What is an (almost) complex manifold in simple words

[gerald V]

TL;DR Summary

What is a complex manifold in simple words? What, in contrast, are manifolds equipped with some unusual metric (skew-symmetric parts, complex entrances)?

I try to understand (almost) complex manifolds and related stuff. Am I right that the condition for almost complexity simply is that the metric locally can be written in terms of the complex coordinates $z$, i.e. $g = g(z_1, ... z_m)$ (complex conjugate coordinates must not appear)? These relations may involve complex constants?

If one starts from a space with real coordinates and a real metric it is not sure that the degrees of freedom can be grouped appropriately, but the other way round it is trivial, even globally (complex manifolds)?

What if I start from a space of even number of dimensions (the coordinates being real numbers), and equip it with any crazy metric which can have skew-symmetric parts and/or complex entrances? Do these manifolds have names? Can this lead to self-contradicitons? Are there examples where such structures are used?

Sorry if I may have confused a lot. But I am asking because this entire topic for me is hard to comprehend.

 

[Infrared]

As you probably already know, a (smooth) manifold is a space locally homeomorphic to open subsets of $\mathbb{R}^n$ with smooth transition functions (and Hausdorff, second countable).

A complex manifold is defined the same way, except that it is locally homeomorphic to open subsets of $\mathbb{C}^n$ and the transition functions are holomorphic (complex differentiable). Every complex manifold can be viewed as a (real) smooth manifold, but not vice-versa even for even dimensions, since being holomorphic is a much stronger condition than being smooth.

One important property of complex manifolds is that their tangent spaces are naturally vector spaces over $\mathbb{C}$. Suppose we want to replicate this feature in real manifolds, meaning that we want a way to view the tangent spaces of a real manifold as complex vector spaces. To view a real vector space $V$ as complex vector space, we need to specify how to scale a vector by $i$; that is, we need to specify a linear map $J:V\to V$ such that $J^2=-1$. Then $V$ would be a vector space with the scalar multiplication law $(a+bi)v=av+bJv.$ For (necessarily even-dimensional) manifolds, an almost complex structure is an endomorphism of the tangent bundle that squares to $-1.$ This makes each tangent space into a vector space over $\mathbb{C}.$ All complex manifolds are of course also almost-complex, but the converse is not true. Not every almost complex structure comes from a complex structure. The Nijenhuis tensor associated to an almost complex structure has to vanish for this to happen, and it is a difficult theorem that this is actually sufficient.

 

[mathwonk]

As infrared said, an "almost complex" manifold is a real manifold whose tangent bundle at least looks like the tangent bundle of a complex manifold. In particular the tangent bundle, and hence also the manifold, has even real dimension. One might think this structure would be enough to give the manifold itself a complex structure, since you should already be able to tell which smooth functions are holomorphic, since their differentials should be complex linear functionals on the tangent bundle, and this should be visible from the complex structure on the tangent bundle. But there is, as infrared said, another condition, called "integrability", that let's you go backwards, i.e. integrate, from the complex tangent bundle structure, to a complex manifold structure. You might google: Newlander - Nirenberg, if memory serves, but I have forgotten what the condition is.

Ah yes, (source: wikipedia) one uses the complex structure on the tangent bundle to decompose the exterior derivative d into two summands ∂ and ∂bar. (holomorphic functions would then be those with ∂bar = 0.) Then the integrability condition seems to be that (∂bar)^2 = 0. (But I am not expert and there seems to be some contradictory looking information on the wikipedia page, stating that just d = ∂ + ∂bar is also equivalent??) But I used to be (in the 1960's) a student of some experts in this stuff and (∂bar)^2 = 0 rings a bell as the condition they always gave. This of course is the condition that let's one define Dolbeault cohomology for complex manifolds.

 

[Infrared]

Ah yes, (source: wikipedia) one uses the complex structure on the tangent bundle to decompose the exterior derivative d into two summands ∂ and ∂bar. (holomorphic functions would then be those with ∂bar = 0.) Then the integrability condition seems to be that (∂bar)^2 = 0. (But I am not expert and there seems to be some contradictory looking information on the wikipedia page, stating that just d = ∂ + ∂bar is also equivalent??)

It's slightly more complicated than that. You first complexify the tangent space $TM$, and then given $J$, extend it to be $\mathbb{C}$-linear and decompose $(TM)^\mathbb{C}$ as the direct sum of the $i$ and $-i$ eigenspaces for $J$ (these are the holomorphic and anti-holomorphic tangent spaces $T^{1,0}M$ and $T^{0,1}M$, respectively, and they would be spanned by $\frac{\partial}{\partial z_i}$ and $\frac{\partial}{\partial \overline{z}_i}$ if they in fact came from a complex structure.)

This induces a decomposition $\left(\Lambda^k M\right)^\mathbb{C}=\bigoplus_{p+q=k} \Lambda^{p,q}(M)$ and given $\alpha\in\Lambda^{p,q}$, you can define $\partial\alpha$ to be the part of $d\alpha$ that lives in $\Lambda^{p+1,q}$ and similarly for $\overline{\partial}.$ If $J$ is induced by a complex structure, then we'll have $d=\partial+\overline{\partial}$ since $d (f dz^{i_1}\wedge\ldots\wedge dz^{i_p}\wedge d\overline{z}^{j_1}\wedge\ldots\wedge d \overline{z} ^{j_q})$ will only have terms of types $(p+1,q)$ and $(p,q+1).$ But if $J$ is not integrable, there is no such guarantee and there will be terms in $d\alpha$ not appearing in $\partial\alpha+\overline{\partial}\alpha.$

Now if $d=\partial+\overline{\partial}$, then $\overline{\partial}^2\alpha$ should be the $(p,q+2)$ component of $d^2\alpha,$ so $\overline{\partial}^2=0.$ I think the wikipedia page actually looks pretty good on this point.

 

[mathwonk]

Thank you! That helps a lot. (I didn't actually read the wikipedia page, only skimmed it looking for the familiar form, to me, of the integrability condition.)