I have the following question: Let $(M,J_{M}, \omega_{M})$ and $(N,J_{N}, \omega_{N})$ be two Kähler manifolds. I consider $M \times N$. Can I introduce on $M \times N$ a complex structure ? I think one can define the complex structure by $J_{M} \oplus J_{N}$. Is this right? But then I do not understand why the Nijenhuis-tensor vanishes. Can someone explain this to me please. Thanks in advance.
Complex structure on the product of two complex Kähler manifolds
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differential-geometry
manifolds
riemannian-geometry
complex-geometry
kahler-manifolds
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0Yes, this is right. I don't think there's any deep reason why two integrable almost complex structures give rise to an integrable one, we just have to plough through the calculations to show it. If calculating the Nijenhuis tensor of $J_M \oplus J_N$ seems tiring, perhaps you could try showing the equivalent condition that $\bar \partial_{J_M \oplus J_N}^2 = 0$. – 2012-11-01
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0how could one check such a thing? – 2012-11-01
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0For any form $\alpha$ and almost complex structure $J$ we have (by definition) $$\bar \partial_J \alpha = \tfrac 12 (d\alpha \pm i d J \alpha),$$ where $d$ is the exterior derivative and I can never remember whether there should be a $+$ or $-$ in the place of $\pm$ (one gives $\partial$, the other gives $\bar \partial$). Integrability of $J$ (i.e. the vanishing of the Nijenhuis tensor) is equivalent to either $\partial^2 = 0$ or $\bar \partial^2 = 0$. – 2012-11-01
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0By the way, your question is independent of the Kahler condition. – 2012-11-01