MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Using the chern character, it can be shown that there is no complex structure on $S^n$ if $n > 6$: See May's book: if $S^{2n}$ has a complex structure, let $\tau$ be the tangent bundle. $c_n(\tau) = \chi(S^{2n}) = 2$ must be divisible by $(n-1)!$ by Husemöller, Fibre bundles, chapter 20, Theorem 9.8. So the only case left is $S^4$ and $S^6$.

Is there a complex structure on $S^4$ or $S^6$?

share|cite|improve this question
Half of your question is a duplicate:… – Reid Barton Jan 13 '10 at 16:27
up vote 25 down vote accepted

It is known that $S^4$ doesn't even have an almost complex structure, and the case for $S^6$ is open. The lack of almost complex structure can be proved a number of ways, one way is by showing that an almost complex, compact, four manifold with $\dim_{\mathbb{Q}}H^2(X,\mathbb{Q})=0$ has $\chi(X)=0$, but the four sphere doesn't. (It follows from the index theorem, here's a quick reference, first result.)

share|cite|improve this answer
A clarifying comment: $S^6$ DEFINITELY has an almost complex structure, but it's not integrable. It is open if there is a complex structure, and for more details, see the other question that Reid linked to. – Charles Siegel Jan 13 '10 at 17:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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