Currently learning about complex and Kähler manifolds and I'm reading this article about Kähler forms.
Specifically I'm trying to understand how do they conclude that on $\mathbb{C}^2$ the Kähler form is given by $$\omega = -\frac{i}{2} dz_1 \wedge d\overline{z}_1 + dz_2 \wedge d\overline{z}_2$$ but there are some minor holes in my understanding.
In the article they state that $M$ is a complex manifold and then suddenly $M$ has a Hermitian metric $h=g-i\omega$. I don't think that complex manifolds naturally carry such metric or am I wrong? What I'm trying to get at that $M$ cannot simply be a complex manifold in order for us to have this kind of Hermitian metric. Do we need to assume that $M$ also has a Riemannian metric?
Following this they suddenly jump to the conclusion that $M$ also has an almost complex structure $J$. This is the second issue I have since I do not know whether every complex manifold naturally has such a structure?
So if anyone could help me figure out how $\omega$ is derived an give some insights on what kind of assumptions we need for $M$ I would be very happy.
The article states that if $M$ is a complex manifold, a Kähler form is a two form $\omega$ such that there is a hermitian structure $h$ on $M$ for which $\omega$ is the imaginary part, i.e. $h=g-i\omega$ such that $d\omega=0$. Every complex manifold (and in fact every complex vector bundle over a paracompact space) admits a hermitian structure by the following argument.
Let $\{U_{\alpha}\}_{\alpha\in A}$ be a locally finite cover of $M$ with a subordinate partition of unity $\{(V_{\alpha}, \phi_{\alpha})\}$ such that $TM$ trivializes over each $U_{\alpha}$, with given trivialization $\psi_{\alpha}: \pi^{-1}(U_{\alpha})\to \mathbb{C}^n\times U_{\alpha}$. Over each $U_{\alpha}$ we can endow $\pi^{-1}(U_{\alpha})$ a Hermitian metric by pulling back the standard hermitian metric from $\mathbb{C}^n\times U_{\alpha}$. Call this metric $h_{\alpha}$. We can define a global sesquilinear form on $TM$ by $h:=\sum \phi_{\alpha}h_{\alpha}$. This is well defined because $\{U_{\alpha}\}$ is locally finite. Because each of the $\phi_{\alpha}$'s is non-negative, and the sum at any given point is nonvanishing, $h_{p}=\sum\phi_{\alpha}(p)h_{\alpha}(p)$ is a hermitian form on $T_pM$ (Exercise: verify that for $a,b>0$, $h$ and $h'$ hermitian forms on $V$ a complex vector space, $ah+bh'$ is again a hermitian form on $V$) and hence $h$ defines a global hermitian form on $M$. The same argument tells us that any smooth manifold admits a Riemannian metric.
The question regarding the existence almost complex structures on complex manifolds is either tautologically true or true by a small exercise depending on your definition of a complex manifold.
In analogy to smooth manifolds, one may define a complex manifold as a topological space $M$ equipped with a cover of charts $\phi_{\alpha}: U_{\alpha}\to \mathbb{C}^n$ where each of the overlaps $\phi_{\alpha}\circ \phi_{\beta}^{-1}$ is a holomorphic map between open sets of $\mathbb{C}^n$, i.e. the Jacobian at every point is a complex linear transformation. Because the transition maps are complex differentiable, we can give $M$ an almost complex structure by defining multiplication by $i$ in charts.
The other definition of a complex manifold, is a smooth manifold $M$ admitting an endomorphism $J: TM\to TM$ with $J^2=-1$ (this makes each $T_pM$ into a complex vector space) such that $J$ is integrable, that is the Nijenhuis tensor of $J$ vanishes. In this case $M$ is already an almost complex manifold by definition.
It is important to note that not every complex manifold admits a Kähler form, and there are cohomological obstructions which help to quantify this.
The fact that the Kähler structure on $\mathbb{C}^2$ takes this particular form is really part of a more general fact about hermitian forms on complex vector spaces. If $V$ is a complex vector space and $h$ is a hermitian form thereof (i.e. non-degenerate sesquilinear form on $V$), $\mathrm{Im} (h)$ is a Kähler form on $V$. This means that to find the Kähler form on $\mathbb{C}^2$ it suffices to calculate the imaginary part of the standard hermitian form.
Many of the previous results are of the first things one learns when studying complex geometry, and not having a strong grasp on them means that continuing will become increasingly difficult. I might suggest taking a deep dive in a complex geometry text before going forward.
EDIT: The first expression you give $h(u,v)=u_1\bar{w}_1+u_2\bar{w}_2$ is sufficient to see that $h=dz^1\otimes d\bar{z}^1+dz^2\otimes d\bar{z}^2$ since $h(e^i,e^j)=\delta^{ij}$. Then to go about calculating the corresponding Kähler form, we see that since $Im(z)=-\frac{i}{2}(z-\bar{z})$, \begin{align}\mathrm{Im}(h)&=\mathrm{Im}(dz^1\otimes d\bar{z}^1)+\mathrm{Im}(dz^2\otimes d\bar{z}^2)\\ &=-\frac{i}{2}( dz^1\otimes d\bar{z}^1-d\bar{z}^1\otimes d z^1+ dz^2\otimes d\bar{z}^2-d\bar{z}^2\otimes d z^2)\\ &=-\frac{i}{2}(dz^1\wedge d\bar{z}^1+dz^2\wedge d\bar{z}^2)\end{align} As desired.