpha} \rho_{\alpha}\, \int_{U_{\alpha}} \phi_{\alpha}^{*}\frac{\omega^{k}}{k!} \\ &= \sum_{\alpha} \rho_{\alpha}\, \int_{U_{\alpha}} \phi_{\alpha}^{*}\frac{\omega^{k}(\dd_{1}, \dots, \dd_{2k})}{k!}\, dV \\ &\leq \sum_{\alpha} \rho_{\alpha}\, \int_{U_{\alpha}} dV \\ &= \sum_{\alpha} \rho_{\alpha}\, \operatorname{Vol}(U_{\alpha}) = \operatorname{Vol}(M), \end{align*} with equality if and only if $T_{p}M$ is a complex subspace of $T_{p}\Cpx^{n}$ at each point, i.e., if $M$ is an immersed complex submanifold of $\Cpx^{n}$.

To address the questions about Ballmann's notes, let $(M, J, g)$ be a complex manifold equipped with a complex structure $J$ and a $J$-compatible Riemannian matric $g$. The $2$-tensor $\omega$ defined by $\omega(X, Y) = g(JX, Y)$ is skew-symmetric since $$\omega(Y, X) = g(JY, X) = g(JJY, JX) = g(-Y, JX) = -g(JX, Y) = -\omega(X, Y). \tag{2}$$

Moreover, the complexified tangent bundle of $M$, $TM \otimes \Cpx$, splits into eigenbundles of $J$, i.e., into sub-bundles of $(1, 0)$- and $(0, 1)$-vectors. If $X$ is a tangent vector to $M$, define complex vectors $$X^{(1, 0)} = \tfrac{1}{2}(X - iJX),\qquad X^{(0, 1)} = \tfrac{1}{2}(X + iJX).$$ It's easy to check that $X = X^{(1, 0)} + X^{(0, 1)}$, that $J(X^{(1, 0)}) = iX^{(1, 0)}$, and that $J(X^{(0, 1)}) = -iX^{(0, 1)}$.

The complex vector bundle $(TM, J)$ is isomorphic to the complex vector bundle $(T^{(1, 0)}M, i)$ under the mapping $X \mapsto X^{(1, 0)}$. The Riemannian metric extends uniquely to a complex bilinear pairing $H$ on $TM \otimes \Cpx$. Since $$H(X, Y) = g(X, Y) = H(JX, JY),\qquad H(JX, Y) = g(JX, Y) = \omega(X, Y) = -H(JY, X),$$ equation (2) implies the Hermitian metric $h$ defined by $h(X, Y) = 2H(X^{(1, 0)}, \overline{Y^{(1, 0)}})$ satisfies \begin{align*} h(X, Y) = 2H(X^{(1, 0)}, Y^{(0, 1)}) &= \tfrac{1}{2} H(X - iJX, Y + iJY) \\ &= \tfrac{1}{2}\bigl(H(X, Y) + H(JX, JY) - iH(JX, Y) + iH(X, JY)\bigr) \\ &= g(X, Y) - i \omega(X, Y). \end{align*}

It's meaningful to consider Riemannian metrics not compatible with a complex structure $J$, but:

• The preceding algebraic yoga only works out nicely if $g$ is $J$-compatible;

• If $g_{0}$ is an arbitrary Riemannian metric, then $$g(X, Y) = \tfrac{1}{2}\bigl(g_{0}(X, Y) + g_{0}(JX, JY)\bigr)$$ (the resulting of averaging $g_{0}$ over the action of $J$) is $J$-compatible. That is, assuming $J$-compatibility is not especially intrusive.