19
$\begingroup$

I'm told that $\overline{\mathbb{C}P^2}$, i.e. $\mathbb{C}P^2$ with reverse orientation, is not a complex manifold. But for example, $\overline{\mathbb{C}}$ is still a complex manifold and biholomorphic to $\mathbb{C}$.

This makes me wonder, if $X$ is complex manifold is there a general criterion for when $\overline{X}$ also has a complex structure? For example, it seems that if $X$ is an affine variety than simply replacing $i$ with $-i$ gives $\overline{X}$ a complex structure and $X, \overline{X}$ are biholomorphic.

EDIT: the last claim is wrong; see BCnrd's comments below and Dmitri's example. Also, as explained by Dmitri and BCnrd, $X$ should be taken to have even complex dimension.

Another question: if $X$ and $\overline{X}$ both have complex structures, are they necessarily biholomorphic? Edit: No per Dmitri's answer below.

$\endgroup$
4
  • 1
    $\begingroup$ Is there a simple reason for why $\overline{\mathbb{CP}^2}$ is not a complex manifold? $\endgroup$
    – J.C. Ottem
    Nov 30, 2010 at 22:31
  • 5
    $\begingroup$ @J.C. Ottern: Any almost complex structure compatible with the orientation on a closed 4-manifold $X$ satisfies $c_1^2[X]=2\chi+3\sigma$ ($\chi$=Euler char, $\sigma$=signature). This is by Hirzebruch's signature theorem. $\endgroup$
    – Tim Perutz
    Nov 30, 2010 at 22:40
  • 4
    $\begingroup$ Fix an alg. closure $\mathbf{C}$ of $\mathbf{R}$, equipped with unique abs. value extending the one on $\mathbf{R}$, complex analysis is developed without needing a preferred $\sqrt{-1}$. The complex structure has no reliance on any orientation. The so-called canonical orientation on complex manifolds is just the functorial one arising from a choice of $\sqrt{-1}$; can make either choice, complex structure can't tell! Likewise, the analytification functor on locally finite type $\mathbf{C}$-scheme has nothing to do with any such choice. Note $p$-adic analysis goes the same way. $\endgroup$
    – BCnrd
    Nov 30, 2010 at 22:48
  • 5
    $\begingroup$ What is canonical is that even-dim'l C-manifolds have an intrinsic orientation determined by C-structure: an orientation of $\mathbf{C}$ endows all C-manifolds with functorial orientation, and changing initial choice affects the orientation on $n$-dimensional C-manifolds by $(-1)^n$. So for even $n$ the question is well-posed. This has nothing to do with changing $i$ and $-i$, and your impression in the affine case is wrong. In any dim., can "twist" structure sheaf by C-conj. to get a new C-manifold (modelled on $\overline{f}(\overline{z})$), but that's a different beast. $\endgroup$
    – BCnrd
    Nov 30, 2010 at 23:13

2 Answers 2

14
$\begingroup$

If you take an odd dimensional complex manifold $X$ with holomorphic structure $J$ then $-J$ defines on $X$ a holomorphic structure as well. And, of course, $J$ and $-J$ induce on $X$ opposite orientations. In general it is not true that these two complex manifolds are biholomorphic. Indeed, if $X$ is a complex curve, then $(X,J)$ is biholomorphic to $(X,-J)$ only if $X$ admits an anti-holomorphic involution (this will be the case for example if $X$ is given by an equation with real coefficients).

Starting from this example on can construct a (singular) affine variety $Y$ of dimension $3$, such that $(Y,J)$ is not biholomorphic to $(Y,-J)$. Namely, let $C$ be a compact complex curve that does not admit an anti-holomorphic involution say of genus $g=2$. Consider the rank two bundle over it, equal to the sum $TC\oplus TC$ ($TC$ is the tangent bundle to $C$). Contract the zero section of the total space of this bundle, this gives you desired singular $Y$.

$\endgroup$
2
  • $\begingroup$ Nice example but I wonder -and this is not so important, just curious if there's a nice answer- if $C$ is a genus 2 curve and I describe it as a degree 2 cover of $\mathbb{P}^1$: $y^2 = (x - a_1)\cdots (x - a_6)$, is there simple condition on the $a_i$ that guarantees that it does not have an antiholomorphic involution? $\endgroup$
    – solbap
    Dec 1, 2010 at 16:23
  • 2
    $\begingroup$ Yes, there should be a relatively simple criterion, one should check when the configuration of points $a_1,...,a_6$ is (not) invariant under any anti-holomorphic involution of $\mathbb CP^1$. For example, in the case of elliptic curve $y^2=(x-a_1)...(x-a_4)$ the necessary and sufficient condition for having anti-holomorphic involution is that the double ratio of $\frac{a_1-a_2}{a_3-a_4}$ is real. If all double ratios of $a_1,...,a_6$ are real, then again we have an anti-holomorphic involution. But this is not necessary there are two more cases (like $(x^2+a)(x^2+b)(x^2+c)$ $a,b,c>0$)... $\endgroup$ Dec 1, 2010 at 17:51
7
$\begingroup$

It seems to me that you could be interested in the following (I haven't checked the paper in detail, but I think theorems of this "style" could be helpful for you):

Theorem Let $X$ be a compact complex surface admitting a complex structure for $\bar{X}$. Then $X$ (and $\bar{X}$) satisfies one of the following:

  1. $X$ is geometrically ruled, or
  2. the Chern numbers $c_1^2$ and $c_2$ of $X$ vanish, or
  3. $X$ is uniformised by the polydisk.

In particular, the signature of $X$ vanishes.

Other material that could be helpful is:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.