# Proof that the Nijenhuis tensor vanishes in a complex manifold

I'm in trouble proving that if $(M,J)$ is a complex manifold with $J$ a compatible almost complex structure then the Nijenhuis tensor of $J$ vanishes: in other words I would like to find that for any two vector fields $X,Y$ one has $$J[X,Y]=J[X,JY]+J[JX,Y]+[X,Y]$$ I tried applying all the definitions of commutator I actually know, but I can't manage it...

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Edit: Am I using the wrong definition of commutator? Can you please confirm me that $[X,Y]$ is defined for $X=X_i\partial_i$, $Y=Y_j\partial_j$ to be the vector field $$\sum_{i=1}^N\Big(\sum_{j=1}^N X_j\frac{\partial Y_i}{\partial x_j}-Y_j\frac{\partial X_i}{\partial x_j}\Big)\partial_i$$

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Doesn't this identity simply say that the commutator bracket is complex linear on complex manifolds? Also, it is more familiar to me in the form $\frac{1}{2}[J,J] = 0$ where $\frac{1}{2}[J,J](X,Y) = [JX,JY] - [X,Y]-J[X,JY] - J[JX,Y]$. – t.b. Jul 22 '11 at 12:31
Also, are you sure you've got your identity right? – t.b. Jul 22 '11 at 12:43
@Theo: Yes, and yes. The resul I want to prove is part of the theorem by Newlander e Nirenberg saying that $M$ is a complex manifold iff $J$ is integrable, iff the Nijenhuis is zero. – Fosco Loregian Jul 22 '11 at 15:52
In the sign conventions I know this result says that $\frac{1}{2}[J,J] = 0$ is equivalent to being a complex manifold. But somehow our identities do not look the same (we seem to disagree about one $J$ here and I'm sure about my version, too). Unfortunately, I don't know a good (= readable) reference for Newlander-Nirenberg. – t.b. Jul 22 '11 at 16:07
Kobayashi-Nomizu's Foundations states the result in the form of theorem 2.5 page 124, and they define the Nijenhuis tensor in "my" way, upto a factor 2... – Fosco Loregian Jul 22 '11 at 16:16

Yes, you're right. But even writing it in this way, the only thing I can get is that $[X,JY]+[JX,Y]=0$ so a couple of term cancels out, but now.. How can $[JX,JY]=[X,Y]$ be true? – Fosco Loregian Jul 22 '11 at 17:05
@tetrapharmakon: replace $X$ by $JX$ then you get a minus sign on the left and you're reduced to the previous identity. – t.b. Jul 22 '11 at 17:16