Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $J$ be an almost complex structure on an algebraic variety $V$. As we all know, $J$ comes from a complex structure if the Nijenhuis tensor of $J$ vanishes. What I would like to know is if there exists a simpler characterisation of integrability than this for varieties (as opposed to general manifolds).

share|improve this question
4  
When you say that $V$ is an algebraic variety, are you assuming that it is complex-algebraic (i.e., already has a complex structure), and that $J$ is an additional almost-complex structure? –  S. Carnahan Feb 21 '11 at 16:48
1  
@Janos: If you want equivalent conditions to the Nijenhuis tensor vanishing then one is that the induced $\bar \partial$ operator defines a complex, i.e. that $\bar \partial^2 = 0$. Another one is that the exterior derivative decomposes as $d = \partial + \bar \partial$. If you want explicit examples of almost complex manifolds that are not complex, that's going to be more difficult, see the answers to tinyurl.com/4zrkar6 –  Gunnar Magnusson Feb 21 '11 at 18:03
1  
No, wait, do you want to hit two birds with one stone and get integrability for the almost complex structure and projectivity of the resulting complex manifold in one swoop? I'm not sure that's going to be doable... just look at two-dimensional tori. You can realize any complex torus as an integrable complex structure on $M = \mathbb R^2/\mathbb Z^2$, but not all tori are projective. Given an arbitrary integrable complex structure on $M$, I don't know how to link its properties to the question of projectivity of the torus. –  Gunnar Magnusson Feb 21 '11 at 18:10
1  
There is a simpler criterion that does integrability and Kählerness in one step: if there is both a Riemann metric $g$ and an almost complex structure $J$ and $\nabla J =0$ (i.e., $J$ is parallel w.r.t. the Levi-Civita connection of the Riemann metric), then $J$ is automatically integrable and $g$ is a kähler metric. –  Johannes Ebert Feb 23 '11 at 22:13
1  
@Gunnar: complex tori of complex dimension $1$ are automatically projective (they are smooth elliptic curves). In general, compact Riemann surfaces are smooth projective algebraic curves... –  Qfwfq Feb 23 '11 at 23:50
show 10 more comments

1 Answer 1

up vote 3 down vote accepted

If you want equivalent conditions to the Nijenhuis tensor vanishing then one is that the induced $\bar \partial$ operator defines a complex, i.e. that $\bar \partial^2 = 0$. Another one is that the exterior derivative decomposes as $d = \partial + \bar \partial$. If you want explicit examples of almost complex manifolds that are not complex, that's going to be more difficult, see the answers to tinyurl.com/4zrkar6

To find the $\bar \partial$ operator associated to an almost complex structure $J$ on a smooth manifold $M$, one needs to note that $J$ induces a splitting $T_M \otimes \mathbb C = T^{1,0} \oplus T^{0,1}$ of the tangent bundle into $i$ and $-i$ eigenvectors (the eigenspaces being marked with $(1,0)$ and $(0,1)$, respectively).

The same thing happens on the level of 1-forms (and indeed on the level of $k$-forms): they split into $(p,q)$-forms like on complex manifolds. If $\pi^{p,q} : \bigwedge^k T_M \to \bigwedge^{p,q} T_M$ is the projection onto the space of $(p,q)$-forms, then the $\bar \partial : \bigwedge^{p,q} T_M \to \bigwedge^{p,q+1} T_M$ operator associated to $J$ is $\bar \partial_J = \pi^{p,q+1} \circ d$.

Once one does the calculations this comes out to $$ \bar \partial \alpha = \frac 1 2 \left( d \alpha + i d J \alpha \right) $$ for a $(p,q)$-form $\alpha$. A similar formula holds for the $\partial$ operator, you just have to change $i$ to $-i$. (I may have confounded the signs here.)

A very good reference for the linear algebra parts (i.e. most of this) is Chapter 2 of Huybrecht's "Complex geometry".

For the conditions equivalent to the vanishing of the Nijenhuis tensor I seem to remember the first chapter of http://tinyurl.com/4cqspu7 Moroianu's notes on Kahler geometry being very helpful when I went through this a couple of months ago.

Finally, on why the vanishing of the Nijenhuis tensor implies that we indeed have a complex structure I recommend Demailly's book - Chapter 8, section 11 (page 396) has all the details: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf

share|improve this answer
    
Great, thanks a lot! –  Janos Erdmann Feb 21 '11 at 19:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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