TITLE: Confusion about basic Kahler geometry
QUESTION [6 upvotes]: I am really struggling to understand the basics of Kahler geometry and hope someone can give me some guidance.
Suppose we have a complex manifold with some complex structure $J$ and let $g$ be a Hermitian metric. That is, $g$ is a Riemannian metric and satisfies the extra condition $g(JX,JY)=g(X,Y)$. So we have a Riemannian manifold $(M^{2n},g)$. Why do we all of a sudden start considering the complexified tangent space? What good does that do? The metric $g$ is of the form $$g=\sum_{i,j=1}^{2n}g_{ij}E^i\otimes E^j,$$ where $\{E^1,\dots, E^{2n}\}=\{dx^1,dy^1,\dots, dx^n, dy^n\}$. We introduce $\frac{\partial}{\partial z_k}$ and $\frac{\partial}{\partial \overline z_k}$ and $dz_k$ and $d\overline z_k$ and then extend the metric to $TM^\mathbb{C}$ and start doing all these computations. But by doing this we've completely changed the domain of the metric. One metric is in $\Omega^2_\mathbb{C}(M)$ and the other is in $\Omega^2(M)$. How is this new metric telling us stuff about the original one? What is the relationship between the matrix $g_{k\overline j}$, where $1\leq j,k\leq n$ and $g_{jk}$, where $1\leq j,k\leq 2n$? We introduce the 'extended' connection and curvature but again, I don't see how this new information tells us something about the original manifold. For example, apparently the 'extended' definition of Ricci curvature on $TM^\mathbb{C}$ is the same as the Ricci curvature on $(M^{2n},g)$. How can these two things be the same, they are functions on completely different spaces?? Another example I am struggling with is how the Kahler form $\omega$ satisfies $\frac{\omega^n}{n!}=\text{vol}_M$. We have $\omega=\sqrt{-1}g_{j\overline k}dz^j\wedge d\overline z^k$ an element of $\Omega^2_\mathbb{C}(M)$. How can this be raised to powers to give an element of
$\Omega^{2n}(M)$??
Really any help would be greatly appreciated!

REPLY [2 votes]: Customary we write Kahler metric $$ ds^2=2\sum_{a,b} \
g_{a\overline{b}}dz_a\otimes d\overline{z}_b
$$
Here $g$ is a Riemannian metric. If $E_a=\frac{\partial }{\partial
x_a},\ E_{n+a}=\frac{\partial }{\partial x_{n+a}}$ are coordinate
field, then we have dual $dE_a,\ dE_{n+a}$.
In further, define $$ \frac{\partial
}{\partial z_a} =\frac{\partial }{\partial x_a} -i\frac{\partial
}{\partial x_{n+a}} $$ so that its dual is $dz_a=\frac{1}{2}\{ dE_a+
idE_{n+a} \}$.
If $g_{a\overline{b}} :=g(\frac{\partial }{\partial
z_a},\overline{\frac{\partial }{\partial z_b}} )$, then by a direct
computation considering $J$, we have $$ g=2{\rm Re}\ \sum_{a,b} \
g_{a\overline{b}}dz_a\otimes d\overline{z}_b = {\rm Re}\ ds^2 $$
In further, fundamental form is $\Phi (X,Y)=g(X,JY)$. By
computation, we have $\Phi = {\rm Im}\ ds^2 $.