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Assume we have $X$ a complex manifold and $Y = Y^{\alpha} \frac{\partial}{\partial z^{\alpha}}$ and $Z = Z^{\alpha} \frac{\partial}{\partial z^{\alpha}}$ two vector fields on $X$. Let $\nabla$ be the covariant derivative. What is the computation of $\nabla_{Z}Y$ (i.e. what is $\nabla_{\frac{\partial}{\partial z^{\beta}}}\frac{\partial}{\partial z^{\alpha}}$, i think its $0$ but why ?)?

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You don't have a Levi-Civita covariant derivative unless you also impose a Riemannian metric. It does not come for free on a complex manifold. Your question has nothing to do with complex manifolds at all. The answer is in absolutely any text on Riemannian geometry. –  Spiro Karigiannis Jun 12 '11 at 11:51
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The again perhaps the OP does not mean the Levi-Civita connection, but only a complex connection: one relative to which the complex structure is parallel. –  José Figueroa-O'Farrill Jun 12 '11 at 12:09
    
ok sorry the without levi civita. but does it mean that: for some complex connection one has $\nabla_{\frac{\partial}{\partial z^{\beta}}}\frac{\partial}{\partial z^{alpha}} = 0$ ? du you know some literature where i can find this ? –  gregor Jun 12 '11 at 13:22
    
No. If $z^{\alpha}$ and $z^{\beta}$ are local holomorphic coordinates on the manifold, as I assume you mean, then you can choose a connection such that in that single coordinate chart, this condition holds, but you cannot always ensure this holds in every single coordinate chart. Because if it did, this would be a "flat" connection, and not all complex manifolds admit flat connections. –  Spiro Karigiannis Jun 12 '11 at 13:32
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Any of the three following references should give you enough background to work this out, though the answer to your question may not be spelled out in detail: [1] "Complex differential geometry" by F. Zheng, [2] "Lectures on Kahler geometry" by A. Moroianu, [3] "Complex analytic and differential geometry" by J-P. Demailly (tinyurl.com/2dcnycm). –  Gunnar Magnusson Jun 12 '11 at 15:41

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