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 $C^n \supset B^{2n}$.

  • $\begingroup$ why did you made the discussion on $S^{2n}$ before this (apparently) independent question? $\endgroup$ Sep 28, 2014 at 14:02
  • 2
    $\begingroup$ @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. $\endgroup$ Sep 28, 2014 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 $\endgroup$ Sep 28, 2014 at 15:02

1 Answer 1


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.

  • $\begingroup$ Thank you! I will to find and read that paper. Zhou Jianwei $\endgroup$ Sep 28, 2014 at 22:45
  • 1
    $\begingroup$ 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 $\endgroup$ Sep 29, 2014 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.