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Apr 22, 2011 at 3:26 comment added Joel Fine No access to Gray's paper where I am, so no idea if he does the following, but here's a short proof that $h^{0,1}\geq 1$ for those who are curious. The argument holds whenever both integer cohomology in degree 2 and $b_n$ vanish on a complex $n$-fold. Since $H^2(Z)=0$, the exponential sequence shows $H^{0,1}$ surjects onto the group of line bundles. So it's enough to find a non-trivial holomorphic line bundle. Now the canonical bundle is nontrivial, for if not a holomorphic $n$-form $\Omega$ would have $\int \Omega \wedge \bar{\Omega} >0$ so $[\Omega]$ would be non-zero contradicting $b_n=0$.
Apr 22, 2011 at 2:12 history edited YangMills CC BY-SA 3.0
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Apr 22, 2011 at 2:04 history edited YangMills CC BY-SA 3.0
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Apr 21, 2011 at 18:43 history answered YangMills CC BY-SA 3.0