# Structure on a complex manifold and the Kähler form on $\mathbb{C}^2$.

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.

• Thanks again for the nice answer. So in order to calculate the Kähler form it would suffice to calculate $\Im(h)$ for $h(u,w)= u_1\overline{w}_1 + u_2\overline{w}_2$ and $u,v\in \mathbb{C}^2$. If we let $u_j=x_j+iy_j$ and $w_k=a_k+ib_k$, then $$h(u,w)=(x_1a_1+b_1y_1)+i(-b_1x_1+y_1a_1)+(x_2a_2+b_2y_2)+i(-b_2x_2+y_2x_2)$$ and now I guess that there is a way to express this in the form $$h=h_{ij} dz_i \otimes d\overline{z}_j$$ which I don't see immediately? @j-v-gaiter Commented Jun 12, 2023 at 21:37
• I have added a comment which addresses this. Commented Jun 13, 2023 at 18:26