# When can you reverse the orientation of a complex manifold and still get a complex manifold?

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.

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Is there a simple reason for why $\overline{\mathbb{CP}^2}$ is not a complex manifold? –  J.C. Ottem Nov 30 '10 at 22:31
@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. –  Tim Perutz Nov 30 '10 at 22:40
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. –  BCnrd Nov 30 '10 at 22:48
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. –  BCnrd Nov 30 '10 at 23:13

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 biholomprphic. Indeed, if $X$ is a complex curve, then $(X,J)$ is biholomorphic to $(X,-J)$ only if $X$ admits and 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 byholomorphic 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 $Y$.

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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? –  solbap Dec 1 '10 at 16:23
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 $(x^2+a)(x^2+b)(x^2+c)$ $a,b,c>0$)... –  Dmitri Dec 1 '10 at 17:51
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.