Confusion about basic Kahler geometry 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!
 A: Kähler manifolds are particularly nice examples of manifolds where the complex structure and the Riemannian structure coalesce perfectly. Every complex manifold may be viewed as a Riemannian manifold (since holomorphic charts are automatically smooth), but it is not necessarily the case that these two structures will be compatible. This compatibility can be expressed in a number of ways:


*

*We have a notion of parallel transport on Riemannian manifolds. A certain compatibility criterion between the complex and Riemannian structures would be to require that parallel transport commutes with $J$. 

*On the tangent bundle of a complex manifold, we have two choices of canonical connection, the Levi--Civita connection from Riemannian geometry and the Chern connection from complex geometry. These two connections do not necessarily coincide on an arbitrary complex manifold. 

*On a Riemannian manifold we may diagonalise the metric at a point which gives us the Riemannian Normal Coordinates. On an a complex manifold we have these coordinates also (of course), but these coordinates are not necessarily holomorphic.  
PUNCHLINE: On a Kähler manifold all of these conditions are satisfied! In fact, the Kähler condition $d\omega =0$ is equivalent to the three conditions above (any of which may be taken as a definition).    
Closing remarks: It is worth noting that the condition $g(JX, JY) = g(X,Y)$ may be stated as requiring the complex structure $J$ to be an orthogonal transformation on each tangent space. 
Commuting with the complex structure $J$ is desirable for the same reason that holomorphic functions are desirable relative to smooth functions. That is, a function $f : \mathbb{C} \longrightarrow \mathbb{C}$ is holomorphic if its Jacobian matrix commutes with $J = \begin{bmatrix}
0 & -1 \\
1 & 0 
\end{bmatrix}$. 
Notice that this also alludes to why holomorphic functions are such starkly different beasts when compared with smooth functions. This requirement on the derivative is quite rigid. 
References: Griffiths and Harris, Principles of Algebraic Geometry, Andrei Moroianu Lecture on Kähler Geometry, Werner Ballmann Lectures on Kähler manifolds, Gabór Szekelyhidi Introduction to Extremal Metrics.
This is also nice: Key differences between almost complex manifolds and complex manifolds
Updated: I have recently been made aware of Daniele Angella's text Cohomological Aspects in complex Non-Kähler geometry. As the name suggests, the book focuses on complex manifolds which are not Kähler, but I would highly recommend this well-written text. It has a nice treatment of almost complex manifolds, and it is often insightful to look at the behaviour of the objects which do not satisfy such a nice condition as being Kähler. 
A: 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 $.
