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Assuming that there is a complex structure on $S^{2n}$ and it becomes a complex manifold, also assuming there are complex coordinate $z, w$ on $U, V$ respectively, where $U, V$ are open cover of $S^{2n}$. Let $\rho_U, \rho_V=1-\rho_U$ be partition of unity. Then we have a connection on the bundle $T^{(1,0)}S^{2n}$ defined by $$\nabla= \rho_U \nabla^U+\rho_V \nabla^V, $$ where $\nabla^U, \nabla^V$ are defined by $$\nabla^U\frac {\partial }{\partial z^i}=0; \ \ \ \nabla^V\frac {\partial }{\partial w^i}=0, \ \ i=1,\cdots, n.$$ Denote $A= (\frac {\partial w^j}{\partial z^i}), \ \omega_0=dA\cdot A^{-1}= \partial A\cdot A^{-1}$ and $(\frac {\partial }{\partial z^i})=(\frac {\partial }{\partial z^1},\cdots, \frac {\partial }{\partial z^n})^t=A(\frac {\partial }{\partial w^j})$, then we have $$\nabla(\frac {\partial }{\partial z^i})= \rho_U \nabla^U (\frac {\partial }{\partial z^i}) +\rho_V \nabla^V(\frac {\partial }{\partial z^i})=\rho_VdA\cdot A^{-1}(\frac {\partial }{\partial z^i}), $$ $$\nabla^2(\frac {\partial }{\partial z^i})=(d\rho_V\wedge\omega_0 +\rho_V(1-\rho_V)\omega_0\wedge\omega_0)(\frac {\partial }{\partial z^i})=\Omega|_U (\frac {\partial }{\partial z^i}).$$

Then it is easy to see the top Chern class of the the bundle $T^{(1,0)}S^{2n}$ is trivial if $n>1$, that is, $$c_n(T^{(1,0)}S^{2n})=\det (\frac {\sqrt{-1}}{2\pi}\Omega)=0.$$ On the other hand, we have $c_n(T^{(1,0)}S^{2n}) = e(TS^{2n})$ the Euler class of tangent bundle of $S^{2n}$, a contradiction.

My question is:

If the open ball $B^{2n}\subset R^{2n}$ is a complex manifold, is there a holomorphic isomorphism from $B^{2n}$ to a open subset of $C^n$? Or equivalently, does complex manifold $B^{2n}$ can be covered by one complex coordinates card?

In 1977, R Hamilton (J. D. G., vol.12, no.1, 1-45) showed that such isomorphism exists if the complex structure on $B^{2n}$ is sufficiently close to the structure on $C^n \supset B^{2n}$.

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  • $\begingroup$ why did you made the discussion on $S^{2n}$ before this (apparently) independent question? $\endgroup$ – Daniele Zuddas Sep 28 '14 at 14:02
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    $\begingroup$ @DanieleZuddas: The OP wants to show that a positive answer to his question would settle the question (in the negative) about whether $S^{6}$ has a complex structure. (Aside from $S^2$, the $6$-sphere is the only sphere that could possibly carry a complex structure.) This seems very dubious. I don't see his 'easy' proof that the top Chern class would vanish if $n>1$, though. Why doesn't this proof show that if the coordinate change is $w^i = 1/z^i$ for $1\le i\le n$, then the top Chern form vanishes? It doesn't if $n=1$, and yet the top form is then just a product of the $1$-dim cases. $\endgroup$ – Robert Bryant Sep 28 '14 at 14:22
  • $\begingroup$ I see, the problem is that a possible complex structure on $S^6$ can be defined by a bunch of holomorphic charts $\endgroup$ – Daniele Zuddas Sep 28 '14 at 15:02
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The answer is not. Calabi and Eckmann ("A class of compact, complex manifolds which are not algebraic", Ann. of Math. 58 (1953) 494-500) proved that $\Bbb R^{2n}$, $n > 1$, has a complex structure which cannot be covered by a single holomorphic chart.

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  • $\begingroup$ Thank you! I will to find and read that paper. Zhou Jianwei $\endgroup$ – Zhou Jianwei Sep 28 '14 at 22:45
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    $\begingroup$ Dear Professor Daniele Zuddas, I have found the paper you suggested. The Theorem VI of the paper can answer my question. Thank you very much! Zhou Jianwei $\endgroup$ – Zhou Jianwei Sep 29 '14 at 11:09

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